Problem:

(a) Radioactive decay can produce neutrinos of either of two varieties, called νe and νμ.
There is considerable interest in the possibility that neutrinos have a small mass m1, m2.
Suppose that |νe> and |νμ>are linear combinations of the ortho-ormal mass (or energy)
eigenstates |ν1> and |ν1>.
μ> = cosθ|ν1> + sinθ|ν2>,  |νe> = -sinθ|ν1> + cosθ|ν2>.
A nuclear reactor emits neutrinos of type νe.
A distance L away a detector is able to record the passage of νe
Assume that m1,2c2 << E1,2, so that E = Pc + m2c3/2P.  Show that the intensity of νe at L relative to that at the source is
I(L) = 1 - sin2(2θ) sin2(c3∆m2L/(4ħE)) = 1 - sin2(2θ) sin2(1.27 L(m) ∆m2c4 (eV2)/E(MeV)).
Here  ∆m2 = (m12 - m22).

(b) Suppose that ∆m2c4 = 1 eV2 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|ν1> = E11>,  H|ν2> = E22>.
U(t,t0) = exp(-iH(t - t0)/ħ) is the evolution operator.
Therefore |ν1(t)> = exp(-iE1t/ħ)|ν1(0)>,  |ν2(t)> = exp(-iE2t/ħ)|ν2(0)>.
A second operator, which we denote by A, has eigenstates |νe>  and |νμ>.  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 νe
Pe(t) = |<νe|Ψ(t)>|2, with |Ψ(0)> = |νe> = -sinθ|ν1> + cosθ|ν2>.
|Ψ(t)> = -sinθ exp(-iE1t/ħ)|ν1> + cosθ exp(-iE2t/ħ)|ν2>.
e|Ψ(t)> = (-sinθ<ν1| + cosθ<ν2|)(-sinθ exp(-iE1t/ħ)|ν1> + cosθ exp(-iE2t/ħ)|ν2>)
= sin2θ exp(-iE1t/ħ) + cos2θ exp(-iE2t/ħ).
|<νe|Ψ(t)>|2 = sin4θ + cos4θ + 2 sin2θ cos2θ cos((E1 - E2)t/ħ)
= (3 + 4cos2θ + cos4θ)/8 + (3 - 4cos2θ + cos4θ)/8
+ ½(1 + cos2θ)(1 - cos2θ)cos((E1 - E2)t/ħ)
= 6/8 + (2/8)cos4θ + ½(1 - cos22θ) cos((E1 - E2)t/ħ)
= 6/8 + (2/8)(1 - 2sin22θ) + ½sin22θ cos((E1 - E2)t/ħ)
= 1 - ½sin22θ + ½sin22θ cos((E1 - E2)t/ħ)
= 1 - sin22θ[½ + ½cos((E1 - E2)t/ħ)]
= 1 - sin22θ)sin2(E1 - E2)t/(2ħ)).
We have
E = Pc + m2c3/2P.    [E = (p2c2 + m2c4)1/2 ≈ pc(1 + m2c2/2p2)]
E1 - E2 = (m12 - m22)c3/(2P) =  ∆m2c3/(2P).
For t we have t = L/v ≈ L/c.
Therefore
(E1 - E2)t/(2ħ) = ∆m2c3L/(4ħPc) ≈ ∆m2c3L/(4ħE) since E ≈ Pc.
(E1 - E2)t/(2ħ) = ∆m2c4(eV2) (10-6 MeV/eV) L(m)/(4E(MeV)(6.6*10-16 eV s)(3*108 m/s)
= 1.27 L(m) ∆m2c4 (eV2)/E(MeV).

