Consider a composite system made of two spin ½ particles. For t < 0 the
Hamiltonian does not depend on time and can be taken to be zero. For t > 0
the Hamiltonian is given by H = (4∆/ħ^{2})**S**_{1}∙**S**_{2}, where ∆ is a constant. Suppose that the system is in the state |+-> for t
≤ 0. Find, as a function of time, the probability for being in each of the
states |++>, |+->, |-+>, and |-->,

(a) by solving the problem exactly, using |Ψ(t)> = U(t, t_{0})Ψ(t_{0})>
and

(b) by solving the problem assuming the validity of first-order
time-dependent perturbation theory with H as a perturbation which is switched on
a t = 0.

(b) Under what conditions does the perturbation calculation disagree with
the exact solution and why?

Solution:

- Concepts:

The state space of two spin ½ particles, time-dependent perturbation theory. - Reasoning:

**S**_{1}∙**S**_{2}= ½(S^{2}- S_{1}^{2}- S_{2}^{2}). The eigenstates of**S**_{1}∙**S**_{2}are the singlet state and the triplet states, {|S, M_{s}>}, S = 0, 1. These are common eigenfunctions of S^{2}and S_{z}. - Details of the calculation:

(a) The common eigenfunctions of S^{2}and S_{z}are

|1,1> = |++>,

|1,0> = ½^{½}(|+-> + |-+>),

|1,-1> =|-->,

|0, 0> = ½^{½}(|+-> - |-+>).

Therefore |+-> = ½^{½}(|0,0> + |1,0>), |-+> = ½^{½}(|0,0> - |1,0>).

For the singlet state we have**S**_{1}∙**S**_{2}|0,0> = -(3/4)ħ^{2}, H|0,0> = -3∆|0,0>.

For the triplet states we have**S**_{1}∙**S**_{2}|1, M_{s}> = (1/4)ħ^{2}, H|1, M_{s}> = ∆|1, M_{s}>.

|ψ(0)> = |+-> = ½^{½}(|0,0> + |1,0>),

|ψ(t)> = U(t, 0)|+-> = 2^{-½}(exp(i3∆*t/ħ)|0,0> + exp(-i∆*t/ħ)|1,0>).

P(t) of being in the state |++> = |1 1> = 0.

P(t) of being in the state |--> = |1 -1> = 0.

P(t) of being in the state |+-> = |<+-| ψ(t)>|^{2}.

<+-|ψ(t)> = ½(<0,0| + <1,0|)(exp(i3∆*t/ħ)|0,0> + exp(-i∆*t/ħ)|1,0>)

= ½(exp(i3∆*t/ ħ) + exp(-i∆*t/ħ))

= ½exp(i∆*t/ ħ) (exp(i2∆*t/ħ) + exp(-i2∆*t/ħ)) = exp(i∆*t/ħ) cos(2∆*t/ħ).

P(t) of being in the state |+-> = cos^{2}(2∆*t/ħ).

For 2∆*t/ħ << 1, P(t) ~ 1.

P(t) of being in the state |-+> = sin^{2}(2∆*t/ħ), since the total probability of being in one of the 4 basis states must be equal to 1.

For 2∆*t/ħ << 1, P(t) ~ (2∆*t/ħ)^{2}.

(b) Time-dependent perturbation theory

P_{if}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{fi}t')W_{fi}(t')dt'|^{2}.

The initial and the final state must be different for this formula to be valid. For i = f, P(t) ~ 1.

W_{fi}= <f|H|i>. Here W_{fi}= <f|H|+->.

W_{fi}= 0 for |f> = |++> and |f> = |-->, so P_{if}(t) = 0 for these states.

For |f> = |-+> we have W_{fi}= 2^{-½}<-+|H|(|00> + |10>)

= 2^{-½}[-3∆<-+| 0, 0> + ∆<-+| 1, 0>] = ½(3∆ + ∆) = 2∆.

ω_{fi}= (E_{f}- E_{i})/ω_{fi}= 0 for the unperturbed states.

P_{if}(t) = (1/ħ^{2})|∫_{0}^{t}(2∆)dt'|^{2}= 4∆^{2}t^{2}/ħ^{2}.

(c) The time-dependent perturbation theory calculation result agrees with the exact result as long as x = 2∆*t/ħ << 1, so that cos(x) = 1 and sin(x) = x are very good approximations. For large x (or t) the probability of finding the particle in the state |-+> oscillates between 0 and 1.

In a one-dimensional structure a particle of mass m is in the ground state of a
potential energy function U(x) = ½kx^{2}. A phase transition occurs,
and the effective spring constant suddenly doubles. What is the probability that
the particle will be found in an exited state after the phase transition? Give a
numerical answer.

Solution:

- Concepts:

The QM harmonic oscillator, the sudden approximation - Reasoning:

The sudden approximation can be used to calculate transition probabilities when the Hamiltonian changes rapidly. The reaction time is so short that the transition amplitude <β|U(t_{2},t_{1})|α> is simply given by the overlap <β|α>. The transition probability is |<β|α>|^{2}. Here |α> is the eigenstate of the Hamiltonian before the transition and |β> is the eigenstate of the Hamiltonian after the transition. - Details of the calculation:

The normalized ground-state wave function of the 1D harmonic oscillator for t < 0 is

Φ_{0i}(x) = (mω/(πħ))^{¼}exp(-½mωx^{2}/ħ), ω = (k/m)^{½}.

The normalized ground-state wave function for t > 0 is

Φ_{0f}(x) = (m√(2)ω/(πħ))^{¼}exp(-½m√(2)ωx^{2}/ħ).

The probability that the oscillator remains in the ground state is

P = |<Φ_{0i}|Φ_{0f}>|^{2}= |∫_{-∞}^{∞}dx Φ_{0i}*(x) Φ_{0f}(x)|^{2 }= 2^{5/4}/(1 + √2) = 0.985.

The probability that the particle will be found in an exited state after the phase transition is

1 - 0.985 = 0.015.

In one dimension, consider a spinless particle trapped in a delta-function potential U(x) = -Cδ(x), C > 0.

At t = 0, a time dependent perturbation W(t) = Wcosωt is turned on.

Assume ω >> mC^{2}/(2ħ^{3}) so that the particle can be ejected
from the trap. Use perturbation theory to find the transition rate.

You can assume that the free particles will be in a box of size L, L =
very large.

Solution:

- Concepts:

Fermi's golden rule, the delta-function potential - Reasoning:

Only one bound state exist in the delta-function potential. This is the initial state. The final state of the particle is a continuum state. - Details of the calculation:

(a) Let H_{0}= p^{2}/(2m) - Cδ(x).

Normalized eigenfunction of H_{0}: Φ(x) = ρ^{½}exp(-ρ|x|),

ρ^{2}= -2mE_{i}/ħ^{2}, E_{i}= -mC^{2}/(2ħ^{2}), ρ = mC/ħ^{2}.

To find the transition rate we use Fermi's golden rule.

If W(t) = Wcos(ωt), then the transition probability per unit time is given by

w(i,E) = (π/(2ħ))ρ(E)|W_{Ei}|^{2}δ_{E-Ei,ħω}, where W_{Ei}= <Φ_{E}|W|Φ_{i}>.

Since ħω >> -mC^{2}/(2ħ^{2}), we can assume that the ejected particle is a nearly free particle with energy E = ħω + E_{i}and k^{2}= 2mE/ħ^{2}.

We assume that the particle is confined to a region of width L >> a and use periodic boundary conditions.

Then Φ_{k}(x) = L^{-½}exp(ikx), with k = 2πn/L, n = 0, ±1, ±2, ... is the final state wave function.

The number of states with wave vectors whose magnitudes lie between k and k + dk is

dN = 2 dk/(2π/L) = Ldk/π. (The particle can move towards the left or to the right.)

dN/dk = L/π.

The density of states is dN/dE = (dN/dk)(dk/dE).

With E = ħ^{2}k^{2}/(2m) we have ρ(E) = dN/dE = Lm/(πkħ^{2}).

W_{Ei}= (ρ/L)^{½}W[∫_{0}^{∞}dx e^{ikx}e^{-ρx}+ ∫_{-∞}^{0}dx e^{ikx}e^{ρx}]

= (ρ/L)^{½}W 2ρ/(ρ^{2}+ k^{2}).

w(i,E) = (π/(2ħ))(Lm/(πkħ^{2}))(ρ/L)W^{2}4ρ^{2}/(ρ^{2}+ k^{2})^{2 }= (2mW^{2}/ħ^{3}) ρ^{3}/[k(ρ^{2}+ k^{2})^{2}].

