Problems Chapter 7

Quantum Mechanics
                       K. Konishi, G. Paffuti

Problem

Solve the Heisenberg equations of motion for a free particle and for a particle in an external uniform field.

Solution

The Heisenberg equations for

H = ((p^2)/(2m)),

read

(d/dt)(Subscript[q, H])[t] =- (( i)/ℏ)[(Subscript[q, H])[t],H] = (((Subscript[p, H])[t])/m);   (d/dt)(Subscript[p, H])[t] =- (( i)/ℏ)[(Subscript[p, H])[t],H]=0 .

From the second equation we get (Subscript[p, H])[t]=(Subscript[p, H])[0]= p. The solution of the first equation is

(Subscript[q, H])[t]= q + (p/m)t.

For a uniform external field

H = ((p^2)/(2m))- F q.

The Heisenberg equations are

(d/dt)(Subscript[q, H])[t] =- (( i)/ℏ)[(Subscript[q, H])[t],H] = (((Subscript[p, H])[t])/m);   (d/dt)(Subscript[p, H])[t] =- (( i)/ℏ)[(Subscript[p, H])[t],H]=F .

The second equation has solution

(Subscript[p, H])[t] = p + F t.

By substitution in the first equation and time integration we find

(Subscript[q, H])[t] = q + (p/m)t + (1/2)(F/m)(t^2).

Problem

Solve the Heisenberg equations of motion for a harmonic oscillator in a uniform constant external field. Generalize the solution for a uniform time dependent force F[t].

Solution

The Hamiltonian is

H = ((p^2)/(2m))+ (1/2)m (ω^2)(q^2)- F q.

The Heisenberg equations of motion

(d/dt)(Subscript[q, H])[t] =- (( i)/ℏ)[(Subscript[q, H])[t],H] = (((Subscript[p, H])[t])/m);    (d/dt)(Subscript[p, H])[t] =- (( i)/ℏ)[(Subscript[p, H])[t],H]=- m (ω^2)q + F .

The change of variables

q = Q + (F/(m (ω^2)));   p = P;

is a canonical transformation which leaves commutators invariant and transform the equations in

(d/dt)(Subscript[Q, H])[t]=(((Subscript[P, H])[t])/m) ;  (d/dt)(Subscript[P, H])[t] =- m (ω^2) (Subscript[Q, H])[t].

These are the usual harmonic oscillator equation, the solution has been found in the text and is easily checked to be

(Subscript[Q, H])[t] = Q[0] Cos[ω t] + ((P[0])/(m ω))Sin[ω t];   (Subscript[P, H])[t]= P[0]Cos[ω t] - m ω Q[0] Sin[ω t].

In terms of p and q :

(Subscript[q, H])[t] = (F/(m (ω^2)))+(q- (F/(m (ω^2)))) Cos[ω t] + (p/(m ω))Sin[ω t];  
 (Subscript[P, H])[t]= p Cos[ω t] - m ω (q- (F/(m (ω^2)))) Sin[ω t].

The procedure can be generalized to the case of a time dependent force. The equations are the same:

Let us make the canonical (unitary) transformation

q = Q + (Subscript[f, q])[t];   p = P + (Subscript[f, p])[t];

The equations of motion become

(d/dt)(Subscript[Q, H])[t]=(((Subscript[P, H])[t])/m)  + ((d/dt)(Subscript[f, q]) - (1/m)(Subscript[f, p]));

(d/dt)(Subscript[P, H])[t] =- m (ω^2) (Subscript[Q, H])[t] + ((d/dt)(Subscript[f, p]) + m (ω^2)(Subscript[f, q]) - F[t])

If we choose (Subscript[f, p])and (Subscript[f, q]) as a particular solution of the classical equations of motion, the equations for P and Q reduce to the usual harmonic oscillator equations, then we can write at once the solution for q and p:

(Subscript[q, H])[t] = (Subscript[f, q])[t]+(q- (Subscript[f, q])[t]) Cos[ω t] + (((p - (Subscript[f, p])[t]))/(m ω))Sin[ω t];  
 (Subscript[P, H])[t]= (Subscript[f, p])[t]+(p - (Subscript[f, p])[t]) Cos[ω t] - m ω (q- (Subscript[f, q])[t]) Sin[ω t].

A particular solution of the classical equations of motion can be easily find using the method of  "variation of constants". Write

(Subscript[f, q])[t] = α[t] Cos[ω t] + ((β[t])/(m ω))Sin[ω t];   (Subscript[f, p])[t]= β[t]Cos[ω t] - m ω α[t] Sin[ω t];

Substitution in the equation of motion give

((dα)/dt)= -(( dβ)/dt) ((Sin[ω t])/(m ω)) ;     (( dβ)/dt) Cos[ω t] - m ω ((dα)/dt) Sin[ω t] = F[t].

Substituting in the second equation the first one get, as  a particular solution (with β[0] = 0):

β[t] = (∫_0^t)F[τ] Cos[ω τ]dτ

and from the first equation:

α[t] = -(1/(m ω))(∫_0^t)F[τ] Sin[ω τ]dτ.

Problem

Suppose that the system described by the wave function (Subscript[ψ, S]) at the instant t=0  is an eigenstate of the operator f, with eigenvalue,  (Subscript[f, 0]).  Show that the wave function at time t  is an eigenstate of the Heisenberg  operator (Subscript[f, H])[-t], with the same eigenvalue.

Solution

We have

(Subscript[f, H])[-t]ψ[t] = (e^(-i H t))f[0](e^(i H t)) (e^(-i H t))ψ[0]= (e^(-i H t))f[0]ψ[0]= (Subscript[f, 0]) (e^(-i H t))ψ[0]= (Subscript[f, 0])ψ[t].