*392*
*74*
*1MB*

*English*
*Pages 302
[90]*
*Year 2005*

- Author / Uploaded
- Charles E. Burkhardt
- Jacob J. Leventhal

TOPICS IN ATOMIC PHYSICS SOLUTIONS TO PROBLEMS

C. E. Burkhardt Department of Physics St. Louis Community College St. Louis, MO 63135 & J. J. Leventhal Department of Physics University of Missouri - St. Louis St. Louis, MO 63121

TABLE OF CONTENTS

CHAPTER 1 - BACKGROUND.................................................................................................... 2 CHAPTER 2 - ANGULAR MOMENTUM ................................................................................... 4 CHAPTER 3 - ANGULAR MOMENTUM - TWO SOURCES .................................................. 15 CHAPTER 4 - THE QUANTUM MECHANICAL HYDROGEN ATOM ................................. 20 CHAPTER 5 - THE CLASSICAL HYDROGEN ATOM ........................................................... 31 CHAPTER 6 - THE LENZ VECTOR AND THE ACCIDENTAL DEGENERACY ................. 36 CHAPTER 7 - BREAKING THE ACCIDENTAL DEGENERACY .......................................... 42 CHAPTER 8 - THE HYDROGEN ATOM IN EXTERNAL FIELDS ........................................ 49 CHAPTER 9 - THE HELIUM ATOM ......................................................................................... 63 CHAPTER 10 - MULTIELECTRON ATOMS............................................................................ 69 CHAPTER 11 - THE QUANTUM DEFECT............................................................................... 73 CHAPTER 12 - MULTIELECTRON ATOMS IN EXTERNAL FIELDS.................................. 77 CHAPTER 13 - INTERACTION OF ATOMS WITH RADIATION ......................................... 83

1

CHAPTER 1 - BACKGROUND 1.1 Bohr's theory of the atom pre-dated deBroglie's hypothesis of what we now call the deBroglie wavelength, λ D . Suppose, however, that Bohr was aware of this hypothesis and, rather than using the correspondence principle stated: Assumption III: The orbits of the stationary states are defined by the condition that only integral multiples of λ D = h / p "fit" into the orbital circumference. Show that this is identical to the "assumption" that the angular momentum is quantized in units of , and, thus, it would lead to the correct energies and the same atomic parameters that he derived using the correspondence principle. Solution: In order to fit into the circumference of the nth orbit we must have h h 2π rn = nλD = n ⋅ = n ⋅ p me vn But, the angular momentum of the electron in the nth circular orbit is given by L = me v n rn so h =n L = me vn rn = n ⋅ 2π from which the other parameters follow. 1.2 Positronium is a bound state of a positron (anti-electron) and an electron. It is essentially a hydrogen atom with the proton replaced by a positron. What is the ground state energy of positronium? Solution: me c 2 α 2 RH ( 0) Hydrogen energies are given by En = 2 = . 2n 2 n In fact, the mass of the electron is really the reduced mass μ of the proton-electron system. For hydrogen μ ≈ me because the proton is so massive. For positronium, however, the reduced mass is given by m 1 1 1 = + ⇒ μ= e μ me me 2 Therefore, the ground state energy of positronium is one-half the ground state energy of hydrogen, −6.8 eV.

(

)

1.3 Obtain the expression for the energy levels of a particle in an infinite square well of length L, Equation (1.29), by fitting the deBroglie wave in the box. Note that in this special case the deBroglie wavelength for each level is constant throughout the box because there is no potential energy, i.e. the kinetic energy (and hence the momentum) is constant for each level. Solution: To fit deBroglie waves in the box the wavelength must be λn = ( 2 L ) / n . A short table shows that this is the correct. n λn

1 2 3

2L L (2/3) L

2

( 2L ) = h ∴ λn = n

2

⇒

p

En = T + V =

2

n 2π 2 ⎛ nh ⎞ ⎛ 2π ⎞ p =⎜ ⎟ ⎜ ⎟ = L2 ⎝ 2 L ⎠ ⎝ 2π ⎠ 2

p2 1 n 2π 2 +0= 2m 2m L2

2

2

1.4 What is the ionization energy of ground state He+? In other words, what is the the ground state energy of He+? Solution: Since the energy of a one-electron atom scales as Z 2 the ionization energy of He+ is 22 ×13.6 eV=54.4 eV 1.5 A diatomic molecule for which the reduced mass of the nuclei is μ undergoes vibrational motion. a) Use the uncertainty principle to minimize the energy to show that an approximate value for / 2 μω . the amplitude of vibration A is A = b) Show that, using A from part a), the minimum energy is the correct zero point energy of a harmonic oscillator. c) The spacing between the quantized vibrational energy levels of a diatomic hydrogen molecule is roughly 0.4 eV. Find the approximate amplitude of vibration for this molecule. Solution: p2 1 a) E = + μω 2 x 2 2μ 2 Take Δx ≈ A = amplitude of vibration . 1 1 ∴ Δp = p = = 2 Δx 2 A 2

1 ⎛ ⎞ 1 2 2 ⎜ ⎟ + μω A 2μ ⎝ 2 A ⎠ 2 Find the amplitude that gives minimum energy. 2 2 dE 2 4 =− + μω = 0 ⇒ = ⇒ A= A A 4 μ A3 4 μ 2ω 2 dA

In terms of the amplitude then E =

/ 2 μω

2

⎞ 1 1 ⎛ ⎞ ⎛ 2 μω ⎞ 1 2⎛ ω b) E = ⎟= ⎜ ⎟ ⎜ ⎟ + μω ⎜ 2μ ⎝ 2 ⎠ ⎝ ⎠ 2 ⎝ 2 μω ⎠ 2

c) A = 2

2μω

=

2

2 ( m proton / 2 ) ( ω )

=

2

1836me ( ω )

Use atomic units. A2 =

2

1836me ( ω )

=

12

⎡⎛ 2 ⎞ ⎛ 1a.u. ⎞ ⎤ 1836 (1) ⎢⎜ eV ⎟ ⎜ ⎟⎥ ⎣⎝ 5 ⎠ ⎝ 27.2 eV ⎠ ⎦ ( a0 is the a.u. of length)

So A ≈ 0.2 a0 3

=

5 ⋅ 27.2 a.u. = 0.037 a0 2 2 ⋅1836

CHAPTER 2 - ANGULAR MOMENTUM

2.1 Consider a particle of mass μ subjected to a two-dimensional harmonic oscillator potential in which the force constants for the x- and y- components of motion are kx = ky = k. Show that the Schrödinger equation is separable in Cartesian coordinates. How many quantum numbers are there? Find an expression for the energy of this oscillator and discuss degeneracy. Do you think that the problem is separable in any other coordinate system? If so, which one? Why? Solution: The Hamiltonian is – 2 ⎛ ∂2 ∂2 ⎞ 1 ⎜⎜ 2 + 2 ⎟⎟ + (x 2 + y 2 ) Hˆ = − 2 μ ⎝ ∂x ∂y ⎠ 2

= Hˆ x ( x ) + Hˆ y ( y ) When the Hamiltonian can be written as a sum of terms, each consisting of one variable then the wave function can be written as a product of eigenfunctions of each constituent Hamiltonian and the energy is the sum of the eigenfunctions.

Letting Ψ ( x, y ) = X ( x ) Y ( y ) where X and Y are one-dimensional harmonic oscillator eigenfunctions the Schrödinger equation is separable and the energy eigenvalues are given by the sum of the energy eigenvalues corresponding to each. Therefore, there are two quantum numbers and we have E nm = E n + E m 1⎞ 1⎞ ⎛ ⎛ = ⎜n + ⎟ ω + ⎜m + ⎟ ω 2⎠ 2⎠ ⎝ ⎝ = (n + m + 1) ω

where ω 2 =

k

μ

Clearly there is a degeneracy, e.g. n = 1, m = 0 gives the same energy as n = 0, m = 1, etc. Note that the ground state is non-degenerate. The degeneracy suggests that the problem will be separable in some other coordinate system. Because of the cylindrical symmetry the obvious choice is polar (cylindrical coordinates). 2.2 Consider a particle of mass μ confined to a two dimensional circular "box" of radius a such that V (ρ) = 0 ρ < a = ∞ otherwise Show that the Schrödinger equation is separable in polar coordinates ρ,φ. Find and solve the equations for the radial and the angular motion. Indicate how to obtain the energy eigenvalues. Solution: The Schrödinger equation inside the "box", that is, for ρ < a is ⎛ ∂2 1 ∂ 1 ∂2 ⎞ ⎟ Ψ ( ρ , φ ) = EΨ ( ρ , φ ) ⎜⎜ 2 + + 2 2 μ ⎝ ∂ρ ρ ∂ρ ρ ∂φ 2 ⎟⎠ Letting Ψ (ρ , φ ) = R (ρ )Φ (φ ) the equation is separable, i.e. −

2

ρ 2 d 2 R 2 μ E 2 ρ dR 1 d 2Φ + + = − ρ 2 R dρ2 R dρ Φ dφ 2 4

This equation can be satisfied only if each side is a constant, choose −m2 . We get two equations: d 2Φ = −m 2 Φ Equation for Φ: 2 dφ The solutions are Φ (φ ) = e imφ where m = 0, ± 1, ± 2 ... Equation for R: ρ 2 d 2 R 2μ E 2 ρ dR + 2 ρ + = m2 2 R dρ R dρ Making the substitution ρ =

2

2μ E

r so that

dR dR dr dR 2 μ E = ⋅ = ⋅ 2 d ρ dr d ρ dr

and

d 2R d ⎛ dR ⎞ d ⎛ dR ⎞ dr d ⎪⎧ dR 2μ E ⎪⎫ 2μ E 2μ E d 2 R = = ⋅ = ⋅ = 2 ⎨ ⋅ ⎜ ⎟ ⎜ ⎟ 2 ⎬ 2 d ρ 2 d ρ ⎝ d ρ ⎠ dr ⎝ d ρ ⎠ d ρ dr ⎪⎩ dr dr 2 ⎪⎭ we have ρ 2 d 2 R 2μ E 2 ρ dR + 2 ρ + = m2 R dρ2 R dρ 2

2

2 2 2 ⎛ ⎞ 1 ⎧ 2μ E d 2 R ⎫ 2μ E ⎛ ⎞ ⎛ ⎞ 1 ⎧⎪ 2μ E dR ⎫⎪ 2 + ⋅ + r⎟ r r⎟ ⎨ ⎜ ⎜ ⎟ ⎜ ⎨ ⎬ ⎬=m 2 2 2 2 ⎜ 2μ E ⎟ R ⎩ ⎜ ⎟ ⎜ ⎟ dr ⎭ dr ⎪⎭ ⎝ ⎠ ⎝ 2μ E ⎠ ⎝ 2μ E ⎠ R ⎪⎩ and the equation for ρ (in terms of r) becomes r 2 d 2 R 2 r dR +r + = m2 2 R dr R dr 2 d R 1 dR ⎛ m 2 ⎞ + + ⎜1 − 2 ⎟ R dr 2 r dr ⎝ r ⎠ which is the Bessel equation of order m. Because R ( r ) must be finite at the origin, only the Bessel functions of the first kind, J m ( ρ ) are solutions. The Bessel functions of the second kind blow up at the origin. ⎛ 2μ E ⎞ We have R ( r ) = const ⋅ J m ⎜⎜ ρ ⎟⎟ , and the full wave function is given by 2 ⎝ ⎠ ⎛ 2μE ⎞ imφ Ψ ( ρ , φ ) = const ⋅ J m ⎜⎜ ρ ⎟⎟ ⋅ e where m = 0, ± 1, ± 2 .... 2 ⎝ ⎠ Now, to find the energies we use the remaining boundary condition. Since V ( ρ ) = ∞ for ρ > a we must have R ( r = a ) = 0 . The energies are given by the condition ⎛ 2μE ⎞ J m ⎜⎜ a ⎟⎟ = 0 2 ⎝ ⎠ for a given value of the quantum number m which determines the order of the Bessel function in the energy eigenfunctions. That it, every time the Bessel function goes through a zero (with ρ = a ) there will be an allowed energy. Since the Bessel functions wiggle as shown below we will have many zeros of the argument.

5

The energy eigenvalues may be written Emn =

2

rmn 2μ a 2 where rmn is the nth root (zero) of the mth Bessel function. These can be obtained from any tabulation of Bessel functions. Several are given below and are seen to be consistent with the three Bessel functions plotted above.

ith zero of the mth Bessel function J m ( r ) i 1 2 3 4 5 6 7 8 9

m=0 2.405 5.520 8.654 11.792 14.931 18.071 21.212 24.353 27.494

m =1 3.832 7.016 10.173 13.323 16.470 19.616 22.760 25.903 29.047

m=2 5.135 8.147 11.620 14.796 17.960 21.117 24.270 27.421 30.571

m=3 6.379 9.760 13.017 16.224 19.410 22.583 25.749 28.909 32.050

m=4 7.586 11.064 14.373 17.616 20.827 24.018 27.200 30.371 33.512

m=5 8.780 12.339 15.700 18.982 22.220 25.431 28.628 31.813 34.983

We see that there are two quantum numbers and, because of the circular symmetry, there is a degeneracy.

6

2.3 Discuss modifications of the energies and wave functions obtained in Problem 2.2 if the particle were in a three dimension cylindrical box of length L such that V (ρ) = 0 ρ < a

= ∞ otherwise V ( z) = 0

z .

ψ = A 11 + B 10 + C 1 − 1 where A, B and C are complex constants. Calculate the expectation value of Lˆ x . Solution: Lˆ x = ψ Lˆ x ψ 1 ψ Lˆ + + Lˆ − ψ 2 Applying the raising and lowering operators to each term we get Lˆ+ 11 = 0 =

Lˆ+ 10 = 2 11 Lˆ+ 1 − 1 = 2 10 Lˆ− 11 = 2 10 Lˆ− 10 = 2 1 − 1 Lˆ− 1 − 1 = 0

Then

(

⎛1⎞ Lˆ x ψ = ⎜ ⎟ 0 + B 2 11 + C 2 10 + A 2 10 + B 2 1 − 1 + 0 ⎝2⎠ =

2

( B 11 + C 10

+ A 10 + B 1 − 1

)

)

and Lˆx =

2

( A* B + B *C + B * A + C * B)

2.8 a) Find the eigenvectors of the operator Sˆ y in terms of α and β , the eigenvectors of Sˆ z . Express them as spinors. b) Suppose that an electron is in the spin state 1 ⎛ 2⎞ ⎜ ⎟ 5 ⎜⎝ − 1⎟⎠ with the Sˆ z eigenvectors as the basis. If we measure the y-component of the spin, what is the

probability for finding a value of +(1/2) ? Solution: a) ⎛0 − i⎞ σ y = ⎜⎜ ⎟⎟ ⎝i 0 ⎠ so that σˆ y α

y

= α

y

becomes, in matrix form

⎛ 0 − i ⎞⎛ a ⎞ ⎛a⎞ ⎛ − ib ⎞ ⎛ a ⎞ ⇒ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = +⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ i 0 ⎠⎝ b ⎠ ⎝b⎠ ⎝ ia ⎠ ⎝ b ⎠ This equation can be true only if a = –ib. Therefore 9

α

y

⎛ − ib ⎞ = ⎜⎜ ⎟⎟ ⎝ b ⎠ ⎛ 1⎞ = −ib⎜⎜ ⎟⎟ ⎝i⎠ 2

2

Since a + b = 1 we can choose − ib = Similarly ⎛ 0 − i ⎞⎛ a ⎞ ⎛a⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = −⎜⎜ ⎟⎟ ⎝ i 0 ⎠⎝ b ⎠ ⎝b⎠ so that ia = –b and 1 ⎛1⎞ β y= ⎜ ⎟ 2 ⎜⎝ − i ⎟⎠

⇒

1 2

and α

y

=

1 ⎛1⎞ ⎜ ⎟ 2 ⎜⎝ i ⎟⎠

⎛ − ib ⎞ ⎛a⎞ ⎜⎜ ⎟⎟ = − ⎜⎜ ⎟⎟ ⎝ ia ⎠ ⎝b⎠

b) To find the probability of measuring + (1/ 2 ) function in terms of the eigenfunctions of Sˆ .

for Sˆ y we must express the given wave

y

1 ⎛ 2⎞ 1 ⎛ 1⎞ 1 ⎛1⎞ ⎜⎜ ⎟⎟ = A ⎜⎜ ⎟⎟ + B ⎜ ⎟ 5 ⎝ − 1⎠ 2 ⎝i⎠ 2 ⎜⎝ − i ⎟⎠ 1 ⎛ A+ B ⎞ = ⎜ ⎟ 2 ⎜⎝ i ( A − B )⎟⎠ 2 where A is the desired probability. Therefore

2 1 ( A + B ) and − 1 = i ( A − B ) = 5 2 5 2 which may be written 2 2 = ( A + B ) and i = (A − B) 2 5 5 2 (2 + i ) = 2 A so that A = 1 ⋅ 2 ⋅ (2 + i ) . 5 2 5 1 2 2 2 1 2 ⎛ 1 ⎞ The probability of measuring Sˆ y ⎜ + ⎟ = A = ⋅ ⋅ (2 + 1 ) = 4 5 2 ⎝ 2 ⎠

Adding, we get

2.9 Use the spin 1 Sˆ z eigenstates as a basis to angular momentum operators. Answer: ⎛ 0 1 0⎞ ⎛0 ⎜ ⎟ ⎜ Sˆ x = ⎜ 1 0 1 ⎟ ; Sˆ y = ⎜i 2⎜ 2⎜ ⎟ ⎝ 0 1 0⎠ ⎝0

form the matrix representations of the

−i 0 i

0⎞ ⎟ − i ⎟ ; Sˆ z = 0 ⎟⎠

⎛1 0 0 ⎞ ⎜ ⎟ ⎜0 0 0 ⎟ 2⎜ ⎟ ⎝ 0 0 − 1⎠

Solution: Since the representation is in the Sˆ z basis, Sˆ z must have the eigenvalues along the diagonal. These eigenvalues are the values of m j , which in this case is ms . Further, because this is spin 1,

10

i.e. j = s = 1 , the possible values of m j are +1, 0, –1. Therefore, Sˆ z is given by ⎛1 0 0 ⎞ ⎜ ⎟ Sˆ z = ⎜0 0 0 ⎟ 2⎜ ⎟ ⎝ 0 0 − 1⎠ An easy way to find Sˆ x and Sˆ y is to find Sˆ+ and Sˆ− because we know the result of action by the raising and lowering operators. Then use the fact that 1 1 ˆ Sˆ x = Sˆ+ + Sˆ− and Sˆ y = S + − Sˆ− 2 2i The action of Sˆ+ and Sˆ− is Sˆ s m = s(s + 1) − m(m + 1) s (m + 1)

(

)

(

)

+

and Sˆ− s m = s(s + 1) − m(m − 1) s (m − 1) where we have let m = ms . Now, the matrix representation of Sˆ+ is given by

⎛ 11 Sˆ+ 11 11 Sˆ+ 10 11 Sˆ+ 1 − 1 ⎞ ⎜ ⎟ Sˆ+ = ⎜ 10 Sˆ+ 11 10 Sˆ+ 10 10 Sˆ+ 1 − 1 ⎟ ⎜ ⎟ ⎜ 1 − 1 Sˆ+ 11 1 − 1 Sˆ+ 10 1 − 1 Sˆ+ 1 − 1 ⎟ ⎝ ⎠ All diagonal elements clearly vanish because they have the same kets and Sˆ+ shifts one away from the other. Also, all first column elements vanish because Sˆ 11 = 0 . Similarly, all bottom +

row elements vanish because Sˆ+ does not shift the kets on the right to be the same as the kets on the left. The same is true for the upper right hand element. The only ones that are non-zero are 11 Sˆ+ 10 = 1 ⋅ (1 + 1) − 0 ⋅ (0 + 1) = 2 and 10 Sˆ+ 1 − 1 = 1 ⋅ (1 + 1) − (− 1) ⋅ (− 1 + 1) = 2 This gives ⎛ 0 1 0⎞ ⎜ ⎟ Sˆ+ = 2 ⎜ 0 0 1 ⎟ ⎜ 0 0 0⎟ ⎝ ⎠ Now, we could use the same procedure to find Sˆ− but we know that Sˆ− is the hermetian conjugate of Sˆ . Therefore, we can obtain Sˆ by taking the conjugate transpose of Sˆ which is +

