Quantum Physics III: Lecture Notes [3]

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Chapter 1

Perturbation theory c B. Zwiebach

It is often the case that the Hamiltonian of a system differs slightly from a Hamiltonian that is well studied and completely understood. This is a situation where perturbation theory can be useful. Perturbation theory allows us to make statements about the Hamiltonian of the system using what we know about the well studied Hamiltonian. The well studied Hamiltonian could be the that of the simple harmonic oscillator in one, two, or three dimensions. In a diatomic molecule, for example, the potential that controls the vibrations is not exactly quadratic; it has extra terms that make the vibrations slightly anharmonic. In that situation the extra terms in the potential represent perturbations of the Hamiltonian. The hydrogen atom Hamiltonian is also a well understood system. If we place the atom inside a weak external magnetic field or electric field, the situation is described by adding some small terms to the hydrogen Hamiltonian. Similarly, the interaction between the magnetic moments of the proton and the electron can be incorporated by modifying the original hydrogen atom Hamiltonian. The interaction between two neutral hydrogen atoms at a distance, leading to the van der Waals force can be studied in perturbation theory by thinking of the two atoms as electric dipoles. The Hamiltonian of interest is written as the understood, original Hamiltonian H (0) , plus a perturbation δH: H (0) + δH . (1.0.1) Since H (0) is Hermitian and the sum must be a Hermitian Hamiltonian, the perturbation operator δH must also be Hermitian. It is convenient to introduce a unit-free constant λ ∈ [0, 1] and to consider, instead, a λ-dependent Hamiltonian H(λ) that takes the form H(λ) = H (0) + λ δH .

(1.0.2)

When λ = 1 we have the Hamiltonian of interest, but λ allows us to consider a family of Hamiltonians that interpolate from H (0) , when λ is equal to zero, to the Hamiltonian of interest for λ equal to one. In many cases perturbations can be turned on and off; think, 1

2

CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY

for example, of an atom in an external magnetic field that can be varied continuously. In that case we can view λ as the parameter that allows us to turn on the perturbation by letting λ 6= 0. The parameter λ is also useful in organizing the perturbation analysis, as we will see below. We spoke of a Hamiltonian that differs slightly from H (0) . In order to use perturbation theory we need λδH to be a ‘small’ perturbation of the Hamiltonian H (0) . We will have to deal with the meaning of small. At first sight we may imagine that small means that, viewed as matrices, the largest entries in λδH are smaller than the largest entries in H (0) . While this is necessary, more is needed, as we will see in our analysis. An additional advantage of using λ is that by taking it to be sufficiently small we can surely make λδH small. We assume that the Hamiltonian H (0) is understood, namely, we know the eigenstates and eigenvalues of H (0) . We want to know the eigenstates and eigenvalues of H(λ). One may be able to calculate those exactly, but this is seldom a realistic possibility. Diagonalizing δH is seldom useful, since δH and H do not generally commute and therefore δH eigenstates are not eigenstates of H(λ). In perturbation theory the key assumption is that the eigenvalues and eigenvectors of H(λ) can be found as series expansions in powers of λ. We hope, of course, that there are some values of λ for which the series converges, or at least gives useful information.

Figure 1.1: The energy eigenvalues of H(λ) as λ varies from zero to one. On the λ = 0 vertical axis the H (0) eigenstates are represented by heavy dots. By the time λ = 1 the dots have shifted. In Figure 1.1 we illustrate some of the phenomena that we may see in the spectrum of a system with Hamiltonian H(λ). We show how the energies of the various states may change as the parameter λ is increased from zero. The two lowest energy eigenstates are non-degenerate and their energies can go up and down as λ varies. Next up in energy we have two degenerate states of H (0) (the Hamiltonian as λ = 0). The perturbation splits the

3

1.1. NONDEGENERATE PERTURBATION THEORY

two levels, and that happens generically. In the figure, the perturbation splits the levels to first order in λ, as shown by the different slopes of the two curves that meet at λ = 0. In other words, viewed as power series in λ the energies of the two states have different linear terms in λ. The last level shown corresponds to four degenerate states. The perturbation to first order in λ splits the states into a group of three states and a fourth. To second order in λ the three states split further. A single Hamiltonian can exhibit behavior like this, with many possible variations. To analyze the evolution of states and energies as functions of λ we have two possible cases: (i) we are following a non-degenerate state or, (ii) we are following a collection of degenerate states. The challenges are quite different and therefore we must analyze them separately. Clearly both situations can occur for a single Hamiltonian, depending on the spectrum of H (0) . To follow a non-degenerate state we use non-degenerate perturbation theory. To follow a set of degenerate states we use degenerate perturbation theory. Since Hamiltonians H (0) generally have both non-degenerate and degenerate states we need to consider both types of perturbation theory. We begin with non-degenerate perturbation theory.

1.1

Nondegenerate perturbation theory

We begin by describing the original Hamiltonian H (0) . We assume this Hamiltonian has a discrete spectrum with an orthonormal basis |k(0) i of energy eigenstates, where k ∈ Z is a label that ranges over a possibly infinite set of values: (0)

H (0) |k(0) i = Ek |k(0) i ,

hk(0) |l(0) i = δkl .

(1.1.1)

We will let k = 0 denote the ground state and we order the states so that the energies generally increase as the value of the label increases, so that (0)

E0

(0)

≤ E1

(0)

≤ E2

(0)

≤ E3

≤ ...

(1.1.2)

The equal signs are needed because some states may be degenerate. In this section we focus on a non-degenerate state |n(0) i with fixed n. This means that |n(0) i is a single state that is separated by some finite energy from all the states with more energy and from all the states with less energy. In other words the following must be part of the sequence of inequalities in (1.1.2) (0)

(0)

. . . ≤ En−1 < En(0) < En+1 ≤ . . . (0)

(1.1.3) (0)

If the chosen state is the ground state, we have instead E0 < E1 . As the perturbation is turned on by making λ different from zero, the energy eigenstate |n(0) i of H (0) will evolve into some energy eigenstate |niλ of H(λ) with energy En (λ): H(λ)|niλ = En (λ) |niλ ,

(1.1.4)

4

CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY

where |niλ=0 = |n(0) i ,

and En (λ = 0) = En(0) .

(1.1.5)

As we said, the solution is assumed to take the form of a regular power series expansion in λ. To make this clear consider a function f (λ) such that its derivatives to all orders exist for λ = 0. In that case we have a Taylor expansion ∞ X 1 (n) f (0) λn = f (0) + f ′ (0) λ + 12 f ′′ (0) λ2 + f (λ) = n!

3 1 ′′′ 3! f (0) λ

n=0

+ ···

(1.1.6)

The expansion is a power series in λ, with coefficients f (0), f ′ (0), etc, that are λ independent and reflect the value of the function and its derivatives at λ = 0. For our problem we note the values of |niλ and En (λ) for λ = 0 (1.1.5) and write: |niλ = |n(0) i + λ|n(1) i + λ2 |n(2) i + λ3 |n(3) i + . . . , En (λ) = En(0) + λ En(1) + λ2 En(2) + λ3 En(3) + . . .

(1.1.7)

The superscripts on the states and energies denote the power of λ that accompanies them in the above expressions. The above equations are a natural assumption; they state that the perturbed states and energies, being functions of λ, admit a Taylor expansion around λ = 0. Our aim is to calculate the states |n(1) i , |n(2) i , |n(3) i , . . .

(1.1.8)

En(1) , En(2) , En(3) , . . .

(1.1.9)

and the energies Note that all these states and energies are, by definition, λ independent. Here |n(1) i is the (1) leading correction to the state |n(0) i as we turn on λ, and En is the leading correction to the energy as we turn on λ. We will not impose the requirement that |niλ is normalized. It suffices that |niλ is normalizable, which it will be for sufficiently small perturbations. For λ = 1 we would find the solution for H(1) = H (0) + δH in the form |ni ≡ |ni1 = |n(0) i + |n(1) i + |n(2) i + |n(3) i + . . . , En ≡ En (1) = En(0) + En(1) + En(2) + En(3) + . . .

(1.1.10)

Substituting the ansatz (1.1.7) into the Schr¨ odinger equation (1.1.4) we will find the conditions for such solution to exist:  H (0) + λδH − En (λ) |niλ = 0 , (1.1.11)

which more explicitly takes the form   (H (0) − En(0) ) − λ(En(1) − δH) − λ2 En(2) − λ3 En(3) − . . . − λk En(k) + . . .   |n(0) i + λ|n(1) i + λ2 |n(2) i + λ3 |n(3) i + . . . + λk |n(k) i + . . . = 0 .

(1.1.12)

1.1. NONDEGENERATE PERTURBATION THEORY

5

Multiplying out we get a series in λ with coefficients λ-independent vectors in the state space of the theory. If this is to vanish for all values of λ those coefficients must be zero. Collecting the coefficients for each power of λ we λ0 :

(H (0) − En(0) ) |n(0) i = 0 ,

λ1 :

(H (0) − En(0) ) |n(1) i = (En(1) − δH)|n(0) i ,

λ2 :

(H (0) − En(0) ) |n(2) i = (En(1) − δH)|n(1) i + En(2) |n(0) i ,

λ3 :

(H (0) − En(0) ) |n(3) i = (En(1) − δH)|n(2) i + En(2) |n(1) i + En(3) |n(0) i ,

.. . λk :

.. .

.. .

(H (0) − En(0) ) |n(k) i = (En(1) − δH)|n(k−1) i + En(2) |n(k−2) i + . . . + En(k) |n(0) i .

(1.1.13) Each equation is the condition that the coefficient multiplying the power of λ indicated to the left vanishes. That power is reflected as the sum of superscripts on each term, counting δH as having superscript one. This gives a simple consistency check on our equations. These (1) (2) are equations for the kets |n(1) i, |n(2) i, . . . as well as the energy corrections En , En , . . .. Note again that λ does not enter into the equations, and thus the kets and energy corrections are λ independent. The first equation, corresponding to λ0 , is satisfied by construction. The second equation, corresponding to λ1 , should allow us to solve for the first correction |n(1) i to the state (1) and the first correction En to the energy. Once these are known, the equation corre(2) sponding to λ2 involves only the unknowns |n(2) i and En , and should determine them. At each stage each equation has only two unknowns: a state correction |n(k) i and an energy (k) correction En . A useful choice. We now claim that without loss of generality we can assume that all the state corrections |nk i, with k ≥ 1 contain no vector along |n(0) i. Explicitly: 0 = hn(0) |n(1) i = hn(0) |n(2) i = hn(0) |n(3) i = . . . .

(1.1.14)

To show this we explain how we can manipulate a solution that does not have this property into one that does. Suppose you have solution in which the state corrections |n(k) i have components along |n(0) i: |n(k) i = |n(k) i′ − ak |n(0) i,

k ≥ 1,

(1.1.15)

with some constants ak and with |n(k) i′ orthogonal to |n(0) i. Then the solution for the full

6

CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY

corrected state is   |niλ = |n(0) i + λ |n(1) i′ − a1 |n(0) i + λ2 |n(2) i′ − a2 |n(0) i + . . .  = 1 − a1 λ − a2 λ2 − . . . |n(0) i + λ|n(1) i′ + λ2 |n(2) i′ + . . .

(1.1.16)

Since this is an eigenstate of the Hamiltonian H(λ), it will still be an eigenstate if we change its normalization by dividing it by any function of λ. Dividing by the coefficient of |n(0) i we have the physically identical solution |ni′λ given by |ni′λ = |n(0) i +

  1  λ|n(1) i′ + λ2 |n(2) i′ + . . . 1 − a1 λ − a2 λ2 − . . .

(1.1.17)

We can expand the denominator so that we get

 |ni′λ = |n(0) i + λ|n(1) i′ + λ2 |n(2) i′ + a1 |n(1) i′ + . . .

(1.1.18)

The explicit expressions do not matter, the key point, actually visible in (1.1.17), is that we have a physically identical solution of the same equation in which the state corrections are all orthogonal to |n(0) i. This shows that we can impose the conditions (1.1.14) without loss of generality. Solving the equations. Let us finally begin solving equations (1.1.13). For this we note that the Schrodinger equation for the ket |n(0) i implies that for the bra we have hn(0) |(H (0) − En(0) ) = 0 .

(1.1.19)

This means that acting with hn(0) | on the left-hand side of any of the equations in (1.1.13) will give zero. Consistency requires that acting with hn(0) | on the right-hand side of any of the equations in (1.1.13) also give zero, and presumably some interesting information. For the λ-equation this gives: 0 = hn(0) |(En(1) − δH)|n(0) i . (1.1.20) (1)

Since |n(0) i is normalized and En is a number, this means that En(1) = hn(0) |δH|n(0) i .

(1.1.21)

This is the most famous result in perturbation theory: the first correction to the energy of an energy eigenstate is simply the expectation value of the correction to the Hamiltonian in the uncorrected state. You need not know the correction to the state to determine the first correction to the energy! Note that the hermicity of δH implies the required reality of the energy correction. We can actually find some interesting formulae (but not yet fully explicit!) for the higher energy corrections. For the λ2 equation, acting with hn(0) | on the right-hand side gives   (1.1.22) 0 = hn(0) | (En(1) − δH)|n(1) i + En(2) |n(0) i .

1.1. NONDEGENERATE PERTURBATION THEORY

7 (1)

Recalling our orthogonality assumption, we have hn(0) |n(1) i = 0 and the term with En drops out. We get En(2) = hn(0) |δH|n(1) i , (1.1.23) which states that the second correction to the energy is determined if we have the first correction |n(1) i to the state. Note that this expression is not explicit enough to make it (2) manifest that En is real. This and the earlier result for the first correction to the energy have a simple generalization. Acting with hn(0) | on the last equation of (1.1.13) we get   0 = hn(0) | (En(1) − δH)|n(k−1) i + En(2) |n(k−2) i + . . . + En(k) |n(0) i . (1.1.24) Using the orthogonality of |n(0) i and all the state corrections, we have 0 = −hn(0) |δH)|nk−1 i + En(k) ,

(1.1.25)

En(k) = hn(0) | δH |n(k−1) i .

(1.1.26)

and therefore we have

At any stage of the recursive solution, the energy at a fixed order is known if the state correction is known to previous order. So it is time to calculate the corrections to the states! Let us solve for the first correction |n(1) i to the state. This state must be some particular superposition of the original energy eigenstates |k(0) i. For this we look at the equation (H (0) − En(0) ) |n(1) i = (En(1) − δH)|n(0) i .

(1.1.27)

This is a vector equation: the left-hand side vector set equal to the right-hand side vector. As in any vector equation, we can check it using a basis set of vectors. Forming the inner product of each and every basis vector with both the left-hand side and the right-hand side, we must get equal numbers. We already acted on the above equation with hn(0) | to figure (1) out En . The remaining information in this equation can be obtained by acting with all the states hk(0) | with k 6= n: hk(0) |(H (0) − En(0) ) |n(1) i = hk(0) |(En(1) − δH)|n(0) i .

(1.1.28)

On the left-hand side we can let H (0) act on the bra. On the right-hand side we note that (1) with k 6= n the term with En vanishes (0)

(Ek − En(0) ) hk(0) | n(1) i = −hk(0) |δH|n(0) i .

(1.1.29)

To simplify notation we define the matrix elements of δH in the original basis δHmn ≡ hm(0) |δH|n(0) i .

(1.1.30)

8

CHAPTER 1. TIME INDEPENDENT PERTURBATION THEORY

Note that the Hermiticity of δH implies that δHnm = (δHmn )∗ .

(1.1.31)

With this notation, equation (1.1.29) gives hk(0) | n(1) i = −

δHkn (0) Ek

(0)

− En

,

k 6= n .

(1.1.32)

Since we now know the overlap of |n(1) i with all basis states, this means that the state has been determined. Indeed we can use the completeness of the basis to write X X |n(1) i = |k(0) ihk(0) |n(1) i = |k(0) ihk(0) |n(1) i , (1.1.33) k

k6=n

since the term with k = n does not contribute because of the orthogonality assumption. Using the overlaps (1.1.32) we now get |n(1) i = −

X |k(0) iδHkn (0)

k6=n

(0)

Ek − En

.

(1.1.34)

This shows that the first correction |n(1) i can have components along all basis states, except |n(0) i. The component along a state |k(0) i vanishes if the perturbation δH does not couple |n(0) i to |k(0) i, namely, if δHkn vanishes. Note that the assumption of non-degeneracy is needed here. We are summing over all states |k(0) i that are not |n(0) i and if any such state has the same H (0) energy as |n(0) i the energy denominator will vanish causing trouble! Now that we have the first order correction to the states we can compute the second order correction to the energy. Using (1.1.23) we have X hn(0) |δH|k(0) iδHkn . (1.1.35) En(2) = hn(0) |δH|n(1) i = − (0) (0) Ek − En k6=n

In the last numerator we have hn(0) |δH|k(0) i = δHnk = (δHkn )∗ and therefore X |δHkn |2 En(2) = − . (0) (0) E − E n k6=n k

(1.1.36) (2)

This is the second-order energy correction. This explicit formula makes the reality of En manifest.

In summary, going back to (1.1.7), we have that the states and energies for H(λ) = H (0) + λδH are, to this order, |niλ = |n(0) i − λ En (λ) =

En(0)

X

δHkn

(0) k6=n Ek

(0)

− En

+ λ δHnn − λ

2

X

k6=n

|k(0) i + O(λ2 ) , |δHkn |2

(0)

(0)

Ek − En

(1.1.37) 3

+ O(λ ) ,

9

1.1. NONDEGENERATE PERTURBATION THEORY Remarks:

1. The first order corrected energy of the (non-degenerate) ground state overstates the true exact ground state energy. To see this consider the first order corrected ground (0) (1) state energy E0 + λE0 . Writing this in terms of expectation values, with |0(0) i denoting the unperturbed ground state, we have (0)

(1)

= h0(0) |H (0) |0(0) i + λh0(0) |δH|0(0) i

E0 + λE0

= h0(0) |(H (0) + λδH)|0(0) i

(1.1.38)

= h0(0) |H(λ)|0(0) i .

By the variational principle, the expectation value of the Hamiltonian on an arbitrary (normalized) state is larger than the ground state energy E0 (λ), therefore (0)

(1)

E0 + λE0

= h0(0) |H(λ)|0(0) i ≥ E0 (λ) ,

(1.1.39)

which is what we wanted to prove. Given this overestimate at first order, the second order correction to the ground state energy is always negative. Indeed, − λ2

X k6=0

|δHk0 |2

(0)

(0)

Ek − E0

,

(1.1.40) (0)

and each term is negative because the unperturbed excited state energies Ek (k 6= 0) (0) exceed the unperturbed ground state energy E0 . 2. The second order correction to the energy of the |n(0) i eigenstate exhibits level repulsion: the levels with k > n push the state down and the levels with k < n push the state up. Indeed, − λ2

X

k6=n

|δHkn |2

(0)

(0)

Ek − En

= − λ2

X

k>n

|δHkn |2

(0)

(0)

Ek − En

+ λ2

X

k 0 all energy shifts are down. Indeed n j+

1 2

n jmax +

1 2

=

n ℓmax + 12 +

1 2

=

n =1 n

n j+

1 2

3 4

1 4

.

(2.4.64)

5. For a given fixed n, states with lower values of j get pushed further down. As n increases splittings fall off like n−3 . A table of values of Sn,j is given here below n

j

Sn,j

1 2

1 2 1 2

1 8 5 128

3

3 2 1 2

1 128 1 72

3 2

1 216

5 2

1 648

The energy diagram for states up to n = 3 is given here (not to scale)

.. . n=3

S ℓ=0 .. .

P ℓ=1 .. .

3S1/2

D ℓ=2 .. .

3P3/2 3P1/2

3D5/2 3D3/2

Fine Structure Spectrum

n=2

2S1/2

n=1

1S1/2

2P3/2 2P1/2

45

2.5. ZEEMAN EFFECT

For the record, the total energy of the hydrogen states is the zeroth contribution plus the fine structure contribution. Together they give Enℓjmj = −

e2 1 h α2  n 1 + 2a0 n2 n2 j +

1 2

3 4

i

.

(2.4.65)

This is the fine structure of hydrogen! There are, of course, finer corrections. The socalled Lamb shift, for example, breaks the degeneracy between 2S1/2 and 2P1/2 and is of order α5 . There is also hyperfine splitting, which arises from the coupling of the magnetic moment of the proton to the magnetic moment of the electron. Such coupling leads to a splitting that is a factor me /mp smaller than fine structure.

2.5

Zeeman effect

In remarkable experiment done in 1896, the Dutch physicist Pieter Zeeman (1865-1943) discovered that atomic spectral lines are split in the presence of an external magnetic field. For this work Zeeman was awarded the Nobel Prize in 1902. The proper understanding of this phenomenon had to wait for Quantum Mechanics. The splitting of atomic energy levels by a constant, uniform, external magnetic field, the Zeeman effect, has been used as a tool to measure inaccessible magnetic fields. In observing the solar spectrum, a single atomic line, as seen from light emerging from outside a sunspot, splits into various lines inside the sunspot. We have learned that magnetic fields inside a sunspot typically reach 3, 000 gauss. Sunspots are a bit darker and have lower temperature than the rest of the solar surface. They can last from hours to months, and their magnetic energy can turn into powerful solar flares. The external magnetic field interacts with the total magnetic moment of the electron. The electron has magnetic moment due to its orbital angular momentum and one due to its spin e e L , µs = − S, (2.5.1) µℓ = − 2mc mc where we included the g = 2 factor in the spin contribution. The Zeeman Hamiltonian is thus given by e (L + 2S) · B . (2.5.2) δHZeeman = −(µℓ + µs ) · B = 2mc Conventionally, we align the magnetic field with the positive z axis so that B = B z and thus get eB ˆ δHZeeman = (Lz + 2Sˆz ) . (2.5.3) 2mc When we consider the Zeeman effect on Hydrogen we must not forget fine structure. The full Hamiltonian to be considered is H = H (0) + δHfs + δHZeeman .

(2.5.4)

46

CHAPTER 2. HYDROGEN ATOM FINE STRUCTURE

Recall that in fine structure, there is an internal magnetic field Bint associated with spinorbit coupling. This is the magnetic field seen by the electron as it goes around the proton. We have therefore two extreme possibilities concerning the external magnetic field B of the Zeeman effect: (1) Weak-field Zeeman effect: B ≪ Bint . In this case the Zeeman effect is small compared with fine structure effects. Accordingly, the original Hamiltonian H (0) together with ˜ (0) , and the fine structure Hamiltonian Hfs are thought as the “known” Hamiltonian H the Zeeman Hamiltonian is the perturbation: H = H (0) + δHfs +δHZeeman . {z } |

(2.5.5)

H = H (0) + δHZeeman +δHfs . {z } |

(2.5.6)

˜ (0) H

(2) Strong-field Zeeman effect: B ≫ Bint . In this case the Zeeman effect is much larger than fine structure effects. Accordingly, the original Hamiltonian H (0) together with ˇ (0) and the fine the Zeeman Hamiltonian are thought as the “known” Hamiltonian H structure Hamiltonian Hfs is viewed as the perturbation:

ˇ (0) H

You may thing that H (0) + δHZeeman does not qualify as known, but happily, as we will confirm soon, this is actually a very simple Hamiltonian. When the Zeeman magnetic field is neither weak nor strong, we must take the sum of the Zeeman and fine structure Hamiltonians as the perturbation. No simplification is possible and one must diagonalize the perturbation. ˜ (0) are the coupled states Weak-field Zeeman effect. The approximate eigenstates of H |nℓjmj i that exhibit fine structure corrections and whose energies are a function of n and j, as shown in the Fine Structure diagram. Degeneracies in this spectrum occur for different values of ℓ and different values of mj . To figure out the effect of the Zeeman interaction on this spectrum we consider the matrix elements: hnℓjmj |δHZeeman |nℓ′ jm′j i . (2.5.7) Since δHZeeman ∼ Lz + 2Sz we see that δHZeeman commutes with L2 and with Jˆz . The matrix element thus vanishes unless ℓ′ = ℓ and m′j = mj and the Zeeman perturbation is diagonal in the degenerate fine structure eigenspaces. The energy corrections are therefore (1)

Enℓjmj =

e~ ˆ z + 2Sˆz )|nℓjmj i 1 , B hnℓjmj |(L 2mc ~

(2.5.8)

where we multiplied and divided by ~ to make the units of the result manifest. The result of the evaluation of the matrix element will show a remarkable feature: a linear dependence

47

2.5. ZEEMAN EFFECT

E (1) ∼ ~mj on the azimuthal quantum numbers. The states in each j multiplet split into equally separated energy levels! We will try to understand this result as a property of ˆ z + 2Sˆz = Jˆz + Sˆz and matrix elements of vector operators. First, however, note that L therefore the matrix element of interest in the above equation satisfies ˆ z + 2Sˆz )|nℓjmj i = ~m + hnℓjmj |Sˆz |nℓjmj i . hnℓjmj |(L

(2.5.9)

It follows that we only need to concern ourselves with Sˆz matrix elements. ˆ is said to be a vector operator Let’s talk about vector operators. The operator V ˆ if the following commutator holds for all values of under an angular momentum operator J i, j = 1, 2, 3:   Jˆi , Vˆj = i~ ǫijk Vˆk . (2.5.10) ˆ commutators that J ˆ is a vector operator under J. ˆ Additionally, It follows from the familiar J ˆ is a vector operator it has a standard commutation relation with J ˆ 2 that can be quickly if V confirmed:  ˆ 2 , V] ˆ = 2i~ V ˆ ×J ˆ − i~ V ˆ . [J (2.5.11)

ˆ is chosen to be J ˆ the left-hand side vanishes by the standard property of J ˆ 2 and the If V ˆ commutation relations can be written as J× ˆ J ˆ = i~J. ˆ right-hand side vanishes because the J Finally, by repeated use of the above identities you will show (homework) that the following formula holds  1 h ˆ2 ˆ2 ˆ i ˆ +V ˆ Jˆ2 . ˆ · J) ˆ J ˆ − 1 Jˆ2 V (2.5.12) J , [J , V] = (V 2 2 (2i~) ˆ 2 , Jˆz ) eigenstates |k; jmj i where k stands for other quantum number that bear Consider (J no relation to angular momentum. The matrix elements of the left-hand side of (2.5.12) on such eigenstates is necessarily zero: h i ˆ 2 , [J ˆ 2 , V] ˆ |k; jmj i = 0 , hk′ ; jm′j | J (2.5.13)

ˆ 2 gives the same as can be seen by expanding the outer commutator and noticing that J eigenvalue when acting on the bra and on the ket. Therefore the matrix elements of the right-hand side gives ˆ · J) ˆ J|k; ˆ jmj i = ~2 j(j + 1) hk′ ; jm′ |V|k; ˆ hk′ ; jm′j |(V jmj i , j

(2.5.14)

which implies that ′

hk ;

ˆ jm′j |V|k;

ˆ · J) ˆ J|k; ˆ jmj i hk′ ; jm′j |(V . jmj i = ~2 j(j + 1)

(2.5.15)

This is the main identity we wanted to establish. Using the less explicit notation h· · · i for the matrix elements we have found that ˆ ˆ ˆ ˆ = h(V · J) J i . (2.5.16) hVi ˆ2i hJ

48

CHAPTER 2. HYDROGEN ATOM FINE STRUCTURE

ˆ This is sometimes called the projection lemma: the matrix elements of a vector operator V ˆ onto J. ˆ Recall that the projection of a vector are those of the conventional projection of V 2 v along the vector j is (v · j)j/j . Let us now return to the question of interest; the computation of the the expectation ˆ is a vector operator under J ˆ we can use (2.5.15). Specializing value of Sˆz in (2.5.20). Since S to the z-component ˆ ·J ˆ |nℓ jmj i ~mj hnℓ jmj |S hnℓ jmj |Sˆz |nℓ jmj i = . ~2 j(j + 1)

(2.5.17)

We already see the appearance of the predicted ~mj factor. The matrix element in the ˆ ·J ˆ is a numerator is still to be calculated but it will introduce no mj dependence. In fact S ˆ scalar operator (it commutes with all Ji ) and therefore it is diagonal in mj . But even more is true; the expectation value of a scalar operator is in fact independent of mj ! We will not ˆ =J ˆ−S ˆ we have show this here, but will just confirm it by direct computation. Since L  ˆ2 + S ˆ2 − L ˆ2 , ˆ·J ˆ = 1 J (2.5.18) S 2 and therefore

hnℓ jmj |Sˆz |nℓ jmj i =

 ~mj j(j + 1) − ℓ(ℓ + 1) + 43 . 2j(j + 1)

(2.5.19)

Indeed, no further mj dependence has appeared. Back now to (2.5.20) we get 3  ˆ z + 2Sˆz )|nℓjmj i = ~mj 1 + j(j + 1) − ℓ(ℓ + 1) + 4 . hnℓjmj |(L 2j(j + 1)

(2.5.20)

The constant of proportionality in parenthesis is called the Lande g-factor gJ (ℓ): gJ (ℓ) ≡ 1 +

j(j + 1) − ℓ(ℓ + 1) + 2j(j + 1)

3 4

(2.5.21)

We finally have for the Zeeman energy shifts in (2.5.8) (1)

Enℓjmj =

e~ B gJ (ℓ) mj . 2mc

(2.5.22)

e~ Here the Bohr magneton 2mc ≃ 5.79 × 10−9 eV/gauss. This is our final result for the weakfield Zeeman energy corrections to the fine structure energy levels. Since all degeneracies within j multiplets are broken and j multiplets with different ℓ split differently due to the ℓ dependence of gJ (ℓ), the weak-field Zeeman effect removes all degeneracies!

