Problems Chapter 2

Quantum Mechanics
                      K. Konishi, G. Paffuti

Problem

(δ^3)(r-(Subscript[r, 0]))= (1/(r^2))δ(r-(Subscript[r, 0]))δ(Cos[θ]-Cos[(Subscript[θ, 0])])δ(φ-(Subscript[φ, 0])).

⟨ A (SuperDagger[A]) ⟩ ≥ 0.

Exp[ζ A] B Exp[-ζ A] = B + c ζ.

Solution

The proof of the points 1 - 4 is left to the reader.

5

By definition of the adjoint (we assume that there are no problems related to the domains):

⟨ψ| A (SuperDagger[A])|ψ⟩ = ⟨A ψ| A |ψ⟩ = |A|ψ⟩((|)^2) ≥ 0 .

6

Let O[ζ] be given by

O[ζ] = Exp[ζ A] B Exp[-ζ A] .

A differentiation with respect to ζ gives:

(d/(dζ))O[ζ]= A (e^(ζ A)) B (e^(-ζ A)) - (e^(ζ A)) B Exp (e^(-ζ A)) A= (e^(ζ A)) (A B - B A)(e^(-ζ A)) = (e^(ζ A)) c (e^(-ζ A)) = c.

Integrating this from 0 to ζ

O[ζ] - O[0] = c ζ  ⇒ O[ζ] = B + c ζ.

Problem

Consider a two-state system  with H = (({{(Subscript[E, 0]), -η}, {-η, (Subscript[E, 0])}})).

Solution

1

No. η must be real for the Hamiltonian to be hermitian.

Let us note that in the general case

H = (({{(Subscript[E, 0]), β}, {(SuperStar[β]), (Subscript[E, 0])}})),

the off diagonal matrix elements can be made real by a redefinition of the phase of vector |2⟩ (or |1⟩ )

|2⟩ → ±|2'⟩(e^(i Arg[β])) ⇒  ⟨1|H|2'⟩ = ±⟨1|H|2⟩(e^(-i Arg[β]))= ±|β|.

2

|a⟩ = (1/(Sqrt[2]))(|1⟩+ |2⟩);  eig. (Subscript[E, 0])-η;     |b⟩ = (1/(Sqrt[2]))(|1⟩- |2⟩);  eig. (Subscript[E, 0])+η;

If η > 0, |a⟩ is the ground state and |b⟩ the excited state.

3

We have, inverting previous relations

|1⟩ = (1/(Sqrt[2]))(|a⟩+ |b⟩);   |2⟩ = (1/(Sqrt[2]))(|a⟩- |b⟩);

Then

|ψ[0]⟩ = (1/2)((|a⟩+ |b⟩) + i (|a⟩- |b⟩)) = (1/2)((1+i)|a⟩ + (1-i) |b⟩).

The time evolution is

|ψ[t]⟩ = (e^(-i (Subscript[E, 0])t/ℏ)) (1/2){(1+i)|a⟩ (e^(-i ηt/ℏ))+ (1-i) (e^(i ηt/ℏ))|b⟩}= 
                 =  (e^(-i (Subscript[E, 0])t/ℏ)) (1/(2(Sqrt[2]))){(1+i)(|1⟩+ |2⟩) (e^(-i ηt/ℏ))+ (1-i) (e^(i ηt/ℏ))(|1⟩- |2⟩)}= 
                  = ((e^(-i (Subscript[E, 0])t/ℏ))/(Sqrt[2])){ (Cos[((η t)/ℏ)]+ Sin[((η t)/ℏ))|1⟩ + i(Cos[((η t)/ℏ)]- Sin[((η t)/ℏ))|2⟩}.

Then the probabilities are

(((((Subscript[P, 1])[t] = (1/2)(Cos[((η t)/ℏ)]+ Sin[((η t)/ℏ)))^2)= ((1+ Sin[((2 η t)/ℏ)])/2);   (Subscript[P, 2])[t] = (1/2)(Cos[((η t)/ℏ)]- Sin[((η t)/ℏ)))^2)= ((1- Sin[((2 η t)/ℏ)])/2).

Problem

A two-level system is described by the Hamiltonian,

H = (Subscript[E, 0])(({{1, 0}, {0, -1}}))

We consider in this system the measurement of two variables described by the operators A and B,

A =(({{0, -i}, {i, 0}}));  B = (({{2, -(Sqrt[2])i}, {(Sqrt[2])i, 1}})).

