## Assignment 9, solutions

#### Problem 1:

The wave function of a system in the ground state is given as

ψ(r,t) = [exp(-iωt)/(πa03)½] exp(-r/a0).

(a)  Sketch the probability density in coordinate space as function of r/a0.
(b)  Find the momentum space wave function Φ(p,t).
Hint:  Use spherical coordinates for evaluation of the integral transform.
(c)  Find the probability density function in momentum space.  Sketch it as a function of pa0/ħ.

Solution:

• Concepts:
Change of representation, the Fourier transform
• Reasoning:
We are asked to change from coordinate space to momentum space representation.
• Details of the calculation:
Let  ψ(r,t) = ψ(r) exp(-iωt), then Φ(p,t) = Φ(p) exp(-iωt).
(a)  ψ(r) = ψ(r), we have spherical symmetry.
Probability density:
|ψ(r)|2 = exp(-2r/a0)(πa03) = constant*exp(-2x), where x = r/a0.

(b) ψ(r) = (2πħ)-3/2∫d3p Φ(p)exp(ipr/ħ),
Φ(p) = (2πħ)-3/2∫d3r ψ(r)exp(-ipr/ħ).
ψ(r) and Φ(p) are Fourier transforms of each other.

Symmetry demands that Φ(p) = Φ(p).
Let us choose a convenient direction for p.  Let p = pk.  Then
Φ(p) = (2πħ)-3/20dφ∫0πsinθ dθ∫0r2dr exp(-iprcosθ/ħ) (πa03)exp(-r/a0)
= (2πħ)-3/2(πa03)2π∫-11dcosθ∫0r2dr exp(-(ipcosθ/ħ + 1/a0)r).
0r2dr exp(-(iprcosθ/ħ + /a0)r) = 2/(ipcosθ/ħ + 1/a0)3
using ∫0xn exp(-x) dx = n!.
Φ(p) = a-3/2ħ3/2(√2)(-ip3π)-1-11dx (x + ħ/(ia0p))-3
= (a0ħ)-3/2ħ3(√2)(-ip3π)-1(2ħ/(ia0p))(1 + ħ2/(a0p)2)-2.
Φ(p) = (1/π) (2a0/ħ)3/2 [1/(1 + a02p22)2] = Φ(p).

(c)  Probability density:
| Φ(p)|2=  (1/π2) (2a0/ħ)3[1/(1 + a02p22)2]4 = constant*[1/(1 + x2)]4.
where x =  a0p/ħ.

#### Problem 2:

An exotic atom consists of a Helium nucleus (Z = 2) and an electron and an antiproton p(bar) both in n = 2 states.  Take the mass of the p(bar) to be 2000 electron masses and that of the helium nucleus to be 8000 me.  For an electron in the n = 1 state of hydrogen E = -13.6 eV.
(a)  How much energy is required to remove the electron from this atom?
(b)  How much energy is required to remove the p(bar) from this atom?
(c)  Assume both the p(bar) and the electron are in 2p states.  Then each can de-excite to their ground state.  It is observed that radiation always accompanies those transitions when the electron jumps first, but when the p(bar) jumps first there is often no photon emitted.  Explain!

Solution:

• Concepts:
Hydrogenic atoms
• Reasoning:
To find the eigenfunctions and eigenvalues of the Hamiltonian of a hydrogenic atom we replace in the eigenfunctions of the Hamiltonian of the hydrogen atom a0 by
a0' = ħ2/(μ'Ze2) = a0(μ/μ')(1/Z),
and in the eigenvalues of the Hamiltonian of the hydrogen atom we replace EI by
EI' = μ'Z2e4/(2ħ2) = EI(μ'/μ)Z2.
Here μ is the reduced mass for the hydrogen atom.
• Details of the calculation:
(a)  For the p(bar) we have μ' = 2000*8000/10000 me = 1600 me.
For the p(bar) we have a0' = a0/3200.
The electron therefore move in a field that is approximately that of a nucleus with Z = 1.
E2 = -13.6 eV*/4 = -3.4 eV.  3.4 eV is required to remove the electron.
(b)  For the p(bar) we have μ' = 1600, Z = 2.
E2 = -13.6 eV*4*1600/4 = -21760 eV.  21760 eV is required to remove the p(bar).
(c)  When the electron jumps first, (13.6 - 3.4) = 10.2 eV of energy is released.  This energy can only be given to a photon, since at least 13.6*4*1600(1/4 - 1/9) eV = 12088 eV is needed to excite the p(bar) to the n = 3 level.
When the p(bar) jumps first, enough energy is released to remove the electron from the atom.  Radiation-less transitions, in which the energy is given to the electron, are now possible.

#### Problem 3:

The potential energy of the nuclei of a diatomic molecule as a function of their separation r is given by
U(r) = -2D[a0/r - a02/r2].
Here D is a constant with units of energy.
Approximate this potential energy function near its minimum by a harmonic oscillator potential energy function and determine the vibrational energy levels of the molecule with zero angular momentum.