−

+

⎛ 0 0 0⎞ ⎜ ⎟ Sˆ− = 2 ⎜ 1 0 0 ⎟ ⎜ 0 1 0⎟ ⎝ ⎠ Then we have

(

)

1 Sˆ x = Sˆ+ + Sˆ − = 2

⎛ 0 1 0⎞ ⎜ ⎟ 1 ˆ S + − Sˆ − = ⎜ 1 0 1 ⎟ and Sˆ y = i 2 2⎜ ⎟ ⎝ 0 1 0⎠

(

11

)

⎛0 − i 0 ⎞ ⎜ ⎟ ⎜ i 0 − i⎟ 2⎜ 0 ⎟⎠ ⎝0 i

2.10 An unpolarized beam of neutral spin-1 particles is passed through a Stern-Gerlach device with magnetic field in the y-direction, an SGy device. The beam of emerging particles having S y = is then passed through an SGz device. What fraction of the particles with S y = will be

found to have S z = ? Solution: To answer this question we must find the eigenfunction of Sˆ y corresponding to the eigenvalue + , call it 11 y , in terms of the eigenfunctions of Sˆ z , i.e. use the latter as a basis set. Therefore, we must first solve the eigenvalue equation for Sˆ z . From a previous problem we know that, on the Sˆ basis z

⎛0 − i 0 ⎞ ⎜ ⎟ Sˆ y = ⎜ i 0 − i⎟ 2⎜ 0 ⎟⎠ ⎝0 i so the eigenvalue equation Sˆ y 1μ y = μ 1μ y becomes ⎛ 0 − i 0 ⎞⎛ a ⎞ ⎜ ⎟⎜ ⎟ ⎜ i 0 − i ⎟⎜ b ⎟ = μ 2⎜ 0 ⎟⎠⎜⎝ c ⎟⎠ ⎝0 i

⎛a⎞ ⎜ ⎟ ⎜b⎟ ⎜c⎟ ⎝ ⎠

or −i ⎛ ⎞ 0 ⎟ ⎜− μ 2 ⎜ ⎟⎛ a ⎞ − i ⎟⎜ ⎟ ⎜ i ⎜b⎟ = 0 ⎜ 2 −μ 2 ⎟⎜ ⎟ ⎜ ⎟⎝ c ⎠ i ⎜⎜ 0 − μ ⎟⎟ 2 ⎝ ⎠ This equation can have a non-trivial solution only if the determinant of the coefficients vanishes 1 ⎞ ⎛ i ⎞⎛ − iμ ⎞ ⎛ − μ⎜ μ 2 − ⎟ + ⎜ ⎟⎜ ⎟ = 0 ⇒ μ = 0, ± 1 as expected. 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ Now, we are only interested in the eigenvalue +1, so we find that eigenvector, i.e. |1,1>. Return now to the eigenvalue equation, but put in µ = + . 1 2

⎛ 0 − i 0 ⎞⎛ a ⎞ ⎛ a ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ − i 0 i ⎟⎜ b ⎟ = ⎜ b ⎟ ⎜0 i 0 ⎟⎠⎜⎝ c ⎟⎠ ⎜⎝ c ⎟⎠ ⎝

or ⎛ − ib ⎞ ⎛ a ⎞ ⎟ ⎜ ⎟ 1 ⎜ ⎜ i (a − c )⎟ = ⎜ b ⎟ 2⎜ ⎟ ⎜ ⎟ ⎝ b ⎠ ⎝c⎠ from which we see that −ib = 2a ; − ia − ic = 2b ; b = 2c 12

From the first and last of these equations −ib = 2a = −i 2c or c = − a . Also, b = i 2a . Then ⎛ a ⎞ ⎜ ⎟ 11 y = ⎜ i 2a ⎟ ⎜ −a ⎟ ⎝ ⎠ We must require the state to be normalized so that ⎛ a ⎞ ⎟ ⎜ 2 a * − i 2a * − a * ⎜ i 2a ⎟ = 4 a = 1 ⎜ −a ⎟ ⎠ ⎝ Therefore, up to an overall phase, we may choose a = 1/2 giving ⎛ 1 ⎞ ⎟ ⎛ 1 ⎞⎜ 11 y = ⎜ ⎟⎜ i 2 ⎟ ⎝ 2 ⎠⎜ ⎟ ⎝ −1 ⎠

(

)

1 2 1 11 + i 10 − 1 − 1 . 2 2 2 Note that the absence of subscript on the kets means eigenfunction of Sˆ z . or, expressed in terms of kets 11

y

=

We see now that the square of the amplitude of the + eigenfunction is (1/2)2 = 1/4 so that 25% of the particles in the +

beam from the SGy device have Sz = + . Note that 50% have 0 and

25% have – . Schematic representation:

(

)

2.11 Show that the operators Lˆ z and Hˆ = pˆ 2 / 2m + V ( r ) commute for a spherically

symmetric potential. Solution: For a spherically symmetric potential Hˆ = pˆ 2 / 2m + V (r ) ˆj iˆ kˆ

(

Lˆ = r × pˆ = x pˆ x

y pˆ y

)

z so Lˆ z = xpˆ y − ypˆ x pˆ z

Then

⎡ Lˆz , Hˆ ⎤ = 1 ⎡( xpˆ y − ypˆ x ) , pˆ x 2 + pˆ y 2 + pˆ z 2 ⎤ + ⎡ Lˆ z ,V ( r ) ⎤ ⎣ ⎦ 2m ⎣ ⎦ ⎦ ⎣ ∂ But ⎡⎣ Lˆz ,V ( r ) ⎤⎦ = 0 because Lˆ z = −i so we have ∂φ

(

)

13

(

)

2m ⎡⎣ Lˆz , Hˆ ⎤⎦ = ⎡⎣( xpˆ y − ypˆ x ) , pˆ x 2 + pˆ y 2 + pˆ z 2 ⎤⎦ = ⎡⎣ xpˆ y , pˆ x 2 ⎤⎦ + ⎡⎣ xpˆ y , pˆ y 2 ⎤⎦ + ⎡⎣ xpˆ y , pˆ z 2 ⎤⎦ − ⎡⎣ ypˆ x , pˆ x 2 ⎤⎦ − ⎡⎣ ypˆ x , pˆ y 2 ⎤⎦ − ⎡⎣ ypˆ x , pˆ z 2 ⎤⎦

= pˆ y ⎡⎣ x, pˆ x 2 ⎤⎦ + 0 + 0 − 0 − pˆ x ⎡⎣ y, pˆ y 2 ⎤⎦ − 0 We can evaluate the non-zero commutators in this expression using the identity ˆ ˆ ⎤ = ⎡ Aˆ , Bˆ ⎤ Cˆ + Bˆ ⎡ Aˆ , Cˆ ⎤ ⎡ Aˆ , BC ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⎡⎣ x j , pˆ j 2 ⎤⎦ = ⎡⎣ x j , pˆ j ⎤⎦ pˆ j + pˆ j ⎡⎣ x j , pˆ j ⎤⎦ = 2i pˆ j so that 2m ⎡⎣ Lˆz , Hˆ ⎤⎦ = pˆ y ⎡⎣ x, pˆ x 2 ⎤⎦ − pˆ x ⎡⎣ y, pˆ y 2 ⎤⎦ = pˆ y ( 2i pˆ x ) − pˆ x ( 2i pˆ y ) =0

14

CHAPTER 3 - ANGULAR MOMENTUM - TWO SOURCES 3.1 An electron in an atom is in an uncoupled spin and orbital eigenstate given by m ; ms = 1 − 1;α where we have omitted s = 1/ 2 from the ket because all electrons have s = 1/ 2 . What is the probability that a measurement of the total angular momentum will yield the value j = 3 / 2 ? j = 1/ 2 ? j = 5 / 2 ? Clebsch-Gordan coefficients for j2 = 1/ 2

1 j1 , ; m1 , m2 j m j 2 m2 = +1 / 2 = m s = α 2

j

( j +m

j1 + 1/ 2

j1 − 1/ 2

1

−

j

m 2 = −1 / 2 = m s = β 2

+ 1/ 2 )

1

j

+ 1/ 2 )

+ 1/ 2 )

( j +m

j

+ 1/ 2 )

1

( 2 j1 + 1)

Solution: We need to find the wave function

j

1

( 2 j1 + 1)

( j −m

( j −m

= 1, m = −1, α

( 2 j1 + 1)

( 2 j1 + 1)

in terms of the coupled basis set

1 1 = − (recall that m is given in 2 2 the problem and ms = 1/ 2 ). Therefore, for the coefficient of the j = 3 / 2 coupled state, we need the top coefficient in the first column with j1 = 1 and m j = −1/ 2 . That is, the wave j , m j . In our case j1 = = 1 and m j = m + ms = −1 +

function that we have, 1 − 1;α , can be written as a linear combination of coupled eigenfunctions j , m j : m ;α = 1 − 1;α

3 1 1 1 = A j = :m j = − + B j = ; mj = − 2 2 2 2 We do not need any β 's because our eigenstate is one of α . It is also one of m j , namely m j = −1/ 2 so we don't need any other values of m j in our expansion. From the table

A=

⎡ ⎛ 1 ⎞ 1⎤ ⎢1 + ⎜ − 2 ⎟ + 2 ⎥ ⎠ ⎣ ⎝ ⎦ = 1 ⎡⎣ 2 (1) + 1⎤⎦ 3 2

Probability of finding the electron with j = 3 / 2 = A = 1/ 3 2

2

Probability of finding the electron with j = 1/ 2 = B = 1 − A = 2 / 3 Probability of finding the electron with j = 5 / 2 = 0 Note that the other coefficient, B could have been evaluated from the 15

table. ⎡ ⎛ 1 ⎞ 1⎤ ⎢1 − ⎜ − 2 ⎟ + 2 ⎥ ⎝ ⎠ ⎦ =− 2 B=− ⎣ 3 ⎡⎣ 2 (1) + 1⎤⎦ 2

2

and we have A + B = 1 as it must. 1 1 1 − ;1 . What is 2 2 2 the probability that a measurement of the z-component of the orbital angular momentum will yield the value m = 0 ? m = +1 ? m = −1 ? Solution: We must expand the coupled eigenket as a linear combination of uncoupled kets 3.2 An electron in an atom is in an coupled eigenstate j m j ; s =

1 2

m ; s ms . In our case these are the 1 m ; ms . We use the table of Clebsch-Gordan

coefficients in the previous problem. Since j1 = = 1 and the state that we must expand has j = 1/ 2 , we need the bottom row, i.e. the row that gives j = 1/ 2 . Further, m j = −1/ 2 , as given in the coupled ket in the problem. The coefficients are ⎛ 1⎞ 1 1− ⎜ − ⎟ + ⎝ 2⎠ 2 = − 2 − 2 (1) + 1 3

⎛ 1⎞ 1 1+ ⎜ − ⎟ + ⎝ 2⎠ 2 = 1 and − 2 (1) + 1 3

so that the coupled ket is given by 1 1 1 2 11 1 1 1 1 m = −1; 1 m = 0; − − ;1 =− + 2 2 2 3 22 3 2 2 Note that the values m = −1 and m = 0 in each of the uncoupled kets are consequences of the first one being spin up, +1/2, so that m = −1 is required to make m j = −1/ 2 . The second one is spin down, –1/2, so that m = 0 is required to make m j = −1/ 2 . We read off the probabilities: 2

⎛ 1 ⎞ 1 Probability of m = 0 : ⎜ ⎟ = ⎝ 3⎠ 3 Probability of m = +1 : zero 2

2 ⎛ 2 ⎞ Probability of m = −1 : ⎜ − ⎟ =3 3⎠ ⎝ ⎛1⎞ 3.3 Show that Jˆ1 • Jˆ 2 = Jˆ1z Jˆ2 z + ⎜ ⎟ Jˆ1+ Jˆ2− + Jˆ1− Jˆ2+ ⎝ 2⎠ Solution: Jˆ • Jˆ = Jˆ Jˆ + Jˆ Jˆ + Jˆ Jˆ

(

1

2

1x

(

2x

1y

But Jˆ x = ⎛⎜ 1 ⎞⎟ Jˆ+ + Jˆ− ⎝ 2⎠

)

2y

1z

2z

(

)

)

and Jˆ y = ⎛⎜ 1 ⎞⎟ Jˆ+ − Jˆ− . Therefore ⎝ 2i ⎠

16

( = ⎛⎜ 1 ⎞⎟ ( Jˆ ⎝ 4⎠ = ⎛⎜ 1 ⎞⎟ ( Jˆ ⎝ 2⎠

Jˆ1 • Jˆ 2 = ⎛⎜ 1 ⎞⎟ Jˆ1+ + Jˆ1− ⎝ 2⎠

) ⎛⎜⎝ 12 ⎞⎟⎠ ( Jˆ

2+

)

(

+ Jˆ2− + ⎛⎜ 1 ⎞⎟ Jˆ1+ − Jˆ1− ⎝ 2i ⎠

) ⎛⎜⎝ 21i ⎞⎟⎠ ( Jˆ

2+

)

− Jˆ2− + Jˆ1z Jˆ2 z

)

1+

Jˆ2+ + Jˆ1+ Jˆ2− + Jˆ1− Jˆ2+ + Jˆ1− Jˆ2− − Jˆ1+ Jˆ2+ + Jˆ1+ Jˆ2− + Jˆ1− Jˆ2+ − Jˆ1− Jˆ2− + Jˆ1z Jˆ2 z

1+

Jˆ2− + Jˆ1− Jˆ2+ + Jˆ1z Jˆ2 z

)

3.4 An eigenfunction of Jˆ 2 and Jˆ z (total angular momentum and its z-component) is given

in terms of the uncoupled eigenfunctions j1 m j1 ; j2 m j 2 as 3 3 1 2 3 3 − ;1 − 1 + − ;10 5 2 2 5 2 2 Using the operator identity of the previous problem, find the value of the total angular momentum and its z-component. Solution: We must operate on ψ with Jˆ 2 and Jˆ z and find the eigenvalues.

ψ =

(

Jˆ 2 first: Jˆ 2 = Jˆ1 + Jˆ 2

)

2

= Jˆ12 + Jˆ22 + 2Jˆ1 • Jˆ 2

From the previous problem we write the last term in terms of the raising and lowering operators so that 2 ⎡⎛ 1 ⎞ ⎤ Jˆ 2 = Jˆ1 + Jˆ 2 = Jˆ12 + Jˆ22 + 2 ⎢⎜ ⎟ Jˆ1+ Jˆ2− + Jˆ1− Jˆ2+ + Jˆ1z Jˆ2 z ⎥ ⎣⎝ 2 ⎠ ⎦ Now evaluate each of these terms separately and add them up at the end. 15 2 ⎛ 3 ⎞⎛ 3 ⎞ ψ and Jˆ12 ψ = ⎜ ⎟ ⎜ + 1⎟ 2 ψ = 4 ⎝ 2 ⎠⎝ 2 ⎠ Jˆ 2 ψ = (1)(1 + 1) 2 ψ = 2 2 ψ

(

)

(

)

2

Note that ψ is an eigenfunction of both Jˆ12 and Jˆ22 since each of the constituent basis kets has the same value of j1 and j2. 3 3 1 2 3 3 Jˆ1+ Jˆ 2− ψ = Jˆ1+ Jˆ 2− − ;1 − 1 + Jˆ1+ Jˆ 2− − ;1 0 5 2 2 5 2 2 The first term vanishes because Jˆ lowers it out of existence. We have 2−

2ˆ 3 3 2 3 3 − ;1 − 1 1(1 + 1) − ( 0 )(..) Jˆ1+ Jˆ2− ψ = Jˆ1+ J 2− − ;10 = Jˆ1+ 5 2 2 5 2 2 =2

1 5

2

3 ⎛ 3 ⎞ ⎛ 3 ⎞⎛ 3 ⎞ 3 1 3 ⎜ + 1⎟ − ⎜ − ⎟⎜ − + 1⎟ − ;1 − 1 = 2 2 ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ 2 2 5

2

3 1 − ;1 − 1 2 2

3 3 1 2 3 3 Jˆ1− Jˆ 2+ ψ = Jˆ1− Jˆ 2+ − ;1 − 1 + Jˆ1− Jˆ 2+ − ;1 0 5 2 2 5 2 2 The second term vanishes because Jˆ1− lowers it out of existence. We have

17

3ˆ 3 1 Jˆ1− Jˆ2+ ψ = Jˆ1− J 2+ − ;1 − 1 5 2 2 3 3 1 = Jˆ1− − ;10 1(1 + 1) − ( −1)( 0 ) 5 2 2 6 5

= =3

2 5

3 ⎛ 3 ⎞ ⎛ 1 ⎞⎛ 1 ⎞ 3 3 ⎜ + 1⎟ − ⎜ − ⎟⎜ − − 1⎟ − ;10 2 ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ 2 2

2

2

3 3 − ;10 2 2

3 3 1 2 3 3 2 Jˆ1z Jˆ 2 z ψ = 2 Jˆ1z Jˆ 2 z − ;1 − 1 + 2 Jˆ1z Jˆ 2 z − ;1 0 5 2 2 5 2 2 The second term vanishes because the eigenvalue of Jˆ is 0. 2z

3 3 1 3⎛ 1 ⎞ 2 Jˆ1z Jˆ2 z ψ = 2 Jˆ1z Jˆ2 z − ;1 − 1 = 2 ⎜ − ⎟ (− 5 2 2 5⎝ 2 ⎠

3 1 3 − ;1 − 1 = 2 2 5

)

2

3 1 − ;1 − 1 2 2

Now, to evaluate Jˆ 2 ψ add up all these terms. 15 Jˆ 2 ψ = 4

3 5

2

ψ +2

2

ψ +2

2

ψ +2

2

ψ + 3⎢

3 1 2 − ;1 − 1 + 3 2 2 5

2

2

3 3 3 − ;1 0 + 2 2 5

2

3 1 − ;1 − 1 2 2

or 15 Jˆ 2 ψ = 4

⎡ 3 ⎣ 5

2

3 1 2 − ;1 − 1 + 2 2 5

2

⎤ 3 3 − ;10 ⎥ 2 2 ⎦

35 2 ⎛ 15 ⎞ ⎛ 5 ⎞⎛ 5 ⎞ ψ = ⎜ ⎟ ⎜ + 1⎟ 2 ψ = ⎜ + 2 + 3⎟ 2 ψ = 4 ⎝ 4 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ Therefore, this is the eigenket corresponding to j = 5 / 2 . To obtain the value of m j we may operate with Jˆ z = Jˆ z1 + Jˆ z 2 or it is much simpler to notice that the sums of the individual z-components add up to –3/2 for each of the 1 3 3 3 and − + 0 = − . Therefore, the uncoupled kets that compose ψ , i.e. − − 1 = − 2 2 2 2 eigenvalue of Jˆ z for this ψ is −3/2. 3.5 An eigenfunction of Jˆ 2 and Jˆ z (total angular momentum and its z-component) is given

in terms of the uncoupled eigenfunctions j1 m j1 ; j2 m j 2 as 2 33 1 31 8 3 1 − ;11 ;1 − 1 + ;10 − 5 22 15 2 2 15 2 2 Using the table of Clebsch-Gordan coefficients below, find the value of the total angular momentum and its z-component.