Solution

1

The eigenvalues and eigenvectors of A are:

eig. -1: (Subscript[vA, 1])  ={(i/(Sqrt[2])),(1/(Sqrt[2]))},  eig.+1:  (Subscript[vA, 2])= {-(i/(Sqrt[2])),(1/(Sqrt[2]))} .

The eigenvalues and eigenvectors of B are:

eig. 3: (Subscript[vB, 1])= {-i (Sqrt[(2/3)]),(1/(Sqrt[3]))},  eig.0: (Subscript[vB, 2])= {(i/(Sqrt[3])),(Sqrt[(2/3)])} .

2

The commutator of A and B is

[A,B] = (({{0, i}, {i, 0}})).

The operators are not compatible. They are not conserved, as

[A,H]= (Subscript[E, 0])(({{i , i }, {i , -i }})) ;     [B, H] = (Subscript[E, 0])(({{i (Sqrt[2]) , i (Sqrt[2]) }, {1+i (Sqrt[2]) , -i (Sqrt[2]) }})).

3

The possible results for a measurement of B are 3 and 0, the eigenvalues of B. If the measurement of A has given 1, the state just after that is (Subscript[vA, 2]). In this state

P[3,(Subscript[vA, 2])]= |⟨(Subscript[vB, 1])|(Subscript[vA, 2])⟩((|)^2)= (1/6) (3+2 (Sqrt[2]));    P[0,(Subscript[vA, 2])]= |⟨(Subscript[vB, 1])|(Subscript[vA, 2])⟩((|)^2)= (1/6) (3-2 (Sqrt[2])).

The sum of probabilities is 1, as it should be.

4

(Subscript[vA, 2])[t] = {-(i/(Sqrt[2]))(e^(-i (Subscript[E, 0])t/ℏ)),(1/(Sqrt[2]))(e^(+i (Subscript[E, 0])t/ℏ))}.

P[A=1,(Subscript[vA, 2])[t]] = |⟨(Subscript[vA, 2])|(Subscript[vA, 2])[t]⟩((|)^2) = ((Cos[((t (Subscript[E, 0]))/ℏ)])^2).

P[A=1,(Subscript[vB, 1])] = |⟨(Subscript[vA, 2])|(Subscript[vB, 1])⟩((|)^2) =(1/6) (3+2 (Sqrt[2])).

P[A=1,(Subscript[vB, 1])]P[3,(Subscript[vA, 2])] =|⟨(Subscript[vA, 2])|(Subscript[vB, 1])⟩((|)^2)  |⟨(Subscript[vB, 1])|(Subscript[vA, 2])⟩((|)^2) =(17/36)+((Sqrt[2])/3).

|⟨(Subscript[vB, 1])|(Subscript[vA, 2])⟩((|)^2)|⟨(Subscript[vA, 2])|(Subscript[vB, 1])⟩((|)^2)+ |⟨(Subscript[vB, 2])|(Subscript[vA, 2])⟩((|)^2)|⟨(Subscript[vA, 2])|(Subscript[vB, 2])⟩((|)^2) = (17/18).

The probability is not 1, this is a crucial point.

If we do not measure B the state ψ obtained after the first measurement of A, i.e. (Subscript[vA, 2]), can always be analyzed in terms of eigenstates of B

probA =|<ψ | ψ >((|)^2)=1= |<ψ |b1><b1 ψ > + <ψ |b2><b2 ψ >((|)^2)

we have interference between amplitudes.

If instead we measure B the interference is destroyed. The measurement of B gives the two results  (exclusive) with relative probabilities

probb1 =|<b1 | ψ >((|)^2);  probb2 =|<b2 | ψ >((|)^2);

The state after measurement is either |b1> or |b2>, with no interference, and the (relative) probability of measuring again A=1 in the two distinct situations is

probA[b1] =|<ψ | b1 >((|)^2);  probA[b2] =|<ψ | b2 >((|)^2);

If we do not distinguish between the two cases the total probability is

probA =probb1*probA[b1]+probb2 * probA[b2] ;

this is the expression computed above.

Problem

A three states system is described by an Hamiltonian

H = (Subscript[H, 0])+(Subscript[H, 1]) = (({{(Subscript[E, 0]), 0, 0}, {0, (Subscript[E, 0]), 0}, {0, 0, (Subscript[E, 0])}})) - (ϵ/(Sqrt[2]))(({{0, 1, 0}, {1, 0, 1}, {0, 1, 0}})).

|1⟩ = (({{1}, {0}, {0}})) ;   |2⟩ = (({{0}, {1}, {0}})) ;  |3⟩ = (({{0}, {0}, {1}})) .

