- #1

isochore

- 12

- 1

- Homework Statement
- Prove that ##\frac{d^2}{dt^2} \langle \psi (t) \vert X^2 \vert \psi (t) \rangle = 4 \omega^2 \langle \psi (t) \vert X^2 \vert \psi (t) \rangle + \frac{1}{m}\langle \psi (t) \vert H \vert \psi (t) \rangle## for the simple harmonic oscillator.

- Relevant Equations
- ##\frac{i}{\hbar}[H, X] = \frac{1}{m} P##

##\frac{i}{\hbar}[H, P] = -m\omega^2X##

##\frac{d}{dt} \langle \psi (t) \vert A \vert \psi (t) \rangle = \frac{i}{\hbar} \langle \psi (t) \vert [H,A] \vert \psi (t) \rangle## if A is time-independent.

I can show that ##\frac{d}{dt} \langle \psi (t) \vert X^2 \vert \psi (t) \rangle = \frac{1}{m} \langle \psi (t) \vert PX+XP \vert \psi (t) \rangle##.

Taking another derivative with respect to time of this, I get ##\frac{d^2}{dt^2} \langle \psi (t) \vert X^2 \vert \psi (t) \rangle = \frac{i}{m \hbar} \langle \psi (t) \vert [H, PX+XP] \vert \psi (t) \rangle##.

However, I'm not sure how to manipulate this into something that matches the right side of the equation.

I can expand the commutator in the middle to get

##[H, PX + XP] = [H, PX] + [H, XP] = [H, P]X + P[H, X] + [H, X]P + X[H,P]##

I can then substitute in the commutator values known from above (ignoring the factor of ##\frac{i}{\hbar}## momentarily):

##[H, PX + XP] = (-m\omega^2 X)X + P(\frac{1}{m}P) + (\frac{1}{m}P)P + X(-m\omega^2 X) = -2m\omega^2 X^2 + \frac{2}{m}P^2##

I see a way to get from ##P^2## to ##H## using the time-independent Schrödinger equation, but I took derivatives with respect to time to arrive at this equation. Is there another path besides using the time-independent Schrödinger equation, or is it okay to use the time-independent form here?

Taking another derivative with respect to time of this, I get ##\frac{d^2}{dt^2} \langle \psi (t) \vert X^2 \vert \psi (t) \rangle = \frac{i}{m \hbar} \langle \psi (t) \vert [H, PX+XP] \vert \psi (t) \rangle##.

However, I'm not sure how to manipulate this into something that matches the right side of the equation.

I can expand the commutator in the middle to get

##[H, PX + XP] = [H, PX] + [H, XP] = [H, P]X + P[H, X] + [H, X]P + X[H,P]##

I can then substitute in the commutator values known from above (ignoring the factor of ##\frac{i}{\hbar}## momentarily):

##[H, PX + XP] = (-m\omega^2 X)X + P(\frac{1}{m}P) + (\frac{1}{m}P)P + X(-m\omega^2 X) = -2m\omega^2 X^2 + \frac{2}{m}P^2##

I see a way to get from ##P^2## to ##H## using the time-independent Schrödinger equation, but I took derivatives with respect to time to arrive at this equation. Is there another path besides using the time-independent Schrödinger equation, or is it okay to use the time-independent form here?

Last edited: