(a)
Radioactive decay can produce neutrinos of either of two varieties, called ν_{e} and
ν_{μ}.

There is
considerable interest in the possibility that neutrinos have a small mass m_{1},
m_{2}.

Suppose that |ν_{e}> and |ν_{μ}>are linear combinations of the
ortho-ormal mass (or energy)

eigenstates |ν_{1}> and |ν_{1}>.

|ν_{μ}> = cosθ|ν_{1}> + sinθ|ν_{2}>, |ν_{e}>
= -sinθ|ν_{1}> + cosθ|ν_{2}>.

A nuclear reactor emits neutrinos of type ν_{e}.

A distance L away a detector is able to record the passage of ν_{e}.

Assume that m_{1,2}c^{2} << E_{1,2},
so that E = Pc + m^{2}c^{3}/2P. Show that the intensity of
ν_{e} at L relative to that at the source is

I(L) = 1 - sin^{2}(2θ) sin^{2}(c^{3}∆m^{2}L/(4ħE))
= 1 - sin^{2}(2θ) sin^{2}(1.27 L(m) ∆m^{2}c^{4}
(eV^{2})/E(MeV)).

Here ∆m^{2} = (m_{1}^{2} - m_{2}^{2}).

(b) Suppose that ∆m^{2}c^{4} = 1 eV^{2} and that a high energy
accelerator is emitting neutrinos of energy 1 GeV.

What is the optimum distance to
place the detector to observe the effect of oscillations?

(c) The accelerator energy is reduced such that neutrinos emerge with only 1/45 of the
energy that they had previously.

How does the optimal distance change?

Solution:

(a) Equating mass and energy we assume that the mass eigenstates are the eigenstates of
the Hamiltonian, which we assume does not explicitly depend on time.

H|ν_{1}> = E_{1}|ν_{1}>, H|ν_{2}> = E_{2}|ν_{2}>.

U(t,t_{0}) = exp(-iH(t - t_{0})/ħ) is the evolution operator.

Therefore |ν_{1}(t)> = exp(-iE_{1}t/ħ)|ν_{1}(0)>,
|ν_{2}(t)> = exp(-iE_{2}t/ħ)|ν_{2}(0)>.

A second operator, which we denote by A,
has eigenstates |ν_{e}>
and |ν_{μ}>.
It is the operator which
distinguishes between the two varieties of neutrinos.

What is the probability that at time t a measurement of the observable A
yields a neutrino of type ν_{e}?

P_{e}(t) = |<ν_{e}|Ψ(t)>|^{2}, with |Ψ(0)>
= |ν_{e}> = -sinθ|ν_{1}> + cosθ|ν_{2}>.

|Ψ(t)> = -sinθ exp(-iE_{1}t/ħ)|ν_{1}> + cosθ exp(-iE_{2}t/ħ)|ν_{2}>.

<ν_{e}|Ψ(t)> = (-sinθ<ν_{1}| + cosθ<ν_{2}|)(-sinθ
exp(-iE_{1}t/ħ)|ν_{1}> + cosθ exp(-iE_{2}t/ħ)|ν_{2}>)

= sin^{2}θ exp(-iE_{1}t/ħ) + cos^{2}θ exp(-iE_{2}t/ħ).

|<ν_{e}|Ψ(t)>|^{2} = sin^{4}θ + cos^{4}θ
+ 2 sin^{2}θ cos^{2}θ cos((E_{1} - E_{2})t/ħ)

= (3 + 4cos2θ + cos4θ)/8 + (3 - 4cos2θ + cos4θ)/8

+ ½(1 + cos2θ)(1 - cos2θ)cos((E_{1} - E_{2})t/ħ)

= 6/8 + (2/8)cos4θ + ½(1 - cos^{2}2θ) cos((E_{1} - E_{2})t/ħ)

= 6/8 + (2/8)(1 - 2sin^{2}2θ) + ½sin^{2}2θ cos((E_{1}
- E_{2})t/ħ)

= 1 - ½sin^{2}2θ + ½sin^{2}2θ cos((E_{1} - E_{2})t/ħ)

= 1 - sin^{2}2θ[½ + ½cos((E_{1} - E_{2})t/ħ)]

= 1 - sin^{2}2θ)sin^{2}(E_{1} - E_{2})t/(2ħ)).

We have

E = Pc + m^{2}c^{3}/2P. [E = (p^{2}c^{2}
+ m^{2}c^{4})^{1/2} ≈ pc(1 + m^{2}c^{2}/2p^{2})]

E_{1} - E_{2} = (m_{1}^{2} - m_{2}^{2})c^{3}/(2P)
=^{ } ∆m^{2}c^{3}/(2P).

For t we have t = L/v ≈ L/c.

Therefore

(E_{1} - E_{2})t/(2ħ) = ∆m^{2}c^{3}L/(4ħPc)
≈ ∆m^{2}c^{3}L/(4ħE) since E ≈ Pc.