Π = (({{0, 0, 1}, {0, 1, 0}, {1, 0, 0}}));       (Π^2)= 1.

Solution

1

The eigenvalues are

(Subscript[E, 0])-ϵ;  (Subscript[E, 0]); (Subscript[E, 0])+ ϵ .

The eigenvectors

|a⟩ = (({{1/2}, {1/(Sqrt[2])}, {1/2}}))= ((|1⟩+(Sqrt[2])|2⟩+|3⟩)/2); 
|b⟩ = (({{1/(Sqrt[2])}, {0}, {-1/(Sqrt[2])}}))= ((|1⟩-|3⟩)/2); 
|c⟩ = (({{1/2}, {-1/(Sqrt[2])}, {1/2}}))= ((|1⟩-(Sqrt[2])|2⟩+|3⟩)/2);

2

It is easy to check that [H , Π] = 0. The eigenvalues of Π are ± 1 and, as the eigenvalues of H are non degenerate, the eigenstates of H must be also eigenstates of Π. In effect |a⟩and |c⟩ are even states (eigenvalue + 1) and |b⟩is an odd state (eigenvalue -1).

3

Inverting the relations above

|1⟩ =  ((|a⟩+(Sqrt[2])|b⟩+|c⟩)/2); |2⟩ =  ((|a⟩-|c⟩)/2); |3⟩ = ((|a⟩-(Sqrt[2])|b⟩+|c⟩)/2);

4

We have

|ψ[t]⟩ = (e^(-i (Subscript[E, 0])t/ℏ))(((|a⟩(e^(i ϵ t/ℏ))+(Sqrt[2])|b⟩+|c⟩(e^(-i ϵ t/ℏ)))/2)) = 
                  = ((1/2)+(1/2) Cos[((t ϵ)/ℏ)])|1⟩+ ((i  Sin[((t ϵ)/ℏ)])/(Sqrt[2]))|1⟩+ (-(1/2)+(1/2) Cos[((t ϵ)/ℏ)])|3⟩.

(Subscript[P, 2])[t] is given by

(Subscript[P, 2])[t] = (1/2)((Sin[((t ϵ)/ℏ)])^2).

Here is a plot of the probabilities as functions of time.

The Hamiltonian H can be considered as an elementary model for an electron moving in a triangular triatomic molecule. ϵ is the amplitude of tunneling between sites. If ϵ=0 the electron has three equivalent equilibrium positions, with energy (Subscript[E, 0]).

Problem

Let us consider a three states system. Let us suppose that the Hamiltonian has non degenerate levels. In the base of Hamiltonian eigenstates two hermitian operators A and B are given by

A = (({{0, 0, -i a}, {0, 0, 0}, {i a, 0, 0}}));    B = (({{b, 0, 0}, {0, 0, b}, {0, b, 0}}));

with  a, b real and positive.

Solution

1

The eigenvalues of A are ± a, 0 with eigenvectors (we write vectors as row for simplicity):

|A,a⟩= (1/(Sqrt[2]))(1,0,i);  |A,-a⟩= (1/(Sqrt[2]))(1,0,-i); |A,0⟩= (0,1,0).

2

The eigenvalues of B are b, b, -b with eigenvectors

|B,b,1⟩= (1,0,0);  |B,b,2⟩= (1/(Sqrt[2]))(0,1,1); |B,-b⟩=(1/(Sqrt[2])) (0,1,-1) .

3

After the measurement the system is in the state |ψ⟩=|A,a⟩. The probability to measure the value b just after this measurement is

P[b] = |⟨B,b,1|ψ⟩((|)^2) + |⟨B,b,2|ψ⟩((|)^2) = (1/2) + (1/4)= (3/4).

The probability to find - b is

P[-b] = |⟨B,-b|ψ⟩((|)^2)  =  (1/4).

4

At a generic time t the evolution of the state ψ is

ψ[t] = (1/(Sqrt[2]))((e^(-i (Subscript[E, 1])t/ℏ)),0,i (e^(-i (Subscript[E, 3])t/ℏ))).

The probability to find b is always the same

P[b,t] = |⟨B,b,1|ψ[t]⟩((|)^2) + |⟨B,b,2|ψ[t]⟩((|)^2) = (3/4).

This can be best understood by noting that the projection of ψ[t] on  |B,-b⟩changes just by a phase, then the probability P[-b] is constant, so must be constant also P[b].