## Assignment 11, solutions

#### Problem 1:

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∆/ħ2)S1S2, 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, t0)Ψ(t0)> 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:
S1S2 = ½(S2 - S12 - S22).  The eigenstates of S1S2 are the singlet state and the triplet states, {|S, Ms>}, S = 0, 1.  These are common eigenfunctions of S2 and Sz.
• Details of the calculation:
(a)  The common eigenfunctions of S2 and Sz are
|1,1> = |++>,
|1,0> = ½½(|+-> + |-+>),
|1,-1> =|-->,
|0, 0> = ½½(|+-> - |-+>).
Therefore |+-> = ½½(|0,0> + |1,0>), |-+> = ½½(|0,0> - |1,0>).
For the singlet state we have S1S2|0,0> = -(3/4)ħ2,  H|0,0> = -3∆|0,0>.
For the triplet states we have S1S2|1, Ms> = (1/4)ħ2,  H|1, Ms> = ∆|1, Ms>.
|ψ(0)> = |+-> = ½½(|0,0> + |1,0>),
|ψ(t)> = U(t, 0)|+-> = 2(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 |+-> = cos2(2∆*t/ħ).
For 2∆*t/ħ << 1, P(t) ~ 1.
P(t) of being in the state |-+> = sin2(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
Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2.
The initial and the final state must be different for this formula to be valid.  For i = f, P(t) ~ 1.
Wfi = <f|H|i>.  Here Wfi = <f|H|+->.
Wfi = 0 for |f> = |++> and |f> = |-->, so Pif(t) = 0 for these states.
For |f> = |-+> we have Wfi = 2<-+|H|(|00> + |10>)
= 2[-3∆<-+| 0, 0> + ∆<-+| 1, 0>] = ½(3∆ + ∆) = 2∆.
ωfi = (Ef - Ei)/ωfi = 0 for the unperturbed states.
Pif(t) = (1/ħ2)|∫0t(2∆)dt'|2 = 4∆2t22.

(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.

#### Problem 2:

In a one-dimensional structure a particle of mass m is in the ground state of a potential energy function U(x) = ½kx2.  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(t2,t1)|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
Φ0i(x) = (mω/(πħ))¼exp(-½mωx2/ħ),  ω = (k/m)½.
The normalized ground-state wave function for t > 0 is
Φ0f(x) = (m√(2)ω/(πħ))¼exp(-½m√(2)ωx2/ħ).
The probability that the oscillator remains in the ground state is
P = |<Φ0i0f>|2 =  |∫-∞dx Φ0i*(x) Φ0f(x)|2
= 25/4/(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.

#### Problem 3:

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 >> mC2/(2ħ3) 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 H0 = p2/(2m) - Cδ(x).
Normalized eigenfunction of H0:  Φ(x) = ρ½ exp(-ρ|x|),
ρ2 = -2mEi2,  Ei = -mC2/(2ħ2),  ρ = mC/ħ2.
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)|WEi|2δE-Ei,ħω, where WEi = <ΦE|W|Φi>.
Since ħω >> -mC2/(2ħ2), we can assume that the ejected particle is a nearly free particle with energy E = ħω + Ei and k2 = 2mE/ħ2.
We assume that the particle is confined to a region of width L >> a and  use periodic boundary conditions.
Then Φk(x) = Lexp(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 = ħ2k2/(2m) we have ρ(E) = dN/dE = Lm/(πkħ2).
WEi = (ρ/L)½W[∫0dx eikxe-ρx  + ∫-∞0dx eikxeρx]
= (ρ/L)½W 2ρ/(ρ2 + k2).

w(i,E) = (π/(2ħ))(Lm/(πkħ2))(ρ/L)W22/(ρ2 + k2)2
= (2mW23) ρ3/[k(ρ2 + k2)2].
2 + k2)2 = 4m2ω22, ρ3/[(ρ2 + k2)2] = mC3/(4ħ4ω2).
w(i,E) = (2mW23)mC3/(4ħ4ω2)(2mE)½/ħ.
w(i,E) = (2mE)½ 2m2W2C3/(4ω2ħ8).
This is the transition probability per unit time.

#### Problem 4:

A hydrogen atom with Hamiltonian H0(r) is placed in a time-dependent electric field E = E(t) k. The perturbed Hamiltonian is H(r,t) = H0(r) + H'(r,t).
(a)  Show that H'(r,t) = qeE(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 -∫0zpqeE(t)dz = -qeE(t)zp.
The energy of the electron in the field is ∫0zeqe E(t)dz = qeE(t)ze.
The potential energy of both particles is
qeE(t)(ze - zp) = qeE(t)z = qeE(t) r cos(θ).

(b)  For a dipole transition to be allowed, we need <Φf|H'(r,t)|Φi> to be nonzero.
Φnlm(r,θ,φ) = Rnl(r)Ylm(θ,φ).
The ground state wave function is Φ100(r,θ,φ)  =  πa0-3/2 exp(-r/a0).
The wave functions of the quadruply degenerate first excited states are
Φ200(r,θ,φ)  = (4π) (2a0)-3/2 (2 - r/a0) exp(-r/(2a0)),
Φ211(r,θ,φ)  = (8π) (2a0)-3/2 (r/a0) exp(-r/(2a0)) sinθ e,
Φ210(r,θ,φ)  = (4π) (2a0)-3/2 (r/a0) exp(-r/(2a0)) cosθ,
Φ21-1(r,θ,φ) = (8π) (2a0)-3/2 (r/a0) exp(-r/(2a0)) sinθ e-iφ.