ψ =

18

j1 =

3 2

j1 =

j=

5 2

j=

3 2

j=

1 2

m m1

m2

5 2

3 2 3 2 3 2 1 2 1 2 1 2

1

1

0

3 2

1 2

0

1 2

1

−

1 2

0

−

1 2

−1

−

3 2

1

−

3 2

0

−

3 2

−1

−

3 2

−

5 2

3 2

1 2

−

1 2

−

3 2

1 2

−

1 2

3/ 5

1/10

2/5

1/ 2

1/15

− 1/ 3

− 2/5

3/ 5 3/ 5

−1

−

1 2

2/5

−1

1

−

3/10

8 /15

1/ 6

− 8 /15

3/10

1/ 6

− 1/15

3/ 5

3/ 5

− 1/ 3

2/5

− 2/5

1/10

1/ 2

− 3/ 5

2/5

1

Solution: Since it is given that ψ is an eigenfunction of Jˆ 2 and Jˆ z it must be a linear combination

of the uncoupled kets as given in a column in the table. This particular ψ is that in the column for j = 3 / 2 ; m = 1/ 2 .

19

CHAPTER 4 - THE QUANTUM MECHANICAL HYDROGEN ATOM 4.1 The wave function for a hydrogen atom is given by ψ ( r ,θ , φ ) ∝ r 2 e − r / 3a0 3cos 2 θ − 1

(

)

a) What are the quantum numbers n, , m for this atom? No calculations allowed. b) What is the most probable value of r for the electron in this state? What is significant about the result? Solution: a) The n quantum numbers is 3 because the denominator in the exponential in the radial parts of the wave functions is na0 . The function 3cos 2 θ − 1 is proportional to Y20 (θ , φ ) . Note that

(

)

there is no φ dependence which confirms that m = 0 . Answer: n = 3; = 2; m = 0 b) The most probable value of r is the position of the maximum in the probability distribution 2 r 2 ψ . The max will be the same as the max of rψ so, taking the derivative with respect to r: and setting it equal to zero: ⎛ 1 3 ⎞ d r ir 2 e − r / 3a0 = e − r / 3a0 ⎜ − r + 3r 2 ⎟ = 0 ⇒ r0 = 32 a0 dr ⎝ 3a0 ⎠

(

)

This is the third Bohr radius. It turns out that for the state of maximum for a given n the most probable value of r is always the Bohr radius. 4.2 The wave function for a hydrogen atom is given by ψ (r, θ , φ ) = R54 (r )Y m (θ , φ ) + R53 (r )Y30 (θ , φ ) + R52 (r )Y21 (θ , φ )

For what values of and m (= m ) will ψ be an eigenfunction of the Hamiltonian? What is the energy in electron volts? in Rydbergs? in a.u.? in cm-1? Solution: Since the radial parts of the wave function all have n = 5, ψ (r , θ , φ ) is already an eigenfunction of the Hamiltonian. Recall that the energy for a hydrogen atom doesn't depend on either or m . Therefore the only requirements on or m are that they be consistent with n = 5 and, of course, each other. Since the radial part of the first term has = 4, so too must the spherical harmonic. Further, m must be such that –4 ≤ m ≤ +4. The energy for n = 5 is: 13.6 eV = −0.544 eV E5 = − 52 1 = − 2 Ry 5 1 1 = − ⋅ 2 a.u. 2 5 1.097373 ×10−5 cm −1 =− 52 = −4389.492 cm −1

20

4.3 An electron is in the ground state of tritium, i.e a hydrogen atom with two neutrons the symbol for which is 3H. A nuclear reaction instantaneously changes the nucleus to 3He which retains the single electron from the tritium atom. That is, the product is a singly ionized 3He which is a one-electron atom. Calculate the probability that the atomic electron remains in the ground state of the 3He ion. Solution: We may expand the wave function for the "old" atom, the tritium, on the complete set of eigenfunctions of the Hamiltonian for the "new" one-electron atom, the 3He+. ∞

tritium ψ 100 = ∑∑∑ An m ψ nHem n =1

m

But, we are only interested in the probability that the 3He is in the ground state so we need only A100 because the desired probability is | A100 |2. Furthermore, the spherical harmonics will be the same in the new and old wave functions. In fact, Y00 (θ , φ ) = 1/ 4π so when we integrate the square of Y00 over all angles we get 1. Therefore, we need only work with the radial parts of the two wave functions. These wave functions are identical except that the tritium wave functions have Z = 1 while the He wave functions have Z = 2. He tritium Formally, we take the inner product ψ 100 ψ 100 = A100 Now we must perform the integral in coordinate space. The wave functions are 3

⎛ Z ⎞2 ⎛ Z Z ψ 100 = 2⎜⎜ ⎟⎟ exp⎜⎜ − ⎝ a0 ⎠ ⎝ a0 so that

⎞ r ⎟⎟ ⎠ 3

ψ

He 100

ψ

tritium 100

2

and A100 =

∞

= ∫ r drR 0

2

Z =2 10

(r ) R

Z =1 10

⎛ 3r ⎞ 4 32 ∞ 2 22 ( r ) = 2 ∫0 r dr exp ⎜ − ⎟ = 8 3 a0 3 ⎝ a0 ⎠

64 ⋅ 8 = 0.70 36

4.4 An electron in the Coulomb field of a proton is described by the normalized wave function

{

ψ ( r ) = 16 4 ψ 100 ( r ) + 3 ψ 211 ( r ) − ψ 210 ( r ) + 10 ψ 21−1 ( r )

}

where the ψ (r ) are spherical (orbital) eigenfunctions of the hydrogen atom Hamiltonian. Find: a) The expectation value of the energy in a.u. b) The expectation value of Lˆ2 . Keep the 's in the answer. c) The expectation value of Lˆ z . Solution:

(

)

1 1 2 a) Hˆ = ψ Hˆ ψ = 2 42 E1 + 32 E2 + ( −1) E2 + 10 E2 = ( 4 E1 + 5 E2 ) 6 9 1⎡ ⎛ 1 1 ⎞ 21 ⎛ 1 1 ⎞⎤ = ⎢ 4 ⋅ ⎜ − ⋅ 2 ⎟ + 5 ⋅ ⎜ − ⋅ 2 ⎟ ⎥ a.u. = − a.u. 9⎣ ⎝ 2 1 ⎠ 72 ⎝ 2 2 ⎠⎦ (1⋅ 2 ) 1 10 2 b) Lˆ2 = 2 ⎡0 + 32 ⋅ (1⋅ 2 ) + ( −1) ⋅ (1⋅ 2 ) + 10 ⋅ (1 ⋅ 2 ) ⎤ 2 = 2 ( 9 + 1 + 10 ) = ⎦ 6 ⎣ 6 9

21

2

1 c) Lˆ z = 2 ⎡0 + 32 ⋅ ( 6 ⎣

) + ( −1) ⋅ ( 0 ) + 10 ⋅ ( − )⎤⎦ = − 2

1 36

4.5 A hydrogen atom is in an eigenstate of L2 and Lz . Show that, although neither Lx nor L y

are well defined, their sum {L2x + L2y } is well defined. Find the value of {L2x + L2y } in terms of and m . Ignore spin. Solution:

{L

2 x

Therefore

{L

2 x

+ L2y } = L2 − L2z

+ L2y } n m = {L2 − L2z } n m =

[(

+ 1) − m

2

]⋅

2

n m

which does indeed have a definite value. 4.6 Show that the integrals

z nn (m−1)m = n ( − 1)m z n m ∞

= ∫ r 2 drRn (r )Rn ( 0

π

−1) (r )rdr ∫ 0

P m (θ )P(

−1)m

(θ )cosθ ⋅ sin θdθ

and znn (m+1)m = n ( + 1) m z n m π

∞

= ∫ r 2 Rn ( r ) Rn ( 0

+1)

( r ) rdr ∫0

P m (θ ) P(

+1) m

(θ ) cosθ ⋅ sin θ dθ

are given in atomic units by z nn (m−1)m =

3 n 2

(

)(

− m2 n2 − 2 (2 + 1)(2 − 1) 2

)

and z nn (m+1)m =

3 n 2

[(

][

+ 1) − m 2 n 2 − ( + 1) (2 + 3)(2 + 1) 2

2

]

Solution:

To evaluate the θ-integral use the following relation for associated Legendre functions: P m (θ ) cosθ =

(

+ 1) − m 2 P (2 + 3)(2 + 1) ( 2

22

− m2 P +1) m (θ ) + (2 + 1)(2 − 1) ( 2

−1) m

(θ )

and the orthogonality relation

∫

π

0

⎛ 2⎞ Rn ( r ) = − ⎜ ⎟ ⎝n⎠

P m (θ )P 'm (θ )sin θdθ = δ ' . Evaluate the r-integral using

3/ 2

1/ 2

⎡ ( n − − 1)!⎤ ⎢ ⎥ ⎢⎣ 2n ( n + ) ⎥⎦

⎛ 2r ⎞ ⎛ r ⎞ 2 +1 ⎛ 2r ⎞ ⎜ ⎟ ⋅ exp ⎜ − ⎟ ⋅ Ln − −1 ⎜ ⎟ ⎝ n ⎠ ⎝ n⎠ ⎝ n ⎠

and the recursion relation for associated Laguerre functions Lαβ (x ) = Lαβ +1 (x ) − Lαβ −+11 (x ) which is applicable only to the associated Laguerre polynomials as defined in the radial wave function given above, i.e. Definition 2. You will also need the orthogonality relation

∫

∞

0

By letting

→ + 1 obtain Rnln (

ρ p e − ρ Lqp ( ρ )Lqp' ( ρ )dρ = + )1

=

(q + p )! δ q!

qq '

3 2 n n 2 − ( + 1) . 2

The θ-integral: Replace P m (θ ) cosθ with the identity given. For z nn (m−1)m the only term that survives is the one with P m−1 (θ ) . Therefore π

∫

0

P m (θ )P(

⎛ ⎜ ⎜ ⎝

π

−1)m

(θ ) cosθ ⋅ sin θdθ = ∫0

− m2 P (2 + 1)(2 − 1) (

⎞

2

−1) m

(θ )⎟⎟ P( −1)m (θ )sin θdθ ⎠

− m2 (2 + 1)(2 − 1) 2

=

For z nn (m+1)m the only term that survives is the one with P m+1 (θ ) . Therefore

∫

π

0

P m (θ )P(

π

+1)m (θ ) cos θ ⋅ sin θdθ = ∫ 0

=

⎛ ⎜ ⎜ ⎝

(

(

+ 1) − m 2 P (2 + 3)(2 + 1) ( 2

+ 1) − m 2 (2 + 3)(2 + 1) 2

The r-integral:

23

⎞

⎟ +1) m (θ ) P( +1)m (θ ) sin θdθ ⎟ ⎠

Rnn (

− )1

∞

= ∫ r 2 dr Rn (r )Rn ( 0

−1)

(r )rdr

⎧⎪⎛ 2 ⎞ 3 / 2 ⎡ (n − − 1)!⎤ 1 / 2 ⎫⎪ = ⎨⎜ ⎟ ⎢ ⎥ ⎬⋅ ⎪⎩⎝ n ⎠ ⎣ 2n(n + )! ⎦ ⎪⎭ ⎧⎪⎛ 2 ⎞ 3 / 2 ⎡ (n − )! ⎤ 1 / 2 ⎫⎪ ∞ − 2 r / n ⎛ 2r ⎞ 2 −1 2 +1 ⎛ 2r ⎞ 2 −1 ⎛ 2r ⎞ 3 ⎜ ⎟ Ln − −1 ⎜ ⎟ Ln − ⎜ ⎟r dr ⎨⎜ ⎟ ⎢ ⎥ ⎬∫ e ⎝ n⎠ ⎝ n⎠ ⎝ n⎠ ⎪⎩⎝ n ⎠ ⎣ 2n(n + − 1)!⎦ ⎪⎭ 0

n dx . We get 2

Now let x = 2r / n so dr = Rnn (

− )1

∞

= ∫ r 2 dr Rn (r )Rn ( 0

=

−1)

1 (n − − 1)!(n − 4 (n + − 1)!(n +

(r )rdr

)1 / 2 ∞ e − x x 2 + 2 L2 +1 (x )L2 −1 (x )dx n − −1 n− )1 / 2 ∫0

To use the orthogonality integral we must use the recursion relation to raise the upper index to match the power of x, i.e. to 2 + 2 . In the recursion relation Lαβ ( x ) = Lαβ +1 ( x ) − Lαβ −+11 ( x ) we identify α = n − − 1 and β = 2 + 1 in the first associated Laguerre function under the integral. We have then L2n −+1−1 (x ) = L2n −+ 2−1 ( x ) − L2n −+ 2− 2 ( x ) Now we have to do the same thing with the second Laguerre function. α = n −

and β = 2 − 1 .

Therefore, L2n −−1 ( x ) = L2n − ( x ) − L2n −

−1

(x )

Clearly the recursion relation must be applied twice more.

[

] [

[

] [

]

L2n −−1 ( x ) = L2n −+1 ( x ) − L2n −+1−1 ( x ) − L2n −+1−1 ( x ) − L2n −+1− 2 ( x ) = L2n −+1 ( x ) − 2 L2n −+1−1 ( x ) + L2n −+1− 2 ( x )

] [

]

= L2n −+ 2 ( x ) − L2n −+ 2−1 ( x ) − 2 L2n −+ 2−1 ( x ) − L2n −+ 2− 2 ( x ) + L2n −+ 2− 2 ( x ) − L2n −+ 2−3 ( x )

= L2n −+1 ( x ) − 3L2n −+1−1 ( x ) + 3L2n −+ 2− 2 ( x ) − L2n −+ 2−3 ( x )

Now, in order for the integral to be non-zero we must have both associated Laguerre polynomials the same. Therefore, the only terms in the product that will be non-zero are

[

]

[

]

− 3 L2n −+ 2−1 ( x ) − 3 L2n −+ 2− 2 ( x ) 2

2

24

We have then

∫

∞

0

e −x x 2

+2

∞

0

e −x x 2

+2

0

and using

∫

∞

L2n −+1−1 ( x )L2n −−1 ( x )d = −3∫ e − x x 2

+2

∫

∞

0

ρ p e − ρ Lqp ( ρ )Lqp' ( ρ )dρ =

[L

2 +2 n − −1

(q + p )! δ q!

∞

L2n −+1−1 ( x )L2n −−1 (x )d = −3∫ e − x x 2 0

+2

(x )]2 dx − 3∫0

e−x x 2

+2

[L

(x )]2 dx

(x )]2 dx − 3∫0

e −x x 2

+2

[L

(x )]2 dx

∞

qq '

[L

2 +2 n − −1

∞

2 +2 n− −2

⎡ (n − − 1 + 2 + 2)! (n − − 2 + 2 + 2)!⎤ = −3⎢ + (n − − 1)! (n − − 2)! ⎥⎦ ⎣

⎡ (n + + 1)! (n + )! ⎤ = −3⎢ + ⎥ ⎣ (n − − 1)! (n − − 2 )!⎦

⎡ (n + + 1)(n + )! (n + )! ⎤ = −3⎢ + ⎥ ⎣ (n − − 1)(n − − 2)! (n − − 2)!⎦ (n + )! ⎡ n + + 1 + 1⎤ = −3 (n − − 2)! ⎢⎣ n − − 1 ⎥⎦ (n + )! 2n = −3 (n − − 2)! (n − − 1) (n + )! n = −6 (n − − 1)! We have then

R

2 +2 n− −2

n ( − )1 n

1 (n − − 1)!(n − =− 4 (n + − 1)!(n +

)1 / 2 ∞ e − x x 2 + 2 L2 +1 (x )L2 −1 (x )dx n − −1 n− )1 / 2 ∫0 1/ 2 (n + )! n⎤ 1 (n − − 1)!(n − ) ⎡ =− −6 1/ 2 ⎢ 4 (n + − 1)!(n + ) ⎣ (n − − 1)! ⎥⎦ 1/ 2 1 (n − − 1)!(n − ) ⎡ (n + − 1)!(n + ) ⎤ = n 6 (n − − 1)! ⎥⎦ 4 (n + − 1)!(n + )1 / 2 ⎢⎣ 1/ 2 3 (n − ) (n + )n = 2 (n + )1 / 2

3 n n2 − 2 2 If we let → + 1 we obtain Rnn ( =

+1)

Rnln (

+ )1

=

3 2 n n 2 − ( + 1) 2

We have then

25

z

n ( −1)m n m

=

− m2 3 ⋅ n n2 − (2 + 1)(2 − 1) 2 2

3 = n 2 z nn (m+1)m =

(

(

)(

− m2 n2 − 2 (2 + 1)(2 − 1) 2

2

)

+ 1) − m 2 3 2 ⋅ n n 2 − ( + 1) (2 + 3)(2 + 1) 2

3 = n 2

2

[(

][

+ 1) − m 2 n 2 − ( + 1) (2 + 3)(2 + 1) 2

2

]

4.7 A collection of hydrogen atoms is in a parabolic eigenstate ψ given by

1 1 200 + 210 2 2 where the kets on the right are the spherical eigenkets n m .

ψ =

a) What is the energy eigenvalue, in atomic units, corresponding to ψ ? b) What are the possible sets of parabolic quantum numbers that characterize ψ ? Virtually no computation is required. Merely list the combinations that are possible from inspection of the wave function given. Solution: a) Since n = 2 the eigenvalue is: 1 1 1 E2 = − 2 = − = − a.u. 2 2n 2⋅2 8 b) The parabolic quantum numbers n, n1 , n2 , m are related by: n = n1 + n 2 + m + 1 Since n = 2 and m = 0 (from the wave function above), we must have n1 + n 2 = 2 − 1 = 1 Since n1 and n2 can take on only integral values, the possible combinations for parabolic quantum numbers are: {n1n2 m} = {100}

or

{n1n2 m} = {010} 4.8 The normalized energy eigenfunctions for the hydrogen atom in parabolic coordinates are given by m + 3/ 2 exp ( imφ ) n1 ! n2 !ε m /2 m m exp ⎡⎣ −ε (ξ + η ) ⎤⎦ (ξη ) Ln1 + m ( εξ ) Ln2 + m ( εη ) ψ n1 n2 m (ξ ,η , φ ) = 3/ 2 3/ 2 π n ( n1 + m ) ( n2 + m )

where ε = − 2 E and the L's are associated Laguerre polynomials. a) Find the normalized hydrogen atom energy eigenfunctions in parabolic coordinates for n = 2, n1 = 1, n2 = 0 = m in terms of the hydrogen atom energy eigenfunctions in spherical 26

coordinates. b) - h) What are the probabilities of measuring the following: b) the energy of the ground state. c) the energy of the first excited state. d) total angular momentum zero e) total angular momentum 2 . f) z-component of angular momentum zero. g) z-component of angular momentum . h) Find the normalized ground state eigenfunction in parabolic coordinates in terms of the spherical eigenfunctions. What is noteworthy about this eigenfunction and why? Solution: 1 1! 0 ! 0 ⎡ 1 ⎤ a) ψ 100 (ξ ,η , φ ) = ⋅ 3/ 2 3/ 2 ε 3/ 2 exp ⎢ − ε (ξ + η ) ⎥ ⋅ (ξη ) ⋅ L1 ( εξ ) ⋅ L0 (εη ) 2π 1! 0! ⎣ 2 ⎦ but L0 ( x ) = 1 and L1 (x ) = 1 − x so that 1 ⎡ 1 ⎤ ⋅ ε 3/ 2 exp ⎢ − ε (ξ + η ) ⎥ ⋅ (1 − εξ ) 2π ⎣ 2 ⎦ The defining relations for parabolic coordinates are ξ = r + z = r (1 + cosθ ) η = r − z = r (1 − cosθ ) y φ = φ = tan −1 x from which we find x = ξη cos φ

ψ 100 (ξ ,η , φ ) =

y = ξη sin φ 1 (ξ − η ) 2 1 r = (ξ + η ) 2 ξ =r+z η =r−z z=

⎛ 1 ⎞ 1 Also, ε = − 2 E = − 2⎜ − 2 ⎟ = which in this case is ε = 1/ 2 because n = 2 . ⎝ 2n ⎠ n We have then 3/ 2 ⎤ 1 ⎛1⎞ ⎛ r ⎞ ⎡ ⎛1⎞ ψ par ( r ,θ , φ ) = ⎜ ⎟ exp ⎜ − ⎟ ⋅ ⎢1 − ⎜ ⎟ ( r + z ) ⎥ 2π ⎝ 2 ⎠ ⎝ 2⎠ ⎣ ⎝ 2⎠ ⎦

