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

ψ(**r**,t) = [exp(-iωt)/(πa_{0}^{3})^{½}]
exp(-r/a_{0}).

(a) Sketch the probability density in coordinate space as
function of r/a_{0.
}(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 pa_{0}/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/a_{0})(πa_{0}^{3}) = constant*exp(-2x), where x = r/a_{0}.(b) ψ(

**r**) = (2πħ)^{-3/2}∫d^{3}p Φ(**p**)exp(i**p**∙**r**/ħ),

Φ(**p**) = (2πħ)^{-3/2}∫d^{3}r ψ(**r**)exp(-i**p**∙**r**/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**= p**k**. Then

Φ(**p**) = (2πħ)^{-3/2}∫_{0}^{2π}dφ∫_{0}^{π}sinθ dθ∫_{0}^{∞}r^{2}dr exp(-iprcosθ/ħ) (πa_{0}^{3})^{-½}exp(-r/a_{0})

= (2πħ)^{-3/2}(πa_{0}^{3})^{-½}2π∫_{-1}^{1}dcosθ∫_{0}^{∞}r^{2}dr exp(-(ipcosθ/ħ + 1/a_{0})r).

∫_{0}^{∞}r^{2}dr exp(-(iprcosθ/ħ + /a_{0})r) = 2/(ipcosθ/ħ + 1/a_{0})^{3}

using ∫_{0}^{∞}x^{n }exp(-x) dx = n!.

Φ(**p**) = a^{-3/2}ħ^{3/2}(√2)(-ip^{3}π)^{-1}∫_{-1}^{1}dx (x + ħ/(ia_{0}p))^{-3 }= (a_{0}ħ)^{-3/2}ħ^{3}(√2)(-ip^{3}π)^{-1}(2ħ/(ia_{0}p))(1 + ħ^{2}/(a_{0}p)^{2})^{-2}.

Φ(**p**) = (1/π) (2a_{0}/ħ)^{3/2 }[1/(1 + a_{0}^{2}p^{2}/ħ^{2})^{2}] = Φ(p).(c) Probability density:

| Φ(p)|^{2}= (1/π^{2}) (2a_{0}/ħ)^{3}[1/(1 + a_{0}^{2}p^{2}/ħ^{2})^{2}]^{4}= constant*[1/(1 + x^{2})]^{4}.

where x = a_{0}p/ħ.

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 m_{e}. 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 a_{0}by

a_{0}' = ħ^{2}/(μ'Ze^{2}) = a_{0}(μ/μ')(1/Z),

and in the eigenvalues of the Hamiltonian of the hydrogen atom we replace E_{I}by

E_{I}' = μ'Z^{2}e^{4}/(2ħ^{2}) = E_{I}(μ'/μ)Z^{2}.

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 m_{e}= 1600 m_{e}.

For the p(bar) we have a_{0}' = a_{0}/3200.

The electron therefore move in a field that is approximately that of a nucleus with Z = 1.

E_{2}= -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.

E_{2}= -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.

The potential energy of the nuclei of a diatomic molecule as a function of
their separation r is given by

U(r) = -2D[a_{0}/r - a_{0}^{2}/r^{2}].

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 ψ_{klm}(r,θ,φ) = R_{kl}(r)Y_{lm}(θ,φ) = [u_{kl}(r)/r]Y_{lm}(θ,φ) is a product of a radial function R_{kl}(r) and the spherical harmonic Y_{lm}(θ,φ). The differential equation for u_{kl}(r) is

[-(ħ^{2}/(2μ))(∂^{2}/∂r^{2}) + ħ^{2}l(l + 1)/(2μr^{2}) + U(r)]u_{kl}(r) = E_{kl}u_{kl}(r).

