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>,|b_a> } be an othonormal eigenbasis of B.
A|a_i> = a_i|a_i>,  B|b_i> = b_i|b_i>.
|b₁> = cos(θ)|a₁> + sin(θ)e^iφ|a₂>,  |b> = -sin(θ)|a₁> + cos(θ)e^iφ|a₂>
is the most general expansion of the |b_i> in terms of the |a_i>.
<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 a_n is given by
  P(a_n) = ∑_i=0^gn|<u_nⁱ|Ψ>|²,  where {|u_nⁱ>} (i=1,2,...,g_n) is an orthonormal basis in the eigensubspace Ε_n associated with the eigenvalue a_n.
• 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?
  P_a1(b₁) = cos²(θ) (system is now in state |b₁>), P_b1(a₁) = cos²(θ),
  P_a1(b₁) P_b1(a₁)  = cos⁴(θ),
  P_a1(b₂) = sin²(θ) (system is now in state |b₂>), P_b2(a₁) = sin²(θ),
  P_a1(b₂) P_b2(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?
  P_a2(b₁) = sin²(θ) (system is now in state |b₁>), P_b1(a₂) = sin²(θ),
  P_a2(b₁) P_b1(a₂)  = sin⁴(θ),
  P_a2(b₂) = cos²(θ) (system is now in state |b₂>), P_b2(a₂) = cos²(θ),
  P_a2(b₂) P_b2(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
ψ_n(x) = (2/a)^1/2sin(nπx/a)  with eigenvalues  E_n = n²π²ħ²/(2ma²).
<x> = ∫₀^a dx Ψ*(x) x Ψ*(x)
<x> = (2/a∫₀^a dx  x sin²⁽nπx/L) = a/2.
Δx = <(<x²> - <x>²)>^1/2.
<x²> = (2/a∫₀^a 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 x_avg = a/2 for a large number of particles with arbitrary initial positions.
x²_rms = (1/a)∫₀^a dx (x - a/2)² =(x³/3 - ax²/2 + a²x/4)|₀^a = 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 |ψ_i>.
  The probability of finding the system in the state |ψ_f> (f ≠ i)  at time t is
  P_if(t) = (1/ħ²)|∫₀^texp(iω_fit')W_fi(t')dt'|²,
  with  ω_fi = (E_f - E_i)/ħ and W_fi(t) = <ψ_f|W(t)|ψ_i>, with W(t) = Wcosωt.
  Here i = 1 and f = 2.
  We have to consider a sinusoidal perturbation starting at t = 0.
  ∫₀^t exp(iω_fit')W_fi(t')dt' = (W_fi/2)∫₀^t [exp(i(ω_fi+ω)t' + exp(i(ω_fi-ω)t']dt'
  = (W_fi/2)[(1 - exp(i(ω_fi+ω)t)/(ω_fi+ω) + (1 - exp(i(ω_fi-ω)t)/(ω_fi-ω)].
  P_if(t) =[|W_fi|²/(4ħ²)]|(1 - exp(i(ω_fi+ω)t)/(ω_fi+ω) + (1 - exp(i(ω_fi-ω)t)/(ω_fi-ω)|²,
  with W_fi = <ψ_f|W|ψ_i>.
  The expression
  (1 - exp(i(ω_fi+ω)t)/(ω_fi+ω) + (1 - exp(i(ω_fi-ω)t)/(ω_fi-ω)
  has an appreciable amplitude only if the denominator of one of the two terms is approximately zero, i.e. if ω_fi = ±ω, or E_f = E_i ± 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 E_f > E_i, and only the second term needs to be considered.  Then
  P_if(t) = [|W_fi|²/(4ħ²)] sin²((ω_fi - ω)t/2)/((ω_fi - ω)/2)².

Consider a one-dimensional system, with momentum operator p and position operator q.
(a) Show that [q,pⁿ] = iħ n p^n-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]  =iħnp^n-1, then [q,pⁿ⁺¹] = piħnp^n-1 + iħpⁿ = iħ(n+1)pⁿ.
  [q,pⁿ] = iħnp^n-1 holds for n = 1 and n = 2.  Therefore it holds for all n.
  (b)  [q,F(p)] =  [q,∑_nf_npⁿ] = ∑_nf_n[q,pⁿ] = ∑_nf_nniħp^n-1] = iħ∂f(q)/∂q.
  Similarly,  [p,F(q)] =  [p,∑_nf_nqⁿ] = ∑_nf_n[p,qⁿ] = -∑_nf_nniħq^n-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 = μ∙B and U₂ = -μ∙B where μ ≈ μ_n = -μ_n. 
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>.

|ψ₊> 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 P_n(t) = |<n₂|ψ(t)>|².
<n₂|ψ(t)> = cos(θ/2) sin(θ/2)[exp(-iE₊t/ħ) - exp(-iE_-t/ħ)].
P_n(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, μ∙B = μ_zB. 
Then E₁ = m + μ_zB and E₂ = m - μ_zB, ½(E₁ + E₂) = m,  ½(E₁ - E₂) = μ_zB.
tanθ = α/μ_zB,  1/cos²θ = 1 + (α/μ_zB)²,  sin²θ = α²/(μ_z²B² + α²).
E_± = m ± (μ_z²B² + α²)^1/2, E₊ - E_- = 2(μ_z²B² + α²)^1/2.
P_n(t) = ½[α²/(μ_z²B² + α²)] [1 - cos(2(μ_z²B² + 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.