⎤ 1 1 ⎛ r ⎞ ⎡ ⎛1⎞ exp ⎜ − ⎟ ⋅ ⎢1 − ⎜ ⎟ r (1 + cos θ ) ⎥ 2π 2 2 ⎝ 2⎠ ⎣ ⎝ 2⎠ ⎦ where the subscript par refers to the parabolic quantum numbers of the problem, n = 2, n1 = 1, n2 = 0 = m . =

Now Y00 (θ , φ ) =

1 4π

and Y10 (θ , φ ) =

3 1 ⎛ r⎞ cosθ while R20 (r ) = exp⎜ − ⎟ and 4π 2 ⎝ 2⎠ 27

⎛ r⎞ r exp⎜ − ⎟ so that we can rewrite u100 in a more transparent way. 2 6 ⎝ 2⎠ 1 1 1 ⎛ r ⎞⎛ r ⎞ 1 1 ⎛ r⎞ exp ⎜ − ⎟ ⎜1 − ⎟ − ⋅ r ⋅ exp ⎜ − ⎟ cos θ ψ 100 = 4π 2 4π ⎝ 2 ⎠⎝ 2 ⎠ 2 2 ⎝ 2⎠ 1 1 1 1 Y00 (θ , φ ) R20 ( r ) − 2 6 R21 ( r ) ⋅ Y10 (θ , φ ) = 2 2 3 2 1 1 or ψ par ( r ,θ , φ ) = R20 ( r ) Y00 (θ , φ ) − R21 ( r ) Y10 (θ , φ ) 2 2 b) The energy depends only on the radial part of the wavefunction, Rn (r ) . Since the only principal quantum number that appears in ψ 100 is n = 2, the probability of measuring E1, the ground state energy is zero. c) n = 2 corresponds to the first excited state energy so that the probability of measuring E2 is unity. ( + 1) where appears in the spherical d) The total angular momentum is given by R21 (r )

1

harmonic Y m (θ , φ ) . Therefore, the probability of measuring

= 0 is given by the square of the 2

1 ⎛ 1 ⎞ coefficient of the spherical harmonic that contains = 0, i.e. ⎜ ⎟ = 2 ⎝ 2⎠ 2

1 ⎛ 1 ⎞ e) Probability of =1: ⎜ − ⎟ = 2 2⎠ ⎝ f) Since both spherical harmonics in ψ 100 have m = 0 the probability of measuring m = 0 is unity. A better way of saying this is that m is a good quantum number in both spherical and parabolic coordinates and m = 0 for this parabolic eigenstate. g) In view of f) the answer is zero. h) Parabolic quantum numbers for the ground state are: n = 1, n1 = n2 = m = 0. From the expression given for the wavefunction in parabolic coordinates we have 1 ⎡ 1 ⎤ exp ⎢ − (ξ + η ) ⎥ ψ 000 (ξ ,η , φ ) = π ⎣ 2 ⎦ 1 ⎡ 1 ⎤ exp ⎢ − r ⎥ = π ⎣ 2 ⎦

To convert this to spherical coordinates we use Equations (4.32) ξ = r − z = r (1 − cosθ ) ; η = r + z = r (1 + cosθ ) ; φ = φ Adding the first two together we get ξ + η = 2r so in spherical coordinates this wave function is 1 ⎡ 1 ⎤ exp ⎢ − r ⎥ which is ψ 100 ( r ,θ , φ ) . Thus, the ground state parabolic wave function and the π ⎣ 2 ⎦ ground state spherical eigenfunctions are the same. The reason these wave functions are the same is that the ground state is non-degenerate. 4.9 A hydrogen atom is in an eigenstate that is characterized by the parabolic quantum numbers n1 = 1 ; n2 = 0 = m . The wave function in spherical coordinates in atomic units is:

28

3/ 2

[

]

1 ⎛1⎞ 1 –r / 2 un1n2m = u100 = ⎜ ⎟ – 1 + 2 r (1 + cos θ ) ⋅ e 2π ⎝ 2 ⎠ That is, you have been spared the trouble of writing the wave function in parabolic coordinates and converting to spherical coordinates. Some selected spherical harmonics and radial hydrogen atom wave functions are: 1 3 3 ;Y10 = cosθ ;Y11 = sin θe iφ Y00 = 4π 8π 4π 1 1 1 – 12 r e – r / 2 ; R21 = R10 = 2e – r ; R20 = re – r / 2 2 2 6 a) If a measurement of the energy of the atom in this state is made what are the possible values of the energy that could be measured? Give your answer in atomic units. What probabilities are associated with each value? b) If a measurement of the total orbital angular momentum of the atom in this state is made what are the possible values that could be measured? What probabilities are associated with each possible angular momentum? c) If a measurement of the z-component of the angular momentum of the atom in this state is made what are the possible values that could be measured? What probabilities are associated with each possible z-component ? Solution: a) Since n1 = 1 ; n2 = 0 = m we have n = n1 + n2 + m + 1 = 1 + 0 + 0 + 1 = 2 . Therefore, the eigenstate is characterized by n = 2 and the only possible result of the measurement is 1 1 E2 = – = – au 2 2⋅2 8 b) This is the only part that requires any calculation. We must cast u100 u110 in terms of the spherical harmonics and the radial wave functions. From the spherical harmonics and radial wave functions given, we have: 1 – 12 r ⋅ e – r / 2 = 2 R20 ; re – r / 2 = 2 6 R21

(

)

(

)

1 = 4π Y00 ; cos θ =

u100 =

4π Y10 3

1 ⎡ 4π ⎤ 1 Y10 ⎥ ⎢ – 2 R20 4π Y00 + 2 2 6 R21 3 4 π ⎣ ⎦ 8π

1 [– R20Y00 + R21Y10 ] 4 π 2 From the last form we see that = 1 and 0 are possible. Since the expansion coefficients are the same, these values are equally probable. c) The system is in an eigenstate of Lz, i.e m = 0. Therefore, the only possible value of the zcompontent of angular momentum that could be measured is 0. This is also clear from the last form of the wave function since both spherical harmonics have m = 0. =

[– R20Y00 + R21Y10 ] =

4.10 Show that the expectation value of the electric dipole moment of an atom having N electrons in a state of well-defined parity vanishes. Use atomic units. Suppose a hydrogen atom is in a spherical eigenstate. What is the dipole moment? How about a parabolic eigenstate? Solution: The dipole moment is given by 29

⎛ N ⎞ pˆ = ⎜ ∑ rq ⎟ ⎝ q =1 ⎠ so the expectation value of the dipole moment is ⎛ N ⎞ pˆ = ψ ⎜ ∑ rq ⎟ ψ ⎝ q =1 ⎠ where, of course, the integral on the rhs is over all space. Since rq = xq iˆ + yq ˆj + zq kˆ the dipole operator has definite parity – odd. If the state vector ψ has definite parity as specified in the statement of the problem, then the integrand on the rhs vanishes because the product ψ *ψ is necessarily even. Spherical eigenstates have definite parity so a hydrogen atom in a spherical eigenstate has no dipole moment. Parabolic eigenstates do not necessarily have definite parity so a hydrogen atom in such an eigenstate can have a permanent dipole moment. Of course, those states for which the spherical and parabolic eigenfunctions are the same cannot have dipole moments.

30

CHAPTER 5 - THE CLASSICAL HYDROGEN ATOM 5.1 Using SI units, find the period Tn of an electron in a hydrogen Bohr orbit of principal quantum number n. Show that this period is consistent with Kepler's third law. Show that is the same as Tc the period deduced from the correspondence principal. Hint: Eliminate the 's so the

constants match. Show that the result reduces to T = 2π n3 in atomic units. Solution: 2

⎛ 1 ⎞ e2 ⎛ 1 ⎞ e2 ⎛ 2π ⎞ 2 2 m r = ⇒ = =⎜ From electrostatics F = ⎜ ω ω ⎟ 2 ⎜ ⎟ ⎟ e n n n 3 ⎝ 4πε 0 ⎠ rn ⎝ 4πε 0 ⎠ me rn ⎝ Tn ⎠ ( 4πε 0 ) me r 3/ 2 = ⎡⎢ 2π n3 ( 4πε 0 ) me ⎤⎥ a 3/ 2 Solving for Tn we have Tn = 2π ( ) e2 n 0 e2 ⎢⎣ ⎥⎦ which, incidentally, is Kepler's 3rd law. Note that this result can also be obtained as follows. 2π rn 2π n 2 a0 4πε Tn = = 2 = 2 0 ⋅ 2π n3 a0 because rn and vn were obtained using Coulomb's vn e e / 4πε 0 n

(

)

law. This doesn't look the same as the above result though because we must eliminate the . To do this use a0 = ( 4πε 0 ) ⋅

2

me e 2

⇒

=

me e 2 a0 so that 4πε 0

⎤ 4πε 0 me 3/ 2 a0 ⎥ a0 = 2π n3 e2 ⎥⎦ Note that in atomic units everything in Tn is unity except 2π n3 . Now find the frequency from the correspondence principle according to which the frequency of the motion should be the same as the difference between adjacent energies divided by . Since Tn =

⎡ m e2 a 4πε 0 3 0 e π ⋅ 2 n ⎢ 2 e ⎢⎣ 4πε 0

(

)

(

)

2 2 me c 2 α 2 1 me c α 2π ⇒ ΔE = Δn = ω c = En = − 2 3 2 n n Tc where the subscript "c" means "correspondence". But, for adjacent levels Δn = 1 so we have 2π n3 . To compare this expression with that for Tn we must replace α. Tc = me c 2 α 2

(

)

One way to do this is to write α in terms of the Bohr radius, see Table 1.2. 2

⎞ ⎞ 1⎛ 1⎛ 2π n3 ⎡ ⎛ me c ⎞ ⎤ 3 2 ⎛ me ⎞ a a0 = ⎜ ⎟ ⇒ α= ⎜ ⎟ so Tc = ⎟ ⎥ = 2π n a0 ⎜ ⎟ 2 ⎢ 0⎜ a0 ⎝ me c ⎠ α ⎝ me c ⎠ me c ⎣ ⎝ ⎠⎦ ⎝ ⎠

(

But, we must replace Tc = 2π n3 a0 2 me

)

as above so

4πε 0 4πε 0 me = 2π n3 a03/ 2 = Tn 2 me e a0 e2

5.2 The (classical) Lenz vector (in a.u.) is A = p × L − rˆ where rˆ is the unit vector in the r direction. For a general central potential find a general 31

expression for the time derivative of A, i.e. A and show that A is a constant of the motion for a Coulomb potential. Solution: d A = ( p × L) − rˆ dt d ( p × L) . First work on the term dt d ( p × L) = p × L + p × L = p × L because L = 0 for any central potential. dt Now use Newton's 2nd law for an arbitrary central potential V(r). We have r d dV 1 dV dV dV ⎛ ⎞ ( p × L) = − (rˆ × L) = (r × L) . which gives = −⎜ ⎟ p = − rˆ dt dr r dr dr ⎝ r ⎠ dr But L = r × p = r × mr which is L = r × r in atomic units so d ( p × L) = − 1 dV (r × L) = − 1 dV [r × (r × r )] = − 1 dV r (r • r ) − rr 2 dt r dr r dr r dr 1 d (r • r ) = rr so Now r • r = 2 dt d ( p × L) = − 1 dV rrr − rr 2 dt r dr 1 dV 3 ⎛ r r⎞ r ⎜ 2 r− ⎟ =− r dr ⎝ r r⎠ dV d ⎛ r ⎞ = r2 ⋅ ⎜ ⎟ dr dt ⎝ r ⎠ dV d = r2 ⋅ rˆ dr dt dV = r2 rˆ dr dV ⎞ d dV ⎛ Then A = ( p × L) − rˆ = r 2 rˆ − rˆ = ⎜1 − r 2 ⎟rˆ dt dr dr ⎠ ⎝ 1 1 dV For the Coulomb or Kepler potential we have V (r ) = − in atomic units. Therefore =− 2 r dr r and A = 0 .

[

[

]

]

5.3 a) Derive the equation of a Keplerian ellipse in terms of the Lenz vector A and show that the eccentricity is ε = A . b) Show that

ε = 1−

2

n Note that neither nor n are "quantum numbers" because, classically, they are continuously is the classical angular momentum and n is a measure of the energy. Recall that, in variable. atomic units E = −1/ 2n 2 .

( )

c) By averaging over a period show that the electric dipole moment of a Keplerian hydrogen 32

3 2

atom is < p > = n 2 1 −

2

/ n2 .

d) For what value of the is < p > = 0 ? What is special about these orbits? e) Find the positions (or position) of the maxima (or maximum) in the radial probability density for hydrogen atoms having = n − 1 , the maximum angular momentum. What is special about these states? It may be helpful to recall that dp ⎛ r ⎞ ⎛ 2r ⎞ 2 +1 ⎛ 2r ⎞ q Rn (r ) ∝ exp⎜ − ⎟ ⋅ ⎜ ⎟ ⋅ Ln + ⎜ ⎟ in a.u. and that L p ( ρ ) = Lq ( ρ ) . dρ p ⎝ n⎠ ⎝ n ⎠ ⎝ n ⎠ Solution: a) The Lenz vector is A = p × L − a r . Now take the dot product with r. A • r = p × L • r − a r • r Now permute the cross/dot product as follows: Ar cosθ = r × p • L − r Ar cosθ =

2

= 1 + ε cosθ which is the equation r of an ellipse with A = ε provided ε < 1 and θ is measured from pericenter. or, since L =

2

− r . Solving for r we have

2

b) For a Coulomb/gravitational potential the equation of the orbit is of the form where ε = 1 + 2 E c)

2

(atomic units). But E = −

r

= 1 + ε cosθ

2 1 in au so that ε = 1 − 2 2n n

p = r in atomic units.

By symmetery y = 0 . Also x = r cos θ (ignore the minus sign). x =

1

τ

τ∫

0

1 r cosθ dθ but, inserting the equation of the orbit and conservation of angular

θ

momentum, = r 2θ where A has been replaced by ε we have 5 5 π 1 π cosθ r2 d cos ⋅ = = x = r d d θ θ θ 3 ∫−π 2 3 ∫−π 3 3 2πn 2πn 2πn dε (1 + ε cosθ ) Integrating we get π

sin θ 1 ⎛ 1⎞ d x = − − ⎟ 3 ⎜ 2 2πn ⎝ 2 ⎠ dε 1 − ε (1 + cos θ ) −π 1 − ε 2 5

(

=

5

⎛ 1⎞ d ⎜− ⎟ 2πn ⎝ 2 ⎠ dε 3

Now, at the upper limit, tan

−1

)

(

⎡ 1 ⎤ ⎢− 2 ⎥ ⎣ 1− ε ⎦

(

(∞) = π / 2 .

)

2

( 1− ε ) 2

tan −1

cosθ

⎛ 1⎞

π

∫ π ⎜⎝ − 2 ⎟⎠ (1 + ε cosθ ) −

π

2

dθ

dθ

) ∫ π (1 + cosθ ) −

(1 − ε ) tan θ

π

2

(1 − ε ) 2

−π

Similarly, for the −π -limit. Therefore, the term in

= π and we have 33

5 ⎤ 2 2ε ⎛1⎞ d ⎡ ⎛ 1 ⎞⎛ 3 ⎞ ⋅ π = − ⎟⎜ − ⎟ ⋅ ⎢ ⎥ ⎜ ⎟ 3 3/ 2 5 ⎜ 2 2 5/2 2πn ⎝ 2 ⎠ dε ⎣⎢ 1 − ε ⎦⎥ n ⎝ 2 ⎠⎝ 2 ⎠ 1 − ε Leave the ε in the numerator for now, but replace it in the denominator.

x =

ε = 1−

5

(

2

n

2

(

⇒ 1− ε

)

)

2 5/ 2

(

=

5

n

5

⇒ x =

5

⎛3⎞ ⎜ ⎟ n ⎝2⎠ 3

)

ε

( n) 5

5

⎛3⎞ = ⎜ ⎟n 2 ε ⎝2⎠

2

⎛ 3⎞ p = ⎜ ⎟ n2 1 − 2 n ⎝ 2⎠ p = 0 when = n , that is, the maximum angular momentum. This corresponds to a d) circular orbit. ⎛ 2r ⎞ ⎛ 2r ⎞ e) For = n − 1 the associated Laguerre polynomial is L2n(+n(−n)−11+)1 ⎜ ⎟ = L22nn−−11 ⎜ ⎟ When the ⎝ n ⎠ ⎝ n ⎠ lower index and the upper one are the same, the polynomial is a constant as may be seen from dp L pp ( ρ ) = L p ( ρ ) because the Laguerre polynomial Lp is a pth order polynomial. Therefore, dρ p the radial wave function for = n − 1 is given by ⎛ r⎞ Rn ,n −1 (r ) ∝ r n −1 exp⎜ − ⎟ . The radial probability density, P(r), is given by ⎝ n⎠ 2 ⎛ 2r ⎞ P(r ) ∝ r 2 Rn ,n −1 (r ) ∝ r 2 n ⋅ exp⎜ − ⎟ . The peak in the probability distribution can be found by ⎝ n ⎠ differentiating and setting equal to zero. We get rmax=n2. There is only one maximum in the probability density so this is the equivalent of a Bohr orbit. Notice that this is the same result that one obtains from the Bohr model of the atom, i.e. that rmax = n 2 a o . These states obviously correspond to "circular" states. so finally

5.4 This problem should show you why the conservation of the (classical) Lenz vector, A, implies closed orbits for the Kepler problem. Use A in a.u. so that A = p × L − rˆ a) Express A in terms of r and p alone, i.e. eliminate L. No cross products. b) Show that – ⎛ ⎛ 1 ⎞ 1 ⎞ ⎟⎟ = rmin ⎜⎜ 2 E + ⎟⎟ A = rmax ⎜⎜ 2 E + r r max ⎠ min ⎠ ⎝ ⎝

where rmax and rmin are the maximum and minimum values of r, i. e. apocenter and pericenter. c) Show that A is parallel to rmin and antiparallel to rmax . d) From the answer to c) it is clear that for a circular orbit A = 0. Prove this mathematically from the equations that you derived in b). Solution: a) L = r × p so that A = p × (r × p ) − rˆ = rp 2 − p( p • r ) − rˆ b) We eliminate p2 from this expression using the expression for the total energy.

34

E=

p2 1 − 2 r

and letting rˆ =

⇒

p 2 = 2E +

2 r

r we have r

1⎞ ⎛ A = r ⎜ 2 E + ⎟ − p( p • r ) r⎠ ⎝ Now, at rmin or rmax then p ⊥ r and p⋅r = 0, i.e. –

Then, letting r = rmin or rmax gives the desired result. ⎛ ⎛ 1 ⎞ 1 ⎞ ⎟⎟ = rmin ⎜⎜ 2 E + ⎟ and noting that the total energy is given by c) Using A = rmax ⎜⎜ 2 E + rmax ⎠ rmin ⎟⎠ ⎝ ⎝ 1 E=− where a is the semi-major axis and is such that rmin < a < rmax we see that the quantity 2a in ( ) is positive for rmin and negative for rmax thus establishing the directions. d) The only way that A could be parallel and anti-parallel to two opposite vectors is if A = 0. ⎛ ⎛ 1 ⎞ 1 ⎞ ⎟⎟ = rmin ⎜⎜ 2 E + ⎟ To see this mathematically, we use either of A = rmax ⎜⎜ 2 E + rmax ⎠ rmin ⎟⎠ ⎝ ⎝

where, for a circular orbit, rmax = rmin = R = a. Since E = –1/2a, A = 0 in both equations. Note that this is also consistent with A being the eccentricity.