(ρ^{2}+ k^{2})^{2}= 4m^{2}ω^{2}/ħ^{2}, ρ^{3}/[(ρ^{2}+ k^{2})^{2}] = mC^{3}/(4ħ^{4}ω^{2}).

w(i,E) = (2mW^{2}/ħ^{3})mC^{3}/(4ħ^{4}ω^{2})(2mE)^{½}/ħ.

w(i,E) = (2mE)^{½}2m^{2}W^{2}C^{3}/(4ω^{2}ħ^{8}).

This is the transition probability per unit time.

A hydrogen atom with Hamiltonian H_{0}(r) is placed in a
time-dependent electric field **E** = E(t) **k**. The perturbed
Hamiltonian is H(r,t) = H_{0}(r) + H'(r,t).

(a) Show that H'(r,t) = q_{e}E(t) r cos(θ).

(b) Assuming the electron is
initially in the ground state, and recalling that the first excited state of
hydrogen is quadruply degenerate, to which state of the quadruply degenerate
first excited states is a dipole transition from the ground state possible?
Prove this.

(c) If the electron is in the
ground state at t = 0, find the probability (to first order in perturbation theory) that at time
t the
electron will have made the transition to the state determined in (b), as a
function of E(t).

Solution:

- Concepts:

Time dependent perturbation theory - Reasoning:

We consider the interactions of the atomic electrons with the electric field perturbation to the atomic Hamiltonian. - Details of the calculation:

(a) The uniform electric field changes the energy of the proton and the electron, and therefore perturbs the Hamiltonian.

The energy of the proton in the field is -∫_{0}^{zp}q_{e}E(t)dz = -q_{e}E(t)z_{p}.

The energy of the electron in the field is ∫_{0}^{ze}q_{e}E(t)dz = q_{e}E(t)z_{e}.

The potential energy of both particles is

q_{e}E(t)(z_{e}- z_{p}) = q_{e}E(t)z = q_{e}E(t) r cos(θ).

(b) For a dipole transition to be allowed, we need <Φ_{f}|H'(r,t)|Φ_{i}> to be nonzero.

Φ_{nlm}(r,θ,φ) = R_{nl}(r)Y_{lm}(θ,φ).

The ground state wave function is Φ_{100}(r,θ,φ) = π^{-½}a_{0}^{-3/2 }exp(-r/a_{0}).

The wave functions of the quadruply degenerate first excited states are

Φ_{200}(r,θ,φ) = (4π)^{-½}(2a_{0})^{-3/2}(2 - r/a_{0}) exp(-r/(2a_{0})),

Φ_{211}(r,θ,φ) = (8π)^{-½}(2a_{0})^{-3/2}(r/a_{0}) exp(-r/(2a_{0})) sinθ e^{iφ},

Φ_{210}(r,θ,φ) = (4π)^{-½}(2a_{0})^{-3/2}(r/a_{0}) exp(-r/(2a_{0})) cosθ,

Φ_{21-1}(r,θ,φ) = (8π)^{-½}(2a_{0})^{-3/2}(r/a_{0}) exp(-r/(2a_{0})) sinθ e^{-iφ}.

<Φ_{100}| H'(r,t)|Φ_{2lm}> ∝ ∫_{0}^{∞}r^{3}dr R_{00}^{*}(r)R_{2l}(r) ∫_{0}^{π}sinθ dθ∫_{0}^{2π}dφ Y_{00}^{*}(θ,φ) cosθ Y_{lm}(θ,φ)

∝ ∫_{0}^{π}sinθ dθ∫_{0}^{2π}dφ Y_{00}^{*}(θ,φ)^{ }Y_{10}(θ,φ) Y_{lm}(θ,φ).

We can invoke the properties if the spherical harmonics.

Or we can show explicitly

<Φ_{100}|H'(r,t)|Φ_{200}> ∝ ∫_{0}^{π}sinθ cosθ dθ = 0, <Φ_{100}|H'(r,t)|Φ_{211}> ∝ ∫_{0}^{2π}dφ e^{iφ}= 0,

<Φ_{100}|H'(r,t)|Φ_{21-1}> ∝ ∫_{0}^{2π}dφ e^{-iφ}= 0, <Φ_{100}|H'(r,t)|Φ_{210}> ∝ ∫_{0}^{π}sinθ dθ cos^{2}θ ≠ 0.

The only transition from the ground state to an n = 2 state that is allowed is the transition to the n = 2, l = 1, m = 0 state, Φ_{210}(r,θ,φ).

(c) P_{if}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{fi}t')W_{fi}(t')dt'|^{2},

with ω_{fi}= (E_{f}- E_{i})/ħ and W_{fi}(t) = <Φ_{f}|H'(r,t)|Φ_{i}>, with H'(r,t) = q_{e}E(t) r cos(θ).

