Problem 1:

Let A and B be two observables of a system with a two dimensional state space, and suppose measurements are made of A, B, and A again in quick succession.  Show that the probability that the second measurement of A gives the same result as the first is independent of the initial state of the system.

Hint:
Let {|a1>,|a2> } be an othonormal eigenbasis of A and let {|b1>,|ba> } be an othonormal eigenbasis of B.
A|ai> = ai|ai>,  B|bi> = bi|bi>.
|b1> = cos(θ)|a1> + sin(θ)e|a2>,  |b> = -sin(θ)|a1> + cos(θ)e|a2>
is the most general expansion of the |bi> in terms of the |ai>.
<b1|b2> = 0,  <b1|b1> = <b2|b2> = 1.
The initial state can be written as some linear combination of |a1> and a2>.
|Ψ> = c1|a1> + c2|a2>.

Solution:

• Concepts:
Fundamental assumptions of QM
• Reasoning:
When a physical quantity described by the operator A is measured on a system in a normalized state |Ψ>, the probability of measuring the eigenvalue an is given by
P(an) = ∑i=0gn|<uni|Ψ>|2,  where {|uni>} (i=1,2,...,gn) is an orthonormal basis in the eigensubspace Εn associated with the eigenvalue an.
• Details of the calculation:
The initial state can be written as some linear combination of |a1> and a2>.
|Ψ> = c1|a1> + c2|a2>,  |c1|2 + |c2|2 = 1 for a normalized state.
Branch 1:
Assume the first measurement yields a1.  The probability for this outcome is |c1|2.
What is the probability of measuring a1 again after a measurement of B?
Pa1(b1) = cos2(θ) (system is now in state |b1>), Pb1(a1) = cos2(θ),
Pa1(b1) Pb1(a1)  = cos4(θ),
Pa1(b2) = sin2(θ) (system is now in state |b2>), Pb2(a1) = sin2(θ),
Pa1(b2) Pb2(a1)  = sin4(θ).
Probability of the first path + probability of the second path = cos4(θ) + sin4(θ).
Total probability for this branch:  |c1|2(cos4(θ) + sin4(θ)).

Branch 2:
Assume the first measurement yields a2.  The probability for this outcome is |c2|2.
What is the probability of measuring a2 again after a measurement of B?
Pa2(b1) = sin2(θ) (system is now in state |b1>), Pb1(a2) = sin2(θ),
Pa2(b1) Pb1(a2)  = sin4(θ),
Pa2(b2) = cos2(θ) (system is now in state |b2>), Pb2(a2) = cos2(θ),
Pa2(b2) Pb2(a2)  = cos4(θ).
Probability of the first path + probability of the second path = sin4(θ) + cos4(θ).
Total probability for this branch:  |c2|2(cos4(θ) + sin4(θ)).
Probability of obtaining the same result =  (|c1|2 + |c2|2)(cos4(θ) + sin4(θ))
= cos4(θ) + sin4(θ), independent of the initial state.

Problem 2:

Find <x> and ∆x for the nth stationary state of a free particle in one dimension restricted to the interval 0 < x < a.  Show that as n--> ∞ these become the classical values.

Solution:
The eigenfunctions of the infinite well (U(x) = 0,  0 < x < L,  U(x) = ∞ everywhere else) are
ψn(x) = (2/a)1/2sin(nπx/a)  with eigenvalues  En = n2π2ħ2/(2ma2).
<x> = ∫0a dx Ψ*(x) x Ψ*(x)
<x> = (2/a∫0a dx  x sin2(nπx/L) = a/2.
Δx = <(<x2> - <x>2)>1/2.
<x2> = (2/a∫0a dx  x2 sin2(nπx/L) = a2/3 - a2/(2n2π2).
Δx2 = a2/3 - a2/4 - a2/(2n2π2) =  a2/12 - a2/(2n2π2).
Δx2 -->  a2/12  as n --> ∞.

Classically the particle bounces back and forth.  It spends the same amount of time at each position.
Therefore xavg = a/2 for a large number of particles with arbitrary initial positions.
x2rms = (1/a)∫0a dx (x - a/2)2 =(x3/3 - ax2/2 + a2x/4)|0a = a2/12.