35

CHAPTER 6 - THE LENZ VECTOR AND THE ACCIDENTAL DEGENERACY 6.1 Find the probability of measuring the z-component of the Lenz vector to be –1/2 in atomic units for a hydrogen atom for which n = 2; n1 = 1; n2 = 0 = m (parabolic coordinates). Solution: n − n1 0 − 1 1 For these parabolic quantum numbers Az = 2 = =− n 2 2 Thus, the system is in an eigenstate of Aˆ z with eigenvalue –1/2 so that the probability of measuring Az = −1/ 2 is unity. 6.2 Show that

(

)

Aˆ = ⎛⎜ 1 ⎞⎟ pˆ × Lˆ − Lˆ × pˆ − rˆ ⎝ 2⎠

= pˆ × Lˆ − ipˆ − rˆ Solution: We must show that ⎛1⎞ ⎜ ⎟ ⎝ 2⎠

( pˆ × Lˆ − Lˆ × pˆ ) = pˆ × Lˆ − ipˆ

It is sufficient to prove this by showing it for only the x-component.

( pˆ × Lˆ ) − ( Lˆ × pˆ ) x

x

= pˆ y Lˆ z − pˆ z Lˆ y − Lˆ y pˆ z + Lˆ z pˆ y

Now, add and subtract pˆ z Lˆ y and pˆ y Lˆ z we obtain

(pˆ × Lˆ ) − (Lˆ × pˆ ) x

x

(

)

(

)

= pˆ y Lˆ z + pˆ z Lˆ y − pˆ z Lˆ y − pˆ z Lˆ y + pˆ y Lˆ z − pˆ y Lˆ z − Lˆ y pˆ z + Lˆ z pˆ y

(

)

(

)

= pˆ y Lˆ z − pˆ z Lˆ y − Lˆ y pˆ z − pˆ z Lˆ y − pˆ z Lˆ y + Lˆ z pˆ y − pˆ y Lˆ z + pˆ y Lˆ z

) [ ] [ = 2( pˆ Lˆ − pˆ Lˆ ) − ipˆ + (− ipˆ ) = 2( pˆ × Lˆ ) − 2ipˆ (

]

= pˆ y Lˆ z − pˆ z Lˆ y − Lˆ y , pˆ z − pˆ z Lˆ y + Lˆ z , pˆ y + pˆ y Lˆ z y

z

z

y

x

x

x

x

from which we obtain the desired form of Aˆ .

[

]

6.3 a) Show that the commutator Aˆ + , Lˆ + = 0 . b) Assume that it is known that Aˆ + n = D + n ( + 1) m . That is, it is known that Aˆ + raises

[

]

by unity, but we don't know what it does to m ( = ) . Using the commutator Aˆ + , Lˆ + = 0 , confirm that when operating on n , Aˆ + also raises m ( = ) to ( + 1) . Solution: a)

36

(

)(

)

⎡ Aˆ x + iAˆ y , Lˆx + iLˆ y ⎤ = ⎡ Aˆ x , Lˆx ⎤ + i ⎡ Aˆ x , Lˆ y ⎤ + i ⎡ Aˆ y , Lˆx ⎤ − ⎡ Aˆ y , Lˆ y ⎤ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ = 0 + i iAˆ + i −iAˆ + 0

( ) ( z

z

)

=0

{

} = Aˆ Lˆ

b) Lˆ+ Aˆ + n

+

=0

n

+

because the Lˆ z quantum number already has its maximum value, i.e. m = . Also, assuming that we do not know the action of Aˆ on the m quantum number (which is ) we have

{

Lˆ+ Aˆ + n

}= D

+

+

Lˆ+ n ( + 1) m

= D + ⎡⎣( + 1) − m ⎤⎦ ⎡⎣( + 1) + m + 1⎤⎦ n ( + 1)( m + 1)

[

]

But, the right hand side of the last equation must vanish because Aˆ + , Lˆ + = 0 so we must have m = + 1 because m = −( + 2 ) is not an allowed quantum number.

{

6.4 Show that Aˆ − n m

} is an eigenfunction of

eigenvalue ( + 1)( + 2 ) provided m = − . Solution: Using Lˆ z , Aˆ x = iAˆ y ⇒ Lˆ z Aˆ x = Aˆ x Lˆ z + iAˆ y

[ ] [Lˆ , Aˆ ] = −iAˆ z

y

Lˆ z with eigenvalue (m − 1) and of Lˆ2 with

⇒ Lˆ z Aˆ y = Aˆ y Lˆ z − iAˆ x

x

we have Lˆ z Aˆ − n m = Aˆ x Lˆ z + iAˆ y − iAˆ y Lˆ z − Aˆ x n m

{

} ( ) = {Aˆ (Lˆ − 1) − iAˆ (Lˆ − 1)} n m = (m − 1){Aˆ n m } [Lˆ , Aˆ ] = −iAˆ Lˆ − iLˆ Aˆ + iAˆ Lˆ + iLˆ Aˆ x

z

y

z

−

[

]

Using 2 x & Lˆ2 , Aˆ y = iAˆ z Lˆ x + iLˆ x Aˆ z − iAˆ x Lˆ z − iLˆ z Aˆ x z y y z y z z we find that Lˆ2 , Aˆ − = −iAˆ z Lˆ y − iLˆ y Aˆ z + iAˆ y Lˆ z + iLˆ z Aˆ y − i iAˆ z Lˆ x + iLˆ x Aˆ z − iAˆ x Lˆ z − iLˆ z Aˆ x

[

]

(

)

= −iAˆ z Lˆ y − iLˆ y Aˆ z + iAˆ y Lˆ z + iLˆ z Aˆ y + Aˆ z Lˆ x + Lˆ x Aˆ z − Aˆ x Lˆ z − Lˆ z Aˆ x = Aˆ z Lˆ − + Lˆ − Aˆ z − Aˆ − Lˆ z − Lˆ z Aˆ − Lˆ z , Aˆ − = Lˆ z , Aˆ x − i Lˆ z , Aˆ y = iAˆ y − i − iAˆ x = − Aˆ −

[

] [

] [

]

(

)

[Lˆ , Aˆ ] = [Lˆ , Aˆ ]− i[Lˆ , Aˆ ] = −iAˆ − i(iAˆ ) = Aˆ Then [Lˆ , Aˆ ] = Aˆ Lˆ + Aˆ + Aˆ Lˆ − Aˆ Lˆ + Aˆ − Aˆ Lˆ −

z

x

2

−

z

y

z

−

z

−

y

z

−

−

x

−

z

−

Then

37

−

z

= 2 Aˆ z Lˆ − + 2 Aˆ − − 2 Aˆ − Lˆ z

{

} (

)

Lˆ2 Aˆ − n m = Aˆ − Lˆ2 + 2 Aˆ z Lˆ − + 2 Aˆ − − 2 Aˆ − Lˆ z n m =

(

{

}

+ 1) Aˆ − n m + 2 Aˆ z

(

)

(

+ m )( − m + 1) n (m − 1)

+ 2 Aˆ − − 2 Aˆ − m n m Thus Lˆ2 Aˆ − n m = { ( + 1) − 2(m − 1)} Aˆ − n m + 2 Aˆ z ( + m )( − m + 1) n ; ; (m − 1) so that Aˆ − n m is an eigenfunction of Lˆ2 with eigenvalue ( + 1)( + 2 ) if m = − . Note that this is the only way to make the term containing Aˆ vanish because m = + 1 is prohibited.

{

{

}

{

}

}

z

We have Lˆ2 Aˆ − n

{

}= { (

) − 1]}{Aˆ − = ( ( + 1) + 2( + 1)){Aˆ − n = ( + 1)( + 2 ){Aˆ + n } + 1) − 2[(−

n

}

}

We have thus proved that Aˆ − n ; ; (− ) = D −− n ; ( + 1); − ( + 1) 6.5 Show that

⎡ n 2 − ( + 1)2 ⎤ ( ± m + 1)( ± m + 2 ) 1 ⎣ ⎦ Aˆ ± n m = ∓ n( + 1)( m ± 1) n ( 2 + 1)( 2 + 3) ±

2 2 1 ⎡⎣ n − ⎤⎦ ( ∓ m )( ∓ m − 1) n( − 1)( m ± 1) n ( 2 − 1)( 2 + 1)

using the known action of Aˆ z on n m , Equation (6.59), and the commutation relation ⎡ Aˆ z , Lˆ± ⎤ = ± Aˆ ± (see Table 2.2). ⎣ ⎦ Solution: We will work it out for Aˆ + . From the commutation relation

(

)

Aˆ + n m = Aˆ z Lˆ+ − Lˆ+ Aˆ z n m = Aˆ z

(

− m )( + m + 1) n ( m + 1)

1 − Lˆ+ n 1 − Lˆ+ n

(

− m + 1)( + m + 1)

(2

+ 1)( 2 + 3)

⎡ n 2 − ( + 1)2 ⎤ n ( + 1) m ⎣ ⎦

( − m )( + m ) ⎡ 2 n − ( 2 − 1)( 2 + 1) ⎣

2

⎤⎦ n ( − 1) m

where the second and third terms come from applying Aˆ z as per Equation (6.59). Now operate with the remaining operators.

38

Aˆ + n m =

(

− m )( + m + 1)

1 n

(

+

(

− m )( + m + 1)

1 n

(

−

1 n

(

−

1 n

( − m )( + m ) ⎡ n2 − ( 2 − 1)( 2 + 1) ⎣

(2

− m + 1)( + m + 1)

(2

− m )( + m + 2 )

+ 1)( 2 + 3)

+ 1)( 2 + 3)

⎡ n 2 − ( + 1)2 ⎤ n ( + 1)( m + 1) ⎣ ⎦

− m − 1)( + m + 1)

(2

− 1)( 2 + 1)

⎡ n 2 − ( + 1)2 ⎤ ⎣ ⎦ 2

(

⎤⎦

(

⎡⎣ n 2 −

2

⎤⎦ n ( − 1)( m + 1)

+ 1 − m )( + m + 2 ) n ( + 1)( m + 1)

− 1 − m )( + m ) n ( − 1)( m − 1)

where the first two terms come from application of Equation (6.59). Collecting terms 2 2 ⎧ ⎫ ˆA n m = 1 ⎪⎨ ( − m ) ( + m + 1)( + m + 2 ) − ( − m + 1) ( + m + 1)( + m + 2 ) ⎪⎬ + n⎪ ( 2 + 1)( 2 + 3) ( 2 + 1)( 2 + 3) ⎪⎭ ⎩ × n 2 − ( + 1) n ( + 1)( m + 1) 2

⎧ 1⎪ + ⎨ n⎪ ⎩

(

− m)

− m )( − m − 1) ⎫⎪ ⎬ ( 2 − 1)( 2 + 1) ⎪⎭

( − m − 1) − ( ( 2 − 1)( 2 + 1)

× n2 −

2

+ m)

2

2

(

n ( − 1)( m + 1)

1 ⎡( − m ) − ( − m + 1) ⎤⎦ n⎣

(

+ m + 1)( + m + 2 )

1 + ⎡⎣( + m + 1) − ( + m ) ⎤⎦ n

(

− m )( − m − 1)

=

=− +

1 n

(

1 n

(

(2

(2

+ m + 1)( + m + 2 )

(2

+ 1)( 2 + 3)

− m )( − m − 1)

(2

− 1)( 2 + 1)

n2 −

+ 1)( 2 + 3)

− 1)( 2 + 1)

n 2 − ( + 1) n ( + 1)( m + 1)

n2 −

2

2

n ( − 1)( m + 1)

n 2 − ( + 1) n ( + 1)( m + 1) 2

2

n ( − 1)( m + 1)

which is the correct result.

39

6.6 Find all parabolic eigenfunctions for n = 2 as linear combinations of spherical eigenfunctions by applying Lˆ − = Iˆ− + Kˆ − to 211 sp = 2001 par where spherical eigenfunctions

are designated n m

sp

and parabolic eigenfunctions nn1 n2 m

. Solve these simultaneous

par

equations to obtain the parabolic eigenfunctions in terms of the spherical eigenfunctions. Answer: 211 sp = 2001 par 2001 par = 211 sp 210

21 − 1 200

=

sph

2100

2 = 211 − 1

sp

sph

1

=

1 2

1

+

par

2

2010 1

−

par

Therefore, 211 sp = 2001 Also, n (n − 1)(n − 2)

1

=

2

sph

par

=

sph

2010

2

Solution: We know that n (n − 1)(n − 1)

sph

2100

par

par

2100

Therefore, 210

2010

1 2

2100

par

211 − 1

par

= n00 (n − 1)

par

= =

par

1

210

2 1

sph

+

1 2 1

200

200 210 sph − 2 2 = 21 − 1 sp

n 10 (n − 2 ) par

+

1 2

par

1

+

2010

sph

n 01(n − 2)

2 par

=

j ( j + 1) − m(m − 1) n( m − 1)

par

.

THIS IS TWO.

1 2

n 10 (n − 2 )

par

+

1 2

n 01(n − 2)

( j + m )( j − m + 1) n( m − 1) n(n − 1)(n − 2 ) sp = (2n − 3)(n − 1 − n + 2 + 1) n(n − 1)(n − 3) sp = Lˆ − 210

For n = 2.

sp

(2n − 3)2 n(n − 1)(n − 3) sp = 2 21 − 1

sp

Now use Iˆ− nn1 n2 m = (n1 + 1)[n − (n1 + 1)] n (n1 + 1), n2 , (m − 1) And Kˆ − n1 n2 m = (n2 + 1)[n − (n2 + 1)] nn1 , (n2 + 1), (m − 1) to operate on the rhs. First term: Iˆ− + Kˆ − 2100 par = (1 + 1)[2 − (1 + 1)] 220 − 1 par + (0 + 1)[2 − (0 + 1)] 211 − 1 par

(

)

= 0 + 211 − 1

par

This makes sense because n1 ≠ 2 . Second term: Iˆ− + Kˆ − 2010 par = (0 + 1)[2 − (0 + 1)] 211 − 1

(

)

= 211 − 1

par

sph

.

=

Lˆ −

sph

THIS IS ONE.

Now apply Lˆ − = Iˆ− + Kˆ − to n (n − 1)(n − 2) Jˆ − j m =

par

+0

This makes sense because n 2 ≠ 2 .

40

par

+

(1 + 1)[2 − (1 + 1)] 202 − 1

par

par

using

2 21 − 1

We have then Or 21 − 1

sp

= 211 − 1

sp

par

=

1

[

211 − 1 par + 211 − 1 2 THIS IS THREE.

This is three of the four kets. To get the last one 200 orthogonality of the kets. Using 1

=

−

sp

210 200

sp

1

par

sp

]

it is easiest to take advantage of the

= 0 it is clear that

2010 par THIS IS FOUR. 2 2 Adding and subtracting the two parabolic eigenfunctions that consist of two spherical eigenfunctions yield the spherical in terms of the parabolic. 2001 par = 211 sp 200

sph

2100 2010

par

par

211 − 1

= =

par

2100

1 2 1

210

par

sph

+

1 2 1

200

200 210 sph − 2 2 = 21 − 1 sp

sph

sph

41

CHAPTER 7 - BREAKING THE ACCIDENTAL DEGENERACY 7.1 Show that no combination of quantum numbers can conspire to make the fine structure correction to the Bohr energy vanish. Solution: α2 ⎡ n 3⎤ (1) − ⎥ for j = ± 1/ 2 EFS ( n, j ) = En(0) 2 ⎢ n ⎣⎢ ( j + 1/ 2 ) 4 ⎦⎥ To vanish, the quantity in the square bracket must vanish. Therefore, we must have 3 4n − 3 j − ⎡ ⎤ n 3 2 =0 − ⎥=0 ⇒ ⎢ + + j 1/ 2 4 4 j 1/ 2 ( ) ( ) ⎣⎢ ⎦⎥ 3 Since the denominator is manifestly positive we must have 4n − 3 j − . 2 The possible values that j can take on are given by ⎛1⎞ j = n − ( 2k + 1) ⎜ ⎟ where k = 0,1, 2... ( n − 1) . For example, if k = 0 then j = jmax = n − 1/ 2 . If ⎝ 2⎠ k = ( n − 1) then j = jmin = 1/ 2 .

3 = 0 becomes 2 ⎧ ⎛ 1 ⎞⎫ 3 4n − 3 ⎨n − ( 2k + 1) ⎜ ⎟ ⎬ − = 0 ⎝ 2 ⎠⎭ 2 ⎩

Then 4n − 3 j −

3 3 − =0 ⇒ n = −3k 2 2 Since n must be positive this last equation is impossible. n + 3k +

7.2 Suppose the proton is approximated as a spherical shell of radius R ≈ 10−4 nm . Calculate the first order correction to the energy of the ground state, E0(1) , due to the finite size of the nucleus.

Is it positive or negative? Use the fact that R / a0 ≈ 10 −5 to make the approximation e −2 R / a0 ≈ 1 . Estimate the value of E0(1) in terms of the unperturbed ground state energy E0( 0) and compare with the fine structure correction to the ground state. Solution: First we must find the potential energy function to which the electron is subjected. Using Gauss' law we find: 1 e2 ⋅ r≤R V (r ) = − 4πε 0 R

e2 r≥R 4πε 0 r The perturbation is simply the difference between the above potential energy and that for a point nucleus. Thus, the perturbing Hamiltonian is: 1 ⎛ e2 e2 ⎞ ⎟ r≤R ⋅⎜ − Hˆ ' (r ) = 4πε 0 ⎜⎝ r R ⎟⎠ =−

1

⋅

=0 r≥R 42

The first order correction to the energy of the ground state is therefore ψ 100 Hˆ ' ψ 100 , or (1)

E0

1 e2 = ⋅ πa0 3 4πε 0

∫

R

0

⎛ e 2 e 2 ⎞ − 2 r / a0 ⎜⎜ − ⎟⎟ ⋅ e 4πr 2 dr But, e −2 R / a0 ≈ 1 so R⎠ ⎝ r

R⎛1 ⎛ 1 ⎞ 1 1⎞ ⎟⎟ ⋅ ⋅ e 2 ∫ ⎜ − ⎟ ⋅4πr 2 dr E0(1) ≈ ⎜⎜ 3 0 ⎝r R⎠ ⎝ 4πε 0 ⎠ πa0 R

⎛ 1 ⎞ 4 ⎡ r2 r3 ⎤ ⎟⎟ ⋅ 3 ⋅ e 2 ⎢ − = ⎜⎜ ⎥ ⎣ 2 3R ⎦ 0 ⎝ 4πε 0 ⎠ a0 2

⎛ 1 ⎞ 2 e2 ⎛ R ⎞ ⎟⎟ ⋅ ⋅ ⋅ ⎜⎜ ⎟⎟ = ⎜⎜ ⎝ 4πε 0 ⎠ 3 a0 ⎝ a0 ⎠ which is clearly positive. Since the (unperturbed) energy of the hydrogen atom may be written 2

⎛R⎞ ⎛ 1 ⎞ e2 4 ⎟⎟ ⋅ We may write E0(1) in terms of E0(0 ) : E0(1) = E0( 0) ⎜ ⎟ ∼ 10−10 E0( 0) E n = ⎜⎜ 3 ⎝ 4πε 0 ⎠ 2a 0 ⎝ a0 ⎠ 1 () Compared with the magnitude of the fine structure corrections, EFS ( n = 1) ∼ α 2 E0(0) ∼ 10−5 E0( 0) , this is quite small. (0 )

7.3 A particle of rest mass m0 is confined to an infinite one-dimensional potential well, i.e. V ( x ) = ∞ x < 0 and x > L =0

otherwise

(1)

Find the first order correction to E n , the energy of the nth level of this particle-in-a-box, due to the relativistic kinetic energy of the electron. Put your answer in terms of the unperturbed energies and the rest energy of the particle. Under what circumstances will the validity of E n(1) be questionable? Consider two cases, an electron in a 0.1 nm box (an atom) and a proton in a 104 nm box (a nucleus). Solution: The relativistic energy is: 1/ 2

