Problem involving Dirac notation and Ehrenfest's theorem

Question

Ehrenfest's theorem in one guise says (omitting hats for vectors) for suitable operators A

$$\frac{\mathrm d}{\mathrm dt}\langle A\rangle~=~\bigg\langle\frac{\partial A}{\partial t}\bigg\rangle+\bigg\langle\frac{[A,H]}{i\hbar}\bigg\rangle\tag{1}$$

and an "immediate" consequence is that, inserting the position operator x for A in the theorem, using P for the momentum operator,

$$\frac{\mathrm d}{\mathrm dt}\langle x\rangle~=~\bigg\langle\frac{P}{m}\bigg\rangle.\tag{1a}$$

Ignoring the left side of (1) I reasoned (omitting test function in expectation values)

\begin{align} \bigg\langle\frac{\partial x}{\partial t}\bigg\rangle + \bigg\langle\frac{[x,H]}{i\hbar}\bigg\rangle &=~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle + \bigg\langle\frac{xH-Hx}{i\hbar}\bigg\rangle \\ &=~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle + \bigg\langle\frac{xH}{i\hbar}\bigg\rangle~-~\bigg\langle\frac{Hx}{i\hbar}\bigg\rangle \\ &=~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle + \bigg\langle\frac{xH}{i\hbar}\bigg\rangle~-~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle \end{align} and so

$$\frac{\mathrm d}{\mathrm dt}\langle x\rangle~=~ \bigg\langle\frac{xH}{i\hbar}\bigg\rangle,\tag{2}$$

which I can convince myself is true.

On the other hand, using the first relation in this problem set (PDF):

$$[x,H]=\frac{i\hbar}{m}P, $$ we get immediately that

$$\frac{\mathrm d}{\mathrm dt}\langle x\rangle~=~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle+\bigg\langle\frac{[x,H]}{i\hbar}\bigg\rangle$$

$$\frac{\mathrm d}{\mathrm dt}\langle x\rangle~=~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle+\bigg\langle\frac{P}{m}\bigg\rangle\tag{3}$$

but then from (1a) that $\bigg\langle\partial x/\partial t\bigg\rangle~=0,$ which bothers me a little but doesn't seem to contradict anything above directly.

My question is whether the conclusions (2), (3) in these two calculations are right and if not where I went astray.

Answer 1

First, notice that the term: $$\bigg\langle\frac{\partial x}{\partial t} \bigg \rangle$$ Equals zero in the Schrödinger picture (which is the one you seem to be working with) because the operators in the Schrödinger picture are constant. That is, they don't depend explicitly on time. What changes in time is their expected value, and this is represented by: $$\frac{d \langle x \rangle}{dt}$$ So one of your questions is answered (I believe).

Now, notice that: $$[x,H] = \frac{1}{2m}\big(xp^2 - p^2x\big)$$ But $[x,p]=i\hbar$ so $$ \frac{1}{2m}\big(xp^2 - p^2x\big) = \frac{1}{2m}\big(xpp - ppx\big) = \frac{1}{2m}\big((i\hbar+px)p - p(xp-i\hbar)\big)$$ And so $$[x,H] = \frac{1}{2m}\big(i\hbar p + pxp - pxp + i\hbar p\big) = \frac{i\hbar p}{m}$$ It doesn't make much sense saying that $$\bigg\langle\frac{Hx}{i\hbar}\bigg\rangle = \bigg\langle\frac{\partial x}{\partial t}\bigg\rangle $$ As you seem to have done.