Strong-field Zeeman effect. We mentioned earlier that when the Zeeman effect is larger than the fine structure corrections we must take the original hydrogen Hamiltonian together ˇ (0) : with the Zeeman Hamiltonian to form the ‘known’ Hamiltonian H ˆ z + 2Sˆz )B . ˇ (0) = H (0) + e (L (2.5.23) H 2mc

49

2.5. ZEEMAN EFFECT

ˇ (0) is simple because the Zeeman Hamiltonian commutes with the zero-th order Actually H hydrogen Hamiltonian   ˆ z + 2Sˆz , H (0) = 0 . L (2.5.24)

We can thus find eigenstates of both simultaneously. Those are in fact the uncoupled basis states! We have H (0) |nℓmℓ ms i = En(0) |nℓmℓ ms i ˆ z + 2Sˆz )|nℓmℓ ms i = ~(mℓ + 2ms )|nℓmℓ ms i . (L

(2.5.25)

ˇ (0) and have and therefore the uncoupled basis states are the exact energy eigenstates of H energies e~ Enℓmℓ ms = En(0) + B(mℓ + 2ms ) . (2.5.26) 2mc Some of the degeneracy of H (0) has been removed, but some remains. For a fixed principal quantum number n there are degeneracies among ℓ ⊗ 21 states and degeneracies among such multiplets with ℓ and ℓ′ different. This is illustrated in Figure 2.2.

Figure 2.2: Illustrating the degeneracies remaining for ℓ = 0 and ℓ = 1 after the inclusion of Zeeman term in the Hamiltonian. Accounting for the spin of the electron there two degenerate states in the ℓ = 1 multiplet 1 ⊗ 21 and each of the two states in the ℓ = 0 multiplet 0 ⊗ 12 is degenerate with a state in the ℓ = 1 multiplet.

The problem now is to compute the corrections due to δHfs on the non-degenerate and ˇ (0) . The non-degenerate cases are straightforward, but on the degenerate subspaces of H the degenerate cases could involve diagonalization. We must therefore consider the matrix elements (2.5.27) hnℓ′ m′ℓ m′s | δHfs |nℓmℓ ms i , with the condition m′ℓ + 2m′s = mℓ + 2ms ,

(2.5.28)

50

CHAPTER 2. HYDROGEN ATOM FINE STRUCTURE

needed for the two states in the matrix element to belong to a degenerate subspace. Since ˆ 2 commutes with δHfs the matrix elements vanish unless ℓ = ℓ′ and therefore it suffices to L consider the matrix elements hnℓ′ m′ℓ m′s | δHfs |nℓmℓ ms i ,

(2.5.29)

still with condition (2.5.28). Ignoring ℓ = 0 states, we have to re-examine the relativistic correction and spin orbit. The relativistic correction was computed in the uncoupled basis and one can use the result because the states are unchanged and the perturbation was shown to be diagonal in this basis. For spin-orbit the calculation was done in the coupled basis because spin-orbit is not diagonal in the original H (0) degenerate spaces using the uncoupled basis. But happily, it turns out that spin-orbit is diagonal in the more limited degenerate subspaces obtained after the Zeeman effect is included. All these matters are explored in the homework.

MIT OpenCourseWare https://ocw.mit.edu

8.06 Quantum Physics III

Spring 2018

Chapter 3

Semiclassical approximation c B. Zwiebach

3.1

The classical limit

The WKB approximation provides approximate solutions for linear differential equations with coefficients that have slow spatial variation. The acronym WKB stands for Wentzel, Kramers, Brillouin, who independently discovered it in 1926. It was in fact discovered earlier, in 1923 by the mathematician Jeffreys. When applied to quantum mechanics, it is called the semi-classical approximation, since classical physics then illuminates the main features of the quantum wavefunction. The de Broglie wavelength λ of a particle can help us assess if classical physics is relevant to the physical situation. For a particle with momentum p we have λ=

h . p

(3.1.1)

Classical physics provides useful physical insight when λ is much smaller than the relevant length scale in the problem we are investigating. Alternatively if we take h → 0 this will formally make λ → 0, or λ smaller than the length scale of the problem. Being a constant of nature, we cannot really make ~ → 0, so taking this limit is a thought experiment in which we imagine worlds in which h takes smaller and smaller values making classical physics more and more applicable. The semi-classical approximation studied here will be applicable if a suitable generalization of the de Broglie wavelength discussed below is small and slowly varying. The semi-classical approximation will be set up mathematically by thinking of ~ as a parameter that can be taken to be as small as desired. Our discussion in this chapter will focus on one-dimensional problems. Consider therefore a particle of mass m and total energy E moving in a potential V (x). In classical physics E − V (x) is the kinetic energy of the particle at x. This kinetic energy depends on position. 51

52 Since kinetic energy is

CHAPTER 3. WKB AND SEMICLASSICAL APPROXIMATION p2 2m

this suggests the definition of the local momentum p(x): p2 (x) ≡ 2m(E − V (x)) .

(3.1.2)

The local momentum p(x) is the momentum of the classical particle when it is located at x. With a notion of local momentum, we can define a local de Broglie wavelength λ(x) by the familiar relation: h 2π~ λ(x) ≡ (3.1.3) = . p(x) p(x) The time-independent Schr¨ odinger equation −

~2 ∂ 2 ψ(x) = (E − V (x))ψ(x) . 2m ∂x2

(3.1.4)

can be written nicely in terms of the local momentum squared : −~2

∂2 ψ = p2 (x) ψ . ∂x2

(3.1.5)

Using the momentum operator, this equation takes the suggestive form pˆ2 ψ(x) = p2 (x) ψ(x) .

(3.1.6)

This has the flavor of an eigenvalue equation but it is not one: the action of the momentum operator squared on the wavefunction is not really proportional to the wavefunction, it is the wavefunction multiplied by the ‘classical’ position-dependent momentum-squared. A bit of extra notation is useful. If we are in the classically allowed region, E > V (x) and p2 (x) is positive. We then write p2 (x) = 2m(E − V (x)) = ~2 k 2 (x) ,

(3.1.7)

introducing the local, real wavenumber k(x). If we are in the classically forbidden region, V (x) > E and p2 (x) is negative. We then write −p2 (x) = 2m(V (x) − E) = ~2 κ2 (x) ,

(3.1.8)

introducing the local, real κ(x). The wavefunctions we use in the WKB approximation are often expressed in polar form. Just like any complex number z can be written as reiθ , where r and θ are the magnitude and phase of z, respectively, we can write the wavefunction in a similar way: Ψ(x, t) =

i  p ρ(x, t) exp S(x, t) . ~

(3.1.9)

3.2. WKB APPROXIMATION SCHEME

53

We are using here three-dimensional notation for generality. By definition, the functions ρ(x, t) and S(x, t) are real. The function ρ is non-negative and the function S(x, t) as written, has units of ~. The name ρ(x, t) is well motivated, for it is in fact the probability density: ρ(x, t) = |Ψ(x, t)|2 . (3.1.10) Let’s compute the probability current. For this we begin by taking the gradient of the wavefunction 1 ∇ρ iS i ∇Ψ = √ e ~ + ∇S Ψ (3.1.11) 2 ρ ~ We then form:

1 i Ψ∗ ∇Ψ = ∇ρ + ρ∇S 2 ~

(3.1.12)

The current is given by J=

~ Im (Ψ∗ ∇Ψ) . m

(3.1.13)

It follows that J = ρ

∇S . m

(3.1.14)

This formula implies that the probability current J is perpendicular to the surfaces of constant S, the surfaces of constant phase in the wavefunction. In classical physics a fluid with density ρ(x) moving with velocity v(x) has a current p density ρv = ρ m . Comparing with the above expression for the quantum probability current, we deduce that p(x) ' ∇S . (3.1.15) We use the ' because this is an association that is not of general validity. Nevertheless, for ‘basic’ WKB solutions we will see that the gradient of S is indeed the classical local momentum! This association also holds for a free particle: Example. Consider a free particle with momentum p and energy E. Its wavefunction is   ip · x iEt Ψ(x, t) = exp − . (3.1.16) ~ ~ Here we identify the function S as S = p · x − Et and therefore ∇S = p. In this case ∇S is equal to the momentum eigenvalue, a constant.

3.2

WKB approximation scheme

Our aim here is to find approximate solutions for the wavefunction ψ(x) that solves the ¨ time independent Schrdinger equation in one dimension. We wrote before the polar decomposition (3.1.9) of the wavefunction. It will be easier to set the approximation scheme by

54

CHAPTER 3. WKB AND SEMICLASSICAL APPROXIMATION

using a single complex function S(x) to represent the time-independent wavefunction ψ(x). For this we use a pure exponential without any prefactor: i  ψ(x) = exp S(x) , ~

S(x) ∈ C .

(3.2.1)

As before S must have units of ~. This S(x) here is a complex number, because wavefunctions are not in general pure phases. The real part of S, divided by ~, is the phase of the wavefunction. The imaginary part of S(x) determines the magnitude of the wavefunction. Let us plug this into the Schr¨odinger equation (3.1.5): −~2

i d2  i S(x)  e~ = p2 (x)e ~ S(x) . 2 dx

Let us examine the left-hand side and take the two derivatives   00    2  i i i d iS i 0 (S 0 )2 2 2 d S(x) S(x) 2 = −~ −~ = −~ e ~ S(x) . e~ S (x)e ~ − 2 2 dx dx ~ ~ ~ Back into the differential equation and canceling the common exponential  00  iS (S 0 )2 −~2 − 2 = p2 (x) . ~ ~

(3.2.2)

(3.2.3)

(3.2.4)

With minor rearrangements we get our final form: (S 0 (x))2 − i~S 00 (x) = p2 (x) .

(3.2.5)

The presence of an explicit iin the equation tells us that the solution for S, as expected, cannot be real. At first sight one may be baffled: we started with the linear Schr¨ odinger equation for ψ and obtained a nonlinear equation for S! This is actually unavoidable when the variable ψ is parameterized as an exponential and reminds us that sometimes a change of variables can turn a linear equation into a nonlinear one and viceversa. The nonlinear equation (3.2.5) allows us set up an approximation scheme in which ~ is considered small and thus the term involving S 00 (x) is small. We will argue that this is in fact true for slowly varying potentials. Claim: i~S 00 small if V (x) is slowly varying. Indeed, if V (x) = V0 is a constant, then the local momentum p(x) is equal to a constant p0 . Equation (3.2.5) is then solved by taking S 0 = p0 . For this choice S 00 = 0 and the term i~S 00 vanishes identically for constant V . It should therefore be small for slowly varying V (x). Alternatively the term i~S 00 in the differential equation is small as ~ → 0, which makes the local de Broglie wavelength go to zero. In that case, the potential looks constant to the quantum particle.

3.2. WKB APPROXIMATION SCHEME

55

We will thus take ~ to be the small parameter in a systematic expansion of S(x): S(x) = S0 (x) + ~S1 (x) + ~2 S2 (x) + O(~3 ) .

(3.2.6)

Here S0 , just like S, has units of ~. The next correction S1 has no units, and the following, S2 has units of one over ~. Now plug this expansion into our nonlinear equation (3.2.5) S00 + ~S10 + ~2 S20 + . . .

2

− i~ S000 + ~S100 + ~2 S200 + . . .



− p2 (x) = 0 .

(3.2.7)

The left-hand side is a power series expansion in ~. Just as we argued for the parameter λ in perturbation theory, here we want the left-hand side to vanish for all values of ~ and this requires that the coefficient of each power of ~ vanishes. We thus sort the left-hand side terms neglecting terms of order ~2 or higher. We find   (S00 )2 − p2 + ~ 2S00 S10 − iS000 + O ~2 = 0

(3.2.8)

This gives two equations, one for the coefficient of (~)0 and another for the coefficient of ~: (S00 )2 − p2 (x) = 0 ,

(3.2.9)

2S00 S10 − iS000 = 0 . The first equation is easily solved: S00 = ±p(x)

Z →

x

S0 (x) = ±

p(x0 )dx0

(3.2.10)

x0

where x0 is a constant of integration to be adjusted. The next equation allows us to find S1 which is in fact imaginary: S10 =

i S000 i (±p0 (x)) i p0 = = . 2 S00 2 (±p(x)) 2p

(3.2.11)

This is readily solved to give i S1 (x) = − 12 ln p(x) + C 0 . Let us now reconstruct the wavefunction to this order of approximation:     i i 2 ψ(x) = exp (S0 + ~S1 + O(~ )) ' exp S0 exp [iS1 ] ~ ~ Using our results for S0 and S1 we have that the approximate solution is   Z   i x 0 0 ψ(x) = exp ± p(x )dx exp − 12 log p(x) + C 0 . ~ x0

(3.2.12)

(3.2.13)

(3.2.14)

56

CHAPTER 3. WKB AND SEMICLASSICAL APPROXIMATION

We thus have   Z i x 0 0 p(x )dx . ψ(x) = p exp ± ~ x0 p(x) A

(3.2.15)

This is the basic solution in the WKB approximation. We do not attempt to normalize this wavefunction because, in fact, the region of validity of this approximation is still unclear. Observables for the basic solution: i) Probability density: ρ = ψ∗ψ =

|A|2 |A|2 = , p(x) mv(x)

(3.2.16)

where v(x) is the local classical velocity. Note that ρ is higher where v is small as the particle lingers in those regions and is more likely to be found there. This is an intuition we developed long ago and is justified by this result. ii) Probability current: In the language of the polar decomposition (3.1.9) of the wavefunction the basic solution corresponds to Z x S(x) = p(x0 )dx0 (3.2.17) x0

Note that, as anticipated below equation (3.1.15) the gradient of S in the basic solution is the local momentum. Recalling the result (3.1.14) for the current, J = ρ

1 ∂S , m ∂x

(3.2.18)

and therefore, J(x) =

|A|2 p(x) |A|2 = . p(x) m m

(3.2.19)

The fact that the current is a constant should not have taken us by surprise. A position-dependent current for an energy eigenstate is not possible, as it would violate the current conservation equation ∂x J(x) + ∂t ρ = 0, given that ρ is time independent. We con now write the general solutions that can be built from the basic solution and apply to classically allowed and classically forbidden regions. On the allowed region E − V (x) > 0 we write p2 (x) = ~2 k 2 (x), with k(x) > 0 and hence the general solution is a superposition of two basic solutions with waves propagating in opposite directions:  Z x   Z x  B A 0 0 0 0 exp i k(x )dx + p exp −i k(x )dx . (3.2.20) ψ(x) = p k(x) k(x) x0 x0

3.2. WKB APPROXIMATION SCHEME

57

Note that the first wave, with coefficient A moves to the right while the second wave, with coefficient B moves to the left. This can be seen by recalling that the above energy eigenstate ψ(x) is accompanied by the time factor e−iEt/~ when forming a full solution of the Schr¨odinger equation. Moreover, the first phase (accompanying A) grows as x grows, while the second phase becomes more negative as x grows. On the forbidden region p2 = −~2 κ2 (x) so we can take p(x) = iκ(x) in the solution, with κ > 0, to find  Z x   Z x C D 0 0 0 0 ψ(x) = p κ(x )dx (3.2.21) exp κ(x )dx + p exp − κ(x) κ(x) x0 x0 The argument of the first exponential becomes more positive as x grows. Thus the first term, with coefficient C, is an increasing function as x grows. The argument of the second exponential becomes more negative x grows. Thus the second term, with coefficient D, is a decreasing function as x grows.

3.2.1

Validity of the approximation

To understand more concretely the validity of our approximations, we reconsider the expansion (3.2.8) of the differential equation:   (S00 )2 − p2 + ~ 2S00 S10 − iS000 + O ~2 = 0 (3.2.22) We must have that the O (~) terms in the differential equation are much smaller, in magnitude, than the O (1) terms. At each of these orders we have two terms that are set equal to each other by the differential equations. It therefore suffices to check that one of the O (~) terms is much smaller than one of the O (1) terms. Thus, for example, we must have |~S00 S10 |  |S00 |2 .

(3.2.23)

Canceling one factor of |S00 | and recalling that |S00 | = |p| we have |~S10 |  |p| . From (3.2.11) we note that |S10 | ∼ |p0 /p| and therefore we get p0 ~  |p| . p There are two useful ways to think about this relation. First we write it as ~ dp dp  |p| → λ  |p| p dx dx

(3.2.24)

(3.2.25)

(3.2.26)

which tells us that the changes in the local momentum over a distance equal to the de Broglie wavelength are small compared to the momentum. Alternatively we write (3.2.25) as follows: p0 d 1 ~  1 → (3.2.27) 2 ~  1. p dx p

58

CHAPTER 3. WKB AND SEMICLASSICAL APPROXIMATION

This now means dλ  1. dx

(3.2.28)

The de Broglie wavelength must vary slowly. Note the consistency with units, the left-hand side of the inequality being unit free. More intuitive, perhaps, is the version obtained by multiplying the above by λ dλ λ  λ . (3.2.29) dx This tells us that the variation of the de Broglie wavelength λ over a distance λ must be much smaller that λ. It is not hard to figure out what the above constraints tell us about the rate of change of the potential. Finally connect to the potential. Taking one spatial derivative of the equation p2 = 2m(E − V (x)) we get dV dV 0 = 1 |pp0 | . |pp | = m → (3.2.30) dx dx m Multiplying by the absolute value of λ = h/p dV 2π~ 0 p2 = λ(x) |p |  , dx m m

(3.2.31)

where the last inequality follows from (3.2.25). Hence, we find that 2 dV  p . λ(x) dx 2m

(3.2.32)

The change in the potential over a distance equal to the de Broglie wavelength must be much smaller than the kinetic energy. This is the precise meaning of a slowly changing potential in the WKB approximation. The slow variation conditions needed for the basic WKB solutions to be accurate fail near turning points. This could be anticipated since at turning points the local momentum becomes zero and the de Broglie wavelength becomes infinite. Under general conditions, sufficiently near a turning point the potential V (x) is approximately linear, as shown in Figure 3.1. We then have V (x) − E = g(x − a) ,

g > 0.

(3.2.33)

In the allowed region x < a the local momentum is p2 = 2m((E − V (x)) = 2mg(a − x) .

(3.2.34)

3.3. CONNECTION FORMULA

59

Figure 3.1: The slow-varying conditions in WKB are violated at turning points. As a result, the de Broglie wavelength is given by λ(x) =

2π~ 2π~ √ = √ . p 2mg a − x

(3.2.35)

Taking a derivative, we find dλ π~ 1 . = √ dx 2mg (a − x)3/2

(3.2.36)

The right-hand side goes to infinity as x → a and therefore the key condition (3.2.28) is violated as we approach turning points. Our basic WKB solutions can be valid only as long as we remain away from turning points. If we have a turning point, such as x = a in the figure, we need a ‘connection formula’ that tells us how a solution far to the left and a solution far to the right of the turning point are related when they together form a single solution. We will consider a connection formula right below.

3.3

Connection formula

Let us first explain and then use a set of connection formulae, leaving their derivation for next section. The connection formulae refer to solutions away from a turning point x = a separating a classically allowed region to the left and a classically forbidden region to the right. This is the situation illustrated in Figure 3.2. For such turning point at x = a the WKB solutions to the right are exponentials that grow or decay and the WKB solutions to the left are oscillatory functions. They connect

60

CHAPTER 3. WKB AND SEMICLASSICAL APPROXIMATION

Figure 3.2: consistent with the following relations Z a   Z x  π 1 2 0 0 0 0 p cos k(x )dx − ⇐= p exp − κ(x )dx 4 k(x) κ(x) x a 1 −p sin k(x)

Z x

a

k(x0 )dx0 −

π 4



1 =⇒ p exp κ(x)

Z

x

κ(x0 )dx0

(3.3.1)

 (3.3.2)

a

In the first relation the arrow tells us that if the solution is a pure decaying exponential to the right of x = a the solution to the left of x = a is accurately determined and given by the phase-shifted cosine function the arrow points to. The second relation says that if the solution to the left of x = a is of an oscillatory type, the growing part of the solution to the right of x = a is accurately determined and given by the exponential the arrow points to. The decaying part cannot be reliably determined. As we will elaborate upon later, any connection formula is not to be used in the direction that goes against the arrow. Example: Find a quantization condition for the energies of bound states in a monotonically increasing potential V (x) that has a hard wall at x = 0. Assume V (x) increases without bound, as illustrated in Figure 3.3. Let E denote the energy of our searched-for eigenstate. Clearly, the energy and the potential V (x) determine the turning point x = a. The solution for x > a must only have a decaying exponential since the forbidden region extends forever to the right of x = a. The wavefunction for x > a is therefore of the type on the right-side of the connection formula (3.3.1). This means that we have an accurate representation of the wavefunction to the left of x = a. Adjusting the arbitrary normalization, we have Z a  1 0 0 π ψ(x) = p cos k(x )dx − 4 , 0 ≤ x  a . (3.3.3) k(x) x

3.3. CONNECTION FORMULA

61

Figure 3.3: A monotonically increasing potential with a hard wall at x = 0. For an energy eigenstate of energy E the turning point is at x = a.

Since we have a hard wall, the wavefunction must vanish for x = 0. The condition ψ(0) = 0 requires that Z a cos ∆ = 0 , with ∆ = k(x0 )dx0 − π4 . (3.3.4) 0

This is satisfied when Z ∆=

a

k(x0 )dx0 −

0

π 4

π 2

n ∈ Z.

(3.3.5)

π , n = 0, 1, 2, . . . .

(3.3.6)

=

+ nπ ,

The quantization condition is therefore Z 0

a

k(x0 )dx0 = n +

3 4



Negative integers are not allowed because the left-hand side is manifestly positive. The above is easy to use in practice. Using the expression for k(x) in terms of E and V (x) we have Z ar  2m (E − V (x0 )) dx0 = n + 34 π , n = 0, 1, 2, . . . (3.3.7) 2 ~ 0 In some special cases a can be calculated in terms of E and the integral can be done analytically. More generally, this is done numerically, exploring the value of the integral on the left hand side as a function of E and selecting the energies for which it takes the quantized values on the right-hand side.

62

CHAPTER 3. WKB AND SEMICLASSICAL APPROXIMATION Let us rewrite the wavefunction using Z

1

Ra x

=

a

0

Ra 0

− 0

Rx 0

: Z

π 4

ψ(x) = p cos k(x )dx − − k(x) 0   Z x 1 0 0 k(x )dx = p cos ∆ − k(x) 0 Z x  sin ∆ 0 0 = p sin k(x )dx , k(x) 0

x

0

k(x )dx

0



0

(3.3.8)

where we expanded the cosine of a sum of angles and recalled that cos ∆ = 0. In this form it is manifest that the wavefunction vanishes at x = 0. More interestingly, the quantization condition (3.3.6) indicates that the excursion of the phase in the solution from 0 to a is a bit higher than nπ (but less than (n + 1)π). Thus the WKB wavefuntion produces the n nodes the n-th excited state must have, even though this wavefunction is not reliable all to way to x = a.