(E_{1} - E_{2})t/(2ħ) = ∆m^{2}c^{4}(eV^{2})
(10^{-6} MeV/eV) L(m)/(4E(MeV)(6.6*10^{-16} eV s)(3*10^{8}
m/s)

= 1.27 L(m) ∆m^{2}c^{4} (eV^{2})/E(MeV).

(b)
sin^{2}(1.27 L ∆m^{2}c^{4}/E) varies between 0 and 1.
To see
oscillations there is no optimum distance. One must map out the number of
ν_{e} as a function of distance over several periods.

The
first minimum in the number of ν_{e} occurs when
sin^{2}(1.27 L ∆m^{2}c^{4}/E) =
1 for
the first time,
i.e. when 1.27 L ∆m^{2}c^{4}/E = π/2 or when L = πE/(2*1.27
∆m^{2}c^{4}) = 1.24 km.

(c) L = 1.24 km/ 45 = 27 m.

**Problem:**

For the infinite well shown, the wave function for a particle of mass m,
at t = 0, is

ψ(x, t = 0) = (2/a)^{1/2}sin(3πx/a).

(a) Is ψ(x, t = 0) an eigenfunction
of the Hamiltonian?

(b) Calculate <X>, <P_{x}>, and <H> at t = 0.

Solution:

- Concepts:

The eigenfunctions of the infinite square well, the postulates of quantum mechanics. - Reasoning:

We recognize that the given wave function is an eigenfunction of the infinite square well. The mean value of any observable A is <ψ|A|ψ>. - Details of the calculation:

(a) ψ(x, t = 0) = (2/a)^{1/2}sin(3πx/a) is not an eigenfunction of the Hamiltonian.

(b) <x> = (2/a)∫_{0}^{a}dx x sin^{2}(3πx/a) = a/2.

<p_{x}> = (2/a)∫_{0}^{a}dx sin(3πx/a) (ħ/i) ∂sin(3πx/a)/∂x = C ∫_{0}^{a}dx sin(3πx/a) cos(3πx/a) = 0.

<H> = <T> + <U>.

ψ is an eigenfunction of the infinite square well with U = 0 for 0 < x < a.

For the infinite square well we have <H> = E = <T>.

Therefore <T> = n^{2}π^{2}ħ^{2}/(2ma^{2}) = 9π^{2}ħ^{2}/(2ma^{2}).

<U> = ∫_{0}^{a }ψUψ dx = ∫_{0}^{2a/3 }ψU_{0}ψ dx = (2/a)∫_{0}^{2a/3}dx U_{0 }sin^{2}(3πx/a) = 2U_{0}/3.

<H> = 9π^{2}ħ^{2}/(2ma^{2}) + 2U_{0}/3.

**Problem:**

A particle of mass m is inside a one-dimensional infinite well with
walls a distance L apart. One of the walls is suddenly moved by a
distance L so that the wall separation becomes 2L. The wall moves
so suddenly that the particle wave function has no time to change during the
motion. Suppose that the particle is originally in the ground state ψ^{0}.

(a) What is its energy E_{0} and wave function ψ^{0} before the width is doubled?

(b) What are the energy eigenvalues **after** the width is
doubled?

(c) If we measure the energy** after** the width is doubled, what
is the probability that it will not have changed?

(d) If we measure the energy **after** the width is doubled, what
is the probability that the particle will have lost some energy?

(e) What is the expectation value of the energy before and after the doubling
of the width?

Solution:

- Concepts:

A particle in an infinite square well, 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:

For a particle in an infinite well we have

ψ_{n}(x) = (2/L)^{1/2}sin(nπx/L), E_{n}= n^{2}π^{2}ħ^{2}/(2mL^{2}).

(a) ψ^{0}(x) = ψ_{1}(x) = (2/L)^{1/2}sin(πx/L), E_{0}= π^{2}ħ^{2}/(2mL^{2}). Note: the superscript 0 here stands for the ground state, not for the quantum number n.

(b) E_{n}' = n^{2}π^{2}ħ^{2}/(2mL'^{2}) = n^{2}π^{2}ħ^{2}/(8mL^{2}).

(c) For the energy to not have changed the particle must be found in the first excited state of the new potential (n = 2).

P = |<ψ_{f,2}|ψ_{i,1}>|^{2}= |(1/L)^{1/2}(2/L)^{1/2}∫_{0}^{L}dx sin(2πx/(2L))sin(πx/L)|^{2}= 1/2.

(We integrate only from 0 to L since ψ_{i(n=1) }= 0 for x > L.)

(d) If we find the particle in the state n = 1 of the new potential then the particle has lost energy.

(If we do not find it in n = 1 then it has not lost energy.)

