# Hamiltonian differentiation [closed]

$$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2 +V_0 x^3$$ Find $$\frac{d\langle x\rangle}{dt}$$, $$\frac{d\langle p\rangle}{dt}$$ and $$\frac{d\langle H\rangle}{dt}$$.

If this is the hamiltonian given then this should be impossible, as there is no time variable. Am i right? I have trouble evaluating all three.

• Hint: $<x>$ is the average of the operator $x$ on a wave function. In this case, probably it is meant the wave function which solves the time-dependent Schrodinger equation. This is where the time-dependence may come from. Commented May 6, 2020 at 9:52
• so i have to include e^(iwt/hbar)? Commented May 6, 2020 at 10:04
• Can you clarify if you have trouble doing all three of the calculations asked or just the last one (from the question title it sounds like just the last one?) Commented May 6, 2020 at 10:06
• Hi. Welcome to the website. In the future you should use MathJax for for formulas and not attach them as a picture. Commented May 6, 2020 at 10:57
• are you familiar with the Heisenberg picture?
– user245141
Commented May 6, 2020 at 11:35

Assuming that we work in the Schrödinger picture: the time evolution of the system is encoded in its wave function, as determined by the time-dependent Schrödinger equation: $$i\hbar\partial_t\Psi(x, t) = \hat{H}\Psi(x,t).$$ The time-dependent expectation values are then given by $$\langle x(t)\rangle = \langle \Psi(t)|\hat{x}|\Psi(t) = \int dx\Psi(x,t)^*\hat{x}\Psi(x,t), etc.$$ Generally speaking, these are time-dependent and have non-zero time derivatives.
However, if the system is in a stationary state, i.e. $$\hat{H}\psi(x) = E\psi(x),$$ then the time dependence of the wave function is trivial $$\Psi(x,t)=\psi(x)e^{-iEt/\hbar},$$ and the averages in question are time-independent.
Finally, let us note that $$\frac{d\langle x(t)\rangle}{dt}$$ is not the same as $$\left\langle\frac{d x(t)}{dt}\right\rangle$$ - the former is the derivative of an expectation value, whereas the latter is the expectation of the derivative, where the operator of the derivative is defined by $$\hat{\dot{x}} = \frac{d \hat{x}(t)}{dt} = \frac{1}{i\hbar}\left[\hat{x}, \hat{H}\right].$$ Note that, as we are still in the Schrödinger picture, this is definition of the operator of the derivative rather than a time derivative of the operator. In the Heisenberg and interaction pictures the latter is given by the same equation, but with time-dependent operators.