3.4

Airy functions and connection formulae

(3.4.9)

V (x) − E ' g(x − a)

with g > 0 for x ∼ a

(3.4.10)

Solutions A

ψ(x) = p κ(x)

 Z exp −

x

0

κ(x )dx

0



B

Z

x

0

0



+p exp κ(x )dx κ(x) a  Z a   Z a  C D 0 0 0 0 ψ(x) = p exp i k(x )dx + p exp −i k(x )dx k(x) k(x) x x

xa

(3.4.11)

a

x  a (3.4.12)

3.4. AIRY FUNCTIONS AND CONNECTION FORMULAE

63

where k2 ≡

2m 2mg (E − V (x)) ' 2 (a − x) ~2 ~

2mg 2m (V (x) − E) ' 2 (x − a) 2 ~ ~ Idea: Solve Schrodinger equation κ2 ≡

x≤a

(3.4.13)

x≥a

(3.4.14)

~2 00 ψ + (V (x) − E)ψ = 0 2m with the exact linear potential, i.e. −

(3.4.15)

~2 00 ψ + g(x − a)ψ = 0 2m

(3.4.16)

• Solve (3.4.16) exactly. • Match it to the x  a version of (3.4.11) or x  a version of (3.4.12). • Find the required relations between coefficients! Solve (3.4.16). Remove units, find length scale. ~2 = gL =⇒ L = mL2 so define  u=

2mg ~2



~2 mg

 13 (3.4.17)

1 3

(x − a) ≡ η(x − a)

(3.4.18)

thus (3.4.16) becomes d2 ψ − uψ = 0 du2

(3.4.19)

ψ = αAi (u) + βBi (u)

(3.4.20)

with solution where α and β are constants and Ai (u), Bi (u) are Airy functions:  1 u  1,  12 √1π |u|− 4 e−ξ Ai (u) '  √1 |u|− 41 cos ξ − π  u  −1. 4 π

Bi (u) ' with

 1  √1π |u|− 4 eξ

(3.4.21)

u  1,

− √1 |u|− 14 sin ξ − π  4 π 3

ξ ≡ 32 |u| 2 .

(3.4.22) u  −1. (3.4.23)

These are in fact WKB solutions and special cases of connection conditions! (Explain)

64

CHAPTER 3. WKB AND SEMICLASSICAL APPROXIMATION

3.4.1

Take a WKB solution and find the DE it satisfies ψ(u) =

1 u

3

2

1 4

e− 3 |u| 2

(3.4.24)

solves d2 ψ  5 1 − u + ψ=0 du2 16 u2

(3.4.25)

so clearly for u  1 the extra term is negligible. To evaluate (3.4.11) and (3.4.12), determine κ(x) and k(x). Using (3.4.13) and (3.4.14):

Using

0

=

d dx ,

k 2 = η 3 (a − x) = −η 2 u = η 2 |u|

u0

(3.4.27)

the WKB condition is |k 0 |  |k 2 | From ... η3 η3 η2 = = 2|k| 2η|u| 2|u|

|2kk 0 | = η 3 =⇒ |k 0 | =

(3.4.28)

k 2 = η 2 |u| thus |k 0 |  k 2 ,

3 η2 1  η 2 |u| =⇒ |u| 2  =⇒ |u|  0.63 2|u| 2

(3.4.29)

Two integrals to estimate x

Z x>a:

0

0

Z

a

Z

a

x a we can connect it to a solution to the left of x < a using (3.3.1) Z a  2iF eθ 0 0 π cos k(x )dx − 4 . (3.5.7) ψ(x) = − p k(x) x This is a superposition of two waves, a wave ψinc incident on the barrier and a wave ψref reflected from the barrier. The incident part is  Z a  iF eθ 0 0 π ψinc (x) = − p exp −i k(x )dx + i 4 . (3.5.8) k(x) x The sign in front of the integral may seem unusual for a wave moving to the right, but it is correct because the argument x appears in the lower limit of the integration. The transmission coefficient T is the ratio of the transmitted probability current over the incident probability current. Given the result in (3.2.19) we have T =

|F |2 probability current for ψtr = = e−2θ . probability current for ψinc | − iF eθ |2

(3.5.9)

3.5. TUNNELLING THROUGH A BARRIER

69

This is the well-known exponential suppression of the transmission coefficient. Using the earlier definition of θ the result is  Z b  Twkb = exp −2 κ(x0 )dx0 . (3.5.10) a

We added the subscript ‘wkb’ to emphasize that this is the WKB approximation to the exact transmission coefficient. The integral extends in between the two turning points and captures information about the height and the width of the barrier. The integrand, as well as the turning points, depend on the energy as we can display by using the explicit value of κ(x) :  Z b r 2m  0 ) − E) dx0 . Twkb = exp −2 (V (x (3.5.11) ~2 a The WKB approximation only captures the exponentially decaying part of the transmission coefficient. There are corrections that are written as a prefactor to the exponential. These are not determined in this approximation. Example. Use the WKB approximation to estimate the transmission probability T for the rectangular barrier ( V0 for |x| < a , V (x) = (3.5.12) 0 , otherwise. Assume the barrier is large. In terms of the familiar unit-free constant z0 used to characterize square wells, this means that 2mV0 a2 z02 ≡  1. (3.5.13) ~2 Large barrier means large z0 or large V0 a2 . Moreover we assume E smaller than V0 ; we cannot expect the approximation to work as the energy approaches the top of the barrier. Compare the WKB result for T with the exact result. Solution. The WKB estimation, using (3.5.11) is immediate. Since the barrier extends from x = −a to x = a and the potential is constant we have   Z a r 2m 0 Twkb = exp −2 (V − E) dx . (3.5.14) 0 ~2 −a The integral can be done and we find Twkb

 4a p  = exp − 2m(V0 − E) . ~

This is the answer in the WKB approximation. In terms of z0 this reads r  E  Twkb ' exp −4z0 1 − . V0

(3.5.15)

(3.5.16)

70

CHAPTER 3. WKB AND SEMICLASSICAL APPROXIMATION

The validity of the WKB approximation requires the argument in the exponent to be large. This is why we need z0  1 and the energy not to approach V0 . When E is very small compared to V0 , we have T ∼ exp(−4z0 ) . (3.5.17) √ Since z0 ∼ a V0 , this result is the basis for the claim that the exponential suppression of the tunneling probability is proportional to the width of the well and the square root of its height. The exact formula for the tunneling probability of a square barrier has been calculated before, and it is given by  2a p  1 V02 = 1+ sinh2 2m(V0 − E) . T 4E(V0 − E) ~

(3.5.18)

This formula will allow us to confirm the WKB exponential suppression and to find the prefactor. Under the conditions z0  1 and V0 − E finite, the argument of the sinh function is large so this function can be replaced by its growing exponential (sinh x ∼ 12 ex ): 1 V0 ' 1+ T 16E(1 −

E V0 )

exp

  4a p 2m(V0 − E) , ~

(3.5.19)

where we also replaced V0 − E → V0 in the prefactor. The additive unit on the right-hand side can be neglected and we have T ' 16 VE0 (1 −

E V0 )

 4a p  exp − 2m(V0 − E) . ~

(3.5.20)

This result gives the same in exponential suppression as the WKB result (3.5.15). It also gives us a good approximation to the prefactor. The WKB approximation is often used for the estimation of lifetimes. The physical situation is represented in Figure 3.5. We have a particle of mass m and energy E localized between the turning points x = a and x = b of the potential. The classical particle cannot escape because of the energy barrier stretching from x = b to x = c. The quantum particle, however, can tunnel. The state of the quantum particle when localized between a and b is not strictly an energy eigenstate because the wavefunction for any energy eigenstate is nonzero for x > c. Our goal is to determine the lifetime τ of the particle. This is done by using some classical estimates as well as the quantum transmission probability. We first show that the lifetime τ is the inverse of B, the tunneling probability per unit time. To see this we examine the function P (t) that represents the probability to have the particle localized within [a, b] at time t, if it was there at t = 0. The lifetime is said to be τ if P (t) has the time dependence P (t) = e−t/τ . To see that we have such a situation when the tunneling rate B is constant note that  P (t + dt) = P (t) · probability that it does not tunnel in the next dt . (3.5.21)

3.5. TUNNELLING THROUGH A BARRIER

71

Figure 3.5: Estimating the lifetime of a particle with energy E temporarily localized in x ∈ [a, b]. Therefore  P (t + dt) = P (t) · 1 − Bdt

P (t + dt) − P (t) = −BP (t) dt .

(3.5.22)

We thus have the differential equation, and solution, dP = −B P (t) dt from which we identify, as claimed, that

P (t) = e−Bt ,

(3.5.23)

1 . (3.5.24) B The tunneling rate B or tunneling probability per unit time is estimated by counting the number of times nhit the classical particle, bouncing between x = a and x = b, hits the barrier at x = b per unit time, and multiplying this number by the tunneling probability T : τ =

T , ∆t where ∆t is the time the classical particle takes to go from b to a and back to b: Z b Z b dx dx ∆t = 2 = 2m . a v(x) a p(x) B = nhit T =

(3.5.25)

(3.5.26)

We now can put all the pieces together, using the WKB approximation for T . We find Z b  Z c  ∆t dx τ = ' 2m · exp 2 κ(x)dx . (3.5.27) T a p(x) b The smaller the tunneling probability, the larger the lifetime of the localized state. Note that the factor ∆t carries the units of the result. The above result can easily fail to be accurate, as the WKB approximation for T does not give the prefactor multiplying the exponential suppression.

MIT OpenCourseWare https://ocw.mit.edu

8.06 Quantum Physics III

Spring 2018

Chapter 4

Time Dependent Perturbation Theory c B. Zwiebach

4.1

Time dependent perturbations

We will assume that, as before, we have a Hamiltonian H (0) that is known and is time independent. Known means we know the spectrum of energy eigenstates and the energy eigenvalues. This time the perturbation to the Hamiltonian, denoted as δH(t) will be time dependent and, as a result, the full Hamiltonian H(t) is also time dependent H(t) = H (0) + δH(t) .

(4.1.1)

While H (0) has a well-defined spectrum, H(t) does not. Being time dependent, H(t) does not have energy eigenstates. It is important to remember that the existence of energy eigenstates was predicated on the factorization of solutions Ψ(x, t) of the full Schr¨odinger equation into a space-dependent part ψ(x) and a time dependent part that turned out to be e−iEt/~ , with E the energy. Such factorization is not possible when the Hamiltonian is time dependent. Since H(t) does not have energy eigenstates the goal is to find the solutions |Ψ(x, t)i directly. Since we are going to focus on the time dependence, we will suppress the labels associated with space. We simply say we are trying to find the solution |Ψ(t)i to the Schr¨odinger equation  ∂ i~ |Ψ(t)i = H (0) + δH(t) |Ψ(t)i . (4.1.2) ∂t In typical situations the perturbation δH(t) vanishes for t < t0 , it exists for some finite time, and then vanishes for t > tf (see Figure 4.1). The system starts in an eigenstate of H (0) at t < t0 or a linear combination thereof. We usually ask: What is the state of the system for t > tf ? Note that both initial and final states are nicely described in terms of eigenstates of H (0) since this is the Hamiltonian for t < t0 and t > tf . Even during the 73

74

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

Figure 4.1: Time dependent perturbations typically exist for some time interval, here from t0 to tf . time when the perturbation is on we can use the eigenstates of H (0) to describe the system, since these eigenstates form a complete basis, but the time dependence is very nontrivial. Many physical questions can be couched in this language. For example, assume we have a hydrogen atom in its ground state. We turn on EM fields for some time interval. We can then ask: What are the probabilities to find the atom in each of the various excited states after the perturbation turned off?

4.1.1

The interaction picture

In order to solve efficiently for the state |Ψ(t)i we will introduce the Interaction Picture of Quantum Mechanics. This picture uses some elements of the Heisenberg picture and some elements of the Schr¨ odinger picture. We will use the known Hamiltonian H (0) to define some Heisenberg operators and the perturbation δH will be used to write a Schr¨ odinger equation. We begin by recalling some facts from the Heisenberg picture. For any Hamiltonian, time dependent or not, one can determine the unitary operator U(t) that generates time evolution: |Ψ(t)i = U(t)|Ψ(0)i . (4.1.3) The Heisenberg operator AˆH associated with a Schr¨odinger operator Aˆs is obtained by considering a rewriting of expectation values: hΨ(t)|Aˆs |Ψ(t)i = hΨ(0)|U † (t)Aˆs U(t)|Ψ(0)i = hΨ(0)|AˆH |Ψ(0)i ,

(4.1.4)

AH ≡ U † (t)Aˆs U(t) .

(4.1.5)

where This definition applies even for time dependent Schr¨odinger operators. Note that the operator U † brings states to rest: U † (t)|Ψ(t)i = U † (t)U(t)|Ψ(0)i = |Ψ(0)i .

(4.1.6)

In our problem the known Hamiltonian H (0) is time independent and the associated unitary time evolution operator U0 (t) takes the simple form  iH (0) t  U0 (t) = exp − . ~

(4.1.7)

4.1. TIME DEPENDENT PERTURBATIONS

75

The state |Ψ(t)i in our problem evolves through the effects of H (0) plus δH. Motivated e by (4.1.6) we define the auxiliary ket Ψ(t) as the ket |Ψ(t)i partially brought to rest (0) through H :  iH (0) t  e Ψ(t) Ψ(t) . ≡ exp (4.1.8) ~ e Expect that Schr¨ odinger equation for Ψ(t) will be simpler, as the above must have taken (0) e care of the time dependence generated by H . Of cours, if we can determine Ψ(t) we can easily get back the desired state Ψ(t) inverting the above relation to find  (0)  e Ψ(t) = exp − iH t Ψ(t) . ~

(4.1.9)

e Our objective now is to find the Schr¨odinger equation for Ψ(t) . Taking the time derivative of (4.1.8) and using (4.1.2) i~

 iH (0) t  d e e Ψ(t) = −H (0) Ψ(t) (H (0) + δH(t)) Ψ(t) + exp dt ~ " #  iH (0) t   iH (0) t  (0) (0) e Ψ(t) = −H + exp (H + δH(t)) exp − ~ ~  iH (0) t   iH (0) t  e Ψ(t) = exp δH(t) exp − , (4.1.10) ~ ~

where the dependence on H (0) cancelled out. We have thus found the Schr¨odinger equation

i~

d e f Ψ(t) e . Ψ(t) = δH(t) dt

(4.1.11)

f where the operator δH(t) is defined as  iH (0) t   iH (0) t  f δH(t) ≡ exp δH(t) exp − . ~ ~

(4.1.12)

f e Note that as expected the time evolution left in Ψ(t) is generated by δH(t) via a Schr¨odinger f equation. The operator δH(t) is nothing else but the Heisenberg version of δH generated (0) using H ! This is an interaction picture, a mixture of Heisenberg’s and Schr¨odinger pic(0) ture. While we have some Heisenberg operators there is still a time dependent state e Ψ(t) and a Schr¨ odinger equation for it.

76

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

How does it all look in an orthonormal basis? Let |ni be the complete orthonormal basis of states for H (0) : H (0) |ni = En |ni . (4.1.13) This time there is no need for a (0) superscript, since neither the states nor their energies will be corrected (there are no energy eigenstates in the time-dependent theory). We then write an ansatz for our unknown ket: X e Ψ(t) = cn (t)|ni . (4.1.14) n

Here the functions cn (t) are unknown. This expansion is justified since the states |ni form a basis, and thus at all times they can be used to describe the state of the system. The original wavefunction is then:    (0)  X iEn t e Ψ(t) = exp − iH t Ψ(t) |ni . (4.1.15) = cn (t) exp − ~ ~ n (0) is present here. If we had δH = 0 the state Note that the time dependence due to H e Ψ(t) would be a constant, as demanded by the Schr¨odinger equation (4.1.11) and the cn (t) would be constants. The solution above would give the expected time evolution of the states under H (0) . To see what the Schr¨ odinger equation tells us about the functions cn (t) we plug the ansatz (4.1.14) into (4.1.11):

i~

X d X f cm (t)|mi = δH(t) cn (t)|ni dt m n

(4.1.16)

Using dots for time derivatives and introducing a resolution of the identity on the right-hand side we find X X X f i~c˙m (t)|mi = |mihm|δH(t) cn (t)|ni m

m

=

X

n

|mi

m

=

X

X

f hm|δH(t)|nic n (t)

n

f mn (t)cn (t)|mi . δH

(4.1.17)

m,n

Here we have used the familiar matrix element notation f mn (t) ≡ hm|δH(t)|ni f δH . Equating the coefficients of the basis kets |mi in (4.1.17) we get the equations X f mn (t) cn (t) . i~ c˙m (t) = δH n

(4.1.18)

(4.1.19)

4.1. TIME DEPENDENT PERTURBATIONS

77

The Schr¨odinger equation has become an infinite set of coupled first-order differential equations. The matrix elements in the equation can be simplified a bit by passing to un-tilde variables:  (0)   (0) 

f mn (t) = m exp iH t δH(t) exp − iH t n δH ~ ~   i = exp (Em − En )t hm|δH(t)|ni . ~ If we define ωmn ≡

Em − En , ~

(4.1.20)

(4.1.21)

we then have f mn (t) = eiωmn t δHmn (t) . δH

(4.1.22)

The coupled equations (4.1.19) for the functions cn (t) then become i~ c˙m (t) =

X

eiωmn t δHmn (t)cn (t) .

(4.1.23)

n

4.1.2

Example (based on Griffiths Problem 9.3)

Consider a two-state system with basis states |ai and |bi, eigenstates of H (0) with energies Ea and Eb , respectively. Call ωab ≡ (Ea − Eb )/~ . (4.1.24) Now take the perturbation to be a matrix times a delta function at time equal zero. Thus the perturbation only exists for time equal zero:   0 α δ(t) ≡ U δ(t) , (4.1.25) δH(t) = α∗ 0 where α is a complex number. With the basis vectors ordered as |1i = |ai and |2i = |bi we have   Uaa Uab ∗ U = , with Uaa = Ubb = 0 and Uab = Uba = α. (4.1.26) Uba Ubb There is a sudden violent perturbation at t = 0 with off-diagonal elements that should produce transition. Take the system to be in |ai for t = −∞, what is the probability that it is in |bi for t = +∞? Solution: First note that if the system is in |ai at t = −∞ it will remain in state |ai until t = 0− , that is, just before the perturbation turns on. This is because |ai is an energy eigenstate of H (0) . In fact we have, up to a constant phase, |Ψ(t)i = e−iEa t/~ |ai ,

for

− ∞ < t ≤ 0− .

(4.1.27)

78

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

We are asked what will be the probability to find the state in |bi at t = ∞, but in fact, the answer is the same as the probability to find the state in |bi at t = 0+ . This is because the perturbation does not exist anymore and if the state at t = 0+ is |Ψ(0+ )i = γa |ai + γb |bi ,

(4.1.28)

with γ1 and γ2 constants, then the state for any time t > 0 will be |Ψ(t)i = γa |aie−iEa t/~ + γb |bie−iEb t/~ . The probability pb (t) to find the state |bi at time t will be pb (t) = |hb|Ψ(t)i|2 = γb e−iEb t/~ = |γb |2 ,

(4.1.29)

(4.1.30)

and, as expected, this is time independent. It follows that to solve this problem we must just find the state at t = 0+ and determine the constants γ1 and γ2 . Since we have two basis states the unknown tilde state is e Ψ(t) = ca (t)|ai + cb (t)|bi , (4.1.31) and the initial conditions are stating that the system begins in the state |ai are ca (0− ) = 1 ,

cb (0− ) = 0 .

(4.1.32)

The differential equations (4.1.23) take the form i~ c˙a (t) = eiωab t δHab (t)cb (t) , i~ c˙b (t) = eiωba t δHba (t)ca (t) .

(4.1.33)

The couplings are off-diagonal because δHaa = δHbb = 0. Using the form of the δH matrix elements, i~ c˙a (t) = eiωab t α δ(t) cb (t) , i~ c˙b (t) = e−iωab t α∗ δ(t) ca (t) .

(4.1.34)

We know that for functions f continuous at t = 0 we have f (t)δ(t) = f (0)δ(t). We now ask if we are allowed to use such identity for the right-hand side of the above equations. In fact we can use the identity for e±iωab t but not for the functions ca (t) and cb (t) that, are expected to be discontinuous at t = 0. They must be so, because they can only change at t = 0, when the delta function exists. Evaluating the exponentials at t = 0 we then get the simpler equations i~ c˙a (t) = α δ(t) cb (t) , i~ c˙b (t) = α∗ δ(t) ca (t) .

(4.1.35)

4.1. TIME DEPENDENT PERTURBATIONS

79

With such singular right-hand sides, the solution of these equations needs regulation. We will regulate the delta function and solve the problem. A consistency check is that the solution has a well-defined limit as the regulator is removed. We will replace the δ(t) by the function ∆t0 (t), with t0 > 0, defined as follows: ( 1/t0 , for t ∈ [0, t0 ] , (4.1.36) δ(t) → ∆t0 (t) = 0, otherwise . R Note that as appropriate, dt∆t0 (t) = 1, and that ∆t0 (t) in the limit as t0 → 0 approaches a delta function. If we replace the delta functions in (4.1.37) by the regulator we find, for t ∈ [0, t0 ] α For t ∈ [0, t0 ] : i~ c˙a (t) = cb (t) , t0 (4.1.37) α∗ i~ c˙b (t) = ca (t) . t0 For any other times the right-hand side vanishes. By taking a time derivative of the first equation and using the second one we find that   |α| 2 1 α 1 α∗ ca (t) = − ca (t) . (4.1.38) c¨a (t) = i~ t0 i~ t0 ~t0 This is a simple second order differential equation and its general solution is     |α|t |α|t + β1 sin . ca (t) = β0 cos ~t0 ~t0

(4.1.39)

This is accompanied with cb (t) which is readily obtained from the first line in (4.1.37):  |α|t   |α|t  i~t0 i|α|  cb (t) = c˙a (t) = −β0 sin + β1 cos . (4.1.40) α α ~t0 ~t0 The initial conditions (4.1.32) tell us that in this regulated problem ca (0) = 1 and cb (0) = 0. Given the solutions above, the first condition fixes β0 = 1 and the second fixes β1 = 0. Thus, our solutions are:   |α|t For t ∈ [0, t0 ] : ca (t) = cos , (4.1.41) ~t0   |α| |α|t cb (t) = −i sin . (4.1.42) α ~t0 When the (regulated) perturbation is turned off (t = t0 ), ca and cb stop varying and the constant value they take is their values at t = t0 . Hence   |α| ca (t > t0 ) = ca (t0 ) = cos , (4.1.43) ~   |α| |α| cb (t > t0 ) = cb (t0 ) = −i sin . (4.1.44) α ~

80

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

Note that these values are t0 independent. Being regulator independent, we can safely take the limit t0 → 0 to get   |α| ca (t > 0) = cos , (4.1.45) ~   |α| |α| . (4.1.46) sin cb (t > 0) = −i α ~ Our calculation above shows that at t = 0+ the state will be     |α| |α| |α| + + e |Ψ(0 )i = |Ψ(0 )i = cos a − i sin b . ~ α ~ With only free evolution ensuing for t > 0 we have (as anticipated in (4.1.29))     |α| −iEa t/~ |α| |α| −iEb t/~ a e − i b e . |Ψ(t)i = cos sin ~ α ~

(4.1.47)

(4.1.48)

The probabilities to be found in |ai or in |bi are time-independent, we are in a superposition of energy eigenstates! We can now easily calculate the probability pb (t) to find the system in |bi for t > 0 as well as the probability pa (t) to find the system in |ai for t > 0:   |α| pb (t) = |hb|Ψ(t)i|2 = sin2 , ~ (4.1.49)   2 2 |α| pa (t) = |ha|Ψ(t)i| = cos . ~ Note that pa (t)+pb (t) = 1, as required. The above is the exact solution of this system. If we had worked in perturbation theory, we would be taking the strength |α| of the interaction to be small. The answers we would obtain would form the power series expansion of the above formulas in the limit as |α|/~ is small.

4.2

Perturbative solution

In order to set up the perturbative expansion properly we include a unit-free small parameter λ multiplying the perturbation δH in the time-dependent Hamiltonian (4.1.1): H(t) = H (0) + λδH(t) .

(4.2.1)

e With such a replacement the interaction picture Schr¨odinger equation for |Ψ(t)i in (4.1.11) now becomes d e f Ψ(t) e i~ Ψ(t) = λδH(t) . (4.2.2) dt e As we did in time-independent perturbation theory we start by expanding Ψ(t) in powers of the parameter λ: (0) (1) (2)  e e (t) + λ Ψ e (t) + λ2 Ψ e (t) + O λ3 . Ψ(t) = Ψ (4.2.3)

4.2. PERTURBATIVE SOLUTION

81

We now insert this into both sides of the Schr¨odinger equation (4.2.2), and using ∂t for time derivatives, we find (0) (1) (2) (2)  e (t) + λ i~∂t Ψ e (t) + λ2 i~∂t Ψ e (t) + λ3 i~∂t Ψ e (t) + O λ4 i~∂t Ψ (2) (1) (0)  f Ψ f Ψ f Ψ e (t) + O λ4 . e (t) + λ3 δH e (t) + λ2 δH = λ δH

(4.2.4)

The coefficient of each power of λ must vanish, giving us (0) e (t) = 0 , i~∂t Ψ (1) (0) f Ψ e (t) = δH e (t) , i~∂t Ψ (2) (1) f Ψ e (t) = δH e (t) , i~∂t Ψ .. .

=

(4.2.5)

.. .

(n) (n+1) f Ψ e (t) . e i~∂t Ψ (t) = δH The origin of the pattern is clear. Since the Schr¨odinger equation has an explicit λ multif plying the right-hand side, the time derivative of the n-th component is coupled to the δH perturbation acting on the (n − 1)-th component. Let us consider the initial condition in detail. We will assume that the perturbation turns on at t = 0, so that the initial condition is given in terms of |Ψ(0)i. Since our e Schr¨odinger equation is in terms of |Ψ(t)i we use the relation ((4.1.8)) between them to conclude that both tilde and un-tilde wavefunctions are equal at t = 0: e |Ψ(0)i = |Ψ(0)i . Given this, the expansion (4.2.3) evaluated at t = 0 implies that: (0) (1) (2)  e e (0) + λ Ψ e (0) + λ2 Ψ e (0) + O λ3 . Ψ(0) = |Ψ(0)i = Ψ

(4.2.6)

(4.2.7)

This must be viewed, again, as an equation that holds for all values of λ. As a result, the coefficient of each power of λ must vanish and we have (0) e (0) = Ψ(0) , Ψ (n) e (0) = 0 , Ψ n = 1, 2, 3 . . . .

(4.2.8)

These are the relevant initial conditions. (0) e (t) is time independent. Consider now the first equation in (4.2.5). It states that Ψ This is reasonable: if the perturbation vanishes, this is the only equation we get and we e e (0) (t) and the initial should expect Ψ(t) constant. Using the time-independence of Ψ condition we have (0) (0) e (t) = Ψ e (0) = Ψ(0) , Ψ (4.2.9)

82

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

and we have solved the first equation completely: (0) e Ψ = Ψ(0) .