Let Φ_{i}= Φ_{100}and Φ_{f}= Φ_{210}. Then ω_{fi}= ω_{21}= (E_{2}- E_{1})/ħ = (10.2 eV)/ħ and

W_{fi}(t) = <Φ_{210}|H'(r,t)|Φ_{000}> = [q_{e}E(t) /(2^{3/2}a_{0}^{4})] ∫_{0}^{∞}r^{4}dr exp(-3r/(2a_{0}))∫_{0}^{π}sinθ dθ cos^{2}θ

= q_{e}E(t) 2^{½ }2^{7}a_{0}/3^{5}.

Therefore the probability that at time t the electron will have made the transition is

P_{if}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{fi}t')W_{fi}(t')dt'|^{2}= 0.555 q_{e}^{2}a_{0}^{2}|∫_{0}^{t}exp(iω_{21}t')E(t')dt'|^{2}.

Consider a spinless independent electron in the average spherically symmetric
potential due to the nucleus and the other electrons in an atom.

Assume the electron is interacting with a monochromatic plane wave

**E**(r,t)
= (**z**/z)E_{0}cos(ky
- ωt), **B**(r
t) = (**x**/x
)B_{0}cos(ky
- ωt), B_{0}/E_{0} = ω/k
= c.

**E** = -∂**A**/∂t - **∇**Φ,
**B** =
**∇**×**A**.

Choose the Coulomb gauge,
**∇**·**A** = 0, and let Φ = 0,

**A**(r,t)
=
(**z**/z)[A_{0}exp(i(ky
- ωt)) + A_{0}*exp(-i(ky
- ωt))],

with iωA_{0}
= E_{0}/2 and ikA_{0} = B_{0}/2.

The Hamiltonian of the electron interacting with this plane wave is

H = (1/(2m)) (**p** - q_{e}**A**(**r**,t))^{2} +
U(r) = H_{0} + W,

with W = -(q/m)**p**∙**A** + (q_{e}^{2}/(2m))**A**^{2}.

Assume that that the intensity of the wave is low enough so that the term
containing A_{0}^{2} can be neglected compared to terms
containing A_{0}.

(a) Expand
exp(iky) in a Taylor series expansion and evaluate W to zeroth order in ky.

(Since ky is on the order of a_{0}/λ, it is much smaller than 1 in the
visible region of the EM spectrum.)

Write the zeroth order W^{(0)}
in terms of E_{0}. This is the electric dipole Hamiltionian W_{DE}.

(b) Evaluate the commutator [z, H_{0}], and show that the
transition matrix element <Φ_{f}|p_{z}|Φ_{i}>
can be written in terms of the matrix
element <Φ_{f}|z|Φ_{i}>, thus showing that W_{DE}
is equivalent to the form we would get starting with the energy of an electric
dipole in an electric field.

(c) Show that the first order term in the expansion of W is W^{(1)}
= -(q/m)B_{0}cos(ωt)
p_{z}y.

Using
p_{z}y = ½(yp_{z} - zp_{y})
+ ½(yp_{z} + zp_{y})
= ½L_{z}
+ ½(yp_{z} + zp_{y})
write W(1) = W_{DM} + W_{QM}, where W_{DM} is called the
magnetic dipole Hamiltonian and W_{QM} is called the electric
quadrupole Hamiltonian.

For initial and final states for which all transitions are allowed, estimate the
order of magnitude of the ratio of the magnetic
dipole and the electric quadrupole transition probabilities to the electric
dipole transition probability.

(d)
Show that for electric dipole transitions Δl = ±1.

Solution:

- Concepts:

Time dependent perturbation theory - Reasoning:

To find induced transition probabilities, we have to evaluate the matrix elements of W(t) between unperturbed bound states. - Details of the calculation:

(a) exp(iky) = 1 + ky + ... .

W = -(q/m)**p**∙**A**= -(q/m)p_{z}[A_{0}exp(iky)exp(-iωt) + A_{0}*exp(-iky)exp(-iωt)]

W_{DE}≅ -(q/m)p_{z}[A_{0}exp(-iωt) + A_{0}*exp(iωt)]

= -(q/m)p_{z}[(E_{0}/(2iω))exp(-iωt) - (E_{0}/(2iω))exp(iωt))]

= (qE_{0}/(mω)) p_{z}sin(ωt).