⎛ p2 1 p4 p2 ⎞ − ⋅ T = E − m0 c = m0 c + p c − m0 c = m0 c ⎜⎜1 + 2 2 ⎟⎟ − m0 c 2 ≈ 2m0 8 m 0 3 c 2 ⎝ m0 c ⎠ The first term is, of course, the non-relativistic kinetic energy so the second term may be 1 p4 considered the perturbation due to the relativistic motion of the particle, i.e. Hˆ ' = − ⋅ 3 2 . 8 m0 c 2

2

4

2

2

2

Now, the unperturbed energies are: E n(0 ) =

2

π 2 2n2 2m0 L2

. In the well the unperturbed Hamiltonian is

p2 so we may write Hˆ ' in terms of Hˆ 0 , i.e. given by Hˆ 0 = 2m0

(

)

2 1 p4 1 2m0 Hˆ 0 . The first order correction to the energy is thus: Hˆ ' = − ⋅ 3 2 = − 3 2 8 m0 c 8m0 c

43

(1)

(0 )

En = ψ n

Hˆ ' ψ n(0 ) = −

( )

(0 ) 1 (0 ) ˆ ˆ ψ (0 ) = − E n H H ψ | 0 0 n n 2m0 c 2 2m0 c 2

2

Note that E n(1) can also be evaluated by performing the integral, i.e. so that (1)

(0 )

En = ψ n

Hˆ ' ψ n(0 ) = −

1 1 ψ n(0 ) p 4 ψ n(0 ) = − 3 2 3 8m0 c 8m0 c 2

∫

L

0

d ⎞ 2 ⎛ nπx ⎞⎛ sin ⎜ ⎟⎜ ⋅ ⎟ L ⎝ L ⎠⎝ i dx ⎠

4

2 ⎛ nπx ⎞ sin ⎜ ⎟dx L ⎝ L ⎠

4

1 ⎛ nπ ⎞ L 2 ⎛ nπx ⎞ =− ⋅ 4 ⋅⎜ ⎟ ∫ sin ⎜ ⎟dx 3 2 4m0 c L ⎝ L ⎠ 0 ⎝ L ⎠ nπ L Let y = x ; dx = dy ; x = 0 → y = 0 ; x = L → y = nπ . L nπ Therefore, 4 4 1 1 ⎛ L ⎞ nπ ⎛ L 4 ⎛ nπ ⎞ 2 4 ⎛ nπ ⎞ ⋅ ⋅ ⋅ ⋅ = − ⋅ ⋅ E n(1) = − sin y dy ⎜ ⎟ ⎜ ⎟ ∫0 ⎜ ⎟ ⋅⎜ 3 2 3 2 4m 0 c L 4m0 c L ⎝ L ⎠ ⎝ nπ ⎠ ⎝ L ⎠ ⎝ nπ 1 1 =− ⋅ 2 m0 c 2

2

⎞ ⎛ nπ ⎞ ⎟⋅⎜ ⎟ ⎠ ⎝ 2 ⎠

( )

⎡π 2 2 n 2 ⎤ E n(0 ) = − ⎢ 2 ⎥ 2m0 c 2 ⎣ 2m0 L ⎦

2

If m0 is the mass of an electron and L = 0.1 nm, then E n(0 ) ≈ 10 eV , i.e. typical of atomic energy

levels. Since m0 c 2 = 0.51 MeV E n(1) will be small unless n is very large. On the other hand, if a

proton for which m0 c 2 ≈ 1000 MeV is confined to a box with L ≈ 10 −4 nm , the result for E n(1) will be questionable.

7.4 A particle of rest mass m0 is confined to a one-dimensional harmonic oscillator potential V ( x ) = kx 2 . Find the first order correction to E0(1) the energy of the ground state due to the 1 2

relativistic kinetic energy of the electron. Put your answer in terms of the unperturbed energies and the rest energy of the particle. Suppose the vibrations of a diatomic molecule are approximated by the harmonic oscillator. Show that, because the separations between molecular vibrational levels are typically the order of tenths of eV, the correction due to the relativistic motion of the electron is small. There are (at least) three ways to work this problem. Solution: Method #1: The perturbing Hamiltonian is the same as in the previous problem. 1 Hˆ ' = A ⋅ p 4 where A = − . Simply compute the integral 0 H ' 0 using the known 8m03c 2

ground state wave function in coordinate space and making the substitution p → −i ( d / dx ) . Thus, the Gaussian ground state wave function will be differentiated four times and the integral evaluated. Uninteresting! Method #2: ⎞ ⎞ m0ω ⎛ m0ω ⎛ i i p ⎟ and a ≡ p ⎟ so that, as Use the ladder operators: a + ≡ ⎜x− ⎜x+ 2 ⎝ m0ω ⎠ 2 ⎝ m0ω ⎠ usual, a + n =

( n + 1)

n +1

and a n = n n − 1 where the n are the harmonic oscillator

eigenfunctions corresponding to the quantum number n. Subtracting the defining equations of ⎛m ω⎞ the ladder operators we get p = ⎜ 0 ⎟ a − a + . The first order correction that we seek is ⎝ 2 ⎠

(

)

44

2

4 4 ⎛m ω⎞ E0 = A ⎜ 0 ⎟ 0 a − a + 0 . The expansion of a − a + is simplified because a 0 = 0 . ⎝ 2 ⎠ Also, since the kets are orthogonal there must be the same number of raising as lowering operators to keep the integral from vanishing. We have

(1)

(

0 a − a+

)

4

(

( 0 (a

(

)

0 = 0 a 2 − aa + − a + a + a + 2 =

4

)

2

)

0

)

− a 3a + − a 2 a + a + a 2 a + 2 − aa + a 2 + aa + aa + + a + aa + a − aa + 3 0

(

)

0 −a + a 3 + a + a 2 a + + a + aa + a − a + aaa + 2 + a + 2 a 2 − a + 2 aa + − a + 3 a − a + 4 0 Only two of these terms survive the above criteria. We have

(

0 a − a+

)

4

0 = 0 a 2 a +2 0 + 0 aa + aa + 0 = 2 0 a 2 2 + 0 aa + 2 = 2 2 +1 =3

Finally, 3 ( ω) ⎛m ω⎞ E0 = A ⎜ 0 ⎟ ⋅ 3 = − ⋅ 32 m0 c 2 ⎝ 2 ⎠ Method #3: In momentum space

2

2

(1)

∞

E0(1) = 0 H ' 0 = A∫ φ0* ( p ) p 4φ0 ( p ) dp −∞

where φ0 ( p ) is the ground state wave function in

momentum space. We can look this up or calculate it easily as follows. ⎛ ⎞ i a 0 =0 ⇒ ⎜x+ p ⎟ φ0 ( p ) = 0 Making the substitution x → i ( d / dp ) we have m0ω ⎠ ⎝ d φ0 ( p ) ⎛ 1 p2 ⎞ =− p ⇒ φ0 ( p ) = C exp ⎜ − ⎟ . To normalize the wave function we φ0 ( p ) m ω ⎝ 2m0 ω ⎠ compute the integral 1 = C

1= C

2

2

⎛ p2 ⎞ ∫−∞ exp ⎜⎝ − m0 ω ⎟⎠dp Let ∞

m0 ω ∫ exp ( − y 2 )dy . Using the integral ∞

−∞

∫

∞

0

p m0 ω

⇒ dp = m0 ω dy so

⎛ m +1⎞ Γ⎜ ⎟ 2 2 x m e − ax dx = ⎝ ⎛ m +1 ⎞ ⎠ we have ⎜

2a ⎝

⎟ 2 ⎠

1/ 4

⎛ 1 ⎞ C =⎜ ⎟ ⎝ π m0 ω ⎠ E0( ) = A 1

p m0 ω

. Finally, we have

∞ ⎛ 1 p2 ⎞ 4 − exp ⎜ ⎟ p dp . π m0 ω ∫−∞ ⎝ m0 ω ⎠

Making the same substitution as above we have

⇒ dp = m0 ω dy which leads to

45

∞ ⎛ 1 ⎞ 1 2 E0(1) = ⎜ − exp ( − y 2 ) ( m0 ω ) y 4 3 2 ⎟ ∫ −∞ ⎝ 8m0 c ⎠ π m0 ω Γ (5 / 2) ⎛ 1 ⎞ 1 2 = ⎜− ( m0 ω ) 2 ⋅ 3 2 ⎟ 2 ⎝ 8m0 c ⎠ π

(

m0 ω dy

)

⎛ 1 ⎞ 1 2 3 = ⎜− ( m0 ω ) π 3 2 ⎟ 4 ⎝ 8m0 c ⎠ π 3 ( ω) =− ⋅ 32 m0 c 2

2

ω ≈ 10−1 eV the correction for a diatomic molecule will be

Since m0 c 2 ≈ 5 × 105 eV and ∼ 10−8 eV .

7.5 Starting with Equation (7.15), the Hamiltonian for the spin-orbit correction to the energy of the hydrogen atom Hˆ SO = ξ ( r ) ⋅ Sˆ • Lˆ

(

find the energy difference between to the j = ± Solution: We found that Hˆ SO = ξ ( r ) ⋅ Sˆ • Lˆ

(

)

1 levels. 2

so that

(

(1) ESO = Hˆ SO = ξ ( r ) ⋅ Sˆ • Lˆ

= ξ (r ) ⋅

)

2

⎡ j ( j + 1) − 2 ⎣ 2 ⎡ = ξ ( r ) ⋅ ⎢ j ( j + 1) − 2 ⎣

)

(

+ 1) − s ( s + 1) ⎤⎦

(

3⎤ + 1) − ⎥ 4⎦

Then 1⎞ 1⎞ (1) ⎛ (1) ⎛ ΔESO = ESO ⎜ + ⎟ − ESO ⎜ − ⎟ 2⎠ 2⎠ ⎝ ⎝ 2 ⎧ ⎡⎛ 1 ⎞⎛ 3⎞ = ξ ( r ) ⋅ ⎨ ⎢⎜ + ⎟ ⎜ + ⎟ − 2 ⎩ ⎣⎝ 2 ⎠⎝ 2⎠

(

⎤ ⎡⎛ 1 ⎞⎛ 1⎞ + 1) ⎥ − ⎢⎜ − ⎟ ⎜ + ⎟ − 2 ⎠⎝ 2⎠ ⎦ ⎣⎝

(

⎤⎫ + 1)⎥ ⎬ ⎦⎭

2

1 ⎞⎛ 3 1⎞ ⎛ ⎜ + ⎟⎜ + − + ⎟ 2⎝ 2 ⎠⎝ 2 2⎠ 2 1 ⎞⎛ 3 1⎞ ⎛ = ξ (r ) ⋅ ⎜ + ⎟⎜ + − + ⎟ 2⎝ 2 ⎠⎝ 2 2⎠ = ξ (r ) ⋅

= ξ (r ) ⋅

2

2

(2

+ 1)

7.6 An electron of mass me is bound by an isotropic harmonic oscillator potential

46

1 1 1 k x 2 + y 2 + z 2 = kr 2 = meω 2 r 2 2 2 2 Find the corrections to the ground and first two excited state energies due to the spin-orbit 2 1 dV ( r ) ˆ ˆ correction VSO = ⋅ ⋅ L• S . 2 2 dr 2me c r Solution: Because the Hamiltonion is separable, i.e. Hˆ = Hˆ x + Hˆ y + Hˆ z the eigenvalues are the sums of

(

V (r ) =

)

(

)

the eigenvalues for each oscillator. So E n = (n + 3 / 2 ) ω where n = n x + n y + n z . The first three energies are 3 / 2 ω , 5 / 2 ω and 7 / 2 ω .

( ω) . 1 Inserting V(r) into VSO we get VSO = ζ Lˆ • Sˆ = ζ Jˆ 2 − Lˆ2 − Sˆ 2 where ζ = 2 2me c 2 1 The first order correction to the energy is VSO = ζ [ j ( j + 1) − ( + 1) − s (s + 1)] . 2 Now examine the values of for each of the ground and first excited states by comparing the

(

)

(

)

2

degeneracies of the levels recalling that for a given the degeneracy is (2 +1). 3⎞ ⎛ Writing the energy eigenvalues as En = ⎜ nx + n y + nz + ⎟ ω = n ω we have: 2⎠ ⎝ n = 0 : (n x , n y , n z ) = (0,0,0) degeneracy = 1 n = 1 : (n x , n y , n z ) = (1,0,0), (0,1,0), (0,0,1) degeneracy =3

n = 2 : (n x , n y , n z ) = (2,0,0), (0,2,0), (0,0,2 ), (1,1,0), (1,0,1), (0,1,1) degeneracy = 6 1 In general, the degeneracy is ⎡ ( n + 1)( n + 2 ) ⎤ . This may be looked up, but it is relatively easy ⎣2 ⎦ to show as follows: n = nx + n y + nz Since nx = 0,1, 2...n there are ( n + 1) possible values for nx and the sum

Once nx is chosen, there are ( n − nx + 1) ways to choose n y . After both nx

n y + nz = n − nx .

and n y are chosen of course nz is set so the degeneracy is the number of possible ways that

( n − nx + 1)

can be constructed. That is

degeneracy =

n

∑ (n − n

nx = 0

n

n

n

n

nx = 0

nx = 0

nx = 0

nx = 0

x + 1) = n ∑ 1 − ∑ nx + ∑ 1 = n ( n + 1) − ∑ nx + ( n + 1)

1 1 ⎞ 1 2 ⎛ = ( n + 1) − n ( n + 1) = ( n + 1) ⎜ n + 1 − n ⎟ = ( n + 1)( n + 2 ) 2 2 ⎠ 2 ⎝ Note that

n

∑n

nx = 0

x

= n ( n + 1) was evaluated using Gauss' trick. 1 2

The key to determining the allowed values of for each value of n is to recall that for any central potential (such as this isotropic oscillator) the degeneracy for each value of is ( 2 + 1) . Thus, we must determine how to "assemble" the above degeneracies from this fact. Ground state: n = 0, = 0 only because this is the only way to get a degeneracy of 1.

47

Since = 0 and j = s for the ground state, there is no correction to the ground state. First excited state: For n = 1, = 1 only because this is the only way to get a degeneracy of 3. = 1, and j = 1 ± 1/2 = 1/2, 3/2. Therefore, for 1 5 1 j = 3 / 2; VSO = + ζ so E j =3 / 2 = ω+ ζ 2 2 2 5 j = 1 / 2; VSO = −ζ so E j =1 / 2 = ω −ζ 2 Second excited state: For n = 2, = 0 and 2. The degeneracy of 6 is the result of adding 1 + ( 2 ⋅ 2 + 1) . As for the ground state there will be no correction for = 0. For =2, and j = 2 ± 1/2 = 3/2, 5/2. 7 ω +ζ . 2 3 7 3 j = 3 / 2; VSO = − ζ so E j =3 / 2 = ω− ζ 2 2 2 Note: Generalizing, we see that the eigenstates are also eigenstates of the parity operator because even n's have only even 's. j = 5 / 2; VSO = +ζ so E j =5 / 2 =

48

CHAPTER 8 - THE HYDROGEN ATOM IN EXTERNAL FIELDS

(

)

1 8.1 Show that Lˆ • Sˆ = Lˆz Sˆ z + Lˆ+ Sˆ− + Lˆ− Sˆ+ . 2 Solution: Lˆ • Sˆ = Lˆ Sˆ + Lˆ Sˆ + Lˆ Sˆ x

x

y

y

z

z

But Lˆ+ = Lˆx + iLˆ y Lˆ− = Lˆx − iLˆ y

Therefore

⇒

(

)

(

)

1 Lˆ x = Lˆ+ + Lˆ− 2 1 ˆ Lˆ− = L+ − Lˆ− 2i

)(

(

) (

1 1 Lˆ • Sˆ = Lˆ z Sˆz + Lˆ+ + Lˆ− Sˆ+ + Sˆ− − Lˆ+ − Lˆ− 4 4 1 = Lˆ z Sˆz + ⎡⎣ Lˆ+ Sˆ+ + Lˆ+ Sˆ− + Lˆ− Sˆ+ + Lˆ− Sˆ− ⎤⎦ 4 1 − ⎡⎣ Lˆ+ Sˆ+ − Lˆ+ Sˆ− − Lˆ− Sˆ+ + Lˆ− Sˆ− ⎤⎦ 4 1 = Lˆ z Sˆz + ⎡⎣ 2 Lˆ+ Sˆ− + 2 Lˆ− Sˆ+ ⎤⎦ 4 1 = Lˆ z Sˆz + Lˆ+ Sˆ− + Lˆ− Sˆ+ 2

(

) ( Sˆ

+

− Sˆ−

)

)

8.2 Calculate the Zeeman splitting of each of the magnetic sub-levels of hydrogen for n = 2 assuming that the electron has zero spin. Assume that the Zeeman energy and the fine structure splitting are comparable. Solution: E ( n ) = E2( 0) + ΔE ( n ) where ΔE ( n, ) = EFS ( n, ) + EB ( m ) Since the electron has spin zero there is no spin orbit coupling and the energy correction due to fine structure is 2 ⎡ 3 ⎤ α 2 ( 0) n ( 0) α − δ EFS ( n ) = ET ( n ) + ED ( n ) = ⎢ − + E E ⎥ 2 n n n ⎢⎣ 4 ( + 1/ 2 ) ⎥⎦ n ⎧⎪ ⎡ 3 ⎫⎪ α 2 ( 0 ) ⎤ 1 En = ⎨⎢− + ⎥ −δ 0 ⎬ ⎪⎩ ⎣⎢ 4n ( + 1/ 2 ) ⎦⎥ ⎪⎭ n μ μ μ B Hˆ B = B Lˆ + 2 Sˆ • B → Hˆ B = B Lˆ z B ⇒ EB = B m

(

)

Therefore ΔE ( n ) = EFS ( n ) + EB ( n ) ⎧⎪ ⎡ 3 ⎤ 1 = ⎨⎢− + ⎥ −δ ⎩⎪ ⎣⎢ 4n ( + 1/ 2 ) ⎦⎥

⎫⎪ α 2 ( 0) μ B En + B m 0⎬ n ⎭⎪

and 49

0

⎧⎪ ⎡ 3 ⎫⎪ α 2 ( 0) μ B ⎤ 1 En + B m ΔE ( n = 2 ) = ⎨ ⎢ − + − δ ⎥ 0⎬ ⎪⎩ ⎢⎣ 8 ( + 1/ 2 ) ⎥⎦ ⎪⎭ 2 ⎧ ⎡ 3 2 ⎤ ⎫ α 2 ( 0) μ B B ΔE ( n = 2; = 1) = ⎨ ⎢ − + ⎥ ⎬ En + m ⎩⎣ 8 3 ⎦ ⎭ 2 7 2 ( 0) μ B B α En + m 48 For = 0 m = 0 so there is no contribution from the magnetic field, but the Darwin term contributes. ⎧ ⎡ 3 ⎤ ⎫ α 2 ( 0) 5 2 ( 0) ΔE ( n = 2; = 0; m = 0 ) = ⎨ ⎢ − + 2 ⎥ − 1⎬ En = α En 16 ⎩⎣ 8 ⎦ ⎭ 2 =

7 2 ( 0) μ B B α En + 48 7 0 E ( n = 2; = 1; m = 0 ) = α 2 En( ) 48 μ B 7 0 E ( n = 2; = 1; m = 1) = α 2 En( ) − B 48 5 0 E ( n = 2; = 0; m = 0 ) = α 2 En( ) 16 E ( n = 2; = 1; m = 1) =

The only states that are affected by the magnetic field are the two states for which m ≠ 0 . 8.3 Compute the energy shifts when a hydrogen atom in the n = 2 state is immersed in a very strong constant external magnetic field B = Bkˆ . Sketch these energies as a function of μ B B assuming that the (zero-field) fine structure interval is so small compared with the Zeeman energies that it is not noticeable on the sketch. This is the Paschen-Bach effect. How do these energies compare with the asymptotic energies in Table 8.6? Solution: μ B Hˆ B = B Lˆ z + 2 Sˆz