(4.2.10)

Using this result, the O(λ) equation reads: (1) (0) f Ψ f Ψ(0) . e (t) = δH e (t) = δH(t) i~∂t Ψ

(4.2.11)

The solution can be written as an integral: t

Z

(1) e (t) = Ψ

0

f 0) δH(t Ψ(0) dt0 . i~

(4.2.12)

Note that by setting the lower limit of integration at t = 0 we have implemented correctly e (1) (0)i = 0. The next equation, of order λ2 reads: the initial condition |Ψ (2) (1) f Ψ e (t) = δH e (t) , i~∂t Ψ

(4.2.13)

and its solution is (2) e (t) = Ψ

Z 0

t

f 0) δH(t e (1) (t0 ) dt0 , Ψ i~

(4.2.14)

(1) 0 e (t ) we now consistent with the initial condition. Using our previous result to write Ψ have an iterated integral expression: (2) e (t) = Ψ

Z 0

t

f 00 ) f 0 ) Z t0 δH(t δH(t Ψ(0) dt00 . dt0 i~ i~ 0

(4.2.15)

It should (k) be clear from this discussion how to write the iterated integral expression for e (t) , with k > 2. The solution, setting λ = 1 and summarizing is then Ψ !  iH (0) t  e (1) (t)i + |Ψ e (2) (t)i + · · · Ψ(t) = exp − |Ψ(0)i + |Ψ (4.2.16) ~ Let us use our perturbation theory to calculate the probability Pm←n (t) to transition from |ni at t = 0 to |mi, with m 6= n, at time t, under the effect of the perturbation. By definition 2 Pm←n (t) = hm|Ψ(t)i . (4.2.17) Using the tilde wavefunction, for which we know how to write the perturbation, we have 2 2 (0) e e = hm|Ψ(t)i , Pm←n (t) = hm|e−iH t/~ Ψ(t)

(4.2.18)

4.2. PERTURBATIVE SOLUTION

83

since the phase that arises from the action on the bra vanishes upon the calculation of the e norm squared. Now using the perturbation expression for |Ψ(t)i (setting λ = 1) we have   2 (1) (2) e (t) + Ψ e (t) + . . . . Pm←n (t) = hm| Ψ(0) + Ψ

(4.2.19)

Since we are told that |Ψ(0)i = |ni and |ni is orthogonal to |mi we find 2 (2) e (1) (t) + hm Ψ e (t) + . . . . Pm←n (t) = hm Ψ

(4.2.20)

To first order in perturbation theory we only keep the first term in the sum and using our e (1) (t)i we find result for |Ψ (1) Pm←n (t)

2 Z Z t f 0 t hm|δH(t f 0 )|ni δH(t ) 0 |ni dt = dt0 = hm| i~ i~ 0 0

2 .

(4.2.21)

f and those of δH we finally have Recalling the relation between the matrix elements of δH our result for the transition probability to first order in perturbation theory: (1) Pm←n (t)

2 Z t 0 iωmn t0 δH mn (t ) 0 = e dt . i~

(4.2.22)

0

This is a key result and will be very useful in the applications we will consider. Exercise: Prove the remarkable equality of transition probabilities (1) (1) Pm←n (t) = Pn←m (t) ,

(4.2.23)

valid to first order in perturbation theory. It will also be useful to have our results in terms of the time-dependent coefficients cn (t) introduced earlier through the expansion X e Ψ(t) = cn (t)|ni . (4.2.24) n

e Since Ψ(0) = Ψ(0) the initial condition reads X (0) e (0) , Ψ(0) = cn (0)|ni = Ψ

(4.2.25)

n

where we also used (4.2.9). In this notation, the cn (t) functions also have a λ expansion, because we write (k) X (k) e (t) = Ψ cn (t)|ni , k = 0, 1, 2, . . . (4.2.26) n

84

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

and therefore the earlier relation (0) (1) (2)  e e (t) + λ Ψ e (t) + λ2 Ψ e (t) + O λ3 . Ψ(t) = Ψ

(4.2.27)

(1) 2 (2) cn (t) = c(0) n (t) + λcn (t) + λ cn (t) + . . .

(4.2.28)

now gives e (0) (t)i is in fact constant we have Since |Ψ (0) c(0) n (t) = cn (0) = cn (0) ,

(4.2.29)

where we used (4.2.25). The other initial conditions given earlier in (4.2.8) imply that c(k) n (0) = 0 ,

k = 1, 2, 3 . . . .

(4.2.30)

(1) e (t) we have Therefore, using our result (4.2.12) for Ψ (1) X (1) e (t) = Ψ cn (t)|ni =

t

Z 0

n

X f 0) δH(t dt0 cn (0)|ni , i~ n

(4.2.31)

f 0 )|ni hm|δH(t cn (0) dt0 . i~

(4.2.32)

and as a result, c(1) m (t)

(1) X e (t) = ≡ hm Ψ n

Z 0

t

We therefore have c(1) m (t)

=

XZ n

0

t

0

dt0 eiωmn t

δHmn (t0 ) cn (0) . i~

(4.2.33)

The probability Pm (t) to be found in the state |mi at time t is 2 e Pm (t) = |hm|Ψ(t)i|2 = hm|Ψ(t)i = |cm (t)|2 .

(4.2.34)

To first order in perturbation theory the answer would be (with λ = 1) 2 (1) Pm (t) = cm (0) + c(1) m (t) 2 (1) ∗ 2 = cm (0) + cm (0)∗ c(1) m (t) + cm (t) cm (0) + O(δH ) . (1)

(4.2.35)

Note that the |cm (t)|2 term cannot be kept to this order of approximation, since it is of (2) the same order as contributions that would arise from cm (t).

4.2. PERTURBATIVE SOLUTION

4.2.1

85

Example NMR

The Hamiltonian for a particle with a magnetic moment inside a magnetic field can be written in the form H = ω · S, (4.2.36) where S is the spin operator and ω is precession angular velocity vector, itself a function of the magnetic field. Note that the Hamiltonian has properly units of energy (ω~). Let us take the unperturbed Hamiltonian to be H (0) = ω0 Sz =

~ 2

ω0 σz .

(4.2.37)

For NMR applications one has ω0 ≈ 500 Hz and this represents the physics of a magnetic field along the z axis. Let us now consider some possible perturbations. Case 1: Time independent perturbation Let us consider adding at t = 0 a constant perturbation associated with an additional small uniform magnetic field along the x axis: H = H (0) + δH , with δH = ΩSx ,

(4.2.38)

For this to be a small perturbation we will take Ω  ω0 .

(4.2.39)

Hence, for the full Hamiltonian H we have ω = (Ω, 0, ω0 )

H = (Ω, 0, ω0 ) ·(Sx , Sy , Sz ) | {z }

(4.2.40)

ω

The problem is simple enough that an exact solution is possible. The perturbed Hamiltonian ω H has energy eigenstates |n; ±i, spin states that point with n = |ω| with energies ± ~2 |ω|. These eigenstates could be recovered using time-independent non-degenerate perturbation theory using Ω/ω0 as a small parameter and starting with the eigenstates |±i of H (0) . In time-dependent perturbation theory we obtain the time-dependent evolution of initial states as they are affected by the perturbation. Recalling (4.2.12) we have Z t f 0 (1) δH(t ) e (t) = Ψ Ψ(0) dt0 (4.2.41) i~ 0

86

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

f requires a bit of computation. One quickly finds that The calculation of δH h  h  σz i ˆ σz i f ΩSx exp −iω0 t = Ω Sˆx cos ω0 t − Sˆy sin ω0 t . δH(t) = exp iω0 t 2 2

(4.2.42)

As a result we set: Z t  (1) e (t) = Ω Ψ Sˆx cos ω0 t0 − Sˆy sin ω0 t0 Ψ(0) dt0 i~ 0  i 1 Ω hˆ = Sx sin ω0 t + cos(ω0 t0 ) − 1 Sˆy Ψ(0) . i~ ω0

(4.2.43)

As expected, the result is first order in the small parameter Ω/ω0 . Following (4.2.16) the time-dependent state is, to first approximation ! h i h i  σ 1 Ω z Ψ(t) = exp −iω0 t 1+ Sˆx sin ω0 t + cos(ω0 t0 ) − 1 Sˆy |Ψ(0)i 2 i~ ω0 (4.2.44) + O((Ω/ω0 )2 ) . Physically the solution is clear. The original spin state Ψ(0) precesses about the direction of ω with angular velocity ω. Case 2: Time dependent perturbation. Suppose that to the original Hamiltonian we add the effect of a rotating magnetic field, rotating with the same angular velocity ω0 that corresponds to the Larmor frequency of the original Hamiltonian:   δH(t) = Ω Sˆx cos ω0 t + Sˆy sin ω0 t . (4.2.45) f we can use (4.2.42) with t replaced by −t so that the To compute the perturbation δH right-hand side of this equation is proportional to δH: h h   σz i ˆ σz i exp −iω0 t ΩSx exp iω0 t = Ω Sˆx cos ω0 t + Sˆy sin ω0 t . (4.2.46) 2 2 Moving the exponentials on the left-hand side to the right-hand side we find h  h σz i  ˆ σz i ΩSˆx = exp iω0 t Ω Sx cos ω0 t + Sˆy sin ω0 t exp −iω0 t . 2 2

(4.2.47)

f Thus we have shown that The right-hand side is, by definition, δH. f δH(t) = ΩSˆx ,

(4.2.48)

is in fact a time-independent Hamiltonian. This means that the Schr¨odinger equation for e is immediately solved Ψ  f   ˆx t  δHt Ω S e e Ψ(0) Ψ(0) . Ψ(t) = exp −i = exp −i ~ ~

(4.2.49)

4.3. FERMI’S GOLDEN RULE

87

The complete and exact answer is therefore h h h (0) i σz i σx i e Ψ(t) = exp − iH t Ψ(t) exp −iΩt Ψ(0) . = exp −iω0 t ~ 2 2

(4.2.50)

This is a much more non trivial motion! A spin originally aligned along zˆ will spiral into the x, y plane with angular velocity Ω.

4.3

Fermi’s Golden Rule

Let us now consider transitions where the initial state is part of a discrete spectrum but the final state is part of a continuum. The ionization of an atom is perhaps the most familiar example: the initial state may be one of the discrete bound states of the atom while the final state includes a free electron a momentum eigenstate that is part of a continuum of non-normalizable states. As we will see, while the probability of transition between discrete states exhibits periodic dependence in time, if the final state is part of a continuum an integral over final states is needed and the result is a transition probability linear in time. To such probability function we will be able to associate a transition rate. The final answer for the transition rate is given by Fermi’s Golden Rule. We will consider two different cases in full detail: 1. Constant perturbations. In this case the perturbation, called V turns on at t = 0 but it is otherwise time independent: H =

( H (0) , H (0) + V

for t ≤ 0 ,

(4.3.1)

for t > 0 .

This situation is relevant for the phenomenon of auto-ionization, where an internal transition in the atom is accompanied by the ejection of an electron.

88

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY 2. Harmonic perturbations. In this case the time dependence of the perturbation δH is periodic, namely, H(t) = H (0) + δH(t) , (4.3.2) with

( δH(t) =

for t ≤ 0 ,

0,

2H 0 cos ωt ,

(4.3.3)

for t > 0 .

Note the factor of two entering the definition of δH in terms of the time independent H 0 . This situation is relevant to the interaction of electromagnetic fields with atoms. Before starting with the analysis of these two cases let us consider the way to deal with continuum states, as the final states in the transition will belong to a continuum. The fact is that we need to be able to count the states in the continuuum, so we will replace infinite space by a very large cubic box of side length L and we will impose periodic boundary conditions on the wavefunctions. The result will be a discrete spectrum where the separation between the states can be made arbitrarily small, thus simulating accurately a continuum in the limit L → ∞. If the states are energetic enough and the potential is short range, momentum eigenstates are a good representation of the continuum. To count states use a large box, which can be taken to be a cube of side L:

We call L a regulator as it allows us to deal with infinite quantities (like the volume of space or the number of continuum states). At the end of our calculations the value of L must drop out. This is a consistency check. The momentum eigenstates ψ(x take the form 1 ψ(x) = √ eikx x eiky y eikz z , L3 with constant k = (kx , ky , kz ). It is clear that the states are normalized correctly Z Z 1 2 3 |ψ(x)| d x = 3 d3 x = 1 . L box

(4.3.4)

(4.3.5)

The k 0 s are quantized by the periodicity condition on the wavefunction: ψ(x + L, y, z) = ψ(x, y + L, z) = ψ(x, y, z + L) = ψ(x, y, z) .

(4.3.6)

The quantization gives kx L = 2πnx

Ldkx = 2πdnx ,

ky L = 2πny

Ldky = 2πdny ,

kz L = 2πnz

Ldkz = 2πdnz .

(4.3.7)

4.3. FERMI’S GOLDEN RULE

89

Define ∆N as the total number of states within the little cubic volume element d3 k. It follows from (4.3.7) that we have  3 L ∆N ≡ dnx dny dnz = d3 k . (4.3.8) 2π Note that ∆N only depends on d3 k and not on k itself: the density of states is constant in momentum space. Now let d3 k be a volume element determined in spherical coordinates k, θ, φ and by ranges dk, dθ, dφ. Therefore

d3 k = k 2 dk sin θdθ dφ = k 2 dk dΩ

(4.3.9)

We now want to express the density of states as a function of energy. For this we take differentials of the relation between the wavenumber k and the energy E: E= Back in (4.3.9) we have

~2 k 2 2m

kdk =

m dE . ~2

m dE dΩ , ~2  3 L m ∆N = k 2 dΩ dE . 2π ~ d3 k = k

hence

(4.3.10)

(4.3.11)

(4.3.12)

We now equate ∆N = ρ(E)dE ,

(4.3.13)

where ρ(E) is a density of states, more precisely, it is the number of states per unit energy at around energy E and with momentum pointing within the solid angle dΩ. The last two equations determine for us this density: ρ(E) =

L3 m k dΩ . 8π 3 ~2

(4.3.14)

With a very large box, a sum over states can be replaced by an integral as follows Z X ··· → · · · ρ(E)dE (4.3.15) states

where the dots denote an arbitrary function of the momenta k of the states.

90

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

4.3.1

Constant transitions.

The Hamiltonian, as given in (4.3.1) takes the form ( (0) H for t ≤ 0 H= H (0) + V for t > 0 .

(4.3.16)

We thus identify δH(t) = V for t ≥ 0. Recalling the formula (4.2.33) for transition amplitudes to first order in perturbation theory we have XZ t 0 Vmn dt0 eiωmn t c(1) (t) = cn (0) . (4.3.17) m i~ 0 n To represent an initial state i at = 0 we take cn (0) = δn,i . For a transition to a final state f at time t0 we set m = f and we have an integral that is easily performed: Z 0 t0 Vf i eiωf i t 1 t0 (1) iωf i t0 0 Vf i e dt = (4.3.18) cf (t0 ) = i~ 0 i~ iωf i 0

Evaluating the limits and simplifying (1) cf (t0 )

   V eiωf i t0 /2 Vf i  ωf i t0 fi iωf i t0 = 1−e = (−2i) sin . Ef − Ei Ef − Ei 2

(4.3.19) (1)

The transition probability to go from i at t = 0 to f at t = t0 is then |cf (t0 )|2 and is therefore   ω t 4 sin2 f2i 0 Pf ←i (t0 ) = |Vf i |2 . (4.3.20) (Ef − Ei )2 This first order result in perturbation theory is expected to be accurate at time t0 if Pf ←i (t0 )  1. Certainly a large transition probability at first order could not be trusted and would require examination of higher orders. To understand the main features of the result for the transition probability we examine how it behaves for different values of the final energy Ef . If Ef 6= Ei the transition is said to be energy non-conserving. Of course, energy is conserved overall, as it would be supplied by the perturbation. If Ef = Ei we have an energy conserving transition. Both are possible and let us consider them in turn. 1. Ef 6= Ei . In this case the transition probability Pf ←i (t0 ) as a function of t0 is shown in Figure 4.2. The behavior is oscillatory with frequency |ωf i . If the amplitude of the oscillation is much less than one 4|Vf i |2 1 (Ef − Ei )2

(4.3.21)

4.3. FERMI’S GOLDEN RULE

91

Figure 4.2: The transition probability as a function of time for constant perturbations. then this first order transition probability Pf ←i (t0 ) is accurate for all times t0 as it is always small. The amplitude is suppressed as |Ef − Ei | grows, due to the factor in the denominator. This indicates that the larger the energy ‘violation’ the smaller the probability of transition. This is happening because a perturbation that turns on and then remains constant is not an efficient supply of energy. 2. Ef → Ei . In this limit ωf i approaches zero and therefore   (Ef − Ei )2 2 2 ωf i t0 sin ' t0 . 2 4~2

(4.3.22)

It follows from (4.3.20) that lim Pf ←i (t0 ) =

Ef →Ei

|Vf i |2 2 t . ~2 0

(4.3.23)

The probability for an energy-conserving transition grows quadratically in time, and does so without bound! This result, however, can only be trusted for small enough t0 such that Pf ←i (t0 )  1. Note that a quadratic growth of Pf ←i is also visible in the energy non-conserving Ef 6= Ei case for very small times t0 . Indeed, (4.3.20) leads again to limt0 →0 Pf ←i (t0 ) = |Vf i |2 t20 /~2 , while Ef 6= Ei . This behavior can be noted near the origin in Figure 4.2. Our next step is to integrate the transition probability over the now discretized continuum of final states. Remarkably, upon integration the oscillatory and quadratic behaviors of the transition probability as a function of time will conspire to create a linear behavior! The sum of transition probabilities over final states is approximated by an integral, as explained in (4.3.15). We thus have   ω t Z Z sin2 f2i 0 X Pf ←i (t0 ) = Pf ←i (t0 )ρ(Ef )dEf = 4 |Vf i |2 ρ(Ef )dEf . (4.3.24) (Ef − Ei )2 f

92

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

We have noted that Pf ←i in the above integrand is suppressed as |Ef − Ei | becomes large. We therefore expect that the bulk of the contribution to the integral will occur for a narrow range ∆Ef of Ef near Ei . Let us assume now that |Vf i |2 and ρ(Ef ) are slow varying and therefore approximately constant over the narrow interval Ef (we will re-examine this assumption below). If this is the case we can evaluate them for Ef set equal to Ei and take them out of the integrand to find X

Pf ←i (t0 ) =

f

4|Vf i |2 ρ(Ef = Ei ) · I(t0 ) , ~2

where the integral I(t0 ) is given by     Z Z dωf i 1 2 ωf i t0 2 ωf i t0 I(t0 ) ≡ sin dE = ~ sin . f 2 2 2 ωf i ωf2i

(4.3.25)

(4.3.26)

As it stands, the integral extends from minus infinity to plus infinity. It is useful to plot the final integrand for I(t0 ) as a function of the integration variable ωf i . The result is shown in Figure 4.3 and exhibits a main central lobe followed by symmetrically arranged lobes of decreasing amplitude. The lobes are separated by zeroes occurring for ωf i = 2πk/t0 , with k integer. Note that for

Figure 4.3: Plot of

1 ωf2 i

sin2



ωf i t0 2



, the integrand of I.

The largest contribution to I(t0 ) arises from the main lobe −

2π 2π < ωf i < . t0 t0

(4.3.27)

In terms of energies this corresponds to the range Ei −

2π~ 2π~ < Ef < Ei + . t0 t0

(4.3.28)

4.3. FERMI’S GOLDEN RULE

93

We need this range to be a narrow, thus t0 must be sufficiently large. The narrowness is required to justify our taking of the density of states ρ(E) and the matrix element |Vf i |2 out of the integral. If we want to include more lobes this is can be made consistent with a narrow range of energies by making t0 larger. The linear dependence of I(t0 ) as a function of t0 is intuitively appreciated by noticing that the height of the main lobe is proportional to t20 and its width is proportional to 1/t0 . In fact, the linear dependence is a simple mathematical fact made manifest by a change of variables. Letting ωf i t0 t0 u= =⇒ du = dωf i , (4.3.29) 2 2 so that Z Z ~t0 ∞ sin2 u 2~ ∞ sin2 u = , (4.3.30) I(t0 ) = t0 −∞ u2 · t42 2 −∞ u2 0

making the linear dependence in t0 manifest. The remaining integral evaluates to π and we thus get I(t0 ) = π2 ~t0 . (4.3.31) Had restricted the integral to the main lobe, concerned that the density of states and matrix elements would vary over larger ranges, we would have gotten 90% of the total contribution. Including the next two lobes, one to the left and one to the right, brings the result up to 95% of the total contribution. By the time we include ten or more lobes on each side we are getting 99% of the answer. For sufficiently large t0 is is still a narrow range. Having determined the value of the integral I(t0 ) we can substitute back into our expression for the transition probability (4.3.26). Replacing t0 by t X k

Pk←i (t) =

4|Vf i |2 π 2π ρ(Ef ) ~t = |Vf i |2 ρ(Ef ) t . ~2 2 ~

(4.3.32)

This holds, as discussed before, for sufficiently large t. Of course t cannot be too large as it would make the transition probability large and unreliable. The linear dependence of the transition probability implies we can define a transition rate w, or probability of transition per unit time, by dividing the transition probability by t: w≡

1X Pf ←i (t) . t

(4.3.33)

f

This finally gives us Fermi’s golden rule for constant perturbations: Fermi’s golden rule:

w =

2π |Vf i |2 ρ(Ef ) , ~

Ef = Ei .

(4.3.34)

Not only is the density of states evaluated at the energy Ei , the matrix element Vf i is also evaluated at the energy Ei and other observables of the final state, like the momentum. In

94

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

this version of the golden rule the integration over final states has been performed. The units are manifestly right: |Vf i |2 has units of energy squared, ρ has units of one over energy, and with an ~ in the denominator the whole expression has units of one over time, as appropriate for a rate. We can also see that the dependence of w on the size-of-the-box regulator L disappears: The matrix element Vf i = hf |V |ii ∼ L−3/2 ,

(4.3.35)

because the final state wavefunction has such dependence (see (4.3.4)). Then the L dependence in |Vf i |2 ∼ L−3 cancels with the L dependence of the density of states ρ ∼ L3 , noted in (4.3.14). Let us summarize the approximations used to derive the golden rule. We have two conditions that must hold simultaneously: 1. We assumed that t0 is large enough so that the energy range Ei − k

2π~ 2π~ < Ef < Ei + k , t0 t0

(4.3.36)

with k some small integer, is narrow enough that ρ(E) and |Vf i |2 are approximately constant over this range. This allowed us to take them out of the integral simplifying greatly the problem and making a complete evaluation possible. 2. We cannot allow t0 to be arbitrarily large. As we have X Pf ←i (t0 ) = w t0 ,

(4.3.37)

f

we must keep wt0  1 for our first order calculation to be accurate. Can the two conditions on t0 be satisfied? There is no problem if the perturbation can be made small enough: indeed, suppose condition 1 is satisfied for some suitable t0 but condition 2 is not. Then we can make the perturbation V smaller making w small enough that the second condition is satisfied. In practice, in specific problems, one could do the following check. First compute w assuming the golden rule. Then fix t0 such that wt0 is very small, say equal to 0.01. Then check that over the range ∼ ~/t0 the density of states and the matrix elements are roughly constant. If this works out the approximation should be very good! Helium atom and autoionization. The helium has two protons (Z = 2) and two electrons. Let H (0) be the Hamiltonian for this system ignoring the the Coulomb repulsion between the electrons: p21 e2 p2 e2 H (0) = − + 2 − . (4.3.38) 2m r1 2m r2

4.3. FERMI’S GOLDEN RULE

95

Here the labels 1 and 2 refer to each one of the two electrons. The spectrum of this Hamiltonian consists of hydrogenic states, with n1 , n2 the principal quantum numbers for the electrons. The energies are then     1 1 1 1 2 = −(54.4 eV) . (4.3.39) En1 ,n2 = −(13.6 eV)Z + + n21 n22 n21 n22

Figure 4.4: Hydrogenic states in helium and continuum states of negative total energy. For the hydrogen atom we have bound states of negative energy and continuum states of positive energy that can be described to a good approximation as momentum eigenstates of the electron which is no longer bound to the proton. Since we have two electrons in the case of helium, there are continuum states in the spectrum with negative energy. This first happens for n1 = 1 in the limit as n2 → ∞. For n1 = 1 and n2 = ∞ the second electron is essentially free and contributes no energy. Thus a continuum appears for energy E1,∞ = −54.4eV. This (1S)(∞) continuum extends for all E ≥ −54.4eV as the free electron can have arbitrary positive kinetic energy. The state (2S)(∞) is the beginning of a second continuum, also including states of negative energy. In general the state (nS)(∞) with n ≥ 1 marks the beginning of the n-th continuum. In each of these continua one electron is still bound and the other is free. A diagram showing some discrete states and a couple of continua is given in Figure 4.4. Self-ionizing energy-conserving transitions can occur because discrete states can find themselves in the middle of a continuum. The state (2S)2 , for example, with two electrons on the 2S configuration and with energy E2,2 = −27eV is in the middle of the (1S)(∞) continuum. We can view the original (2S)2 hydrogenic state as a t = 0 eigenstate of H (0) and treat the Coulomb repulsion as a perturbation. We thus have a total Hamiltonian H

96

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

that takes the form H = H (0) + V ,

V =

e2 . |r1 − r2 |

(4.3.40)

The perturbation V produces “auto-ionizing” Auger transitions to the continuum. We have a transition from a state with energy E2,2 = −27eV to a state of one bound electron with energy −54.4eV and one free electron with kinetic energy of 27eV. That final state is part of a continuum. Radiative transitions from (2S)2 to 1S2S, with photo-emission, are in fact a lot less likely than auto-ionizing transitions! We will not do the quantitative analysis required to determine lifetime of the (2S)2 state. In general auto-ionization is a process in which an atom or a molecule in an excited state spontaneously emits one of the outer-shell electrons. Auto-ionizing states are in general short lived. Auger transitions are auto-ionization processes in which the filling of an inner shell vacancy is accompanied by the emission of an electron. Our example of the (2S)2 state is an Auger transition. Molecules can have auto ionizing Rydberg states, in which the little energy needed to remove the Rydberg electron is supplied by a vibrational excitation of the molecule.