Solution:

• Concepts:
Motion of a fictitious particle of reduced mass μ in a spherically symmetric potential
• Reasoning:
We have a spherically symmetric potential energy function U(r).
The wave function ψklm(r,θ,φ) = Rkl(r)Ylm(θ,φ) = [ukl(r)/r]Ylm(θ,φ) is a product of a radial function Rkl(r) and the spherical harmonic Ylm(θ,φ).  The differential equation for ukl(r) is
[-(ħ2/(2μ))(∂2/∂r2) + ħ2l(l + 1)/(2μr2) + U(r)]ukl(r) = Eklukl(r).
When l = 0 we have [-(ħ2/(2μ))(∂2/∂r2) + U(r)]ukl(r) = Eklukl(r).
• Details of the calculation:
Find the location of the extrema:  dU/dr = 0,  -2D[-a0/r2 + 2a02/r3] = 0,  r = 2a0 = r0.
There is only one extremum.  As r --> 0 U(r) --> ∞ and as r  --> ∞ U(r) --> 0.  There is only one extremum, it must be a minimum.
Near r0 we write  U(r) ≈ U(r0) + ½ d2U(r)dr2|r0(r - r0)2
d2U(r)dr2|r0 =  - 2D[2a0/(2a0)3 - 6a02/(2a0)4] = -¼D/a02.
U(r) = -U0 + ½ μω2r'2, where r' = r - 2a0, U0 = D/2, and ω2 =  D/(4μa02).
The vibrational energy levels of the molecule with l = 0 are Eν = -U0 + (ν + ½)ħω.
Eν = -D/2 + (ν + ½)ħ(D/(4μa02)½,  ν = 0, 1, 2, ... .

#### Problem 4:

The Rydberg constant, RH = 109737.568525/cm is one of the most accurately known fundamental constants.
(a)  Find the wave number of the Balmer alpha line (n = 3 to n' = 2) in  atomic hydrogen.  Neglect fine structure.
(b)  Is the Balmer alpha line in atomic deuterium shifted towards the blue or towards the red compared to normal hydrogen?
(c)  Calculate the shift in wave number between deuterium and hydrogen.

Solution:

• Concepts:
The hydrogen atoms, reduced mass
• Reasoning:
The deuterium is a hydrogen atom with a slightly different reduced mass than ordinary hydrogen.
• Details of the calculation:
(a)  For ordinary hydrogen:  En = -μe4/(2ħ2n2) with μ = mpme/(mp + me)
En - En' = [μe4/(2ħ2)](1/n'2 - 1/n2) = hν = hc/λ = hcRH(1/n'2 - 1/n2).
So RH = [2πμe4/(2ħ3c)].
For the wave number we have 1/λ = RH(1/32 - ½2) = 15241.328952 cm-1.
[Note: There are unfortunately two different definitions of the wave number, 1/λ or 2π/λ.  Clearly state which definition you are using.]
(d)  For deuterium:  En = -μ'e4/(2ħ2n2) with μ' = (mp + mn)me/(mp + mn + me)
RD = RH μ'/μ > RH.
So 1/λ increases, λ decreases, the line is blue-shifted.
(c)  1/λ' - 1/λ  = (RD - RH)(1/32 - ½2) = (μ'/μ - 1)RH(1/32 - ½2) = 4.15 cm-1.

#### Problem 5:

An electron in the hydrogen atom occupies the combined position and spin state

R21(r)[√⅓ Y10(θ,φ)χ+ + √⅔ Y11(θ,φ)χ-].

(a)  If you measure L2, what value(s) might you get, and with what probability(ies)?
(b)  If you measure Lz, what value(s) might you get, and with what probability(ies)?
(c)  If you measure S2, what value(s) might you get, and with what probability(ies)?
(d)  If you measure Sz, what value(s) might you get, and with what probability(ies)?
(e)  If you measured the position of the electron, what is the probability density for finding the electron at r, θ, φ in terms of the variables given above.
(f)  If you measured both Sz and the distance of the electron from the proton, what is the probability per unit length for finding the particle with spin up a distance r from the proton in terms of the variables given above?

Useful integral:  ∫0πsinθ dθ∫0dφ |Ylm(θ,φ))|2 = 1.

Solution:

• Concepts:
The eigenfunctions of the orbital and spin angular momentum operators, the spherical harmonics, tensor product states
• Reasoning:
The common eigenfunctions of L2 and Lz are the spherical harmonics.
• Details of the calculation:
(a)  L2 = 2ħ2,  probability 1.
(b)  Lz = 0,  probability ⅓,  Lz = ħ,  probability ⅔.
(c)  S2 = (3/4)ħ2,  probability 1.
(d)  Sz = ħ/2,  probability ⅓,  Sz = -ħ/2,  probability ⅔.
(e)  Let P(r,θ,φ) the probability per unit volume of for finding the electron at  r, θ, φ.
P(r,θ,φ) = <ψ|Pr,θ,φ|ψ>, where the projector
Pr,θ,φ = ∑i(|r,θ,φ |χi>)(<r,θ,φ <χi|), and i = +, -.
P(r,θ,φ)
= (√⅓<210|<χ+| + √⅔<211|<χ-|) (|r,θ,φ |χ+>)(<r,θ,φ|<χ+|)(√⅓|210>|χ+> + √⅔|211>|χ->)
+ (√⅓<210|<χ+| + √⅔<211|<χ-|) (|r,θ,> |χ->)(<r,θ,φ|<χ-|)(√⅓|210>|χ+> + √⅔|211>|χ->)
= ⅓|<210|r,θ,φ>|2 + ⅔|<211|r,θ,φ>|2, since <χij|> = δij.
P(r,θ,φ) = |R2l(r)|2[⅓|Y10(θ,φ |2 + ⅔|Y11(θ,φ |2].
(f)  P+(r,θ,φ) = ⅓|R2l(r)|2|Y10(θ,φ)|2.
Probability per unit length P+(r) = ⅓r2|R2l(r)|2.