**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 {|a_{1}>,|a_{2}> } be an othonormal eigenbasis of A
and let {|b_{1}>,|b_{a}> } be an othonormal eigenbasis of
B.

A|a_{i}> = a_{i}|a_{i}>, B|b_{i}> = b_{i}|b_{i}>.

|b_{1}> = cos(θ)|a_{1}> + sin(θ)e^{iφ}|a_{2}>, |b> =
-sin(θ)|a_{1}>
+ cos(θ)e^{iφ}|a_{2}>

is the most general expansion of the |b_{i}> in terms of the |a_{i}>.

<b_{1}|b_{2}> = 0, <b_{1}|b_{1}> = <b_{2}|b_{2}>
= 1.

The initial state can be written as some linear combination of |a_{1}>
and a_{2}>.

|Ψ> = c_{1}|a_{1}> + c_{2}|a_{2}>.

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_{n}^{i}|Ψ>|^{2}, where {|u_{n}^{i}>} (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 |a_{1}> and a_{2}>.

|Ψ> = c_{1}|a_{1}> + c_{2}|a_{2}>, |c_{1}|^{2}+ |c_{2}|^{2}= 1 for a normalized state.

Branch 1:

Assume the first measurement yields a_{1}. The probability for this outcome is |c_{1}|^{2}.

What is the probability of measuring a_{1}again after a measurement of B?

P_{a1}(b_{1}) = cos^{2}(θ) (system is now in state |b_{1}>), P_{b1}(a_{1}) = cos^{2}(θ),

P_{a1}(b_{1}) P_{b1}(a_{1}) = cos^{4}(θ),

P_{a1}(b_{2}) = sin^{2}(θ) (system is now in state |b_{2}>), P_{b2}(a_{1}) = sin^{2}(θ),

P_{a1}(b_{2}) P_{b2}(a_{1}) = sin^{4}(θ).

Probability of the first path + probability of the second path = cos^{4}(θ) + sin^{4}(θ).

Total probability for this branch: |c_{1}|^{2}(cos^{4}(θ) + sin^{4}(θ)).Branch 2:

Assume the first measurement yields a_{2}. The probability for this outcome is |c_{2}|^{2}.

What is the probability of measuring a_{2}again after a measurement of B?

P_{a2}(b_{1}) = sin^{2}(θ) (system is now in state |b_{1}>), P_{b1}(a_{2}) = sin^{2}(θ),

P_{a2}(b_{1}) P_{b1}(a_{2}) = sin^{4}(θ),

P_{a2}(b_{2}) = cos^{2}(θ) (system is now in state |b_{2}>), P_{b2}(a_{2}) = cos^{2}(θ),

P_{a2}(b_{2}) P_{b2}(a_{2}) = cos^{4}(θ).

Probability of the first path + probability of the second path = sin^{4}(θ) + cos^{4}(θ).

Total probability for this branch: |c_{2}|^{2}(cos^{4}(θ) + sin^{4}(θ)).

Probability of obtaining the same result = (|c_{1}|^{2}+ |c_{2}|^{2})(cos^{4}(θ) + sin^{4}(θ))

= cos^{4}(θ) + sin^{4}(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)^{1/2}sin(nπx/a)
with eigenvalues
E_{n} = n^{2}π^{2}ħ^{2}/(2ma^{2}).

<x> = ∫_{0}^{a} dx Ψ*(x) x Ψ*(x)

<x> = (2/a∫_{0}^{a} dx
x sin^{2(}nπx/L) = a/2.

Δx = <(<x^{2}> - <x>^{2})>^{1/2}.

<x^{2}> = (2/a∫_{0}^{a} dx
x^{2} sin^{2(}nπx/L) = a^{2}/3 - a^{2}/(2n^{2}π^{2}).

Δx^{2} =
a^{2}/3 - a^{2}/4 - a^{2}/(2n^{2}π^{2})
=
a^{2}/12 - a^{2}/(2n^{2}π^{2}).

Δx^{2} -->
a^{2}/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.

x^{2}_{rms} = (1/a)∫_{0}^{a} dx (x - a/2)^{2} =(x^{3}/3
- ax^{2}/2 + a^{2}x/4)|_{0}^{a} =
a^{2}/12.

**Problem 3:**

A system makes transitions between eigenstates of H_{0}
under the action of the time dependent Hamiltonian H_{0 }+ Wcosωt, W << H_{0}.

Find an expression for the probability of transition from
||ψ_{1}> to ||ψ_{2}>,
where ||ψ_{1}> and ||ψ_{2}>
are eigenstates of H_{0} with eigenvalues E_{1} and E_{2}.

Show that this probability is small unless E_{2} - E_{1} ≈ 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/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) = <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_{fi}t')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}/(4ħ^{2})]|(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}/(4ħ^{2})] sin^{2}((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,p^{n}] = 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,p^{2}F(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,p^{2}] = p[q,p] + [q,p]p = 2iħp.