(b)  sin2(1.27 L ∆m2c4/E) varies between 0 and 1.  To see oscillations there is no optimum distance.  One must map out the number of νe as a function of distance over several periods.
The first minimum in the number of νe occurs when  sin2(1.27 L ∆m2c4/E)  = 1 for the first time, i.e. when 1.27 L ∆m2c4/E = π/2 or when L = πE/(2*1.27 ∆m2c4) = 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)1/2sin(3πx/a).
(a)  Is ψ(x, t = 0) an eigenfunction of the Hamiltonian?
(b)  Calculate <X>, <Px>, 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)1/2sin(3πx/a) is not an eigenfunction of the Hamiltonian.
(b)  <x> = (2/a)∫0adx  x sin2(3πx/a)  = a/2.
<px> = (2/a)∫0adx sin(3πx/a) (ħ/i) ∂sin(3πx/a)/∂x = C ∫0adx 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> = n2π2ħ2/(2ma2) =  9π2ħ2/(2ma2).
<U> = ∫0a ψUψ dx = ∫02a/3 ψU0ψ dx = (2/a)∫02a/3dx U0 sin2(3πx/a) = 2U0/3.
<H> =  9π2ħ2/(2ma2) + 2U0/3.

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 ψ0.
(a)  What is its energy E0 and wave function ψ0 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(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:
For a particle in an infinite well we have
ψn(x) = (2/L)1/2sin(nπx/L), En = n2π2ħ2/(2mL2).
(a)  ψ0(x) = ψ1(x) = (2/L)1/2sin(πx/L), E0 = π2ħ2/(2mL2).  Note: the superscript 0 here stands for the ground state, not for the quantum number n.
(b)  En' = n2π2ħ2/(2mL'2) = n2π2ħ2/(8mL2).
(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 = |<ψf,2i,1>|2 = |(1/L)1/2(2/L)1/20Ldx sin(2πx/(2L))sin(πx/L)|2 = 1/2.
(We integrate only from 0 to L since ψi(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 = |<ψf,1i,1>|2 = |(1/L)1/2(2/L)1/20Ldx sin(πx/(2L))sin(πx/L)|2
= (2/L2)[(2L/π)]∫0π/2dx' sin(x')sin(2x')|2 = (8/π2)[(sin(x)/2 - sin(3x)/6))|0π/2]2
= (8/π2)(4/6)2 = 0.36.
(e)  <E> = <H> = ∫0Ldx ψi,1(x) (p2/(2m)) ψi,1(x) = E0 before and after the doubling of the width.
Problem:

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 a0 is independent of time.

Solution:
If A commutes with H we can find a common eigenbasis of A and H.  Let {|Eni,amj> } denote a common orthonormal eigenbasis of A and H for the state space E.
H |Eni,amj> = En |Eni,amj>,  A |Eni,amj> = am |Eni,amj>.
Here i and j the degeneracy of the eigenvalues.
|ψ(0)> may be expanded in terms of the basis vectors.
|ψ(0)> = ∑ni,mj Cni,mj |Eni,amj>
Then
|ψ(t)> = ∑ni,mj Cni,mj |Eni,amj> exp(-iEnt/ħ).
The probability of observing the eigenvalue a0 at time t is given by
Pa0(t) = ∑n'i',j' <En'i',a0j'|ψ(t)>|2
The sum is over all the quantum numbers associated with eigenvectors that have eigenvalue a0.
<En'i',a0j'|ψ(t)> = ∑ni,mj Cni,mj <En'i',a0j'|Eni,amj> exp(-iEnt/ħ
= Cn'i',0j'  exp(-iEn't/ħ
|<En'i',a0j'|ψ(t)>|2 = |Cn'i',0j'|2
Pa0(t) = ∑n'i',j'|Cn'i',0j'|2,
which is independent of time.

Problem:

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 = -(ħ2/(2m))∂2/∂x2 + 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(an/n!)dnψ(x)/dxn = T(a)ψ(x)     (Taylor series expansion).
T(a) = ∑n(an/n!)dn/dxn = exp(a d/dx) = exp((ia/ħ) (ħ/i)d/dx) = exp(iap/ħ),
since p = (ħ/i)d/dx.
H = (p2/(2m)) + U(x).
[T(a),H] = (2m)-1[T(a),p2] + [T(a),U(x)].
[T(a),p2] = [exp(iap/ħ),p2] = ∑n(1/n!)(ia/ħ)n[pn,p2] = 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>,