100| H'(r,t)|Φ2lm>  ∝  ∫0r3dr R00*(r)R2l(r)  ∫0πsinθ dθ∫0dφ Y00*(θ,φ) cosθ Ylm(θ,φ)
∝ ∫0πsinθ dθ∫0dφ Y00*(θ,φ) Y10(θ,φ) Ylm(θ,φ).

We can invoke the properties if the spherical harmonics.
Or we can show explicitly
100|H'(r,t)|Φ200> ∝ ∫0πsinθ cosθ dθ = 0,   <Φ100|H'(r,t)|Φ211> ∝ ∫0dφ e = 0,
100|H'(r,t)|Φ21-1> ∝ ∫0dφ e-iφ = 0,   <Φ100|H'(r,t)|Φ210> ∝ ∫0πsinθ dθ cos2θ ≠ 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, Φ210(r,θ,φ).

(c)  Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2,
with  ωfi = (Ef - Ei)/ħ and Wfi(t) = <Φf|H'(r,t)|Φi>, with H'(r,t)  = qeE(t) r cos(θ).
Let Φi = Φ100 and Φf = Φ210.  Then ωfi = ω21 = (E2 - E1)/ħ = (10.2 eV)/ħ and
Wfi(t) = <Φ210|H'(r,t)|Φ000> = [qeE(t) /(23/2a04)] ∫0r4dr exp(-3r/(2a0))∫0πsinθ dθ cos2θ
= qeE(t) 2½ 27 a0/35.
Therefore the probability that at time t the electron will have made the transition is
Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2 = 0.555 qe2a02|∫0texp(iω21t')E(t')dt'|2.

#### Problem 5:

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)E0cos(ky - ωt),  B(r t) = (x/x  )B0cos(ky - ωt),  B0/E0 =  ω/k = c.
E = -∂A/∂t - Φ,  B = ×A.
Choose the Coulomb gauge, ·A = 0, and let Φ = 0,
A(r,t) = (z/z)[A0exp(i(ky - ωt)) + A0*exp(-i(ky - ωt))],
with iωA0 = E0/2 and ikA0 = B0/2.
The Hamiltonian of the electron interacting with this plane wave is
H = (1/(2m)) (p - qeA(r,t))2 + U(r) = H0 + W,
with W = -(q/m)pA + (qe2/(2m))A2.
Assume that that the intensity of the wave is low enough so that the term containing A02 can be neglected compared to terms containing A0.
(a)  Expand exp(iky) in a Taylor series expansion and evaluate W to zeroth order in ky.
(Since ky is on the order of a0/λ, it is much smaller than 1 in the visible region of the EM spectrum.)
Write the zeroth order W(0) in terms of E0.  This is the electric dipole Hamiltionian WDE.
(b)  Evaluate the commutator [z, H0], and show that the transition matrix element <Φf|pzi> can be written in terms of the matrix element <Φf|z|Φi>, thus showing that WDE 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)B0cos(ωt) pzy.
Using pzy = ½(ypz - zpy)  + ½(ypz + zpy) = ½Lz + ½(ypz + zpy) write W(1) = WDM + WQM, where WDM is called the magnetic dipole Hamiltonian and WQM 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)pA = -(q/m)pz[A0exp(iky)exp(-iωt) + A0*exp(-iky)exp(-iωt)]
WDE ≅ -(q/m)pz[A0exp(-iωt) + A0*exp(iωt)]
= -(q/m)pz[(E0/(2iω))exp(-iωt) - (E0/(2iω))exp(iωt))]
= (qE0/(mω)) pzsin(ωt).
(b)  [z, H0] = (2m)-1[z, pz2] = iħpz/m.
(iħ/m)<Φf|pzi> = <Φf|(zH0 - H0z)|Φi> = -(Ef - Ei)<Φf|z|Φi> = -ħωfif|z|Φi>.
f|WDEi> = (qE0/(mω))<Φf|pzi>sin(ωt) = iqE0fi/ω)<Φf|z|Φi>sin(ωt).
(c)  W(1) = -(q/m)pz[A0(iky)exp(-iωt) + A0*(-iky)exp(-iωt)]
= -(q/m)pz[(B0/(2ik))(iky)exp(-iωt) + (B0/(2ik))(iky)exp(-iωt)]
= -(q/m)B0cos(ωt) pzy.
W(1) = -(q/(2m))LzB0cos(ωt) - (q/(2m))B0(ypz + zpy)cos(ωt) = WDM + WQM.
WDM = -(q/(2m))L·B0cos(ωt) = -μ·B0cos(ωt).
W(1)max/W(0)max = ((q/m)B0 pzy)/((qE0/(mω)) pz) = (E0y/c)/(E0/ck) = ky,
which is on the order of a0/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 (a0/λ)2.

(d)  <Φf|WDE(t)|Φi> ∝ <Φf|z|Φi>.
Let Φi(r) = Rni li(r)Yli mi(θ,φ),  Φf(r) = Rnf lf(r)Ylf mf(θ,φ)
With z = r cosθ ∝ r Y10(θ,φ) we have
f|WDE(t)|Φi> ∝ ∫0πsinθ dθ∫0dφ Y*lf mf(θ,φ) Y10(θ,φ) Yli mi(θ,φ).
The integrant is a product of three spherical harmonics.
The integral is zero unless
(i) m1 + m2 + m3 = 0,
(ii) |l1 - l2| ≤ l3 ≤ (l1 + l2),
(iii) l1 + l2 - l3 = even.
Ylm*(θ,φ) = (-1)mYl-m(θ,φ).
We therefore have that <Φf|WDE(t)|Φi> = 0 unless  mf = mi, lf = li ± 1.  If we choose another direction for the polarization of E, i.e.  E = E0i or E0j, then we find  mf = mi ± 1, lf = li ± 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.