4.3.2

Harmonic Perturbation

It is now time to consider the case when the perturbation is harmonic. We will be able to derive a similar looking Fermi golden rule for the transition rate. The most important difference is that now the transitions are of two types. They involve either absorption of energy or a release of energy. In both cases that energy (absorbed or released) is equal to ~ω where ω is the frequency of the perturbation. As indicated in (4.3.3) we have H(t) = H (0) + δH(t) ,

(4.3.41)

where the perturbation δH(t) takes the form ( δH(t) =

for t ≤ 0 ,

0,

2H 0 cos ωt ,

(4.3.42)

for t > 0 .

Here ω > 0 and H 0 is some time independent Hamiltonian. The inclusion of an extra factor of two in the relation between δH and H 0 is convenient because it results in a golden rule does not have additional factors of two compared to the case of constant transitions. We again consider transitions from an initial state i to a final state f . The transition amplitude this time From (4.2.33) (1)

cf (t) ) =

1 i~

Z 0

t0

0

dt0 eiωf i t δHf i (t0 ) .

(4.3.43)

4.3. FERMI’S GOLDEN RULE

97

Using the explicit form of δH the integral can be done explicitly Z 1 t0 iωf i t0 (1) e 2Hf0 i cos ωt0 dt0 cf (t0 ) = i~ 0 Z  Hf0 i t0  i(ω +ω)t0 0 = e fi + ei(ωf i −ω)t dt0 i~ 0 " # Hf0 i ei(ωf i +ω)t0 − 1 ei(ωf i −ω)t0 − 1 =− . + ~ ωf i + ω ωf i − ω

(4.3.44)

Comments: • The amplitude takes the form of a factor multiplying the sum of two terms, each one a fraction. As t0 → 0 each fraction goes to it0 . For finite t0 , which is our case of interest, each numerator is a complex number of bounded absolute value that oscillates in time from zero up to two. In comparing the two terms the relevant one is the one with the smallest denominator.1 • The first term is relevant as ωf i + ω ≈ 0, that is, when there are states at energy Ef = Ei − ~ω. This is “stimulated emission”, the source has stimulated a transition in which energy ~ω is released. • The second term relevant if ωf i − ω ≈ 0, that is, when there are states at energy Ef = Ei + ~ω. Energy is transferred from the perturbation to the system, and we have a process of “absorption”. Both cases are of interest. Let us do the calculations for the case of absorption; the answer for the case of spontaneous emission will be completely analogous. We take i to be a discrete state, possibly bound, and f to be a state in the continuum at the higher energy Ef ≈ Ei + ~ω. Since ωf i ' ω, the second term in the last line of (4.3.44) is much more important than the first. Keeping only the second term we have i

(1) cf (t0 )

= −

Hf0 i e 2 (ωf i −ω)t0 ~

ωf i − ω

 2i sin

ωf i − ω t0 2

 ,

(4.3.45)

and the transition probability is (1)

Pf ←i (t0 ) = |cf (t0 )|2 =

4|Hf0 i |2 ~2



ωf i −ω 2 t0 (ωf i − ω)2

sin2

 (4.3.46)

The transition probability is exactly the same as that for constant perturbations (see (4.3.20)) with V replaced by H 0 and ωf i replaced by ωf i − ω. The analysis that follows is 1 This is like comparing two waves that are being superposed. The one with larger amplitude is more relevant even though at some special times, as it crosses the value of zero, it is smaller than the other wave.

98

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

completely analogous to the previous one so we shall be brief. Summing over final states we have   2 ωf i −ω Z Z t X 0 4|Hf0 i |2 sin 2 Pk←i (t0 ) = Pf ←i ρ(Ef )dEf = ρ(Ef )dEf . (4.3.47) 2 ~ (ωf i − ω)2 k

This time the main contribution comes from the region −

2π 2π < ωf i − ω < t0 t0

(4.3.48)

In terms of the final energy this is the band Ei + ~ω −

2π~ 2π~ < Ef < Ei + ~ω + t0 t0

(4.3.49)

For sufficiently large t0 this is the narrow band of states illustrated in Figure 4.5. Assume

Figure 4.5: In the absorption process we must integrate over a narrow band of states about the final energy Ei + ~ω. that over the band Hf0 i and the density of states is constant so that we get X

Pk←i (t0 ) =

k

4 0 2 |H | ρ(Ei + ~ω) ~ fi



ωf i −ω 2 t0 (ωf i − ω)2

Z sin2

 dωf i .

(4.3.50)

Defining u≡

1 2

(ωf i − ω) t0

du = 12 dωf i t0 ,

(4.3.51)

the integral in (4.3.50) becomes Z

−∞

sin2 u 2 t0  2 du = t0 2 2u t0

Z

−∞

sin2 u t0 π. du = 2 u 2

(4.3.52)

4.4. IONIZATION OF HYDROGEN

99

Finally, the transition probability is X 2π 0 2 Pk←i (t0 ) = |Hf i | ρ(Ei + ~ω) t0 . ~

(4.3.53)

k

The transition rate w is finally given by Fermi’s golden rule:

w =

2π ρ(Ef )|Hf0 i |2 , ~

Ef = Ei + ~ω .

(4.3.54)

Here H 0 (t) = 2H 0 cos ωt. Equation (4.3.54) is known as Fermi’s golden rule for the case of harmonic perturbations. For the case of spontaneous emission the only change required in the above formula is Ef = Ei − ~ω.

4.4

Ionization of hydrogen

We aim to find the ionization rate for hydrogen when hit by the harmonically varying electric field of an electromagnetic wave. We assume the hydrogen atom has its electron on the ground state. In this ionization process a photon ejects the bound electron, which becomes free. Let us first do a few estimates to understand the validity of the approximations that will be required. If the electromagnetic field has frequency ω the incident photons have energy Eγ = ~ω .

(4.4.1)

The energy Ee and the magnitude k of the momentum of the ejected electron are given by Ee =

~2 k 2 = Eγ − Ry , 2m

(4.4.2)

where the Rydberg Ry is the magnitude of the energy of the ground state: 2Ry =

e2 ~2 ~c = =α , 2 a0 a0 ma0

Ry ' 13.6 eV .

(4.4.3)

Inequalities 1. Any electromagnetic wave has spatial dependence. We can ignore the spatial dependence of the wave if the wavelength λ of the photon is much bigger than the Bohr radius a0 : λ  1. (4.4.4) a0 Such a condition puts an upper bound on the photon energy, since the more energetic photon the smaller its wavelength. To bound the energy we first write λ=

2π 2πc 2π~c = = , kγ ω ~ω

(4.4.5)

100

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY and then find Ry 2π ~c 4π Ry λ = = ' 1722 a0 ~ω a0 α ~ω ~ω

(4.4.6)

The inequality (4.4.4) then gives ~ω  1722 Ry ' 23 keV

(4.4.7)

2. For final states we would like to use free particle momentum eigenstates. This requires the ejected electron to be energetic enough not to be much effected by the Coulomb field. As a result, the photon to be energetic enough, and this constraint provides a lower bound for its energy. We have Ee = Eγ − Ry  Ry

~ω  Ry .

(4.4.8)

The two inequalities above imply Ry  ~ω  1722 Ry .

(4.4.9)

If consider that 1  10 we could take 140eV ≤ ~ω ≤ 2.3 keV .

(4.4.10)

Note that even for the upper limit the electron is non relativistic: 2.3 keV  me c2 ' 511keV. Exercise: Determine ka0 in terms of ~ω and Ry . Solution:     ~ω ~2 ~ω ~2 k 2 = ~ω − Ry = Ry −1 = −1 2m Ry 2ma20 Ry s ~ω ~ω k 2 a20 = − 1 → ka0 = − 1. Ry Ry

(4.4.11)

(4.4.12)

This is the desired result. In the range of (4.4.10) we have 3 ≤ ka0 ≤ 13 .

(4.4.13)

Calculating the matrix element. Our perturbation Hamiltonian is δH = −eΦ ,

(4.4.14)

where Φ is the electric scalar potential. Let the electric field at the atom be polarized in the zˆ direction E(t) = E(t)ˆ z = 2E0 cos ωt zˆ (4.4.15)

4.4. IONIZATION OF HYDROGEN

101

The associated scalar potential is Φ = −E(t)z ,

(4.4.16)

and therefore δH = eE(t)z = eE(t)r cos θ = 2 eE0 r cos θ cos ωt ≡ 2H 0 cos ωt | {z }

(4.4.17)

H0

We have thus identified, in our notation, H 0 = eE0 r cos θ .

(4.4.18)

We want to compute the matrix element between an initial state |ii that is the ground state of hydrogen, and a final state |f i which is an electron plane wave with momentum ke . The associated wavefunctions are 1 Final state: uk (x) = 3/2 eik·r (4.4.19) L 1 − r Initial state: ψ0 (r) = p 3 e a0 (4.4.20) πa0 The only physical angle here is that between the ejected electron momentum ke and the electric field polarization E. We expect electron to be ejected maximally along E. This suggests rearranging the axes to have the electron momentum along the z axis and and letting θ be the angle between the electron momentum and the electric field. The integration variable r will have angles θ0 , φ0 , and the angle between the electric field and r is now called θ00 as shown in the figure, so that H 0 = eE0 r cos θ00 : Z 1 1 − r 0 hf |H |ii = d3 x 3/2 eik·r eE0 r cos θ00 p 3 e a0 L πa0 Z eE0 0 − r =p 3 r2 dr sin θ0 dθ0 dφ0 eikr cos θ r cos θ00 e a0 (4.4.21) πa0 L3

102

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

The main complication with doing the integral is the factor cos θ00 . The dot product of unit vectors along E and n is cos θ00 : cos θ00 = (sin θ cos φ)(sin θ0 cos φ0 ) + (sin θ sin φ)(sin θ0 sin φ0 ) + cos θ cos θ0 = sin θ sin θ0 cos(φ − φ0 ) + cos θ cos θ0 Since the following integral vanishes Z

(4.4.22)

dφ0 cos(φ − φ0 ) = 0

(4.4.23)

the first term in the expression for cos θ00 will vanish upon φ0 integration. We therefore can replace cos θ00 → cos θ cos θ0 to find Z eE0 0 − r hf |H 0 |ii = p 3 r3 dr sin θ0 dθ0 dφ0 eikr cos θ cos θ cos θ0 e a0 πa0 L3 Z Z 1 eE0 0 − ar 3 0 = p 3 (cos θ)(2π) r dr e d(cos θ0 ) cos θ0 eikr cos θ (4.4.24) 3 πa0 L −1 Now let r = a0 u and do the radial integration: Z Z 1 eE0 0 4 0 0 0 u3 du e−u(1+ika0 cos θ ) d(cos θ ) cos θ hf |H |ii = p 3 2π cos θa0 3 πa0 L −1 Z 1 eE0 3! = p 3 2π cos θa40 . d(cos θ0 ) cos θ0 (1 + ika0 cos θ0 )4 πa0 L3 −1

(4.4.25)

Writing x = cos θ0 the angular integral is not complicated: Z 1 3! x eE0 2π cos θa40 hf |H 0 |ii = p 3  4 dx 4 3 (ika ) 0 πa0 L −1 x + 1 ika0   1 8 ika 3! eE0 0 2π cos θa40 =p 3 3 4  3 (ika ) 0 πa0 L 3 1 + k21a2 0

32πeE0 a0 ka40 = −i p 3 cos θ πa0 L3 (1 + k 2 a20 )3

(4.4.26)

Hence, with a little rearrangement √ ka4 1 hf |H 0 |ii = −i32 π (eE0 a0 ) p 30 cos θ . 3 a0 L (1 + k 2 a20 )3

(4.4.27)

The above matrix element has the units of (eE0 a), which is energy. This is as expected. Squaring we find k 2 a5 cos2 θ |Hf0 i |2 = 1024 π (eE0 a)2 30 (4.4.28) L (1 + k 2 a20 )6

4.4. IONIZATION OF HYDROGEN

103

At this point, with θ the angle between ke and E, it is more convenient to align E along the z-axis and let ke be defined by the polar angles θ, φ. The density of states was given in (4.3.14)

ρ(Ee ) =

L3 m k dΩ 8π 3 ~2

(4.4.29)

so that using Fermi’s Golden rule, the rate dw to go into final states in the solid angle dΩ is 2π dw = ρ(Ee )|Hf0 i |2 ~ 2 5 cos2 θ 2π L3 m 2 k a0 . (4.4.30) k dΩ 1024π(eE a ) = 0 0 ~ 8π 3 ~2 L3 (1 + k 2 a20 )6 It follows that dw 256 ma20 (eE0 a0 )2 k 3 a30 = cos2 θ . dΩ π ~2 ~ (1 + k 2 a20 )6

(4.4.31)

dw dΩ

is the probability of ionization per unit time and per unit solid angle. All the ejected electrons will have the same momentum ~k. Note that the units have worked out: w ∼ [E]2 1 1 · ∼ ~ T as expected. In here 2E0 is the peak amplitude of the electric field in the [E] wave. The cos2 θ implies that the electron tends to be ejected in the direction of the electric field. The total ionization probability per unit time is obtained by integration over solid angle. Using Z cos2 θ dΩ =

and recalling that

~2 ma20

1 3

4π ,

(4.4.32)

= 2Ry Z w=

dΩ

512 (eE0 a0 )2 k 3 a30 dw = . dΩ 3 ~Ry (1 + k 2 a20 )6

(4.4.33)

For the window of validity (4.4.13) we have 9 ≤ (ka0 )2 ≤ 169, and we neglect the “one” in the denominator to get 512 (eE0 a0 )2 1 w = . (4.4.34) 3 ~Ry (ka0 )9 This is our final answer. For numerical calculations it is useful to note the atomic units in which the answer takes the form 256  Ep 2 1 1 w = . (4.4.35) 3 E∗ t∗ (ka0 )9

104

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

Here Ep = 2E0 is the peak amplitude of the electric field. Moreover, the atomic electric field E∗ and the atomic time t∗ are given by 2Ry e = 2 = 5.14 × 1011 V/m ea0 a0 a0 = = 2.42 × 10−17 sec. αc

E∗ = t∗

(4.4.36)

Note that E∗ is the electric field of a proton at a distance a0 while t∗ is the time it takes the electron to travel a distance a0 at the velocity αc in the ground state. A laser intensity of 3.55 × 1016 W/cm2 has an peak electric field of magnitude E∗ .

4.5 4.5.1

Light and atoms Absorption and stimulated emission

The physical problem we are trying to solve consists of a collection of atoms with two possible energy levels interacting with light at a temperature T . We want to understand the processes of absorption and emission of radiation and how they can produce thermal equilibrium. Let us first consider a single atom and the possible processes. Consider two possible levels of an electron in an atom

Define ωab ≡

Eb − Ea . ~

Imagine now shining light into the atom at a frequency ωab . There are two possibilities depending on the initial state of the atom: (i) Electron initially in |ai: there will be an absorption rate at which a photon is absorbed and the electron goes from |ai → |bi. (ii) Electron initially in |bi: there will be a stimulated emission rate at which the photon field stimulates an electronic transition |bi → |ai with the release of an additional photon.

4.5. LIGHT AND ATOMS

105

These processes are illustrated in the figure below:

A LASER, Light Amplification by Stimulated Emission of Radiation, works by having population inversion, namely most of the atoms are in the excited energy level. Then any small number of photons can trigger larger and larger numbers of stimulated emission processes!

4.5.2

Einstein’s Argument

Atoms in states a or b with Eb > Ea , with populations Na and Nb respectively. The atoms are in equilibrium with a bath of photons all at temperature T . Einstein discovered a number of relations from the condition of equilibrium. Fact 1: Equilibrium values for populations: N˙ a = N˙ b = 0

(4.5.1)

Fact 2: Equilibrium populations governed by thermal distribution e−βEb Na = −βEa = e−β~ωab , Nb e

β≡

1 kb T

(4.5.2)

106

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

Fact 3: For a thermal blackbody, the energy U (ω)dω per unit volume in the frequency range dω is known to be U (ω)dω =

~ ω 3 dω π 2 c3 eβ~ω − 1

(4.5.3)

Process

Rate

Absorption:

|ai → |bi photon absorbed

Bab U (ωba )Na

Stimulated emission:

|bi → |ai photon released

Bba U (ωba )Nb

In the table above we have indicated the two obvious processes: absorption and stimulated emission. The rate indicated is the number of transitions per unit time. For absorption, this rate is proportional to the number Na of atoms in the ground state, capable of absorbing a photon, times U (ωab ) which captures the information about the number of photons available, times a B coefficient Bab that we want to determine. For stimulated emission, the rate is proportional to the number Nb of atoms in the excited state, thus capable of emitting a photon, times U (ωab ) times a B coefficient Bba . The inclusion of U (ωab ) reflects the “stimulated” character of the transition. Can we make this work, that is, can we achieve equilibrium? We will see that we cannot. With the two processes above, the rate of change of Nb is N˙ b = rate absorption − rate stimulated emission

(4.5.4)

At equilibrium N˙ b = 0, hence 0 = N˙ b = Bab U (ωba )Na − Bba U (ωba )Nb = (Bab Na − Bba Nb )U (ωba )   0 = Na Bab − Bba e−β~ωba U (ωba )

(4.5.5)

This is a strange result: in order to have equilibrium we need Bab − Bba e−β~ωba = 0. The B coefficients, however, depend on the electronic configurations in the atom and not on the temperature T . Thus this cancellation is not possible for arbitrary temperature. We also note that the photon energy density does not feature in the equilibrium condition. In conclusion, equilibrium is not possible. What are we missing? Another process: spontaneous emission!!, an emission rate that does not depend on the thermal photons. The rate is proportional to the number of atoms Nb in the excited state, with a coefficient of proportionality called A:

4.5. LIGHT AND ATOMS

107

Process

Rate |bi → |ai photon released

Spontaneous emission:

ANb

Reconsider now the equilibrium condition with this extra contribution to N˙ b : 0 = N˙ b = Bab U (ωba )Na − Bba U (ωba )Nb − ANb   Na =⇒ A = Bab − Bba U (ωba ) Nb A 1 =⇒ U (ωba ) = ba Bab eβ~ωba − B B

(4.5.6)

ab

The equilibrium condition indeed gives some important constraint on U (ωab ). But from Eq. (4.5.3) we know ~ ω3 1 U (ωba ) = 2 ba3 β~ω (4.5.7) π c e ba − 1 Hence comparing the two equations we get Bab = Bba

and

~ω 3 A = 2 ba3 Bab π c

(4.5.8)

As we’ll see, we’ll be able to calculate Bab , but A is harder. Happily, thanks to (4.5.8), we can obtain A from Bab . Spontaneous emission does not care about U (ωba ); the photons flying around. At a deep level one can think of spontaneous emission as stimulated emission due to vacuum fluctuation of the EM field. For a given atomic transition expect stimulated emission rate to dominate at higher temperature, the there are more photons around. Spontaneous emission should dominate at very low temperature. Indeed from (4.5.6)

A = exp Bab U (ωba )



~ ωba kb T

 −1

108

4.5.3

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

Atom/light interaction

 ~ field. Effects of B ~ are weaker by O v ∼ α. For optical frequencies λ ∼ Focus on E c 4000 − 8000 ˚ A and since a0 ' 0.5 ˚ A, we can ignore the spatial dependence of the field (even for ionization frequencies aλ0 ∼ 1700) Electric field at the atom

E(t) = E(t)n = 2E0 cos(ωt)n

where n is the direction of the field. Φ(r, t) = −r · E(t) ,

∇Φ = −E

(4.5.9)

as expected. The perturbation is therefore δH = +qΦ(r) = −qr · E(t)

(4.5.10)

By defining the dipole moment d as d ≡ qr, we can rewrite the perturbation as δH = −d · E(t) = −d · n2E0 cos(ωt) = 2(−d · nE0 ) cos(ωt) .

(4.5.11)

Since we defined δH = 2H 0 cos ωt we can read H 0 = −d · n E0 .

(4.5.12)

Recall that to first order in perturbation theory Pb←a (t) = Pa←b (t) so let’s consider just Pa←b (t), the stimulated emission process, from (4.3.46)   ω −ω 0 |2 sin2 ωba −ω t 4|Hab 4E02 |(d · n)ab |2 sin2 ba2 t 2 Pa←b (t) = = (4.5.13) ~2 (ωba − ω)2 ~2 (ωba − ω)2 Now think in terms of energy. The energy density uE in the electric field E(t) = 2E0 cos(ωt)n is |E(t)|2 4E02 4E02 1 E2 = cos2 ωt =⇒ huE itime = · = 0. (4.5.14) uE = 8π 8π 8π 2 4π In a wave the electric and magnetic energy are the same and therefore huitime = 2huE itime =

E02 =⇒ E02 = 2πhui 2π

(4.5.15)

4.5. LIGHT AND ATOMS

109

The superposition of light is incoherent, so we will add probabilities of transition due to each component of light. For this we must turn a sum of electric field intensities into an integral.

X

E02 (ωi )

Z = 2π

U (ω)dω

(4.5.16)

i

where U (ω) is the energy density per unit frequency. In this way the transition rate in terms of energy reads  Z sin2 ωba2−ω t |(d · n)ab |2 2π U (ω)dω Pa←b (t) = ~2 (ωba − ω)2  Z sin2 ωba2−ω t |(d · n)ab |2 = (2π)U (ωab ) dω ~2 (ωba − ω)2  As usual, take x ≡ ωba2−ω t =⇒ dω = 2 dx t , so that (4.5.17) is sin2 x dx = 2tπ x2

(4.5.18)

4π 2 |(d · n)ab |2 U (ωba ) ~2

(4.5.19)

Z

2 2 ·t t Ra←b =

(4.5.17)

So far we have two vectors: d which depends on the atom geometry and n which is the direction of the electric field. Since light comes in all possible polarization, this amounts to averaging over all possible directions of n D

|dab · n|

2

E

X i 2 D X i ∗  X j E X i ∗ j = dab ni = dab ni dab nj = (dab ) (dab )hni nj i i

i

j

i,j

(4.5.20) Expect hnx nx i = hny ny i = hnz nz i Since

DX

E ni ni = hn2 i = 1

(4.5.21) (4.5.22)

i

then

1 hni nj i = δij 3

(4.5.23)

110

CHAPTER 4. TIME DEPENDENT PERTURBATION THEORY

D E 1 1 |dab · n|2 = d∗ab · dab ≡ |dab |2 3 3 This is the magnitude of a complex vector! Ra←b =

(4.5.24)

4π 2 |dab |2 U (ωba ) 3~2

(4.5.25)

This is the transition probability per unit time for a single atom Bab =

4π 2 |dab |2 3~2

(4.5.26)

thus recalling (4.5.8) 3 3 ~ωba ~ωba 4π 2 B = |dab |2 (4.5.27) ab 2 3 2 3 π c π c 3~2 hence we have that A, the transition probability per unit time for an atom to go from |bi → |ai, is

A=

A = The lifetime for the state is τ =

1 A.

3 4 ωba |dab |2 . 3 ~c3

(4.5.28)

The number N of particles decays exponentially in time

dN = −AN =⇒ N (t) = e−At = N0 e−t/τ dt

(4.5.29)

If there are various decay possibilities A1 , A2 , . . . , the transition rates add Atot. = A1 + A2 + . . .

4.5.4

and

τ=

1 Atot.

(4.5.30)

Selection rules

(to be covered in recitation) Given two states |n`mi and |n0 `0 m0 i, when hn`m|r|n0 `0 m0 i = 6 0? One can learn that hn`m|z|n0 `0 m0 i = 0 0 0

0

0 0

0

hn`m|x|n ` m i = hn`m|y|n ` m i = 0

for m0 6= m

(4.5.31)

0

(4.5.32)

for m 6= m ± 1

Hence no transition unless ∆m ≡ m0 − m = 0, ±1

(4.5.33)

∆` ≡ `0 − ` = ±1

(4.5.34)

MIT OpenCourseWare https://ocw.mit.edu

8.06 Quantum Physics III

Spring 2018

Chapter 6: Adiabatic approximation B. Zwiebach June 2, 2017

Contents 1 Introduction 1.1 Adiabatic approximation in Classical Mechanics . . . . . . . . . . . . . . . . . . . . . . 1.2 Quantum mechanics systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 4

5

3 Systematic approach

7

4 Landau-Zener transitions

9

5 Berry’s phase

12

6 Molecules and Born-Oppenheimer 6.1 Born-Oppenheimer approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Following the direct route . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 The H2+ ion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15 16 19 20

1 1.1

Introduction Adiabatic approximation in Classical Mechanics

Let’s go back to classical mechanics, with a harmonic oscillator performing motion but with ω(t) not constant. The hamiltonian of the system would be H(x, p, ω(t)) =

p2 1 + mω 2 (t)x2 2m 2

where x and p are going to be functions of time. In general dH ∂H ∂H ∂H = x˙ + p˙ + . dt ∂x ∂p ∂t

(1.1)

(1.2)

Hamilton’s equation of motion read ∂H = x˙ , ∂p

∂H = −p˙ . ∂x

1

(1.3)

Even if you are not familiar with these, you probably know well their quantum analogs:     d hxi ∂H d hxi ∂H i~ = h[x, H]i = i~ → = dt ∂p dt ∂p     d hpi ∂H d hpi ∂H i~ = h[p, H]i = −i~ → =− dt ∂x dt ∂x (1.4) At any rate, equations (1.3) imply that the ﬁrst two terms in the right-hand side of (1.2) vanish and we have dH ∂H (1.5) = = mω ωx ˙ 2. dt ∂t We have an adiabatic change if the time scale τ for change is much greater than time scale of the 2π . oscillation, T = ω(t)

What remains roughly constant as ω(t) changes adiabatically? Claim:

I(t) ≡

H(t) ω(t)

is almost constant in adiabatic changes of ω.