Problem 3:

A system makes transitions between eigenstates of H0 under the action of the time dependent Hamiltonian H0 + Wcosωt, W << H0
Find an expression for the probability of transition from ||ψ1> to ||ψ2>, where ||ψ1> and ||ψ2> are eigenstates of H0 with eigenvalues E1 and E2
Show that this probability is small unless E2 - E1 ≈ ħω.
[This shows that a charged particle in an oscillating electric field with angular frequency w will exchange energy with the field only in multiples of E ≈ ħω.]

Solution:

• Concepts:
Time-dependent perturbation theory
• Reasoning:
For small perturbations time-dependent perturbation theory can be used to calculate transition probabilities.
• Details of the calculation:
Assume that Initially the system is in state |ψi>.
The probability of finding the system in the state |ψf> (f ≠ i)  at time t is
Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2,
with  ωfi = (Ef - Ei)/ħ and Wfi(t) = <ψf|W(t)|ψi>, with W(t) = Wcosωt.
Here i = 1 and f = 2.
We have to consider a sinusoidal perturbation starting at t = 0.
0t exp(iωfit')Wfi(t')dt' = (Wfi/2)∫0t [exp(i(ωfi+ω)t' + exp(i(ωfi-ω)t']dt'
= (Wfi/2)[(1 - exp(i(ωfi+ω)t)/(ωfi+ω) + (1 - exp(i(ωfi-ω)t)/(ωfi-ω)].
Pif(t) =[|Wfi|2/(4ħ2)]|(1 - exp(i(ωfi+ω)t)/(ωfi+ω) + (1 - exp(i(ωfi-ω)t)/(ωfi-ω)|2,
with Wfi = <ψf|W|ψi>.
The expression
(1 - exp(i(ωfi+ω)t)/(ωfi+ω) + (1 - exp(i(ωfi-ω)t)/(ωfi-ω)
has an appreciable amplitude only if the denominator of one of the two terms is approximately zero, i.e. if ωfi = ±ω, or Ef = Ei ± ħω.
The first order effect of a perturbation that varies sinusoidally with time is to receive from or transfer to the system a quantum of energy hω.
If the system is initially in the ground state, then Ef > Ei, and only the second term needs to be considered.  Then
Pif(t) = [|Wfi|2/(4ħ2)] sin2((ωfi - ω)t/2)/((ωfi - ω)/2)2.
Problem 4:

Consider a one-dimensional system, with momentum operator p and position operator q.
(a) Show that [q,pn] = iħ n pn-1.
(b) Show that [q,F(p)] = iħ ∂F/∂p, if the function F(p) can be defined by a finite polynomial or convergent power series in the operator p.
(c) Show that [q,p2F(q)] = 2iħ pF(q) if F(q) is some function of q only.

Solution:

• Concepts:
Commutator algebra
• Reasoning:
We are asked to find several commutators.
• Details of the calculation:
(a)  [q,p] = iħ,  [q,p2] = p[q,p] + [q,p]p = 2iħp.
[q,pn+1] = [q,ppn] = p[q,pn] + [q,p]pn.
If [q,pn]  =iħnpn-1, then [q,pn+1] = piħnpn-1 + iħpn = iħ(n+1)pn.
[q,pn] = iħnpn-1 holds for n = 1 and n = 2.  Therefore it holds for all n.
(b)  [q,F(p)] =  [q,∑nfnpn] = ∑nfn[q,pn] = ∑nfnniħpn-1] = iħ∂f(q)/∂q.
Similarly,  [p,F(q)] =  [p,∑nfnqn] = ∑nfn[p,qn] = -∑nfnniħqn-1] = -iħ∂f(q)/∂q.
(c)  [q,p2F(q)] = p2[q,F(q)] + [q,p2]F(q)] = 0 + 2iħ pF(q).
FYI:
These formulas are used to proof the Ehrenfest’s theorem.
(d/dt)<R> = <P>/m.
(d/dt)<P> = -<U(R)>.
md2<R>/dt2 = d<P>/dt = -<U(R)>.

