Heisenberg Picture for 1D Simple Harmonic Oscillator

The Halmiltonian for 1D simple harmonic oscillator is

$$H = \frac{1}{2m}(P^2 + m^2 \omega^2 X^2).$$

Show that in the Heisenberg picture, the sum of expectation

$$\langle X_{t+\pi/2\omega}^2 \rangle + \langle X_t^2 \rangle$$

is constant.

Sorry I am an amateur and I have not taken any formal courses in quamtum mechanics. I am completely stuck.

• Use the equation of motion for the Heisenberg picture operators to find $\frac{d}{dt} X = \frac{i}{\hbar} [H,X] = \frac{i}{\hbar} \frac{1}{2m}[P^2,X] = \frac{i}{\hbar} \frac{1}{2m} (-2 i \hbar P) = \frac{P}{m}$ and similarly $\frac{d}{dt} P = \frac{i}{\hbar} [H,P] = \frac{i}{\hbar} \frac{\omega^2 m}{2}[X^2,P] = \frac{i}{\hbar} \frac{\omega^2 m}{2} (2 i \hbar X) = - \omega^2 m X$. Hence $\frac{d^2}{dt^2} X = \frac{1}{m} \frac{d}{dt}P = - \omega^2 X$. Therefore, the general solution for $X$ is of the form $X = A \cos(\omega t) + B \sin(\omega t)$. – secavara Mar 4 at 19:57
• @secavara That should be an Answer. (If you look in the browser "page source" you can copy your Mathjax.) – Keith McClary Mar 4 at 21:07
• Thank you very much – Bernoulli Mar 4 at 22:46

Use the equation of motion for the Heisenberg picture operators to find

$$\begin{equation} \frac{d}{dt} X = \frac{i}{\hbar} [H,X] = \frac{i}{\hbar} \frac{1}{2m}[P^2,X] = \frac{i}{\hbar} \frac{1}{2m} (-2 i \hbar P) = \frac{P}{m} \, , \end{equation}$$

and similarly

$$\begin{equation} \frac{d}{dt} P = \frac{i}{\hbar} [H,P] = \frac{i}{\hbar} \frac{\omega^2 m}{2}[X^2,P] = \frac{i}{\hbar} \frac{\omega^2 m}{2} (2 i \hbar X) = - \omega^2 m X \,. \end{equation}$$

Hence

$$\begin{equation} \frac{d^2}{dt^2} X = \frac{1}{m} \frac{d}{dt}P = - \omega^2 X \, . \end{equation}$$

Therefore, the general solution for $$X$$ is of the form

$$\begin{equation} X = A \cos(\omega t) + B \sin(\omega t) \, . \end{equation}$$

We can plug this solution in the equation for $$P$$ and find

$$\begin{equation} P = - m \omega A \sin(\omega t) + m \omega B \cos(\omega t) \, . \end{equation}$$

Notice that $$A$$ and $$B$$ are not completely arbitrary: since we must satisfy $$[X,P] = i \hbar$$, we must have $$[A,B] = \frac{i \hbar}{m \omega}$$. Hence, they certainly do not commute. Take this into account when computing $$\langle X^2 \rangle$$.