I(t) is called the adiabatic invariant. Let us see why this is so:  2     dI 1  dH 1  � p 2 2 2 1 − H(t)ω˙ = 2 ω mω ωx ˙ = 2 ω − + mω (t)x ω˙ = dt dt ω 2m 2 ω   ω˙ ω˙ p2 2 2 1 = 2 2 mω (t)x − = 2 (V (t) − K(t)) , 2m ω ω

(1.6)

(1.7)

where V and K are the potential and kinetic energies, respectively. We want to understand why this right-hand side is small. Clearly, ωω˙2 is slowly varying if ω˙ is slowly varying, but the term in parenthesis is actually quickly varying. Indeed, with x = A sin ωt and p = Amω cos ωt then the parenthesis goes like   p2 2 2 1 ∼ 21 mω 2 (sin2 ωt − cos2 ωt) ∼ − 21 mω 2 cos 2ωt , 2 mω x − 2m 2

(1.8)

which is a fast variation. Let us see, however, how much I(t) changes over one period Z I(t + T ) − I(t) = t

t+T

dI 0 dt = dt0

Z t

t+T

 ω˙ 0 � (t ) V (t0 ) − K(t0 ) dt0 2 ω

(1.9)

Since ω(t) is slowly varying, it changes very little over a period T and we can take the factor ω/ω ˙ out of the integral: Z t+T ω˙ I(t + T ) − I(t) ' 2 (t) (V (t) − K(t)) dt0 (1.10) ω t In harmonic motion with ω constant, the average over a period of the potential energy V (t) is exactly equal to the average of the kinetic energy K(t), and the two terms above would cancel exactly. When ω is slowly changing they roughly cancel I(t + T ) − I(t) '

ω(t) ˙ ·0'0 ω2

(1.11)

What is the geometrical interpretation of the adiabatic invariant E/ω? For this we consider the motion of a classical oscillator in phase space (x, p) where periodic oscillations trace the ellipse deﬁned by energy conservation: p2 1 + mω 2 x2 = E (ω, E, constants) 2m 2

We quickly see that the semi-major and semi-minor axes have lengths r √ 2E a= and b = 2mE 2 mω

(1.12)

Therefore Area of the ellipse = πab = 2π

E ω

(1.13)

The adiabatic invariant is proportional to the area of the ellipse. That’s neat! We can rewrite the area as an integral that takes a familiar form: I Area of the ellipse = p dx

3

where we integrate over the whole closed trajectory with the orientation shown in the picture so that the area above the x axis is obtained as we go from left to right and the area under the x axis is obtained as we go from right to left. In this example I p dx = 2πI . (1.14) More generally, for other systems the left-hand side is an adiabatic invariant. The value of the invariant is diﬀerent for diﬀerent systems, for the oscillator it is 2πE/ω .

1.2

Quantum mechanics systems

What does it suggest for quantum mechanics? First, for the harmonic oscillator �  �  ~ω n + 12 E = = ~ n + 12 ω ω

(1.15)

Fixed E ω is ﬁxed quantum number! This suggests that in QM the quantum number doesn’t tend to change under adiabatic changes. Indeed, we get the same intuition from WKB: consider a potential with two turning points a, b. We have the Bohr-Sommerﬁeld quantization condition 1 ~

Z a

I rewritten as

b

�  p(x)dx = n + 12 π

�  p(x)dx = 2π~ n + 12

(1.16)

The left hand side, as noted before is an adiabatic invariant, so in the semiclassical approximation for arbitrary potential we ﬁnd that the quantum number may not change under adiabatic changes! We will see that indeed that is the case. The formula for transitions at ﬁrst order in perturbation theory gives complementary intuition: Z 2 t 0 f iωf i t δHf i (t ) 0 Pf ←i (t) = e dt (1.17) 0 i~ f f i is time-independent and For constant perturbations, δH 2 f f i |2 eiωf i T − 1 2 f f i |2 Z t |δH |δH iωf i t 0 Pf ←i (t) = e dt = . ~2 0 ~2 ωf2i • If the spectrum is discrete, it is hard to jump a big energy gap because of the • For slowly varying perturbations (compared to ωf i ) the

1 ωf2i

(1.18) 1 ωf2i

suppression.

suppression will remain.

So it is diﬃcult in general to change state with constant or slow perturbations, suggesting again that quantum numbers are adiabatic invariants. That is why transitions to the continuum with constant perturbations essentially conserve energy. And why you need oscillatory perturbations for eﬃcient transitions between energy levels. 4

2

Suppose you have found a ψ(t) such that H(t)|ψ(t)i = E(t)|ψ(t)i

(2.1)

we’ll call |ψ(t)i, deﬁned by (2.1), an instantaneous eigenstate. I’d like to emphasize from the start that, in general, an instantaneous eigenstate is not a solution of the time dependent Schr¨odinger equation. As it turns out, it is a useful tool to construct approximate solutions to the Schr¨odinger equation. Let us try to understand the relation between |ψ(t)i and |Ψ(t)i, the solution to the Schr¨odinger equation i~∂t |Ψ(t)i = H(t)|Ψ(t)i . (2.2) Let us try an ansatz, building the solution |Ψ(t)i using the instantaneous eigenstate:  Z t  1 0 0 |Ψ(t)i = c(t) exp E(t )dt |ψ(t)i , i~ 0

(2.3)

with c(t) a function of time to be determined and where we have included a familiar time dependent phase. There is no guarantee this ansatz will work but it seems a good try. The LHS of the Schr¨ odinger equation then looks like   Z t   Z t 1 1 0 0 0 0 i~∂t |Ψ(t)i = c˙ (t) exp E(t )dt |ψ(t)i + E(t)|Ψ(t)i + c(t) exp E(t )dt |ψ˙ (t)i (2.4) i~ 0 i~ 0 For the RHS, using the instantaneous eigenstate equation, we have  Z t  1 0 0 H(t)|Ψ(t)i = c(t) exp E(t )dt H(t)|ψ(t)i = E(t)|Ψ(t)i . i~ 0

(2.5)

Equating the two sides, we get  Z t   Z t  1 1 0 0 0 0 c˙(t) exp E(t )dt |ψ(t)i + c(t) exp E(t )dt |ψ˙ (t)i = 0 i~ 0 i~ 0 and canceling the two exponentials we have c˙(t) |ψ(t)i = −c(t) |ψ˙ (t)i .

(2.6)

Multiply by hψ(t)| we get a diﬀerential equation for c(t): c˙(t) = −c(t) hψ(t)|ψ˙ (t)i , which happily we can solve. Letting c(t = 0) = 1 we can write  Z t  0 ˙ 0 0 c(t) = exp − hψ(t )|ψ (t )idt . 0

5

(2.7)

(2.8)

The above exponential is a phase because the bracket in the integrand is actually purely imaginary: Z Z Z d ∗ dψ ∗ ∗ dψ ˙ = dx (ψ ψ) − dx hψ(t)|ψ (t)i = dx ψ ψ dt dt dt (2.9) Z d ∗ = dx ψ ψ − hψ˙ (t)|ψ(t)i dt Since the wavefunction is normalized we have hψ(t)|ψ˙ (t)i = − hψ˙ (t)|ψ(t)i = −hψ(t)|ψ˙ (t)i∗ showing that indeed hψ(t)|ψ˙ (t)i is purely imaginary. To emphasize this fact we write  Z t  0 ˙ 0 0 c(t) = exp i ihψ(t )|ψ (t )idt

(2.10)

(2.11)

0

Having apparently solved for c(t) we now return to our ansatz (2.3), we get  Z t   Z t  1 0 ˙ 0 0 0 0 |Ψ(t)i ' c(0) exp i ihψ(t )|ψ (t )idt exp E(t )dt |ψ(t)i . i~ 0 0

(2.12)

But there is a mistake in this analysis. We really did not solve the Schr¨odinger equation! That’s why we put a ' instead of an equality. The equation we had to solve, (2.6), is a vector equation, and forming the inner product with hψ(t)| gives a necessary condition for the solution, but not a suﬃcient one. We must check the equation forming the overlap with a full basis set of states. Indeed since |ψ(t)i is known, the equation can only have a solution if the two vectors |ψ˙ (t)i and |ψ(t)i are parallel. This does not happen in general. So we really did not solve equation (2.6). The conclusion is that, ultimately, the ansatz in (2.3) is not good enough! We can see the complication more formally. At t = 0 equation (2.6) reads c˙(0)|ψ(0)i = −c(0)|ψ˙ (0)i .

(2.13)

Using Gram-Schmidt we can construct an orthonormal basis B for the state space with the choice |ψ(0)i for the ﬁrst basis vector: n o B = |1i = |ψ(0)i, |2i, |3i , . . . . (2.14) Equation (2.13) requires hn|ψ˙ (0)i = 0

n = 2, 3, . . .

(2.15)

This will not hold in general. The key insight, however, is that (2.12) is a fairly accurate solution if the Hamiltonian is slowly varying. Making the deﬁnitions: Z Z t 1 t 0 0 ˙ θ(t) ≡ − E(t )dt , ν(t) ≡ ihψ(t)|ψ (t)i , γ(t) ≡ ν(t0 )dt0 (2.16) ~ 0 0 so that θ, ν and γ are all real, the state reads |Ψ(t)i ' c(0) exp (iγ(t)) exp(iθ(t))|ψ(t)i . 6

(2.17)

The phase θ(t) is called the dynamical phase and the phase γ(t) is called the geometric phase. Comment: Instantaneous eigenstates are rather ambiguous. If one has the |ψ(t)i, they can be modiﬁed by the multiplication by an arbitrary phase: |ψ(t)i → |ψ 0 (t)i = e−iχ(t) |ψ(t)i .

3

(2.18)

Systematic approach

Must use a basis to see the error terms and get a more general picture. We will do this now systematically. But ﬁrst let us state precisely the idea in adiabatic approximation. Consider a family of instantaneous eigenstates: (3.1) H(t)|ψn (t)i = En (t)|ψn (t)i with E1 (t) < E2 (t) < . . . , so that there are no degeneracies. Adiabatic approximation: If at t = 0 |Ψ(0)i = |ψn (0)i for some n, then if H(t) is slowly varying for 0 ≤ t ≤ T then at time T we have |ψ(T )i ' |ψn (T )i up to a calculable phase. Key facts: 1. The probability to jump to another state |ψk (t)i with k 6= n is highly suppressed. 2. The phases can sometimes be relevant and have geometric meaning. An overall phase cannot be observed but if we have a superposition |Ψ(0)i = c1 |ψ1 (0)i + c2 |ψ2 (0)i + . . .

(3.2)

the relative phases acquired by |ψ1 (0)i and |ψ2 (0)i after time evolution can matter. Calculation: |Ψ(t)i =

X

cn (t)|ψn (t)i

(3.3)

n

Schr¨odinger equation : i~

 X X cn (t)En (t)|ψn (t)i c˙n |ψn (t)i + cn |ψn˙(t)i = n

(3.4)

n

act with hψk (t)| from the left: i~c˙k = Ek ck − i~

X hψk |ψ˙n icn n



 X hψk |ψ˙n icn i~c˙k = Ek − i~hψk |ψ˙k ick − i~

(3.5)

n6=k

We can relate hψk |ψ˙n i to a matrix element of H(t) in the space of instantaneous eigenvectors H(t)|ψn i = En (t)|ψn i

(3.6)

H ˙(t)|ψn i + H(t)|ψ˙n i = E˙n (t)|ψn i + En (t)|ψ˙n i

(3.7)

take time derivative

7

multiplty by hψk (t)| with k 6= n

hence

hψk (t)|H ˙(t)|ψn i + Ek (t)hψk (t)|ψ˙n i = E˙n (t)hψk (t)|ψ˙n i

(3.8)

hψk (t)|H ˙(t)|ψn i H˙ kn hψk (t)|ψ˙n i = ≡ En − Ek En (t) − Ek (t)

(3.9)

We can use plug (3.9) back in Eq.(3.5) to get   X i~c˙k = Ek − i~hψk |ψ˙k ick − i~ n6=k

Note that if the term proportional to stay in |ψk i. If we ignore the extra term:

H˙ kn En −Ek

H˙ kn cn En − Ek

(3.10)

vanishes, then |ck | = 1, hence if you start in |ψk i you

Z   1 t Ek (t0 ) − i~hψk |ψ˙k i dt0 i~ 0  Z t   Z t  1 0 0 0 ˙ = ck (0) exp Ek (t )dt exp i ihψk |ψk idt i~ 0 0 

ck (t) = ck (0) exp

= ck (0)eiθk (t) eiγk (t)

(3.11)

where 1 θk (t) = − ~

Z

t

Ek (t0 )dt0

(3.12)

0

νk (t) = ihψk (t)|ψ˙k (t)i Z t γ(t)k = νk (t0 )dt0 0

8

(3.13) (3.14)

4

Landau-Zener transitions

Take ψ1 (x; R) and ψ2 (x; R) to be the electronic conﬁgurations of a molecule with ﬁxed nuclei separated at a distance R. We have H(R)ψi (x; R) = Ei (R)ψi (x; R)

(4.1)

Depending on R the states are polar or non polar changing characteristic near R = R0 . If R changes slowly, ψ1 remains ψ1 or ψ2 will remain ψ2 . But if R varies quickly, then ψ1 → ψ2 near R0 . This would be a non-adiabatic transition. Thinking of R as a function of time, we have a time dependent hamiltonian �  H R(t) with instantaneous energy eigenstates �  ψi x; R(t) ,

�  Ei R(t)

we can now plot the energies for constant velocity R˙

We can sometimes think of this as a 2-level problem; if no other eigenstates are relevant. So consider a slightly idealized 2 × 2 system  αt    0 1 2 H(t) = , α>0 |1i = (spin up) αt 0 −2 0   αt αt 0 , E2 (t) = − |2i = (spin down) E1 (t) = 1 2 2

9

(4.2)

The instantaneous eigenstates |1i, |2i are actually time independent. Take     1 0 |ψ1 i = |ψ2 i = 0 1

(4.3)

then    Z 1 t iαt2 0 0 |ψ1 (t)i = exp E1 (t )dt |1i = exp − |1i solves the Schr¨odinger equation (4.4) i~ 0 4~  Z t    1 iαt2 0 0 |ψ2 (t)i = exp E2 (t )dt |2i = exp + |2i solves the Schr¨odinger equation (4.5) i~ 0 4~   1 Time evolution is clear: if |ψi = at t = −∞ it will remain in that state forever. Same for 0   0 |ψi = . 1 Now make the problem more interesting by perturbing the hamiltonian  αt  H12 2 H(t) = (4.6) ∗ − αt H12 2 

with H12 constant. At t = 0 the energy eigenvalues are ±|H12 |

10

H=

αt σz + a. For r < a, however, it does not satisfy the equation; ϕ(r) is a solution everywhere only if the potential vanishes. Given an incident wave we will also have a scattered wave. Could it be an ψ(r) = eikr that propagates radially out?  fails badly for r 6= 0!! (7.1.6) ∇2 + k 2 eikr = 6 0 On the other hand

 eikr =0 for r 6= 0 . (7.1.7) r This is consistent with the radial equation having a solution u(r) = eirk in the region where the potential vanishes. Recall that the full radial solution takes the form u(r)/r. ikr Can the scattered wave therefore be e r ? Yes, but this is not general enough. We need to introduce some angular dependence. Hence our ansatz for the scattered wave is ∇2 + k 2

ψs (r) = fk (θ, φ)

eikr r

( 7.1.8)

We expect the intensity of the scattered wave to depend on direction and the function fk (θ, φ) does that. We will see that ψs is only a solution for r  a, arbitrarily far. Both the incident and the scattered wave must be present, hence the wavefunction is ψ(r) = ψs (r) + ϕ(r) ' eikz + fk (θ, φ)

eikr , r

r  a.

(7.1.9)

CHAPTER 7. SCATTERING

136

As indicated this expression is only true far away from the scattering center. We physically expect fk (θ, φ) to be determined by V (r). fk (θ, φ) is called the scattering amplitude. We now relate fk (θ, φ) to cross section!! h

# particles scattered per unit time into solid angle dΩ about (θ, φ)

i

i dσ = h particles flux of incident particles = # area · time

(7.1.10)

dσ is called differential cross section, it’s the area that removes from the incident beam the particles to be scattered into the solid angle dΩ. Let us calculate the numerator and the denominator. First the denominator, which is the probability current: Incident flux = ~ Im [ψ ∗ ∇ψ] = ~k z ˆ (7.1.11) in eikz m m Intuitively, this can be calculated by multiplying the probability density |eikz |2 = 1, by the p ~k = ~k velocity m m . The result is again an incident flux equal in magnitude to m . To calculate the numerator we first find the number of particles in the little volume of thickness dr and area r2 dΩ

dn = number of particles in this little volume

2 eikr 2 dn = |ψ(r)| d r = fk (θ, φ) r dΩ dr = |fk (θ, φ)|2 dΩ dr r 2

3

(7.1.12)

With velocity v = ~k m all these particles in the little volume will go cross out in time dt = therefore the number of particles per unit time reads dn dΩ dr ~k = |fk (θ, φ)|2 dr = |fk (θ, φ)|2 dΩ dt m v

dr v ,

Back in the formula for the cross section we get ~k  |fk (θ, φ)|2 dΩ dσ = m ~k

 m 

hence dσ = |fk (θ, φ)|2 dΩ Z Z Total cross section: σ = dσ = |fk (θ, φ)|2 dΩ

Differential cross section:

(7.1.13) (7.1.14)

7.2. PHASE SHIFTS

7.2

137

Phase shifts

Assume V (r) = V (r), so that we are dealing with a central potential. First recall the description of a free particle in spherical coordinates. With V = 0,

E=

~2 k 2 , 2m

ψ(r) =

uE` (r) Y`m (Ω) r

the Schr¨odinger equation for uE`   ~2 `(` + 1) ~2 k 2 ~2 d2 + u uE` (r) (r) = − E` 2m dr2 2m r2 2m   d2 `(` + 1) − 2+ uE` (r) = k 2 uE` (r) . dr r2 Now take ρ = kr, then (7.2.2) reads   d2 `(` + 1) − 2+ uE` (ρ) = uE` (ρ) dρ ρ2

(7.2.1)

(7.2.2)

(7.2.3)

Since k2 disappeared from the equation, the energy is not quantized. The solution to (7.2.3) is uE` (ρ) = A` ρj` (ρ) + B` ρ n` (ρ) or uE` (r) = A` rj` (kr) + B` r n` (kr) (7.2.4) where j` (ρ) is the spherical Bessel function

j` (ρ) is non singular at the origin

n` (ρ) is the spherical Neumann function

n` (ρ) is singular at the origin

Both have finite limits as ρ → ∞ 

 `π ρ j` (ρ) → sin ρ − 2   `π ρ n` (ρ) → − cos ρ − 2

(7.2.5) (7.2.6)

Now, with the plane wave a solution, we must have eikz = eikr cos θ =

∞ X

a` P` (cos θ)j` (kr)

(7.2.7)

`=0

for some coefficients a` . Using r 2` + 1 Y`,0 (θ) = P` (cos θ) 4π

and

1 j` (x) = ` 2i

Z

1

−1

eixu P` (u)du

(7.2.8)

CHAPTER 7. SCATTERING

138 it can be shown that ikz

e

=

∞ X √

2` + 1 i` Y`,0 (θ)j` (kr) .

(7.2.9)

`=0

This is an incredible relation in which a plane wave is built by a linear superposition of spherical waves with all possible values of angular momentum! Each ` contribution is a partial wave. Each partial wave is an exact solution when V = 0. We can see the spherical ingoing and outgoing waves in each partial wave by expanding (7.2.9) for large r: " #   `π `π 1 1 ei(kr− 2 ) e−i(kr− 2 ) `π j` (kr) → = sin kr − − kr 2 2ik r r

(7.2.10)

Thus ikz

e

√ `π `π ∞ 4π X √ 1 h ei(kr− 2 ) e−i(kr− 2 ) i ` = 2` + 1 i Y`,0 (θ) − , k 2i | {z r } | {z r }

ra

(7.2.11)

`=0

outgoing

7.2.1

incoming

Calculating the scattering amplitude in terms of phase shifts

Recall 1D case

ϕ(x) = sin(kx) =

 1  ikx e − e|−ikx {z } 2i

solution if V = 0

ingoing

(7.2.12) Exact solution ψ(x) =

 1  ikx 2iδk e e − e|−ikx {z } 2i same

for x > a

(7.2.13)

ingoing wave

where the outgoing wave can only differ from the ingoing one by a phase, so that probability is conserved. Finally we defined ψ(x) = ψs (x) + ϕ(x)

(7.2.14)

So do a similar transformation to write a consistent ansatz for ψ(r). We have from (7.1.9) ψ(r) ' eikz + fk (θ)

eikr , r

r  a.

(7.2.15)

7.2. PHASE SHIFTS

139

The incoming partial waves in the left-hand side must be equal to the incoming partial waves in eikz since the scattered wave is outgoing. Introducing the phase shifts on the outgoing waves of the left-hand side we get √ ψ(r) =

`π `π ∞ 4π X √ 1 h ei(kr− 2 ) e2iδk e−i(kr− 2 ) i eikr ` 2` + 1 i Y`,0 (θ) − = eikz + fk (θ) k 2i | r r r {z } | {z }

`=0

outgoing

incoming

The incoming partial waves in eikz cancel in between the two sides of the equation, and moving the outgoing partial waves in eikz into the other side we get √

∞  eikr e− i`π 2 4π X √ 1  2iδ` eikr ` 2` + 1 i Y`,0 (θ) −1 e = fk (θ) k r r |2i {z } `=0 eiδ` sin δ`

√ =

(7.2.16)

∞ eikr 4π X √ 2` + 1 Y`,0 (θ)eiδ` sin δ` , k r

(7.2.17)

`=0

where we noted that e−

i`π 2

= (−i)` and i` (−i)` = 1. Therefore we get √

fk (θ) =

∞ 4π X √ 2` + 1 Y`,0 (θ)eiδ` sin δ` . k

(7.2.18)

`=0

This is our desired expression for the scattering amplitude in terms of phase shifts. We had (7.2.19) dσ = |fk (θ)|2 dΩ and this differential cross section exhibits θ dependence. On the other hand for the full cross section the angular dependence, which is integrated over, must vanish Z Z 2 σ = |fk (θ)| dΩ = fk∗ (θ)fk (θ)dΩ Z √ 4π X √ −iδ` iδ`0 ∗ 0 = 2 (Ω)Y`0 ,0 (Ω) 2` + 1 2` + 1 e sin δ` e sin δ`0 dΩ Y`,0 k `,`0 | {z } δ``0

Hence σ=

∞ 4π X (2` + 1) sin2 δ` . k2 `=0

Now let us explore the form of f (θ) in the forward direction θ = 0. Given that r r 2` + 1 2` + 1 Y`,0 (θ) = P` (cos θ) =⇒ Y`,0 (θ = 0) = 4π 4π

(7.2.20)

CHAPTER 7. SCATTERING

140 then

√ fk (θ = 0) =

∞ 4π X √ 2` + 1 k

r

`=0

∞ 1X 2` + 1 iδ` e sin δ` = (2` + 1)eiδ` sin δ` 4π k `=0

Hence Im(f (0)) =

∞ 1X 1 k2 (2` + 1)eiδ` sin2 δ` = σ k k 4π

(7.2.21)

`=0

Therefore we have found out a remarkable relation between the total elastic cross section and the imaginary part of the forward (θ = 0) scattering amplitude. This relation is known as the Optical theorem and reads σ=

4π Im(f (0)) . k

Optical Theorem

Let us consider ways in which we can identify the phase shifts δk . Consider a solution, restricted to a fixed ` (7.2.22) ψ(x) = (A` j` (kr) + B` n` (kr)) Y`,0 (θ) x>a `

If B 6= 0 then V = 6 0. As a matter of fact, if V = 0 the solution should be valid everywhere and n` is singular at the origin, thus B` = 0. Now, let’s expand ψ(x)|` for large kr as in (7.2.10)      A` `π B` `π ψ(x) ' sin kr − − cos kr − Y`,0 (θ) (7.2.23) kr 2 kr 2 ` Define

B` ( 7.2.24) A` We must now confirm that this agrees with the postulated definition as a relative phase between outgoing and ingoing spherical waves. We thus calculate      1 `π sin δ` `π ψ(x) ' sin kr − + cos kr − Y`,0 (θ) kr 2 cos δ` 2 `   1 `π sin kr − + δ` Y`,0 (θ) ' (7.2.25) kr 2 tan δ` ≡ −

i `π 1 1 h i(kr− `π +δ` ) 2 − e−i(kr− 2 +δ` ) Y`,0 (θ) e kr 2i i `π 1 1 h i(kr− `π +2δ` ) 2 ' e−iδ e − e−i(kr− 2 ) Y`,0 (θ) kr 2i

=

(7.2.26)

7.2. PHASE SHIFTS

141

This shows that our definition of δk above is indeed consistent the expansion in (7.2.16). Finally we can read the phase shift from (7.2.25)   1 `π ψ(x) ' (7.2.27) sin kr − + δ` Y`,0 (θ) ra kr 2 ` in the partial wave solution. As an aside note that both the outgoing and ingoing components of a partial wave are separately acceptable asymptotic solutions obtained as: • If A` = −iB`      `π Y`,0 (θ) ei(kr− 2 ) `π `π ψ(x) ∼ i sin kr − + cos kr − ∼ Y`,0(θ) 2 2 kr kr ` • If A` = iB`

`π e−i(kr− 2 ) ψ(x) ∼ Y`,0 (θ) kr `

(7.2.28)

(7.2.29)

Each wave is a solution, but not a scattering one.

7.2.2

Example: hard sphere

( ∞ V (r) = 0

u = A` j` (kr) + B` n` (kr) r X ψ(a, θ) = R` (a)P` (cos θ) ≡ 0

R` (r) =

r≤a r>a

(7.2.30)

(7.2.31) (7.2.32)

`

The P` (cos θ) are complete, meaning that 0 = R` (a) = A` j` (ka) + B` n` (ka) ∀ `

(7.2.33)

Recalling tan δ` ≡ −

B` j` (ka) = A` n` (ka)

(7.2.34)

this implies all phase shifts are determined tan δ` =

j` (ka) n` (ka)

(7.2.35)

CHAPTER 7. SCATTERING

142

We can now easily compute the cross section σ, which, recalling (7.2.20) is proportional to sin2 δ` j`2 (ka) tan2 δ` = 2 (7.2.36) sin2 δ` = 2 1 + tan δ` j` (ka) + n2` (ka) hence

∞ ∞ j`2 (ka) 4π X 4π X 2 σ= 2 (2` + 1) sin δ` = 2 (2` + 1) 2 k k j` (ka) + n2` (ka) `=0

(7.2.37)

`=0

Griffiths gives you the low energy ka  1 expansion  ` 4 ∞ 2 `! 4π X 1 σ' 2 (ka)4`+2 k 2` + 1 (2`)!

(7.2.38)

`=0

At low energy the dominant contribution is from ` = 0 σ'

4π (ka)2 = 4πa2 k2

Full area of the sphere! Not the cross section!