Problem 5:

If baryon number were not conserved in nature, the wave functions for neutrons (n) and anti-neutrons (n) would not be stationary mass-energy eigenfunctions.
Rather, the wave functions of such particles would be oscillating, time-dependent superposition of neutron and anti-neutron components, given by

,

where n = 1,  n = 2.

Suppose that such oscillations occur and the Hamiltonian of the system, neglecting all degrees of freedom except for the one corresponding to the oscillations, is

,

where  E1 = m + U1  and  E2 = m + U2  are energies for n and n individually, and α is a real mixing amplitude.
In an external magnetic field B, the potential energies are U1 = μB and U2 = -μB where μμn = -μn.
Suppose that at time t = 0 the initial state is that of a neutron.
(a)  Calculate the time-dependent probability for observing an anti-neutron.  Determine the period of oscillation.
(b)  Describe the effect of increasing the external magnetic field.

Solution:

Let |n1> and |n2> denote the eigenstates of the operator that distinguishes the neutron from the anti-neutron.
In this basis the matrix of H is

.
Let us rewrite H.

H = H1 + H2.  H1 = ½(E1 + E2)I,  H2 = ½(E1 - E2)K.
Define tanθ = 2α/(E1 - E2), then
.
Since every vector is an eigenvector of I, the eigenvectors of K are the eigenvectors of H.
The eigenvalues λ of K are found by solving det(K - λI) = 0.
λ2 = 1 +  tan2θ,  λ± = ±1/cosθ.
For the eigenvectors we have |ψ±> = c1|n1>+  c2|n2>.
For the eigenvector corresponding to λ+:  (1 - 1/cosθ) c1 + tanθ c2 = 0,
c2 = -c1(cosθ/sinθ)- 1/sinθ) = c1tan(θ/2).
|c1|2 + |c2|2 = 1  is satisfied if c2 = sin(θ/2), c1 = cos(θ/2),  |ψ+> = cos(θ/2)|n1> + sin(θ/2)|n2>.
Similarly, for the eigenvector corresponding to λ- we find c1  = -c2tan(θ/2).
|c1|2 + |c2|2 = 1  is satisfied if c1 = -sin(θ/2), c2 = cos(θ/2),  |ψ-> = -sin(θ/2)|n1> + cos(θ/2)|n2>.

+> and |ψ-> are the eigenvectors of H with eigenvalues E± = ½(E1 + E2) ± ½(E1 - E2)/cosθ.
We have |n1> = cos(θ/2)|ψ+> - sin(θ/2)|ψ->, |n2> = sin(θ/2)|ψ+> + cos(θ/2)|ψ->.
At t = 0 we have |ψ(0)> = |n1>.
|ψ(t)> = cos(θ/2)exp(-iE+t/ħ)|ψ+> - sin(θ/2)exp(-iE-t/ħ)|ψ->.
The probability of observing an antineutron at time t is Pn(t) = |<n2|ψ(t)>|2.
<n2|ψ(t)> = cos(θ/2) sin(θ/2)[exp(-iE+t/ħ) - exp(-iE-t/ħ)].
Pn(t) =  ¼sin2θ[2 - exp(i(E+ - E-)t/ħ) + exp(i(E+ - E-)t/ħ)] = ½sin2θ [1 - cos((E+ - E-)t/ħ)].
Period of oscillation:  2π/T = (E+ - E-)/ħ,  T = h/(E+ - E-).
Let B = Bk, μB = μzB.
Then E1 = m + μzB and E2 = m - μzB, ½(E1 + E2) = m,  ½(E1 - E2) = μzB.
tanθ = α/μzB,  1/cos2θ = 1 + (α/μzB)2,  sin2θ = α2/(μz2B2 + α2).
E± = m ± (μz2B2 + α2)1/2, E+ - E- = 2(μz2B2 + α2)1/2.
Pn(t) = ½[α2/(μz2B2 + α2)] [1 - cos(2(μz2B2 + α2)1/2t/ħ)].

(b)  The larger B, the shorter the period of the oscillations.  However, The larger B, the smaller the amplitude of the oscillations.