(b) [z, H_{0}] = (2m)^{-1}[z, p_{z}^{2}] = iħp_{z}/m.

(iħ/m)<Φ_{f}|p_{z}|Φ_{i}> = <Φ_{f}|(zH_{0}- H_{0}z)|Φ_{i}> = -(E_{f}- E_{i})<Φ_{f}|z|Φ_{i}> = -ħω_{fi}<Φ_{f}|z|Φ_{i}>.

<Φ_{f}|W_{DE}|Φ_{i}> = (qE_{0}/(mω))<Φ_{f}|p_{z}|Φ_{i}>sin(ωt) = iqE_{0}(ω_{fi}/ω)<Φ_{f}|z|Φ_{i}>sin(ωt).

(c) W^{(1)}= -(q/m)p_{z}[A_{0}(iky)exp(-iωt) + A_{0}*(-iky)exp(-iωt)]

= -(q/m)p_{z}[(B_{0}/(2ik))(iky)exp(-iωt) + (B_{0}/(2ik))(iky)exp(-iωt)]

= -(q/m)B_{0}cos(ωt) p_{z}y.

W^{(1)}= -(q/(2m))L_{z}B_{0}cos(ωt) - (q/(2m))B_{0}(yp_{z}+ zp_{y})cos(ωt) = W_{DM}+ W_{QM}.

W_{DM}= -(q/(2m))**L**·**B**_{0}cos(ωt) = -**μ**·**B**_{0}cos(ωt).

W^{(1)}_{max}/W^{(0)}_{max}= ((q/m)B_{0}p_{z}y)/((qE_{0}/(mω)) p_{z}) = (E_{0}y/c)/(E_{0}/ck) = ky,

which is on the order of a_{0}/λ.

Transition probabilities are proportional to the square of the matrix elements, so the ratio of the magnetic dipole and the electric quadrupole transition probabilities to the electric dipole transition probability.is on the order of (a

_{0}/λ)^{2}.

(d) <Φ

_{f}|W_{DE}(t)|Φ_{i}> ∝ <Φ_{f}|z|Φ_{i}>.

Let Φ_{i}(r) = R_{ni li}(r)Y_{li mi}(θ,φ), Φ_{f}(r) = R_{nf lf}(r)Y_{lf mf}(θ,φ)

With z = r cosθ ∝ r Y_{10}(θ,φ) we have

<Φ_{f}|W_{DE}(t)|Φ_{i}> ∝ ∫_{0}^{π}sinθ dθ∫_{0}^{2π}dφ Y^{*}_{lf mf}(θ,φ) Y_{10}(θ,φ) Y_{li mi}(θ,φ).

The integrant is a product of three spherical harmonics.

The integral is zero unless

(i) m_{1}+ m_{2}+ m_{3}= 0,

(ii) |l_{1}- l_{2}| ≤ l_{3}≤ (l_{1 }+ l_{2}),

(iii) l_{1}+ l_{2}- l_{3}= even.

Y_{lm}*(θ,φ) = (-1)^{m}Y_{l-m}(θ,φ).

We therefore have that <Φ_{f}|W_{DE}(t)|Φ_{i}> = 0 unless m_{f}= m_{i}, l_{f }= l_{i}± 1. If we choose another direction for the polarization of**E**, i.e.**E**= E_{0}**i**or E_{0}**j**, then we find m_{f }= m_{i}± 1, l_{f }= l_{i}± 1.

The dipole transition selection rules therefore are

Δl = ±1, Δm = 0, ±1.

These selection rules result as a >consequence of the properties of the spherical harmonics.

More information:

The electric dipole transition rules are Δl = ±1, Δm = 0, ±1.

If H_{0}contains a spin orbit coupling term f(r)**L**∙**S**, then the electric dipole selection rules become

Δj = 0, ±1, (except j_{i }= j_{f }= 0), Δl = ±1, Δm_{j }= 0, ±1.If spin is included, the magnetic dipole Hamiltonian becomes W

_{DM}= -(q/(2m))(**L**+ 2**S**)·**B**_{0}cos(ωt).

Transition induced by W_{DM}obey the magnetic dipole transition selection rules Δl = 0, Δm = 0, ±1, Δm_{s}= 0, ±1.

or, if H_{0}includes a spin orbit coupling term, Δl = 0, Δj = 0, ±1, Δm_{j}= 0, ±1.Transition induced by the electric quadrupole Hamiltonian W

_{QM}obey the electric quadrupole transition selection rules