(7.2.39)

One more calculation! i tan δ =

eiδ − e−iδ e2iδ − 1 = eiδ + eiδ e2iδ + 1

Therefore e2iδ` =

e2iδ` =

7.2.3

1 + i nj``(ka) (ka) 1−

i nj``(ka) (ka)

=⇒

=

e2iδ =

1 + i tan δ 1 − i tan δ

n + ij i(j − in) = n − ij −i(j + in)

j` (ka) − in` (ka) , j` (ka) + in` (ka)

Hard sphere phase shifts

(7.2.40)

(7.2.41)

(7.2.42)

General computation of the phase shift

Suppose you have a radial solution R` (r) known for r ≤ a (V (r ≥ a) = 0). This must be matched to the general solution that holds for V = 0, as it holds for r > a:

At r = a must match the function and its derivative R` (a) = A` j` (ka) + B` n` (ka)

(7.2.43)

7.2. PHASE SHIFTS

143   aR`0 (a) = ka A` j`0 (ka) + B` n0` (ka)

(7.2.44)

Let’s form the ratio and define β` as the logarithmic derivative of the radial solution ` 0 j` (ka) + B j` (ka) − tan δ` η`0 (ka) aR`0 (a) A` j` (ka) + B` η`0 (ka) A` η` (ka) = ka = ka = ka ` R` (a) A` j` (ka) + B` n` (ka) j` (ka) − tan δ` n` (ka) j` (ka) + B A` n` (ka) (7.2.45) In principle we can use eq. (7.2.45) to get tan δ`, but let’s push the calculation further to extract the phase shift from e2iδ`   j` eiδ` + e−iδ` + iη`0 eiδ` − e−iδ` j` cos δ` − η`0 sin δ` β` = ka = ka j` cos δ` − n` sin δ` j` (eiδ` + e−iδ` ) + in` (eiδ` − e−iδ` )

β` ≡

= ka

e2iδ` (j` + iη`0 ) + (j` − iη`0 ) eiδ` (j` + iη`0 ) + e−iδ` (j` − iη`0 ) = ka eiδ` (j` + in` ) + e−iδ` (j` − in` ) e2iδ` (j` + in` ) + (j` − in` )

Solve for e2iδ` e2iδ` = −

ka(j` − iη`0 ) − β` (j` − in` ) ka(j` + iη`0 ) − β` (j` + in` )

(7.2.46)

(7.2.47)

which can be rewritten as e2iδ` = −



    β − ka j` −iη`0 j −in j` − in`  `  ` 0`   j +iη j` + in` β` − ka j``+in``

(7.2.48)

which we can also write as  e2iδ` =

2iξ` e|{z}

Hard sphere phase shift

β` − ka



j` −iη`0 j` −in`



β` − ka



j` +iη`0 j` +in`



(7.2.49)

Phase shifts are useful when a few of them dominate the cross section. This happens when ka < 1 with a the range of the potential. So short range and/or low energy. 1. Angular momentum of the incident particle is L = b · p with b impact parameter and p momentum

L ' bp

~` ' b~k

b'

` k

(7.2.50)

CHAPTER 7. SCATTERING

144 When b > a there is no scattering ` >a k

Expect no scattering for ` > ka

(7.2.51)

Thus only values ` ≤ ka contribute to σ. 2. Confirm the impact parameter intuition from partial wave expression. Consider the free partial wave ∼ j` (kr)Y`,0 . The impact parameter b` of such a wave would be estimated to be `/k. Recall that j` (kr) is a solution of the V = 0 radial equation with angular momentum ` and energy ~2 k 2 /(2m). Setting the effective potential equal to the energy: ~2 `(` + 1) ~2 k 2 = 2mr2 2m

(7.2.52)

we find that the turning point for the solution is at k 2 r2 = `(` + 1) . Thus we expect j` (kr) to be exponentially small beyond the turning point p kr ≤ `(` + 1) ' `

(7.2.53)

(7.2.54)

confirming that the wave is negligible for r less than the impact parameter `/k!

Figure 7.2: Plot of ρ2J 2 ` (ρ)

7.3

Integral scattering equation

Some useful approximations can be made when we reformulate the time-independent Schr¨odinger equation as an integral equation. These approximations, as opposed to the partial waves method, allow us to deal with potentials that are not spherically symmetric. We consider the Schr¨ odinger equation   ~2 2 − ∇ + V (r) ψ(r) = Eψ(r) , (7.3.1) 2M

7.3. INTEGRAL SCATTERING EQUATION

145

set the energy equal to that of a plane wave of wavenumber k and rewrite the potential in terms of a rescaled version U (r) that simplifies the units: E=

~2 k 2 2M

and V (r) =

~2 U (r) 2M

(7.3.2)

then the Schr¨odinger equation reads   −∇2 + U (r) ψ(r) = k 2 ψ(r)

(7.3.3)

 ∇2 + k 2 ψ(r) = U (r)ψ(r)

(7.3.4)

rewrite as This is the equation we want to solve. Let us introduce G(r − r0 ), a Green function for the operator ∇2 + k 2 , i.e.  ∇2 + k 2 G(r − r0 ) = δ (3) (r − r0 )

(7.3.5)

Then we claim that any solution of the integral equation Z ψ(r) = ψ0 (r) + d3 r0 G(r − r0 )U (r0 )ψ(r0 )

(7.3.6)

where ψ0 (r) is a solution of the homogeneus equation  ∇2 + k 2 ψ0 (r) = 0

(7.3.7)

is a solution of equation (7.3.4). Let’s check that this is true Z   2 2 2 2 ∇r + k ψ(r) = ∇r + k d3 r0 G(r − r0 )U (r0 )ψ(r0 ) Z = d3 r0 δ (3) (r − r0 )U (r0 )ψ(r0 ) = U (r)ψ(r) To find G we first recall that  ±ikr  Ce for r = 6 0 r G(r) ∼ − 1 for r → 0 4πr

(7.3.8)

X

 e±ikr

as a matter of fact

∇2r + k 2

as a matter of fact

 1 ∇2 − 4πr = δ 3 (r)

r

= 0 ∀ r 6= 0

(7.3.9) Thus try G± (r) = −

1 e±ikr 4π r

(7.3.10)

CHAPTER 7. SCATTERING

146

and we are going to refer to G+ as the outgoing wave Green function and to G− as the ingoing wave Green function.  1 Let’s verify that this works, write G as a product: G± (r) = e±ikr − 4πr , so that           1 1 1 2 2 ±ikr ±ikr 2 ±ikr ~ ~ − ∇ G± (r) = ∇ e + e ∇ − + 2 ∇e ·∇ − 4πr 4πr 4πr (7.3.11) Recall that r ∇2 = ∇ · ∇ , ∇r = , ∇ · (f A) = ∇f · A + f ∇ · A (7.3.12) r Therefore   r ±ikr 2ik ±ikr 2 ±ikr 2 e±ikr ∇e = ±ik e , ∇ e = −k ± (7.3.13) r r then      1 r  r  2ik ±ikr 2 2 e − + e±ikr δ 3 (r) + 2 ±ike±ikr · ∇ G± (r) = −k ± r 4πr r 4πr3     2ik ±ikr 2ik ±ikr = −k 2 G± (r) ∓ e + δ 3 (r) ± e  4πr  4πr = −k 2 G± (r) + δ 3 (r) X

(7.3.14)

We will use ψ0 (r) = eikz

and

G = G+

(7.3.15)

Thus, with these choices, (7.3.6) takes the form Z ψ(r) = eikz + d3 r0 G+ (r − r0 )U (r0 )ψ(r0 ) where

(7.3.16)

0

1 eik|r−r | (7.3.17) 4π |r − r0 | We now want to show that this is consistent with our asymptotic expansion for the energy eigenstates. For that we can make the following approximations G+ (r − r0 ) = −

For the G+ denominator:

0

(7.3.18) 0

|r − r | ' r − n · r

For the G+ numerator: where

n≡

|r − r0 | ' r

r r

(7.3.19)

7.3. INTEGRAL SCATTERING EQUATION

147

In this way G+ (r − r0 ) = − Thus

1 ikr −ikn·r0 e e 4πr

(7.3.20)

 ikr Z 1 e 3 0 −ikn·r0 0 0 ψ(r) = e + − d re U (r )ψ(r ) (7.3.21) 4π r The object in brackets is a function of the unit vector n in the direction of r. This shows that the integral equation, through the choice of G, incorporates both the Schr¨ odinger equation and the asymptotic conditions. By definition, the object in brackets is fk (θ, φ), i.e. Z 1 0 fk (θ, φ) = − (7.3.22) d3 r0 e−ikn·r U (r0 )ψ(r0 ) . 4π ikz



Of course, this does not yet determine fk as the undetermined wavefunction ψ(r) still appears under the integral. The incident wave has the form of eiki ·r with ki the incident wave number, |ki | = k. For the outgoing wave we define ks ≡ nk (in the direction of n), the scattered wave vector. The expression for ψ(r) then becomes   ikr Z 1 e iki ·r 3 0 −iks ·r0 0 0 (7.3.23) ψ(r) = e + − d re U (r )ψ(r ) 4π r We will do better with the Born approximation.

7.3.1

The Born approximation

Consider the original integral expression: Z iki ·r ψ(r) = e + d3 r0 G+ (r − r0 )U (r0 )ψ(r0 )

(7.3.24)

Rewrite by just relabeling r → r0 0

ψ(r0 ) = eiki ·r +

Z

d3 r00 G+ (r0 − r00 )U (r00 )ψ(r00 ) .

(7.3.25)

Now plug (7.3.25) under the integral in (7.3.24) to get Z Z Z iki ·r 0 0 3 0 0 0 iki ·r0 3 0 ψ(r) = e + d r G+ (r−r )U (r )e + d r G+ (r−r )U (r ) d3 r00 G+ (r0 −r00 )U (r00 )ψ(r00 ) (7.3.26) Repeat the trick once more to find Z 0 ψ(r) = eiki ·r + d3 r0 G+ (r − r0 ) U (r0 )eiki ·r Z Z 00 + d3 r0 G+ (r − r0 )U (r0 ) d3 r00 G+ (r0 − r00 ) U (r00 )eiki ·r Z Z Z + d3 r0 G+ (r − r0 )U (r0 ) d3 r00 G+ (r0 − r00 )U (r00 ) d3 r000 G+ (r000 − r00 )U (r000 )ψ(r000 ) (7.3.27)

CHAPTER 7. SCATTERING

148

By iterating this procedure we can form an infinite series which schematically looks like ψ = eiki ·r +

Z

G U eiki r +

Z

Z GU

G U eiki r +

Z

Z GU

Z GU

G U eikir + . . .

(7.3.28)

The approximation in which we keep the first integral in this series and set to zero all others is called the first Born approximation. The preparatory work was done in (7.3.23), so we have now Z  ikr 1 e Born iki ·r 3 0 −iks ·r0 0 iki ·r0 − ψ (r) = e d re U (r )e (7.3.29) 4π r hence fkBorn (θ, φ) = −

1 4π

Z

d3 re−iK·r U (r) .

(7.3.30)

Here we defined the wave-number transfer K:

|ki | = |ks | ,

K ≡ ks − ki

(7.3.31)

Note that we eliminated the primes on the variable of integration – it is a dummy variable after all. The wave-number transfer is the momentum that must be added to the incident one to get the scattered one. We call θ the angle between ki and ks (it is the spherical angle θ if ki is along the positive z-axis). Note that K = |K| = 2k sin 2θ

(7.3.32)

In the Born approximation the scattering amplitude fk (θ, φ) is simply the Fourier transform of U (r) evaluated at the momentum transfer K! fk (θ, φ) captures some information of V (r). If we have a central potential V (r) = V (r), we can simplify the expression for the Born scattering amplitude further by performing the radial integration. We have fkBorn (θ) = −

1 2m 4π ~2

Z

d3 re−iK·r V (r) .

(7.3.33)

By spherical symmetry this integral just depends on the norm K of the vector K. This is why we have a result that only depends on θ: while K is a vector that depends on both θ and φ, its magnitude only depends on θ. To do the integral think of K fixed and let Θ be

7.3. INTEGRAL SCATTERING EQUATION

149

the angle with r

fk (θ) = −

m 2π~2

m =− 2 ~

Z 0

Z

2πdr r2

1

d(cos Θ)e−iKr cos Θ V (r)

−1

0

Z

e−iKr − eiKr m dr r V (r) =− 2 −iKr ~ 2

Z



2

dr r V (r) 0

2 sin(Kr) Kr

 (7.3.34)

hence fk (θ) = −

2m K~2

Z

dr rV (r) sin(Kr)

(7.3.35)

0

with K = 2k sin 2θ . The Born approximation treats the potential as a perturbation of the free particle waves. This wave must therefore have kinetic energies larger than the potential. So most naturally, this is a good high-energy approximation. Example: Yukawa potential

Given

V (r) = V (r) = β

e−µr r

β, µ > 0

(7.3.36)

from (7.3.35), one gets 2mβ fk (θ) = − K~2

Z 0

dr e−µr sin(Kr) = −

2mβ . + K 2)

~2 (µ2

(7.3.37)

We can give a graphical representation of the Born series. Two waves reach the desired point r. The first is the direct incident wave. The second is a secondary wave originating at the scattering “material” at a point r0 . The amplitude of a secondary source at r0 is given by the value of the incident wave times the density U (r0 ) of scattering material at r0 . In the second figure again an incident wave hits r. The secondary wave now takes two steps: the incident wave hits scattering material at r00 which then propagates and hits scattering material at r0 , from which it travels to point r.

150

CHAPTER 7. SCATTERING

Figure 7.3: Pictorial view of Born approximation. Wave at r is the sum of the free incident wave at r, plus an infinite number of waves coming from the secondary source at r0, induced by the incident wave.

Figure 7.4: Pictorial view of second order Born term

Figure 7.5: Given (θ, φ) and k (∼ energy), K is determined.

MIT OpenCourseWare https://ocw.mit.edu

8.06 Quantum Physics III

Spring 2018

Chapter 8

Identical Particles c B. Zwiebach

Two particles are identical if all their intrinsic properties (mass, spin, charge, magnetic moment, etc.) are the same and therefore no experiment can distinguish them. Of course, two identical particles can have different momentum, energy, angular momentum. For example all electrons are identical, all protons, all neutrons, all hydrogen atoms are identical (the possible excitation states viewed as energy, momentum, etc.)

8.1

Identical particles in Classical Mechanics

We can assign a labeling of the particles and follow through. Results are labeling independent. As shown in Fig.8.2 we have two cases: • Case 1: Assume we solve the dynamics and find r1 (t) = r(t) 0

r2 (t) = r (t)

with

r(t0 ) = r0

with

r0 (t0 ) = r00

• Case 2: Since the particles are identical the Hamiltonian must make this clear H(r1 , p1 ; r2 , p2 ) = H(r2 , p2 ; r1 , p1 )

(8.1.1)

This time, when we solve r1 (t) = r0 (t)

with

r0 (t0 ) = r00

r2 (t) = r(t)

with

r(t0 ) = r0

The two descriptions are equivalent. We can follow the particles and the particle that started at {r0 , p0 } will be at r(t), while the one that started at {r00 , p00 } will be at r0 (t). In conclusion, we choose one labeling, just as if the particles were different. Follow through without regard to other possible labeling. 151

CHAPTER 8. IDENTICAL PARTICLES

152

Figure 8.1: Example of two different ways of labeling the initial state

8.2 8.2.1

Identical particles in Quantum Mechanics Exchange degeneracy

If we cannot follow particles once they overlap and/or collide, we can’t know what alternative took place. Question: How to write kets for the initial and final states? Simpler case to see the complications Let there be two spin-1/2 particles. Consider only their spin1 , one is |+i, the other |−i. Recall tensor product |vi i(1) ⊗ |vj i(2) describing particle 1 in state vi and particle 2 in state vj . More briefly |vi i(1) ⊗ |vj i(2) ≡ |vi i ⊗ |vj i with the understanding that the first ket is for particle 1 and the second for particle 2. What is the state of the two spin 1/2 particles? |+i(1) ⊗ |−i(2)

or

|−i(1) ⊗ |+i(2)

(8.2.1)

A priori either one! This, despite the fact that in our conventions for inner products these two states are orthogonal! Can we declare these states to be physically equivalent and thus solve the ambiguity? No. If the two states above are equivalent we would have to admit that even the states (8.2.2) |ψi = α|+i ⊗ |−i + β|−i ⊗ |+i 1 in this approximation we could say that they are static and very close to each other, even on top of each other

8.2. IDENTICAL PARTICLES IN QUANTUM MECHANICS

153

Figure 8.2: Example of two different processes involving identical particles. In principle, in quantum mechanics it is impossible to tell which one happened. with |α|2 + |β|2 = 1 for normalization, are equivalent. This ambiguity in the specification of a state of identical particles is called exchange degeneracy. What is the probability to find these particles in the |+, χi ⊗ |+, χi state? This states is   1  1  |ψ0 i = √ |+i(1) + |−i(1) ⊗ √ |+i(2) + |−i(2) 2 2  1 = |+i(1) ⊗ |+i(2) + |+i(1) ⊗ |−i(2) + |−i(1) ⊗ |+i(2) + |−i(1) ⊗ |−i(2) (8.2.3) 2 the probability is 2 1 2 hψ0 |ψi = (α + β) 2

(8.2.4)

thus a tremendous ambiguity because the chosen values of α, β matter! Three particle degeneracy; 3 different eigenstates |ai, |bi, |ci give the following combinations |ai(1) ⊗ |bi(2) ⊗ |ci(3) ,

|ai(1) ⊗ |ci(2) ⊗ |bi(3) ,

|bi(1) ⊗ |ci(2) ⊗ |ai(3) ,

|bi(1) ⊗ |ai(2) ⊗ |ci(3) ,

|ci(1) ⊗ |ai(2) ⊗ |bi(3) ,

|ci(1) ⊗ |bi(2) ⊗ |ai(3) .

CHAPTER 8. IDENTICAL PARTICLES

154

8.2.2

Permutation operators

Two particle systems Consider the case when the vector space V relevant to the particles is the same for both particles, even though the particles may be distinguishable. Call the particles 1 and 2. Consider the state in which particle one is in state ui and particle 2 in state uj |ui i(1) ⊗ |uj i(2) ∈ V ⊗ V

(8.2.5)

Since the particles are possibly distinguishable note that |ui i(1) ⊗ |uj i(2) 6= |uj i(1) ⊗ |ui i(2) . Define Pˆ21 the linear operator on V ⊗ V such that h i Pˆ21 |ui i(1) ⊗ |uj i(2) ≡ |uj i(1) ⊗ |ui i(2) .

(8.2.6)

(8.2.7)

note that Pˆ21 Pˆ21 = 1 ,

(8.2.8)

† Pˆ21 = Pˆ21 , Hermitian

(8.2.9)

i.e. Pˆ21 is its own inverse. Claim

ˆ the adjoint of O ˆ is defined s.t. Proof: First of all let’s recall that for a generic operator O, ˆ † |βi = hβ|O|αi ˆ hα|O

(8.2.10)

In our case, not writing the subscript labels,   huk | ⊗ hu` |Pˆ21 |ui i ⊗ |uj i = huk | ⊗ hu` | |uj i ⊗ |ui i = δkj δ`i  ∗ h i∗ i∗ h † huk | ⊗ hu` |Pˆ21 |ui i ⊗ |uj i = hui | ⊗ huj |Pˆ21 |uk i ⊗ |u` i = hui | ⊗ huj | |u` i ⊗ |uk i = δi` δjk = δkj δ`i hence † Pˆ21 = Pˆ21

X.

(8.2.11)

Because of (8.2.8) we also have that Pˆ21 is unitary: † ˆ Pˆ21 P21 = Pˆ21 Pˆ21 = 1 .

(8.2.12)

Given a generic state |ψi it is not clear a priori what would it be its behaviour under the action of Pˆ21 , hence to make our life easier we want to rewrite a generic state |ψi in

8.2. IDENTICAL PARTICLES IN QUANTUM MECHANICS

155

terms of eigenstates of Pˆ21 . We can define two eigenstates of Pˆ21 with the following properties Pˆ21 |ψS i = |ψS i Symmetric state Pˆ21 |ψA i = −|ψA i Antisymmetric state and two operators Sˆ and Aˆ 1 Sˆ ≡ (1 + Pˆ21 ) , 2

1 Aˆ ≡ (1 − Pˆ21 ) . 2

Note that Sˆ and Aˆ are such that 1 Pˆ21 Sˆ = (Pˆ21 + Pˆ21 Pˆ21 ) = 2 1 Pˆ21 Aˆ = (Pˆ21 − Pˆ21 Pˆ21 ) = 2 Therefore, given a generic state |ψi we have that ˆ ˆ Pˆ21 S|ψi = S|ψi ˆ ˆ Pˆ21 A|ψi = −A|ψi

(8.2.13)

1 ˆ (P21 + 1) = Sˆ 2 1 ˆ (P21 − 1) = −Aˆ 2 ˆ S|ψi is symmetric ˆ A|ψi is anti-symmetric.

=⇒ =⇒

Because of this, the Hermitian operators Sˆ and Aˆ are called symmetric/antisymmetric projectors. From a mathematical point of view Sˆ and Aˆ are orthogonal projectors2 and satisfy Sˆ2 = Sˆ , Aˆ2 = Aˆ , Sˆ + Aˆ = 1 , SˆAˆ = AˆSˆ = 0 . (8.2.15) Action on operators Let B(n) be an operator acting on the n-th vector space, i.e.   B(1)|ui i(1) ⊗ |uj i(2) = B|ui i ⊗ |uj i(2) (1)   B(2)|ui i(1) ⊗ |uj i(2) = |ui i(1) ⊗ B|uj i (2)

then the action of Pˆ21 on B(1) is † Pˆ21 B(1)Pˆ21 |ui i(1) ⊗ |uj i(2) = Pˆ21 B(1)|uj i(1) ⊗ |ui i(2)     = Pˆ21 B|uj i ⊗ |ui i(2) = |ui i(1) ⊗ B|uj i (1)

(2)

= B(2)|ui i ⊗ |uj i (8.2.16)

2

A projector P : V → U ⊂ V is orthogonal if V = ker P ⊕ range P , with ker P ⊥ range P . Take v ∈ V , then v = |{z} Pv +v − Pv = u + w (8.2.14) | {z } rangeP

ker P

where w ∈ ker P , u = P v ∈ rangeP , then ker P ⊥ rangeP because hw, ui = hw, P vi = hP † w, vi = hP w, vi = h0, vi = 0

CHAPTER 8. IDENTICAL PARTICLES

156 hence

† Pˆ21 B(1)Pˆ21 = B(2)

(8.2.17)

† Pˆ21 B(2)Pˆ21 = B(1)

(8.2.18)

Similarly we have ˆ 2), then and if we consider a generic operator Θ(1, ˆ 2)Pˆ † = Θ(2, ˆ 1) . Pˆ21 Θ(1, 21

(8.2.19)

ˆ 2) = Θ(2, ˆ 1) we say Θ(1, ˆ 2) is symmetric. If an operator is symmetric then Note that if Θ(1, ˆ 2)Pˆ † − Θ(1, ˆ 2) 0 = Pˆ21 Θ(1, 21 ˆ 2) − Θ(1, ˆ 2)Pˆ21 0 = Pˆ21 Θ(1, ˆ 2)] 0 = [Pˆ21 , Θ(1,

(8.2.20)

ˆ 2)] = 0 ⇐⇒ Θ ˆ is symmetric . [Pˆ21 , Θ(1,

(8.2.21)

N particle systems In a N particle system we can define N ! permutation operators Pˆi1 ,...,iN , with Pˆ12...N being the identity. For a 3-particle system, for example, the operator Pnpq acting on a state has the effect of

Pˆnpq

means

n→1

n − th state moved to position 1

p→2

p − th state moved to position 2

q→3

q − th state moved to position 3

e.g. Pˆ231 |ui i(1) ⊗ |uj i(2) ⊗ |uk i(3) = |uj i(1) ⊗ |uk i(2) ⊗ |ui i(3)

(8.2.22)

You should check that its inverse is Pˆ312 , so that Pˆ231 Pˆ312 = 1 .