When l = 0 we have [-(ħ^{2}/(2μ))(∂^{2}/∂r^{2}) + U(r)]u_{kl}(r) = E_{kl}u_{kl}(r). - Details of the calculation:

Find the location of the extrema: dU/dr = 0, -2D[-a_{0}/r^{2}+ 2a_{0}^{2}/r^{3}] = 0, r = 2a_{0}= r_{0}.

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 r_{0}we write U(r) ≈ U(r_{0}) + ½ d^{2}U(r)dr^{2}|_{r0}(r - r_{0})^{2}

d^{2}U(r)dr^{2}|_{r0}= - 2D[2a_{0}/(2a_{0})^{3}- 6a_{0}^{2}/(2a_{0})^{4}] = -¼D/a_{0}^{2}.

U(r) = -U_{0}+ ½ μω^{2}r^{'2}, where r' = r - 2a_{0}, U_{0}= D/2, and ω^{2}= D/(4μa_{0}^{2}).

The vibrational energy levels of the molecule with l = 0 are E_{ν}= -U_{0}+ (ν + ½)ħω.

E_{ν}= -D/2 + (ν + ½)ħ(D/(4μa_{0}^{2})^{½}, ν = 0, 1, 2, ... .

The Rydberg constant, R_{H} = 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: E_{n}= -μe^{4}/(2ħ^{2}n^{2}) with μ = m_{p}m_{e}/(m_{p}+ m_{e})

E_{n}- E_{n'}= [μe^{4}/(2ħ^{2})](1/n'^{2}- 1/n^{2}) = hν = hc/λ = hcR_{H}(1/n'^{2}- 1/n^{2}).

So R_{H}= [2πμe^{4}/(2ħ^{3}c)].

For the wave number we have 1/λ = R_{H}(1/3^{2}- ½^{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: E_{n}= -μ'e^{4}/(2ħ^{2}n^{2}) with μ' = (m_{p}+ m_{n})m_{e}/(m_{p}+ m_{n}+ m_{e})

R_{D}= R_{H}μ'/μ > R_{H}.

So 1/λ increases, λ decreases, the line is blue-shifted.

(c) 1/λ' - 1/λ = (R_{D}- R_{H})(1/3^{2}- ½^{2}) = (μ'/μ - 1)R_{H}(1/3^{2}- ½^{2}) = 4.15 cm^{-1}.

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

R_{21}(r)[√⅓ Y_{10}(θ,φ)χ_{+} + √⅔ Y_{11}(θ,φ)χ_{-}].

(a) If you measure L^{2}, what value(s) might you get, and with what
probability(ies)?

(b) If you measure L_{z}, what value(s) might you get, and with what
probability(ies)?

(c) If you measure S^{2}, what value(s) might you get, and with what
probability(ies)?

(d) If you measure S_{z}, 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 S_{z} 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: ∫_{0}^{π}sinθ dθ∫_{0}^{2π}dφ
|Y_{lm}(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 L^{2}and L_{z}are the spherical harmonics. - Details of the calculation:

(a) L^{2}= 2ħ^{2}, probability 1.

(b) L_{z}= 0, probability ⅓, L_{z}= ħ, probability ⅔.

(c) S^{2}= (3/4)ħ^{2}, probability 1.

(d) S_{z}= ħ/2, probability ⅓, S_{z}= -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|P_{r,θ,φ}|ψ>, where the projector

P_{r,θ,φ}= ∑_{i}(|r,θ,φ |χ_{i}>)(<r,θ,φ <χ_{i}|), 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 <χ_{i}|χ_{j}|> = δ_{ij}.

P(r,θ,φ) = |R_{2l}(r)|^{2}[⅓|Y_{1}^{0}(θ,φ |^{2}+ ⅔|Y_{1}^{1}(θ,φ |^{2}].

(f) P_{+}(r,θ,φ) = ⅓|R_{2l}(r)|^{2}|Y_{1}^{0}(θ,φ)|^{2}.

Probability per unit length P_{+}(r) = ⅓r^{2}|R_{2l}(r)|^{2}.