(

)

If this perturbing Hamiltonian is much greater than that of the fine structure then, since the eigenstates nm ms of the hydrogen atom Hamiltonian are also eigenstates of Hˆ B , we can immediately write the eigenvalues as 1⎞ 1 ⎛ E ⎜ n, , m , ms ± ⎟ = − 2 + μ B B ( m ± 1) 2⎠ 2n ⎝ Thus, for the n = 2 states, the = 0 state, for which m = 0 , the Bohr level is split by ± μ B B . The = 1 state, for which m = 0, ± 1 is unshifted for m = ∓1 , shifted by ±2 μ B B for m = ±1 and shifted by ± μ B B for m = 0 . These five enegies (including zero shift) are the same as the asymptotic energies of Table 8.6. The Zeeman energies are

0, ± μ B B, ± 2μ B B . 50

8.4 Show that EZ(1) for the strong and weak field cases, Equations (8.18) and (8.28), are the same for the top of the ladder state (or the bottom of the ladder state) and that they are the same as Equation (8.32). Solution: General expression for the weak field case is Equation (8.28): ⎛ 2 j +1⎞ α2 ⎡ n 3⎤ − ⎥ + μ B Bm j ⎜ EZ(1) ( n, j , m j ) = En( 0) 2 ⎢ ⎟ for j = + 1/ 2 n ⎣⎢ ( j + 1/ 2 ) 4 ⎦⎥ ⎝ 2j ⎠

⎛ 2 j +1 ⎞ 3⎤ n − ⎥ + μ B Bm j ⎜ ⎢ ⎟ for j = − 1/ 2 n ⎢⎣ ( j + 1/ 2 ) 4 ⎥⎦ ⎝ 2j+2⎠ General expression for the strong field case is Equation (8.18): ⎫⎪ ⎛ α 2 ⎞ ⎧⎪ 3 ⎡ ( + 1) − m ms ⎤ EZ(1) = − En( 0) ⎜ ⎟ ⎨ − ⎢ ⎥ + μ B B ( m + 2ms ) ⎬ ⎝ n ⎠ ⎪⎩ 4n ⎣⎢ ( + 1/ 2 )( + 1) ⎦⎥ ⎪⎭ Equation (8.32) is the top of the ladder state derived from the weak field case: 2 n 3⎤ (1) ( 0) α ⎡ − ⎥ + μ B B ( j + 1/ 2 ) EZ = En 2 ⎢ n ⎢⎣ ( j + 1/ 2 ) 4 ⎥⎦ For the top of the ladder state: m j = j ; m = = j − 1/ 2; ms = +1/ 2 Weak field: Use the first of the weak field equations because = j − 1/ 2 ⇒ j = + 1/ 2 . = En( 0)

EZ ( n, j , m j ) = En (1)

( 0)

= En( 0)

α2 ⎡ 2

α2 ⎡

⎛ 2 j +1⎞ n 3⎤ − ⎢ ⎥ + μ B Bm j ⎜ ⎟ for j = + 1/ 2 2 n ⎣⎢ ( j + 1/ 2 ) 4 ⎦⎥ ⎝ 2j ⎠

α2 ⎡

⎛ 2 j +1⎞ n 3⎤ − ⎥ + μ B Bj ⎜ ⎢ ⎟ n ⎣⎢ ( j + 1/ 2 ) 4 ⎦⎥ ⎝ 2j ⎠ 2

n 3⎤ − ⎥ + μ B B ( j + 1/ 2 ) ⎢ n ⎢⎣ ( j + 1/ 2 ) 4 ⎥⎦ which is the same as Equation (8.32). = En( 0)

α2 ⎡ 2

51

Strong field:

⎛ α 2 ⎞ ⎪⎧ 3 ⎡ ( j − 1/ 2 )( j + 1/ 2 ) − ( j − 1/ 2 )(1/ 2 ) ⎤ ⎪⎫ EZ(1) = − En( 0) ⎜ ⎟ ⎨ − ⎢ ⎥⎬ ( j − 1/ 2 ) j ( j + 1/ 2 ) ⎥⎦ ⎪⎭ ⎝ n ⎠ ⎪⎩ 4n ⎣⎢ + μ B B ⎡⎣( j − 1/ 2 ) + 2 (1/ 2 ) ⎤⎦ ⎤ ⎪⎫ ⎛ α 2 ⎞ ⎪⎧ 3 ⎡ 1 EZ(1) = − En( 0) ⎜ ⎟ ⎨ − ⎢ ⎥ ⎬ + μ B B ( j + 1/ 2 ) ⎝ n ⎠ ⎪⎩ 4n ⎣⎢ ( j + 1/ 2 ) ⎦⎥ ⎪⎭ which is the same as Equation (8.32). 8.5 Consider the effect of application of a constant magnetic field B = Bkˆ on the hyperfine levels of the ground state of hydrogen. Using the notation of Section 3.5 the Hamiltonian is μ 2μ 2κ ⎛ 2μ ⎞ Hˆ = ⎜ B Sˆ1z + N Sˆ2 z ⎟ B + KSˆ1 • Sˆ 2 ≈ B Sˆ1z B + 2 Sˆ1 • Sˆ 2 ⎝ ⎠ where the subscripts 1 and 2 refer to the electron and proton respectively. a) Recast the Hamiltonian in the form

κ Hˆ = μ B Bσˆ1z + σˆ 1 • σˆ 2 2 = ξσˆ1z + Wσˆ 1 • σˆ 2

where W = κ / 2 and ξ = μ B B . b) Find the exact values of the energies for this Hamiltonian. It is not necessary to find the eigenkets. c) Correlate the energies that you found in part b) with the eigenkets for B = 0 from which they emanate. These kets were found in Section 3.5. The eigenkets are not needed for this part either. Draw a graph of the energy E versus B, E vs will do. Solution: 1 a) Sˆ = σˆ where σˆ = σˆ x i + σˆ y j + σˆ x k 2 The Hamiltonian is therefore Hˆ = μ B Bσˆ1z + Wσˆ 1 • σˆ 2

(

)

b)Re-write the Hamiltonian as Hˆ = ξσˆ1z + W σˆ1xσˆ 2 x + σˆ1 y Sˆ2 y + σˆ1zσˆ 2 z . The eigenkets with B = 0 are, from Section 3.5: 11 = triplet 1 = α1 α 2

1 1 α1 β 2 + β1 α 2 2 2 1 − 1 = triplet −1 = β1 β 2 10 = triplet

0

=

1 1 α1 β 2 − β1 α 2 2 2 Use these as the "unperturbed eigenkets". Check to see if the unperturbed kets are eigenkets of the Hamiltonian with B ≠ 0 . No such luck! Only the top of the ladder and bottom of the ladder triplet kets are eigenkets of the full Hamiltonian as is easily seen as follows: singlet =

52

Hˆ 11 = [ξσˆ1z + Wσˆ 1 • σˆ 2 ] α1 α 2 = (ξ + W ) α1 α 2 Hˆ 1-1 = [ξσˆ1z + Wσˆ 1 • σˆ 2 ] β1 β 2 = ( −ξ + W ) β1 β 2 Thus, 11 and 1 − 1 are eigenfunctions of Hˆ with eigenvalues ( ±ξ + W ) respectively. On the

other hand, 10 and 00 are not eigenfunctions so we must diagonalize the remaining portion of the Hamiltonian. We may as well use α1 β 2 and β1 α 2 as the basis kets. Using the notation α1 β 2 = α1 β 2 and β1α 2 = β1 α 2 the matrix is ⎛ α1 β 2 Hˆ α1 β 2 α1β 2 Hˆ β1α 2 ⎞ ⎜ ⎟ ⎜ β α Hˆ α β ˆ βα ⎟ β α H 1 2 1 2 1 2 ⎠ ⎝ 1 2 Now we have to compute these matrix elements. At this point we may drop the subscripts with the understanding that the first of the α and β refers to the electron, particle 1, and the second to the proton. We require the actions of the Pauli spin matrices on α and β . From Chapter 2

σˆ x α = β : σˆ x β = α σˆ y α = i β ; σˆ y β = −i α σˆ z α = α ; σˆ z β = − β we have Hˆ αβ = ξ αβ + W ⎡⎣ βα + i ( −i ) βα − αβ ⎤⎦ = ξ αβ − W αβ + 2W βα = (ξ − W ) αβ + 2W βα Hˆ βα = −ξ βα + W ⎡⎣ αβ + ( −i ) i βα − βα ⎤⎦ = −ξ βα + 2W αβ − W βα

= − (ξ + W ) βα + 2W αβ The matrix is therefore ⎛ αβ Hˆ αβ αβ Hˆ βα ⎞ ⎛ (ξ − W ) 2W ⎞ ⎜ ⎟=⎜ ⎟ ⎜ βα Hˆ αβ − (ξ + W ) ⎠ βα Hˆ βα ⎟⎠ ⎝ 2W ⎝ The secular equation is 2W (ξ − W ) − E = − ⎡⎣(ξ − W ) − E ⎤⎦ ⎡⎣(ξ + W ) + E ⎤⎦ − 4W 2 = 0 − (ξ + W ) − E 2W

Expanding the determinant we have ⎡⎣(ξ − W ) − E ⎤⎦ ⎡⎣(ξ + W ) + E ⎤⎦ + 4W 2 = 0 or

(ξ − W )(ξ + W ) + E (ξ − W ) − E (ξ + W ) − E 2 + 4W 2 = 0 (ξ − W )(ξ + W ) + Eξ − EW − Eξ − EW − E 2 + 4W 2 = 0 (ξ − W )(ξ + W ) − 2 EW − E 2 + 4W 2 = 0 53

E 2 + 2WE − ⎡⎣ 4W 2 + (ξ − W )(ξ + W ) ⎤⎦ = 0

(

)

E 2 + 2WE − ⎡⎣ 4W 2 + ξ 2 − W 2 ⎤⎦ = 0 Using the quadratic formula we have 1 E = ⎡ −2W ± 4W 2 + ⎡⎣16W 2 + 4 ξ 2 − W 2 ⎤⎦ ⎤ ⎥⎦ 2 ⎢⎣ 1 16W 2 + 4ξ 2 = −W ± 2

(

= −W ± 4W 2 + ξ 2 Thus, the four energies are:

)

(

E11 = (W + ξ ) ; E1−1 = (W − ξ ) ; E10 = −W + 4W 2 + ξ 2

(

E00 = −W − 4W 2 + ξ 2

)

)

c) The subscripts on the E's obtained in part b) were chosen to coincide with the answer to this part of the problem which can be done by inspection. To correlate each of these energies with the uperturbed states from which they emanate we begin by noting that the top and bottom of the ladder states are already eigenfunctions so E11 and E1−1 obviously correlate to 11 and 1 − 1 respectively. We next examine the limit of each of these energies as ξ → 0 . It is readily seen that E11 → W ; E1−1 → W ; E10 → W E00 → −3W

Therefore, the energy that we have labeled E00 emanates from the singlet state 00 . The graph is

54

8.6 Consider an electron of mass me confined to an infinite one-dimensional potential well of width L. Show that the polarizability of this "atom" in the ground state is e 2 L4 m ∞ 4q 2 α = 256 6 2 e ∑ 5 2 π q =1 4q − 1

(

)

Note that q is an index, not the electric quantum number. The following integral will be helpful. 4n ⎧ ⎫ , n even ⎪ π 2 2 ⎪− ⎬ ∫0 y sin y sin ( ny ) dy = ⎨ ( n + 1) ( n − 1) ⎪ 0 , n odd ⎭⎪ ⎩ Solution: According to Equation (8.59) the polarizability is given by α d = −d 2 E / dF 2 . We must therefore find the correction to the energy that is proportional to the square of the electric field. We apply a constant electric field F and use perturbation theory to find the correction to the unperturbed energy due to application of the field. We could use a box centered at x = 0 with ends at x = ± L / 2 , but it will be easier to use a box with ends at x = 0 and x = L because the eigenfunctions are all sines. (For the symmetric box the eigenfunctions alternate between cosines and sines.) For a constant electric field F, the perturbing potential is V ' ( x ) = eFx as shown in the drawing.

Letting 1 = the ground state eigenket of the infinite square well the first order correction to the ground state energy is 2 L ⎛π x⎞ E1(1) = 1 eFx 1 = eF ∫ x sin 2 ⎜ ⎟dx L 0 ⎝ L ⎠ L

2 ⎡ ⎛ π x ⎞ ⎤ ⎫⎪ 2 ⎧⎪ x 2 1 ⎛ L ⎞ ⎡ ⎛ π x ⎞⎤ 1 ⎛ L ⎞ = eF ⎨ − x ⎜ ⎟ sin ⎢ 2 ⎜ − ⎟⎥ ⎜ ⎟ cos ⎢ 2 ⎜ ⎟⎥ ⎬ 4 ⎝ π ⎠ ⎣ ⎝ L ⎠⎦ 8 ⎝ π ⎠ L ⎩⎪ 4 ⎣ ⎝ L ⎠ ⎦ ⎭⎪0

= eF

2 ⎧⎪ ⎡ L2 1⎛ L⎞ ⎨⎢ − 0 − ⎜ ⎟ 8⎝π ⎠ L ⎩⎪ ⎢⎣ 4

2

2 ⎤ ⎡ 1 ⎛ L ⎞ ⎤ ⎫⎪ − − − 0 0 ⎥ ⎢ ⎜ ⎟ ⎥⎬ 8 ⎝ π ⎠ ⎥⎦ ⎭⎪ ⎥⎦ ⎢⎣

⎛ eL ⎞ =⎜ ⎟F ⎝ 2 ⎠ Since E1(1) is proportional to the first power of F, the first order correction is not what we are after. The constant of proportionality above is, in fact, the permanent electric dipole moment because it is the integral of the charge distribution, eψ *ψ , times x. Moreover, the first order correction to the energy is simply the energy of orientation of this permanent electric dipole in the field. For V ' ( x ) = eFx as shown above, the permanent electric dipole moment is non-zero. 55

Clearly the permanent electric dipole moment of this box is non-vanishing because the charge is displaced from the origin. Note that if we had used a symmetric box this correction would vanish since 2 L/2 ⎛π x ⎞ E1(1) = 1 eFx 1 = eF ∫ x cos 2 ⎜ ⎟dx ≡ 0 / 2 L − L ⎝ L ⎠ That is, the symmetric box and the perturbation would have the form

and the charge distribution would be symmetric with respect to the origin. It would also have vanished if we had added a constant to V ' ( x ) to make the straight line go through x = L / 2 in the asymmetric box as shown in the drawing below, but this is equivalent to translating the origin and "symmetrizing" the box as shown in the drawing.

Note that symmetrizing the box adds a term to the total energy because it adds a term to the perturbing potential energy. In particular, it add and amount −eLF / 2 which is, of course, exactly the amount necessary to cancel out E1(1) as calculated using V ' ( x ) = eFx . Note also that symmetrizing the box is also equivalent to adding an amount of energy to the total energy so that instead of having the zero of energy at zero and, consequently, unperturbed energy given by 2 2 2 n ( 0) π En = 2mL2 we would have

π 2 2n2

eLF 2mL 2 In other words, symmetrizing the perturbing potential adds energy so there is no contradiction. ( 0)

En =

2

+

56

The fact that the permanent electric dipole moment in this problem is not unique is a consequence of a theorem in Electricity & Magnetism (see e.g. Jackson). The values of the lowest nonvanishing multipole moment of any charge distribution are independent of the choice of origin of coordinates, but all higher multipole moments do in general depend on the location of the origin. In our problem, despite our reference of the electron in the square well as an "atom", there is a monopole moment, namely the charge on the electron. Thus, the permanent electric dipole moment is not unique. It is worth noting that there is a lot of physics that can be learned from the simple particle-in-a-box problem. Now − back to the problem. The permanent electric dipole moment notwithstanding, we seek the correction term that is proportional to F 2 so we calculate the second order correction to the energy which is given by 2 ˆ n H 1 S E1( 2) = ∑ ( 0 ) ( 0) n ≠1 E1 − En where n represents the nth eigenfunction of the infinite square well and En( 0) the corresponding eigenvalue. We can see that E1( 2) will indeed be proportional to F 2 because the matrix element in the numerator is proportional to F, which will be common to all of the squared terms in the summation. This matrix element is 2 L ⎛π x⎞ ⎛ nπ x ⎞ 1 Hˆ S n = ∫ sin ⎜ eFx ) sin ⎜ ( ⎟ ⎟ dx L 0 ⎝ L ⎠ ⎝ L ⎠ To evaluate this integral we again make the substitution πx L ; dx = dy ; x = 0 → y = 0 ; x = L → y = π so y= π L 2 π 2L π ⎛L ⎞ ⎛ L⎞ 1 Hˆ S n = ( eF ) ∫ sin y ⋅ ⎜ y ⎟ ⋅ sin ( ny ) ⋅ ⎜ ⎟ dy = ( eF ) 2 ∫ y sin y sin ( ny ) dy L 0 π 0 ⎝π ⎠ ⎝π ⎠ ⎡ ⎤ 2L ⎡ 2 L ⎢ −4 n −4 n = ( eF ) 2 ⎢ ⎥ = ( eF ) 2 π ⎢⎣ ( n + 1)2 ( n − 1)2 ⎥⎦ π ⎢ n2 − 1 ⎣

(

)

⎤ ⎥ , n even 2 ⎥ ⎦

so 1 Hˆ S n

2

⎡ 4n 2 ⎛ 2L ⎞ = ( eF ) ⎜ 2 ⎟ ⎢ ⎝ π ⎠ ⎢ n2 − 1 ⎣ 2

(

)

⎤ ⎥ 2 ⎥ ⎦

2

, n even

⎛π x⎞ Note that the fact that the integral vanishes for odd n can be seen because sin ⎜ ⎟ is even wrt ⎝ L ⎠ ⎛ nπ x ⎞ x = L / 2 while x is odd wrt to x = L / 2 . Therefore, when sin ⎜ ⎟ is even wrt x = L / 2 (as it ⎝ L ⎠ is for n odd) the integral will vanish. The sum must then indeed be over only even n. We have then

57

E1 = ( eF ) ( 2)

π2

( 0)

where E1 =

2

⎛ 2L ⎞ ⎜ 2⎟ ⎝π ⎠

2

∑

n even

16n 2

(n

2

)

−1

4

i

( 0)

E1

1 1 − n2

(

)

i

1 1 − n2

2

2me L2

so E1 = ( eF ) ( 2)

2

2

2 ⎛ 2 L ⎞ 2me L ⎜ 2⎟ 2 2 ⎝π ⎠ π

⎡ e 2 L4 m = − ⎢128 6 2 e ⎢ π ⎣⎢

∑

n even

∑

n even

16n 2

(n

2

) (

−1

4

)

⎤ ⎥ F2 5⎥ 2 n −1 ⎥ ⎦ n2

(

)

Let q = 2n so ⎡ ⎤ e 2 L4 m ∞ 4q 2 ⎥ 2 2 E1( ) = − ⎢128 6 2 e ∑ F 5 2 ⎢ π q =1 4q − 1 ⎥ ⎣ ⎦ The quantity in brackets is therefore one-half the polarizability so e 2 L4 me ∞ 4q 2 α = 256 6 2 ∑ 5 2 π q =1 4q − 1

(

)

(

)

Alternate solution using well centered on the vertical axis (generally undesirable). To effect this solution we must use the alternating cosine and sine wave functions. The second order correction is 2 ˆ n 1 H S E1( 2) = ∑ ( 0 ) where the matrix element in the numerator is given by ( 0) n ≠1 E1 − En