P = |<ψ_{f,1}|ψ_{i,1}>|^{2}= |(1/L)^{1/2}(2/L)^{1/2}∫_{0}^{L}dx sin(πx/(2L))sin(πx/L)|^{2}

= (2/L^{2})[(2L/π)]∫_{0}^{π/2}dx' sin(x')sin(2x')|^{2}= (8/π^{2})[(sin(x)/2 - sin(3x)/6))|_{0}^{π/2}]^{2}

= (8/π^{2})(4/6)^{2}= 0.36.

(e) <E> = <H> = ∫_{0}^{L}dx ψ_{i,1}(x) (p^{2}/(2m)) ψ_{i,1}(x) = E_{0}before and after the doubling of the width.

Assume
the operator A commutes with the Hamiltonian H of a conservative physical
system.

Prove that in any state |ψ(t)> the probability of observing the eigenvalue a_{0}
is independent of time.

Solution:

If A commutes with H we can find a common eigenbasis of A
and H. Let {|E_{n}^{i},a_{m}^{j}> }
denote a common orthonormal eigenbasis of A and H for the state space E.

H |E_{n}^{i},a_{m}^{j}> = E_{n} |E_{n}^{i},a_{m}^{j}>, A |E_{n}^{i},a_{m}^{j}>
= a_{m} |E_{n}^{i},a_{m}^{j}>.

Here i and j the degeneracy of the eigenvalues.

|ψ(0)> may be expanded in terms of the basis
vectors.

|ψ(0)> = ∑_{ni,mj }C_{ni,mj} |E_{n}^{i},a_{m}^{j}>

Then

|ψ(t)> = ∑_{ni,mj} C_{ni,mj} |E_{n}^{i},a_{m}^{j}>
exp(-iE_{n}t/ħ).

The probability of observing the eigenvalue a_{0} at time t is given by

P_{a0}(t) = ∑_{n'i',j'} <E_{n'}^{i'},a_{0}^{j'}|ψ(t)>|^{2
}The sum is over all the quantum numbers associated with eigenvectors
that have eigenvalue a_{0}.

<E_{n'}^{i'},a_{0}^{j'}|ψ(t)> = ∑_{ni,mj} C_{ni,mj} <E_{n'}^{i'},a_{0}^{j'}|E_{n}^{i},a_{m}^{j}>
exp(-iE_{n}t/ħ

= C_{n'i',0j'}
exp(-iE_{n'}t/ħ

|<E_{n'}^{i'},a_{0}^{j'}|ψ(t)>|^{2}
= |C_{n'i',0j'}|^{2}

P_{a0}(t)
= ∑_{n'i',j'}|C_{n'i',0j'}|^{2},

which is independent of time.

Consider
a one-dimensional problem. Let the translation operator T(a) describe the operation
T(a)ψ(x) = ψ(x + a), where a is a
constant displacement.

(a) Show that this operator commutes with the Hamiltonian, H = -(ħ^{2}/(2m))∂^{2}/∂x^{2}
+ U(x), if the potential
has the periodic property U(x) = U(x + a).

(b) Let ψ(x) be an eigenstate of T(a) with eigenvalue c.
Show
that c is a constant of motion.

Solution:

(a) ψ(x + a) = ∑_{n}(a^{n}/n!)d^{n}ψ(x)/dx^{n}
= T(a)ψ(x) (Taylor series
expansion).

T(a) = ∑_{n}(a^{n}/n!)d^{n}/dx^{n} = exp(a d/dx)
= exp((ia/ħ) (ħ/i)d/dx) = exp(iap/ħ),

since p = (ħ/i)d/dx.

H = (p^{2}/(2m)) + U(x).

[T(a),H] = (2m)^{-1}[T(a),p^{2}] + [T(a),U(x)].

[T(a),p^{2}] = [exp(iap/ħ),p^{2}] = ∑_{n}(1/n!)(ia/ħ)^{n}[p^{n},p^{2}]
= 0.

For any ψ(x) we have

[T(a),U(x)]ψ(x) = T(a)(U(x)ψ(x)) - U(x)(T(a)ψ(x))

= U(x + a)ψ(x + a) - U(x)ψ(x + a) = 0,

since U(x + a) = U(x).

Thus [T(a),H] = 0.

(b) T(a)ψ(x) = cψ(x).
Since T(a)
does not explicitly depend on time we have

(d/dt)<T(a)> = (iħ)^{-1}<[T(a),H]> = 0

Since ψ(x) is an eigenstate of T(a)
we have <T(a)> = c.

Therefore dc/dt = 0, c is a constant
of motion.

ψ(x) represents the projection of |ψ> onto the basis vector |x>.

What does T(a)
do to the basis vectors {|x>} ?

T(a)ψ(x) = ψ(x + a), T(a)<x|ψ> = <x + a|ψ>,

exp(a d/dx) <x|ψ> = <x|exp(iap/ħ)|ψ>

<x|exp(iap/ħ) = <x + a|, exp(-iap/ħ)|x> = |x + a>, exp(iap/ħ)|x> =
|x - a>,