[q,p^{n+1}] = [q,pp^{n}] = p[q,p^{n}] + [q,p]p^{n}.

If [q,p^{n}] =iħnp^{n-1}, then [q,p^{n+1}] = piħnp^{n-1 }+ iħp^{n }= iħ(n+1)p^{n}.

[q,p^{n}] = iħnp^{n-1}holds for n = 1 and n = 2. Therefore it holds for all n.

(b) [q,F(p)] = [q,∑_{n}f_{n}p^{n}] = ∑_{n}f_{n}[q,p^{n}] = ∑_{n}f_{n}niħp^{n-1}] = iħ∂f(q)/∂q.

Similarly, [p,F(q)] = [p,∑_{n}f_{n}q^{n}] = ∑_{n}f_{n}[p,q^{n}] = -∑_{n}f_{n}niħq^{n-1}] = -iħ∂f(q)/∂q.

(c) [q,p^{2}F(q)] = p^{2}[q,F(q)] + [q,p^{2}]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**)>.

md^{2}<**R**>/dt^{2}= 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

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 E_{1} = m + U_{1} and E_{2} = m +
U_{2} are energies for n and
n
individually, and α
is a real mixing amplitude.

In
an external magnetic field **B**, the potential energies are U_{1}
= **μ**∙**B**
and U_{2}
= -**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.

Solution:

Let |n_{1}> and |n_{2}> 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 = H_{1} + H_{2}. H_{1} = ½(E_{1} + E_{2})I,
H_{2} = ½(E_{1} - E_{2})K.

Define tanθ = 2α/(E_{1} - E_{2}), 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.

λ^{2} = 1 + tan^{2}θ, λ_{±} = ±1/cosθ.

For the eigenvectors we have |ψ_{±}> = c_{1}|n_{1}>+
c_{2}|n_{2}>.

For the eigenvector corresponding to λ_{+}: (1 - 1/cosθ) c_{1}
+ tanθ c_{2} = 0,

c_{2} = -c_{1}(cosθ/sinθ)- 1/sinθ) = c_{1}tan(θ/2).

|c_{1}|^{2} + |c_{2}|^{2} = 1 is satisfied
if c_{2} = sin(θ/2), c_{1} = cos(θ/2), |ψ_{+}> =
cos(θ/2)|n_{1}> + sin(θ/2)|n_{2}>.

Similarly, for the eigenvector corresponding to λ_{-} we find c_{1}
= -c_{2}tan(θ/2).

|c_{1}|^{2} + |c_{2}|^{2} = 1 is satisfied
if c_{1} = -sin(θ/2), c_{2} = cos(θ/2), |ψ_{-}> =
-sin(θ/2)|n_{1}> + cos(θ/2)|n_{2}>.

|ψ_{+}> and |ψ_{-}> are the eigenvectors of H with eigenvalues E_{±} = ½(E_{1} + E_{2}) ± ½(E_{1} -
E_{2})/cosθ.

We have |n_{1}> = cos(θ/2)|ψ_{+}> - sin(θ/2)|ψ_{-}>, |n_{2}>
= sin(θ/2)|ψ_{+}> + cos(θ/2)|ψ_{-}>.

At t = 0 we have |ψ(0)> = |n_{1}>.

|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) = |<n_{2}|ψ(t)>|^{2}.

<n_{2}|ψ(t)> = cos(θ/2) sin(θ/2)[exp(-iE_{+}t/ħ) - exp(-iE_{-}t/ħ)].

P_{n}(t) = ¼sin^{2}θ[2 - exp(i(E_{+} - E_{-})t/ħ)
+ exp(i(E_{+} - E_{-})t/ħ)] = ½sin^{2}θ [1 - cos((E_{+}
- E_{-})t/ħ)].

Period of oscillation: 2π/T = (E_{+} - E_{-})/ħ, T =
h/(E_{+} - E_{-}).

Let **B** = B**k**, **μ**∙**B** = μ_{z}B.

Then E_{1} = m + μ_{z}B and E_{2} = m - μ_{z}B,
½(E_{1} + E_{2}) = m, ½(E_{1} - E_{2}) = μ_{z}B.

tanθ = α/μ_{z}B, 1/cos^{2}θ = 1 + (α/μ_{z}B)^{2},
sin^{2}θ = α^{2}/(μ_{z}^{2}B^{2} + α^{2}).

E_{±} = m ± (μ_{z}^{2}B^{2} + α^{2})^{1/2},
E_{+} - E_{-} = 2(μ_{z}^{2}B^{2} + α^{2})^{1/2}.

P_{n}(t) = ½[α^{2}/(μ_{z}^{2}B^{2} + α^{2})] [1 -
cos(2(μ_{z}^{2}B^{2} + α^{2})^{1/2}t/ħ)].

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