(8.2.23)

More formally we can define a permutation of N numbers by the function α that maps the standard ordered integers 1, . . . , N into some arbitrary ordering of them α : [1, 2, . . . , N ] → [α(1), α(2), . . . , α(N )]

(8.2.24)

and associate it with a permutation operator Pˆα ≡ Pˆα(1), α(2), ..., α(N )

(8.2.25)

Pˆα |u1 i(1) ⊗ · · · ⊗ |uN i(N ) = |uα(1) i(1) ⊗ · · · ⊗ |uα(N ) i(N )

(8.2.26)

8.2. IDENTICAL PARTICLES IN QUANTUM MECHANICS

157

For example Pˆ3142 |u1 i(1) ⊗ |u2 i(2) ⊗ |u3 i(3) ⊗ |u4 i(4) = |u3 i(1) ⊗ |u1 i(2) ⊗ |u4 i(3) ⊗ |u2 i(4)

(8.2.27)

Pˆ3142 |ai(1) ⊗ |bi(2) ⊗ |ci(3) ⊗ |di(4) = |ci(1) ⊗ |ai(2) ⊗ |di(3) ⊗ |bi(4)

(8.2.28)

or The set of all permutations of N objects forms the symmetric group SN and it has N ! elements. For S3 we have 6 elements or the 6 permutation operators: cyclic cyclic cyclic cyclic Pˆ123 → Pˆ312 → Pˆ231 , Pˆ132 → Pˆ213 → Pˆ321 |{z} | {z } 1

these are transpositions, a permutation in which only 2 particles are exchanged without affecting the rest

Pˆ132 is a transposition in which the states of the second and third particles are exchanged while the first particle is left unchanged. For transpositions we sometimes use the notation where we just indicate the two labels that are being transposed. Those two labels could be written in any order without risk of confusion, but we will use ascending order: (12) ≡ P213 (13) ≡ P321

(8.2.29)

(23) ≡ P132 While all permutations that are transpositions are Hermitian (see the proof for Pˆ21 that easily generalizes), general permutations are not Hermitian. It is intuitively clear that any permutation can be written as product of transpositions: any set of integers can be reordered into any arbitrary position by transpositions (in fact by using transpositions of consecutive labels). The decomposition of a permutation into a product of transpositions is not unique, but it is unique (mod 2). Hence we have that every permutation is either even or odd. A permutation is said to be even if it is the product of an even number of transpositions, and it is said to be odd if it is the product of an odd number of transpositions. Since transposition operators are and unitary any permutation is a unitary operator. All transpositions are also Hermitian, but an arbitrary product of them is not hermitian because the transpositions do not necessarily commute. In fact, the Hermitian conjugate of a permutation is its inverse, which is a permutation of the same parity. This is clear from writing Pˆα as a product of transpositions Pti : Pˆα = Pˆt1 Pˆt2 . . . Pˆtk Pˆα† = Pˆt†k . . . Pˆt†2 Pˆt†1 = Pˆtk . . . Pˆt2 Pˆt1

(8.2.30)

=⇒ Pˆα Pˆα† = 1

(8.2.31)

and therefore

CHAPTER 8. IDENTICAL PARTICLES

158

Theorem 8.2.1. The number of even permutations is the same as the number of odd permutations Proof. Consider the map that multiplies any permutation by (12) from the left (12) : Peven → Podd so that if σ ∈ Peven then (12)σ ∈ Podd . This map is one to one (12)σ = (12)σ 0 =⇒ σ = σ 0 ,

(8.2.32)

by multiplying from the left by (12), which is the inverse of (12). This map is also surjective or onto:for any β ∈ Podd , we have β = (12) (12)β . | {z } ∈Peven

A\B Pˆ312 Pˆ231

Pˆ312 Pˆ231

(23) (12) (13)

1 (13) (23) (12)

Pˆ231

(23)

(12)

(13)

1 ˆ P312 (12) (13) (23)

(12) (13) 1 ˆ P312 Pˆ231

(13) (23) Pˆ231 1 Pˆ312

(23) (12) Pˆ312 Pˆ231 1

Table 8.1: A · B matrix for S3. Complete symmetrizer and antisymmetrizer Permutation operators do not commute, so we can’t find complete basis of states that are eigenstates of all permutation operators. It is possible, however, to find some states that are simultaneous eigenvectors of all permutation operators. Consider N particles, each with the same vector space. Let Pˆα be an arbitrary permutation, then Symmetric state |ψS i : Antisymmetric state |ψA i : where

( +1 α = −1

if if

Pˆα |ψS i = |ψS i ∀α Pˆα |ψA i = α |ψA i

Pˆα is an even permutation Pˆα is an odd permutation

(8.2.33)

In the total Hilbert space V ⊗N ≡ V ⊗ · · · ⊗ V , we can identify a subspace SymN V ⊂ V ⊗N | {z } N

of symmetric states and a subspace AntiN V ⊂ V ⊗N of antisymmetric states. Can we construct projectors into such subspaces? Yes! 1 Xˆ 1 X ˆ Sˆ ≡ Pα and Aˆ ≡ α P α (8.2.34) N! α N! α

8.2. IDENTICAL PARTICLES IN QUANTUM MECHANICS

159

where we sum over all N ! permutations. Sˆ is called the symmetrizer and Aˆ is called the antisymmetrizer. Aˆ = Aˆ† (8.2.35) Claim: Sˆ = Sˆ† −1 Hermitian conjugation of Pˆα gives Pˆα which is even if Pˆα is even and odd if Pˆα is odd. Thus Hermitian conjugation just rearranges the sums, leaving them invariant. Moreover Pˆα Sˆ = SˆPˆα = Sˆ (8.2.36) 0

0

Pˆα0 Aˆ = AˆPˆα0 = α0 Aˆ

(8.2.37)

Proof. Note that Pˆα0 acting on the list of permutations simply rearranges the list, given two permutations Pγ1 6= Pγ2 then Pˆα0 Pγ1 6= Pˆα0 Pγ2 , hence 1 Xˆ 1 Xˆ ˆ 1 Xˆ Pˆα0 Sˆ = Pˆα0 Pα = P α0 P α = Pβ = Sˆ X ( 8.2.38) N! α N! α N! β

analogously 1 X ˆ ˆ 1 X 1 X ˆ α P α = α P α 0 P α = α α0 α0 Pˆα0 Pˆα Pˆα0 Aˆ = Pˆα0 | {z } N! α N! α N! α =1 α 0 X α 0 X ˆ ˆ ˆ ˆ β P β = α 0 A X α α 0 P α 0 P α = = N! α N!

(8.2.39)

β

Finally they are projectors Sˆ2 = Sˆ ,

Aˆ2 = Aˆ ,

SˆAˆ = AˆSˆ = 0

1 Xˆ ˆ 1 Xˆ 1 Sˆ2 = Pα S = S= N ! Sˆ = Sˆ X N! α N! α N! 1 X ˆ ˆ 1 X 1 X ˆ 1 Aˆ2 = α P α A = α α Aˆ = A= N ! Aˆ = Aˆ N! α N! α N! α N! 1 X ˆ ˆ 1 X ˆ Sˆ X AˆSˆ = α P α S = α S = α = 0 N! α N! α N! α

X

(8.2.40) (8.2.41) X

(8.2.42) (8.2.43)

Since, as explained before there are equal numbers of even and odd permutations, i.e. P  = 0. α α Note that ˆ ˆ ˆ S|ψi ∈ SymN V since Pˆα S|ψi = S|ψi ∀α (8.2.44) and analogously ˆ A|ψi ∈ AntiN V

since

ˆ ˆ Pˆα A|ψi = α A|ψi ∀α

(8.2.45)

Hence, they are, as claimed, projectors into the symmetric and antisymmetric subspaces: Sˆ : V ⊗N → SymN V , Aˆ : V ⊗N → AntiN V (8.2.46)

CHAPTER 8. IDENTICAL PARTICLES

160 Example: N = 3

For S3 we have 6 elements or the 6 permutation operators: Pˆ123 = 1 , Pˆ312 , Pˆ231 , Pˆ132 , Pˆ213 , Pˆ321

In this case the symmetrizer and antisymmetrizer operators Sˆ and Aˆ are Sˆ = 16 (1 + Pˆ312 + Pˆ231 + Pˆ132 + Pˆ213 + Pˆ321 )

(8.2.47)

Aˆ = 61 (1 + Pˆ312 + Pˆ231 − Pˆ132 − Pˆ213 − Pˆ321 )

(8.2.48)

6 1 = Pˆ123 Sˆ + Aˆ = 13 (1 + Pˆ312 + Pˆ231 ) =

(8.2.49)

Note that

This is a manifestation of the fact that in general for N > 2, we have SymN V ⊕ AntiN V ⊂ V ⊗N

(8.2.50)

i.e. in principle, the N -particle Hilbert space is not spanned by purely symmetric or antisymmetric states. We define Θ(1, 2, . . . , N ) to be a completly symmetric observable if [Θ(1, 2, . . . , N ), Pˆα ] = 0

8.3

∀α

(8.2.51)

The symmetrization postulate

In a system with N identical particles the states that are physically realized are not arbitrary states in V ⊗N , but rather they are totally symmetric (i.e. belong to SymN V ), in which case the particles are said to be bosons, or they are totally antisymmetric (i.e. belong to AntiN V ) in which case they are said to be fermions. Comments: 1. The above is a statement of fact in 3D. Alternative possibilities can happen in worlds with 2 spatial dimensions (anyons) 2. The postulate describes the statistical behaviour of bosons and of fermions 3. Spin-statistics theorem from Quantum Field Theory shows that bosons are particles of integer spins (0, 1, 2, . . . ) while fermions are particles of half-integer spin (1/2, 3/2 , ...) 4. The symmetrization postulate for elementary particles lead to a definite character, as bosons or fermions, for composite particles, which in turn obey the symmetrization

8.3. THE SYMMETRIZATION POSTULATE

161

postulate. Take for example two hydrogen atoms

The system is made by 4 particles and its total wavefunction is Ψ(p1 , e1 ; p2 , e2 ). Since the two electrons are identical particles of spin 1/2, the wavefunction must be antisymmetric under the exchange e1 ↔ e2 Ψ(p1 , e2 ; p2 , e1 ) = −Ψ(p1 , e1 ; p2 , e2 ) .

(8.3.1)

Exactly the same argument applies to the protons, Ψ(p2 , e1 ; p1 , e2 ) = −Ψ(p1 , e1 ; p2 , e2 ) .

(8.3.2)

Therefore under the simultaneous exchange Ψ(p2 , e2 ; p1 , e1 ) = +Ψ(p1 , e1 ; p2 , e2 ) .

(8.3.3)

The exchange i n ( 8.3.3) corresponds t o p1 ↔p2 and e 1 ↔e 2, an exchange of t he t wo hydrogen atoms! Since t he wavefunction i s s ymmetric under t his exchange, ( 8.3.3) shows that the hydrogen atom is a boson! 5. The symmetrization postulate solves the exchange degeneracy problem.  Say |ui ∈ V ⊗N represents mathematically a state. Let V|ui ≡ span Pˆα |ui , ∀α . Depending on |ui the dimension of V|ui can go from 1 to N !. This dimensionality, if different from one, is the degeneracy due to exchange. The exchange degeneracy problem (i.e. the ambiguity in finding a representative for the physical state in V|ui ) is solved by the symmetrization postulate by showing that V|ui contains, up to scale, a single ket of SymN V and a single ket of AntiN V . Proof. Suppose we have two states |ψi, |ψ 0 i ∈ V|ui that both happen to be symmetric: |ψi, |ψ 0 i ∈ SymN V . We can write them as X X |ψi = cα Pα |ui and |ψ 0 i = c0α Pα |ui (8.3.4) α

α

CHAPTER 8. IDENTICAL PARTICLES

162

with cα , c0α some coefficients. Then, since |ψi ∈ SymN V X X X X ˆ ˆ α |ui = ˆ ˆ |ψi = S|ψi = Sˆ cα Pα |ui = cα SP cα S|ui = S|ui cα . α

α

α

(8.3.5)

α

Analogously ˆ |ψ 0 i = S|ui

X

c0α .

(8.3.6)

α

Hence |ψi ∝ |ψ 0 i, the states are the same up to scale. 6. Building antisymmetric states. ˆ Constructing the three-particle state A|ui with |ui ∈ V ⊗3 given by |ui = |ϕi(1) ⊗ |χi(2) ⊗ |ωi(3) The claim is that the antisymmetric state is constructed by a determinant: |ϕi(1) |ϕi(2) |ϕi(3) X 1 1 ˆ A|ui = 3! α Pˆα |ϕi(1) ⊗ |χi(2) ⊗ |ωi(3) = 3! |χi(1) |χi(2) |χi(3) α |ωi (1) |ωi(2) |ωi(3)

(8.3.7)

(8.3.8)

When writing the products in the determinant one must reorder each term to have the standard order |· i(1) ⊗ |· i(2) ⊗ |· i(3) . You can confirm you get the right answer. Now do it generally. Recall the formula for the determinant of a matrix X ˆ= det B α Bα(1),1 Bα(2),2 . . . Bα(N ),N (8.3.9) α

Let |ωi be a generic state ∈ V ⊗N |ωi = |ω1 i(1) |ω2 i(2) . . . |ωN i(N )

(8.3.10)

Pˆα |ωi = |ωα(1) i(1) |ωα(2) i(2) . . . |ωα(N ) i(N )

(8.3.11)

then so that 1 X ˆ 1 X ˆ A|ωi = α Pα |ωi = α |ωα(1) i(1) |ωα(2) i(2) . . . |ωα(N ) i(N ) . N! α N! α

(8.3.12)

If we define a matrix then

ωij ≡ |ωi i(j)

(8.3.13)

1 X 1 ˆ A|ωi = α ωα(1),1 ωα(2),2 . . . ωα(N ),N = det(ω) N! α N!

(8.3.14)

8.4. OCCUPATION NUMBERS

163

i.e. |ω1 i(1) |ω1 i(2) . . . |ω2 i(1) |ω2 i(2) . . . 1 ˆ A|ωi = N ! .. . ... ... |ωN i(1)

8.4

|ω1 i(N ) |ω2 i(N ) .. . |ωN i(N )

(8.3.15)

Occupation numbers

Consider a system of N identical particle. Basis states in V ⊗N take the form |ui i(1) ⊗ . . . |up i(N )

(8.4.1)

where the one-particle states form an orthonormal basis of V : V = span{|u1 i, |u2 i, . . . }

(8.4.2)

By applying Sˆ or Aˆ to the full set of states in V ⊗N we obtain all physical states in SymN V and AntiN V . But many different states in V ⊗N can give rise to the same state in SymN V and AntiN V after the application of the projectors. To distinguish basis states in V ⊗N that after application of Sˆ or Aˆ are linearly independent, define the occupation number. We assign a set of occupation numbers to a basis state |·i ⊗ · · · ⊗ |·i. An occupation number is an integer ni ≥ 0 associated with each vector in V : |u1 i , |u2 i , . . . , |uk i , . . . n1

n2

(8.4.3)

nk

We define nk to be the number of times that |uk i appears in the chosen basis state |·i ⊗ · · · ⊗ |·i. Thus, by inspection of the state |·i ⊗ · · · ⊗ |·i we can read all the occupation numbers n1 , n2 , · · · . It should be clear that the action of a permutation operator on a basis state in V ⊗N will not change the occupation numbers. Two basis states in V ⊗N with the same occupation numbers can be mapped into each other by a permutation operator; they lead to the same state in SymN V and to the same state (up to a sign) in AntiN V . Two basis states in V ⊗N with different occupation numbers cannot be mapped into each other by a permutation operator. They must lead to different states in SymN V and to different states in AntiN V , unless they give zero. Given the occupation numbers of a basis state, we denote the associated basis state in SymN V as follows |n1 , n2 , . . . iS ni ≥ 0 (8.4.4)

CHAPTER 8. IDENTICAL PARTICLES

164 Explicitly

|n1 , n2 , . . . iS ≡ cS Sˆ |u1 i . . . |u1 i ⊗ |u2 i . . . |u2 i ⊗ . . . | {z } | {z } n1 times

(8.4.5)

n2 times

where cS is a constant that is used to give the state unit normalization. More briefly we write (8.4.6) |n1 , n2 , . . . iS ≡ cS Sˆ |u1 i⊗n1 ⊗ |u2 i⊗n2 ⊗ . . . where |ui i⊗ni is equal to 1 when ni = 0 These states form an orthonormal basis in SymN V : 0 S hn1 ,

n02 , . . . |n1 , n2 , . . . iS = δn1 ,n01 δn2 ,n02 . . .

The space SymN V relevant to identical bosons is spanned by all the states X |n1 , n2 , . . . i with nk = N

(8.4.7)

(8.4.8)

k

The space AntiN V relevant to identical fermions is spanned by all the states X |n1 , n2 , . . . iA with nk = N and nk ∈ {0, 1} ,

(8.4.9)

k

since occupation numbers cannot be greater than one (any state with an occupation number ˆ We have two or larger is killed by A). |n1 , n2 , . . . iA ≡ cA Aˆ |u1 i⊗n1 ⊗ |u2 i⊗n2 ⊗ . . .

(8.4.10)

where cA is a constant that is used to give the state unit normalization. These states form an orthonormal basis in AntiN V .

8.5 Particles that live in V ⊗ W A particle may have space degrees of freedom, described by a vector space V and spin degree of freedom associated with W . Suppose we have a state than of 2 such particles described in (V ⊗ W )⊗2 for example |ψi = |vi i(1) ⊗ |wi i(1) ⊗ |vj i(2) ⊗ |wj i(2) + . . .

(8.5.1)

This ψ should belongs either to Sym2 (V ⊗ W ) or to Anti2 (V ⊗ W ). The permutation operator here that exchange particles 1 and 2 is   Pˆ3412 |ψi = Pˆ3412 |vi i(1) ⊗|wi i(1) ⊗|vj i(2) ⊗|wj i(2) = |vj i(1) ⊗|wj i(1) ⊗|vi i(2) ⊗|wi i(2) ( 8.5.2) Wanto to express this in terms of Sym2 V , Anti2 V , Sym2 W , Anti2 W . Why? Because it is possible: for any state  1  W W 1 (8.5.3) . . . |ai . . . |bi = . . . |ai . . . |bi + . . . |bi . . . |ai + . . . |ai . . . |bi − . . . |bi . . . |ai 2| {z } 2| {z } ∈ Sym2 W

∈ Anti2 W

8.5. PARTICLES THAT LIVE IN V ⊗ W

165

Thus we can assume we work with simultaneous eigenstates of Pˆ3214 , which exchanges the V states, and of Pˆ1432 , which exchanges the W states. Note that Pˆ3412 = Pˆ3214 Pˆ1432

(8.5.4)

where the order is not important since [Pˆ3214 , Pˆ1432 ] = 0. The eigenvalues are Pˆ3214 1 −1 1 −1

Pˆ1432 1 −1 −1 1

Pˆ3412 1 1 −1 −1

This means that   Sym2 (V ⊗ W ) ' Sym2 V ⊗ Sym2 W ⊕ Anti2 V ⊗ Anti2 W   Anti2 (V ⊗ W ) ' Sym2 V ⊗ Anti2 W ⊕ Anti2 V ⊗ Sym2 W

(8.5.5) (8.5.6)

where with ' we indicate |vi i(1) ⊗ |wi i(1) ⊗ |vj i(2) ⊗ |wj i(2) ' |vi i(1) ⊗ |vj i(1) ⊗ |wi i(2) ⊗ |wj i(2) .

(8.5.7)

The generalization to 2 particle belonging to (U ⊗ V ⊗ W ) is simple, for 3 or more particle is more complicated.

Example.

Two electrons with spin wavefunction Ψ(x1 , m1 ; x1 , m2 ) = φ(x1 , x2 ) · χ(m1 , m2 )

Sˆz = m~

(8.5.8)

χ can be some normalized state in the space spanned by the triplet and the singlet. The probability dP to find one electron in d3 x1 around x1 and in d3 x2 around x2 is dP = |φ(x1 , x2 )|2 d3 x1 d3 x2

(8.5.9)

Assume for simplicity that the electrons are non interacting so that the Schr¨odinger equation   ~2 2 ~2 2 ∇ + V (x1 ) − ∇ + V (x2 ) Ψ = EΨ (8.5.10) − 2m x1 2m x2 is separable, so there is a solution of the form ψA (x1 )ψB (x2 ) with Z Z Z 3 2 3 2 ∗ d x |ψA (x)| = 1 , d x |ψB (x)| = 1 , d3 x ψA (x)ψB (x) = αAB = 6 0 (8.5.11)

CHAPTER 8. IDENTICAL PARTICLES

166

p p By Schwarz’s inequality |hu, vi| ≤ hu, ui hv, vi we have Z p p ∗ d3 x ψA (x)ψB (x) = |hψA , ψB i| ≤ |hψA , ψA i| |hψB , ψB i| =⇒ |αAB | ≤ 1

(8.5.12)

But then must build  N± φ± (x1 , x2 ) = √ ψA (x1 )ψB (x2 ) ± ψA (x2 )ψB (x1 ) 2

(8.5.13)

with N± a real normalization constant. Take the following combination of φ and χ: φ+ · χsinglet

so that the total wavefunction is antisymmetric

φ− · χtriplet

any of the 3 states of the triplet

Again, the probability to find one electron in d3 x1 around x1 and in d3 x2 around x2 is dP± = |φ± (x1 , x2 )|2 d3 x1 d3 x2  ∗ o N± n ∗ (x1 )ψA (x2 )ψB (x2 )ψB (x1 ) d3 x1 d3 x2 = √ |ψA (x1 )ψB (x2 |2 + |ψA (x2 )ψB (x1 )|2 ± 2< ψA | {z } 2 Exchange density

(8.5.14) If we take the case x1 = x2 = x, we get n o dP± = N± |ψA (x)ψB (x)|2 ± |ψA (x)ψB (x)|2 d3 x1 d3 x2 =⇒ dP+ = 2N± |ψA (x)ψB (x)|2 d3 x1 d3 x2 dP− = 0

(8.5.15) (8.5.16)

Recall that P+ is associated with the singlet, while P− with the triplet. Therefore electrons avoid each other in space when they are in the triplet state. In the singlet states there is enhanced probability to be at the same point. Note that normalization requires Z n  ∗ o N±2 ∗ 1= d3 x1 d3 x2 |ψA (x1 )ψB (x2 |2 + |ψA (x2 )ψB (x1 )|2 ± 2< ψA (x1 )ψA (x2 )ψB (x2 )ψB (x1 ) 2  Z  Z N±2 3 ∗ 3 ∗ = d x1 ψA (x1 )ψB (x1 ) d x2 ψB (x2 )ψA (x2 ) 1 + 1 ± 2< 2 n o ∗ = N±2 1 ± < [αAB · αAB ] =

N±2



2

1 ± |αAB |



=⇒ N± = p

1 1 ± |αAB |2

(8.5.17)

8.6. COUNTING STATES AND DISTRIBUTIONS

167

So dP+ =

2 |ψA (x)ψB (x)|2 d3 x1 d3 x2 1 ± |αAB |2

(8.5.18)

to be compared with dPD = |ψA (x)|2 |ψB (x)|2 d3 x1 d3 x2

(8.5.19)

for distinguishable particles. Since from (8.5.12) we have |αAB < 1|, then dP+ ≥ dPD

(8.5.20)

Assume ψA (x) is nonzero only in a region RA , ψB (x) is nonzero only in a region RB and RA ∩ RB = 0, then Z ∗ (8.5.21) αAB = d3 x ψA (x)ψB (x) = 0 since ψA requires x ∈ RA and ψB requires x ∈ RB . Therefore in this case N± = 1. Then the probability to find an electron in d3 x1 around x1 ∈ RA and another in d3 x2 around x2 ∈ RB is n  ∗  ∗ 2 2 1 1 dP± = |ψA (x1 )ψB (x2 |2 + | ψA(x ) ψB(x )|2 ± 2< ψA (x1 ) ψA(x )ψB (x2 ) ψB(x ) d3 x1 d3 x2 = |ψA (x)|2 |ψB (x)|2 d3 x1 d3 x2 = dPD

(8.5.22)

it is the probability density for distinguishable particles. Therefore, there is no need to symmetrize or antisymmetrize the wavefunction of non overlapping localized particles.

8.6

Counting states and distributions

Ei energy of the i-th level, di degeneracy of the i-th level. Assume we have N particles and we want to place N1 particles in level 1, . . . , Ni particles in level i (Ei , di ). Let’s call Q(N1 , N2 , . . . ) the number of ways to do this.

CHAPTER 8. IDENTICAL PARTICLES

168

8.6.1

Distinguishable particles (Maxwell Boltzman) N! N1 !N2 !...

Split N into N1 , N2 , . . . in

ways (imagine putting them on a line |{z} . . . |{z} . . . after N1

N2

drawing one ball at a time). Placing Ni particles in di slots gives (di )Ni ways, therefore Q(N1 , N2 , . . . ) = N !

Y (di )Ni

8.6.2

Maxwell Boltzman

Ni !

i

Identical fermions

Splitting the N states into groups for identical particles can only be done in one way Ni < di how many ways to place them? di choose Ni =

di Ni !(di − Ni )!

=⇒ Q(N1 , N2 , . . . ) =

Y i

8.6.3

di Ni !(di − Ni )!

Identical bosons Ni particles!

Ni balls and (di − 1) bars, for example: How many ways to order the Ni + di − 1 objects? (Ni + di − 1)! but we must divide by the irrelevant permutations, hence Q(N1 , N2 , . . . ) =

Y (Ni + di − 1)! i

(8.6.1)

Ni !(di − 1)!

This is for bosons: Q(N P 1 , N2 , . . . ) counts the number of ways to have N particles and fixed energy E = i Ni Ei . Want to find the values of N1 , N2 , . . . that are most likely. P P Want to maximize Q(N1 , N2 , . . . ) under the constraint that N = i Ni and E = i Ni Ei . Maximizing Q is the same as maximizing ln Q and using lagrange multipliers X X f (N1 , N2 , . . . ) = ln Q(N1 , N2 , . . . ) + α(N − Ni ) + β(E − Ni Ei ) (8.6.2) i

i

8.6. COUNTING STATES AND DISTRIBUTIONS

169

Do the fermion one, using QF from (8.6.1) we have X ln QF = ln di ! − ln Ni ! − ln(di − Ni )!

(8.6.3)

i

Since all these quantities di , Ni are large we can use the Stirling approximation ln n! ' n ln n − n

(8.6.4)

so that ln QF =

X

i −(di −Ni ) ln(di −Ni )+di − i = di ln di −di −Ni ln Ni + N N

i

X

di ln di −Ni ln Ni −(di −Ni ) ln(di −Ni )

i

(8.6.5) Minimizing w.r.t. to Ni without constraints, d ln QF = − ln Ni − 1 + ln(di − Ni ) + 1 = ln dNi



di − Ni Ni

 .

(8.6.6)

so that, adding the lagrange multipliers, we have     ∂f di − Ni di − Ni = ln − α − βEi =⇒ = eα+βEi =⇒ di = Ni eα+βEi + 1 ∂Ni Ni Ni =⇒ Ni =

di α+βE i e

+1 P P α and β can be calculated using the equations i Ni = N and i Ni Ei = E: α≡−

µ(T ) , kB T

β≡

1 kB T

then

di

Ni = e The expected occupation number n ∼  n=

e

E−µ(T ) kB T

Ni di

definition of temperature

Ei −µ(T ) kB T

(8.6.7)

(8.6.8)

(8.6.9) +1

for a single state is

−1 +1

Fermi-Dirac distribution

For bosons we have  E−µ(T ) −1 kB T n= e −1

Bose-Einstein distribution

Since n ≥ 0 we need E > µ for all energy levels µ < Ei ∀i. For an ideal gas µ(T ) < 0 ∀T , i.e. α > 0 For fermions

CHAPTER 8. IDENTICAL PARTICLES

170

Figure 8.3: Occupation number for a state as a function of the energy for a system of identical fermions in the T → 0 limit. µ(T = 0) = EF is called the Fermi energy.

Figure 8.4: Chemical potential µ as a function of the temperature for a system of identical bosons. TC , the critical temperature is defined to be the temperature such that µ(TC ) = 0 and it’s the temperature for Bose-Einstein condensation

Figure 8.5: Chemical potential µ as a function of the temperature for a system of identical fermions.

Aside from statistical mechanics

∂E µ= ∂N S,V

keeping constant entropy may require lowering the energy

(8.6.10)

Suppose we add a particle with no energy to the system. S will increase (more ways to

8.6. COUNTING STATES AND DISTRIBUTIONS

171

divide up the total energy) for this not to happen, must reduce the energy. dE(V, S) = T dS − P dV + µdN = d(T S) − SdT − P dV

(8.6.11)

d (E − T S) = −SdT − P dV + µdN | {z } F (T, V ) = E − T S . µ is an intensive quantity, and in terms of F we have ∂F Adding a particle changes µ= the energy and the entropy ∂N T,V

(8.6.12)

(8.6.13)

MIT OpenCourseWare https://ocw.mit.edu

8.06 Quantum Physics III

Spring 2018