⎧ 2 L/2 ⎫ ⎛ π x ⎞ ⎛ nπ x ⎞ ⎪ L ∫− L / 2 ( eFx ) cos ⎜ L ⎟ sin ⎜ L ⎟ dx , n = 2, 4,..⎪ ⎪ ⎝ ⎠ ⎝ ⎠ ⎪ 1 Hˆ S n = ⎨ ⎬ ⎪ 2 L / 2 ( eFx ) cos ⎛ π x ⎞ cos ⎛ nπ x ⎞ dx , n = 3,5,..⎪ ⎜ ⎟ ⎜ ⎟ ⎪⎩ L ∫− L / 2 ⎪⎭ ⎝ L ⎠ ⎝ L ⎠ The second integral, the one for odd n, vanishes because the integrands are odd over symmetric limits. We are left with 2 L/2 ⎛ π x ⎞ ⎛ nπ x ⎞ 1 Hˆ S n = ∫ ( eFx ) cos ⎜ ⎟ sin ⎜ ⎟ dx , n = 2, 4,.. L −L/ 2 ⎝ L ⎠ ⎝ L ⎠ We can put this in more convenient form by making the substitution πx L z= ; dx = dy ; x = − L / 2 → z = −π / 2 ; x = L / 2 → y = π / 2 L π and we have

58

L/2 2 ⎛ π x ⎞ ⎛ nπ x ⎞ 1 Hˆ S n = ( eF ) ∫ x cos ⎜ ⎟ sin ⎜ ⎟ dx −L/ 2 L ⎝ L ⎠ ⎝ L ⎠

=

π /2 2 L L ( eF ) ∫−π / 2 ⎛⎜ z ⎞⎟ ⋅ cos z ⋅ sin ( nz ) ⎛⎜ dz ⎞⎟ L ⎝π ⎠ ⎝π ⎠

⎛ 2L ⎞ π / 2 n even = ( eF ) ⎜ 2 ⎟ ∫ z ⋅ cos z ⋅ sin ( nz ) dz , ⎝ π ⎠ −π / 2 This definite integral is similar, but not identical to, the integral used in the other calculation. It is given by ⎛n ⎞ π /2 4n ⎜ +1⎟ ⎝2 ⎠ z cos z sin nz dz = − ( ) ( ) 2 2 ∫−π / 2 ( n + 1) ( n − 1) so 2 2 2 π /2 2 ⎛ 2L ⎞ ˆ 1 H S n = ( eF ) ⎜ 2 ⎟ ∫ z ⋅ cos z ⋅ sin ( nz ) dz ⎝ π ⎠ −π / 2 = ( eF )

2

⎛ 2L ⎞ ⎜ 2⎟ ⎝π ⎠

2

⎡ ⎤ 4n ⎢ ⎥ 2 2 ⎢⎣ ( n + 1) ( n − 1) ⎥⎦

2

2

⎡ ⎤ 4n ⎥ 2 ⎛ 2L ⎞ ⎢ = ( eF ) ⎜ 2 ⎟ , n even 2 ⎝ π ⎠ ⎢ n2 − 1 ⎥ ⎣ ⎦ Since this is identical with the square of the matrix element that was obtained using the square well that went from x = 0 → L the polarizability will be the same (thankfully). 2

(

)

8.7 The energy levels in atomic units of a hydrogen atom in a constant electric field are, to first order 1 3 E ( n, n1 , n2 ) = − 2 + n ( n1 − n2 ) F 2n 2 where n, n1 and n2 are parabolic quantum numbers. a) Show that the difference between adjacent levels having the same principal quantum number is ⎛ 3⎞ ΔE = ⎜ ⎟ nF ⎝ 2⎠ b) Show that the difference between the two extreme components for a given principal quantum number and magnetic quantum number is ΔE = 3Fn ( n − 1) Solution: a) The parabolic quantum numbers are related by n = n1 + n2 + m + 1 so that adjacent levels

having the same value of m differ only in their values of the electric quantum number q = n1 − n2 which, from the relation between quantum numbers, changes in steps of 2 for fixed n and |m|. Therefore ⎛ 3⎞ ⎛ 3⎞ ΔE = ⎜ ⎟ nqF − ⎜ ⎟ n ( q − 2 ) F ⎝ 2⎠ ⎝ 2⎠ = 3nF 59

But this is twice the answer. The resolution of the problem comes by noting that for different m's the manifold of states dovetails. For example, suppose that n is odd and m = 0 . Then either n1 or n2 must be odd because their sum must be odd. Since only one is odd then their differencer q = n1 − n2 must be also be odd. On the other hand, if m = 1, then both n1 or n2 must be odd or both must be even. Since both are even or both are odd their difference, q = n1 − n2 , must be even. This will give a different set of levels from the case for m = 0. We can find the ΔE between adjacent levels having adjacent m's as follows: Assume m is positive for convenience n = n1 + n2 + m + 1 so that n1 = n − n2 − m − 1 from which n1 − n2 = n − 2n2 − m − 1 For m = 0 and m = 1 we have ( n1 − n2 )m =0 = n − 2n2 − 1 and ( n1 − n2 )m =1 = n − 2n2 − 2 For the same value of n2, but adjacent values of m, e.g. 0 and 1, we have ⎛ 3⎞ ΔE = ⎜ ⎟ n ⎡⎣( n1 − n2 )m =0 − ( n1 − n2 )m =1 ⎤⎦ F ⎝ 2⎠ ⎛ 3⎞ = ⎜ ⎟ n ⎡⎣( n − 2n2 − 1) − ( n − 2n2 − 2 ) ⎤⎦ F ⎝ 2⎠ ⎛ 3⎞ = ⎜ ⎟ nF ⎝ 2⎠ b) qmax = − qmin so that ⎛ 3⎞ ⎛ 3⎞ ΔE = ⎜ ⎟ nqmax F − ⎜ ⎟ nqmin F ⎝ 2⎠ ⎝ 2⎠ = 3nqmax F qmax = n1* − n2* where n1* = maximum value of n1 n2* = maximum value of n2

since n = n1 + n2 + m + 1 , n1* occurs when m = 0 and n1* = n − 1 and n2* = 0 so that

qmax = n1* − n2*

= ( n − 1) − 0 = n −1

and ΔE = 3n ( n − 1) F 8.8 It can be shown [see Am. J. Phys. 60, 324 (1992)] that the time rates of change of the classical angular momentum and the Lenz vector, in atomic units, for an H-atom subjected to a constant electric field F = Fkˆ are ⎛ 3⎞ ⎛ 3⎞ L = ⎜ ⎟ n2 ( A × F ) A = ⎜ ⎟( L× F ) and ⎝ 2⎠ ⎝ 2⎠ Show that these equations lead to a picture of the Keplerian orbit of the electron rotating about the electric field vector with a frequency ωS = (3/2)nF. Solution: These equations can be uncoupled by differentiating one and substituting in the other. Upon 60

differentiating the A equation we have ⎛ 3⎞ A = ⎜ ⎟( L× F ) ⎝ 2⎠ ⎫ ⎤ ⎛ 3 ⎞ ⎧ ⎡⎛ 3 ⎞ = ⎜ ⎟ ⎨ ⎢⎜ ⎟ n 2 ( A × F ) ⎥ × F ⎬ ⎝ 2 ⎠ ⎩ ⎣⎝ 2 ⎠ ⎦ ⎭ 2

⎛3 ⎞ = − ⎜ n ⎟ ⎡⎣ A ( F • F ) − F ( F • A ) ⎤⎦ ⎝2 ⎠ Since F = Fkˆ we see immediately that Az = 0 . The other components of A have the form 2

⎛3 ⎞ Ai ( t ) = − ⎜ nF ⎟ Ai ( t ) which leads to Ai ( t ) = C1 cos (ω S t ) + C2 sin (ω S t ) ⎝2 ⎠ 3 where C1 and C2 are constants and ω S = nF . Taking Ay ( t = 0 ) = 0 and the amplitudes as Ax0 2 and Ay0, we have Ax ( t ) + Ay ( t ) = Axo cos 2 (ω S t ) + Ay 0 sin 2 (ω S t ) 2

2

which, together with Az ( t ) = cons tan t , shows that the Lenz vector traces an ellipse in a plane perpendicular to the direction of the applied electric field. Further, the frequency of the motion is ωS. The changing shape of the ellipse is due to variations in the length of the minor axis while the length of the major axis, and thus the energy, remains constant. Similar treatment of the angular momentum leads to analogous equations for the motion of L. It too outlines an ellipse with frequency ω S in a plane perpendicular to the z-axis. The major axes of the two ellipses are perpendicular. As required by the general condition A • L = 0 , both vectors precess about the z-axis in the same direction. We have the following picture: The plane of the elliptical orbit of the electron precesses about the electric field vector with frequency ω S and, simultaneously, rotates about the electric field vector,also with a frequency ω S . The fact that the frequency of the rotation is the same as the energy between adjacent (quantum mechanical) Stark states is an illustration of the correspondence principle. 8.9 (This solution to this problem is identical with that for Problem 6.6. The question is, however, phrased in the context of this chapter. If you did not work it in Chapter 6 it is recommended that you work it here.) Use the the fact that the parabolic hydrogen atom eigenfunctions are also eigenfunctions in the presence of a constant electric field to derive the Stark effect wave functions for n = 2 in terms of the spherical eigenfunctions for a hydrogen atom in a constant electric field. Use the formalism of Chapter 6 by applying the operator Lˆ− = Iˆ− + Kˆ − to the spherical eigenfunction n m sp = 211 sp which is identical with the

parabolic eigenfunction nn1n2 m

par

= 2001

par

quantum numbers in the parabolic kets. Solution:

61

. Note that, although redundant, we use four

There will be four states. Two of the parabolic states are simply those that are the same as the spherical states: 21 ± 1 sp = 200 ± 1 par The other two will be linear combinations of the spherical states. Start by lowering 211 sp and 2001

par

Lˆ− 211 sph = 2 210

Also Lˆ− 2 0 01

par

sph

(

)

= Iˆ− + Kˆ − 2 0 01 = 2100

Therefore, 210

sp

=

1 2

( 2100

par

par

par

+ 2 010 + 2 010

par

par

)

is, as expected, a linear combination of the two parabolic states for which m = 0 . There is another m = 0 spherical state for n = 2 , viz. 200 sph . This state must also be a linear combination of 2100 to the state 210

sp

par

and 2 010

par

. This linear combination must, of course, be orthogonal

. This linear combination is

(

)

( (

) )

1 2100 par − 2 010 par as can be verified by taking the inner product. 2 We have now two equations for the two unknowns 2100 par and 2 010 par . 200

=

sp

1 2100 par + 2 010 par 2 1 20 0 sp = 2100 par − 2 010 par 2 Solving these two equations we have 1 210 0 par = 210 sp + 200 sp 2 1 2 010 par = 210 sp − 200 sp 2 The four states are: 210

sp

=

( (

2 0 01

par

2 0 0 −1 210 0 2 010

= 211 sp

par

par

par

) )

= 21 − 1 sp

(

sp

(

sp

1 210 2 1 = 210 2 =

+ 200 − 200

sp

)

sp

)

62

CHAPTER 9 - THE HELIUM ATOM 9.1 Two identical, non-interacting spin-1/2 particles of mass m are in the one-dimensional harmonic oscillator for which the Hamiltonian is p x2 1 p x2 1 Hˆ = 1 + mω 2 x12 + 2 + mω 2 x 22 2m 2 2m 2 a) Determine the ground state and first excited state kets and corresponding energies when the two particles are in a total spin-0 state. What are the lowest energy states and corresponding kets for the particles if they are in a total spin-1 state? b) Suppose the two particles interact with a potential energy of interaction V ( x1 − x 2 ) = −V0 for x1 − x 2 < a =0

elsewhere

Will the energies of part a) be raised or lowered as a result of V ( x1 − x2 ) ? Solution: a) The Hamiltonian is the sum of two parts, one containing only x1 and the other x2 where Hˆ 1 , Hˆ 2 = 0 . Therefore, we can label the eigenstates by n1n2 SM s where SM s is the two-

[

]

particle total spin state. The energy is given by E ( n1 , n2 ) = ( n1 + n2 + 1) ω . The ground state energy corresponds to n1 = 0 = n 2 so that the ground state energy is E0 = E ( 00 ) = ω . The ket corresponding to the ground state is n1n2 SM s = 00 00 . The first excited state energy is E1 = E (10 ) = E (01) = 2 ω with 4 possible kets:

1 1 ⎧ 1 ⎫ ⎧ 1 ⎫ 10 + 01 ⎬ 00 (singlet) and ⎨ 10 − 01 ⎬ S = 1 M s = 0 ± 1 (triplet) ⎨ 2 2 ⎩ 2 ⎭ ⎩ 2 ⎭ b) ) For the interaction potential given, the energy is lowest when the two are close together, i.e. closer than a. Therefore, the states for which the particles come closest will have the lowest energy. The particles come closest when they have anti-symmetric spin states, i.e. the ground state. Therefore, the total spin-0 state will have the energy lowered more than that of the total spin-1 state. 9.2 Make an order-of-magnitude estimate of the the singlet-triplet splitting of the energy levels of helium due to a direct spin-spin interaction of the electrons by comparing with the magnitude of the hyperfine interaction in hydrogen. By comparing this estimate with the observed splitting, ∼ 1eV , what conclusions can be drawn about the relative effects of exchange symmetry and spin-spin interaction on the energy? Solution: Since the average separation between the two electrons in the helium atom is comparable with that between the electron and proton in hydrogen, the size of the spin-spin interaction can be inferred from the size of the hyperfine interaction in hydrogen. We must, however, replace the magnetic moment of the proton with that of the electron. The magnetic moment of the electron is greater than that of the proton by roughly the ratio of the masses m p / me . Therefore, the spin-

spin interaction should be roughly 1836 ×10−5 eV ≈ 10−2 eV which is much smaller than the ~1 eV energy splittings that are observed. We conclude that the exchange symmetry of the spatial wave functions is largely responsible for the splitting.

63

9.3 A helium atom is excited from the ground state to the autoionizing state 2s4p by absorption of ultraviolet light. Assume that the 2s electron moves in the unscreened Coulomb field of the nucleus and the 4p electron in the fully screened Coulomb potential a) Obtain the energy of this autoionizing level and the corresponding wavelength of the ultraviolet light required to effect this excitation. Make an energy level diagram showing this level together with the ground states of neutral, singly ionized and doubly ionized helium atoms. b) Find the velocity of the electron emitted in the autoionizing process in which the 2s4p state decays into a free electron and a He+ ion in the ground state. Solution: 22 1 Z2 a) 2s is unscreened. Therefore, (in atomic units) E 2 s = − 2 = − = − . Since the 4p is 2 2 2n 2⋅2 2 2 1 1 Z fully screened the effective nuclear charge is 1 and E 4 p = − 2 = − = − . Therefore, 2 32 2n 2⋅4 1 1 17 E2s 4 p = − − = − au = −14.5 eV . Graphically, we have 2 32 32

12,400 = 192 Å 64.5 b) The electron kinetic energy is the difference between the autoionizing level and He+(1s): 1 1 v2 KE electron = 54.4 − 14.5 = 39.9 eV so mv 2 = (mc 2 ) 2 = 39.9 eV . Therefore, 2 2 c 2 39.9 eV v = 2⋅ = 1.6 × 10 − 4 so v = 1.6 × 10−4 c = 3.8 × 106 m/s 2 5 c 5.1 × 10 eV Since the photon energy is 64.5 eV, the wavelength is λ =

9.4 Use the variational principle to calculate the ground state energy of a hydrogen atom (in atomic units) using the normalized trial functions:

a) ψ (r ) = β 3 / π ⋅ exp(− βr ) and b) ψ (r ) = 2 2 (2γ 3 / π ) parameters. c) Which gives the most accurate answer? Why? Solution: 1 ⎛1⎞ ⎛1⎞ 1 d ⎛ 2 d ⎞ 1 a) Hˆ = −⎜ ⎟∇ 2 − = −⎜ ⎟ 2 ⎜r ⎟ − and r ⎝ 2 ⎠ r dr ⎝ dr ⎠ r ⎝ 2⎠

1/ 4

64

⋅ exp(− γr 2 ) where β and γ are

β 3 ⎡⎛ 1 ⎞ 1 d ⎛ 2 de − βr ⎞ e − βr ⎤ β 3 ⎛ β − 1 β 2 ⎞ − βr ˆ ⎜r ⎟− ⎜ ⎟e . Therefore, − Hψ = − ⎢⎜ ⎟ ⎥= 2 ⎟⎠ dr ⎟⎠ r ⎦ π ⎣⎝ 2 ⎠ r 2 dr ⎜⎝ π ⎜⎝ r β 3 ∞ ⎛ β − 1 β 2 ⎞ − 2 βr 2 β2 ⎜ ⎟ Hˆ = 4π − e r dr = − β . Now, minimizing wrt β, we have π ∫0 ⎜⎝ r 2 ⎟⎠ 2 d Hˆ dβ

= β − 1 = 0 ⇒ β = 1 which gives, as we should have expected, the exact grounds state

1 1 β2 wave function. The energy is therefore, E 0 = Hˆ = − β = − 1 = − which is the exact 2 2 2 ground state energy. b) The Hamiltonian is exactly the same as for part a). Differentiation and integration using the 3γ 2γ −2 . Minimizing wrt γ gives given wavefunction yields Hˆ = 2 π d Hˆ 3 2 1 8 = − ⋅ 1/ 2 = 0 ⇒ γ = and 2 π γ 9π dβ 3γ 2γ 3 8 2 8 4 1 ⎛ 8 ⎞ ⎛ 1⎞ −2 = ⋅ −2 ⋅ =− = − ⋅ ⎜ ⎟ = 0.848⎜ − ⎟ Hˆ = π π 9π 2 2 9π 3π 2 ⎝ 3π ⎠ ⎝ 2⎠ so that using this wavefunction gives a ~15% error. c) Of course a) gives the most accurate answer because the minimization process produced the exact ground state wavefunction. 9.5 A negative ion is formed when an electron attaches to an atom (or molecule), the result being that the nucleus of charge Z binds ( Z + 1) electrons. The binding energy of the electron to the neutral atom is referred to as the electron affinity of the atom. The electron affinity may also be thought of as the ionization potential of the negative ion. Not surprisingly, the halogen atoms form negative ions most readily. This means that the halogen atoms have the highest electron affinities of all atoms (~3 eV). Hydrogen atoms also form negative ions. The electron affinity of hydrogen is 0.75 eV. Use perturbation theory as it was applied to the helium atom to determine the total energy of the hydrogen atom negative ion. Compare the answer with the actual value. Compare the accuracy of the perturbation theory results for the hydrogen negative ion with the result for the helium atom. Why is perturbation theory more accurate for helium? Note that there are no "new" calculations necessary. Solution: Everything is the same as for helium except that Z = 1 . Thus, the unperturbed energy is merely

twice the ground state energy of hydrogen, that is E1(s01)s = E0( ) = ∫ dr1 dr2 ψ 1s ( r1 ) 1

2

(

−

1 2 1 2

)+(

−

1 2 1 2

) = −1a.u. The integral

2 1 ψ 1s ( r2 ) has already been calculated for an arbitrary value of Z. r1 − r2

5 5 It is E0(1) = + Z = + . Therefore, the total energy of the hydrogen negative ion, as given by 8 8 5 3 perturbation theory is E0 = E0( 0) + E0(1) = −1 + = − = −10.2 eV . (This is unrelated to the 8 8 separation between the ground and first excited state of neutral hydrogen being 10.2 eV.) 65

The experimentally determined total energy is Eexp = −13.6 eV − 0.75eV = −14.35eV so the error is considerable, especially compared with that for helium. Evidently the reason is that the perturbation term, the e-e repulsion, is a larger fraction of the unperturbed energy in hydrogen than it is in helium. E0( )

E0( ) 3/8 = = 0.375 , while for helium ( 0) 1 E0

1

That is, for hydrogen

1

( 0)

E0

H

9.6 Assume the normalized trial wave function ψ = (1 + A2 )

= He

5/8 = 0.156 . 4

−1/ 2

⎡⎣ 100 + A 210 ⎤⎦ , where the kets on the right hand side are spherical hydrogen atom eigenkets, represents a ground state hydrogen atom in a constant electric field F. This wave function represents a state that has "ground state character" and, assuming A