Assignment 9, solutions 

Problem 1:

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:

bulletConcepts:
The eigenfunctions of the orbital and spin angular momentum operators, the spherical harmonics, tensor product states
bulletReasoning:
The common eigenfunctions of L2 and Lz are the spherical harmonics.
bulletDetails of the calculation:
(a)  L2 = 2ħ2,  probability 1.
(b)  Lz = 0,  probability ⅓,  Lz = ħ,  probability ⅔.
(c)  S2 = ħ2,  probability 1.
(d)  Sz = ħ/2,  probability ⅓,  Sz = -ħ/,  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,φ> |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.  

Problem 2:

Is each of the following sets a valid combination of quantum numbers (n, ℓ, m, ms) for the energy eigenstates of hydrogen?  If not, explain why not.
(2, 2, -1, )
(3, 1, +2, -)
(3, 1, 0, )
(4, 1, 1, -3/2)
(2, -1, 1, +1/2)

Solution:

bulletConcepts:
Addition of angular momentum
bulletReasoning:
We are supposed to add the orbital and spin angular momentum of the electron in the hydrogen atom and check if the given combinations are valid.
bulletDetails of the calculation:
The combinations below are not valid.
(2, 2,-1, )  The quantum number ℓ is non-negative and smaller than the quantum number n
(3, 1, +2, -)  The quantum number m can take on all integer values between -ℓ and ℓ.
(4, 1, 1, -3/2)  The quantum number ms for the spin electron is restricted to the values .
(2, -1, 1, +1/2)  The quantum number ℓ is non-negative.

Problem 3:

Find the condition that must be satisfied by the spherically symmetric square well potential
U(r) = -|U0|   for r < a,  U(r) = 0   for r > a,
if it is just barely deep enough to contain one bound state.

Solution:

bulletConcepts:
Three-dimensional square potentials
bulletReasoning:
We have to investigate the properties of the ground state in a three dimensional square potential.
bulletDetails of the calculation:
If the potential only supports one bound state, this state must be an l = 0 state.  The radial Schroedinger equation then is
[(-ħ2/(2μ))(∂2/∂r2)+ U(r)]u10(r) = E10u10(r).
Let k2 = (2μ/ħ2)(E + U0), ρ2 = (2μ/ħ2)(-E), and k02 = (2μ/ħ2)U0.
Let r < a define region 1 and r > a define region 2.  The coordinate r is never negative.  Therefore
Φ1(r) = A sin(kr),  Φ1(0) = 0,
Φ2(r) = Bexp(-ρr),  Φ2(∞) = 0.
At r = a we need hat Φ(a) and (∂/∂r)Φ(r)|a are continuous.
A sin(ka) = Bexp(-ρa).
kA cos(ka) = -ρBexp(-ρa).
Therefore cot(ka) = -ρ/k.
1/sin2(ka) = 1 + cot2(ka) = (k2 + ρ2)/k2  = k02/k2.
We can find a graphical solution by plotting |sin(ka)| and k/k0 versus k.  The intersections of the two plots in regions where cot(ka) < 0 gives the values of k for which a solution exist.
We need  nπ/2 < ka < (n + 1)π/2, n = odd.
To only support one bound state we need π/2 < k0a < 3π/2.
With E near zero we have k ~ k0 and  π/2 ~(2μU02)1/2a.
The smallest possible value for U0 is U0 = π2ħ2/(8μa2).

Problem 4:

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:

bulletConcepts:
Hydrogenic atoms
bulletReasoning:
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.
bullet

Details of the calculation:
(a)  For the p(bar) we have μ' = 2000*8000/10000 me = 1600 me.  For the p(bar) 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 5:

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:

bulletConcepts:
The hydrogen atoms, reduced mass
bulletReasoning:
The deuterium is a hydrogen atom with a slightly different reduced mass than ordinary hydrogen.
bulletDetails 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 - 2) = 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 - 2) = (μ'/μ 1)RH(1/32 - 2) = -4.15 cm-1.