0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Perhaps it would have been clearer if you did keep the operator hats on. $\partial \hat{x} / \partial t \equiv \hat{0}$ because the operator $\hat{x}$ has no explicit time dependence. $\endgroup$ – jacob1729 Jun 19 '18 at 11:00
  • $\begingroup$ @jacob1729: I will go back and edit if it's too confusing. x is the operator throughout. $\endgroup$ – daniel Jun 19 '18 at 11:04
  • 2
    $\begingroup$ Note that you can use \langle and \rangle to get better looking bras and kets, instead of a bunch of less-than & greater-than symbols floating about. $\endgroup$ – Kyle Kanos Jun 19 '18 at 11:06
  • $\begingroup$ @daniel My point was more that maybe its confusing you. $\hat{x}$ is a constant operator, not a dynamical variable (its not the position of anything) and so it doesn't have a partial time derivative. $\endgroup$ – jacob1729 Jun 19 '18 at 11:12
  • 1
    $\begingroup$ I've edited the question adding right and left angles au lieu of greater than and smaller than signs. $\endgroup$ – Vitor C Goergen Jun 19 '18 at 12:33
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I will study this but in the four lines in which I break up [x,H] can you tell me which one is wrong? (thanks) $\endgroup$ – daniel Jun 19 '18 at 12:59
  • $\begingroup$ It is the assumption you made from the 3rd to the 4th line, which I have stated in my answer (last equation). $\endgroup$ – Vitor C Goergen Jun 19 '18 at 13:28
  • $\begingroup$ The third full line is $\frac{d}{dt}<x>=<\partial x/\partial t>-<\partial x/\partial t>+<xH/ih> \implies \frac{d}{dt}<x> =~<xH/ih>$. If it's right, why doesn't the 4th line follow? $\endgroup$ – daniel Jun 19 '18 at 13:43
  • $\begingroup$ That’s not right. $\bigg\langle \frac{Hx}{i\hbar \bigg\rangle$ is not equal to the expectation value of the derivative of x wrt time. $\endgroup$ – Vitor C Goergen Jun 19 '18 at 15:20
  • $\begingroup$ Understood. So there must be a mistake in one of the previous lines. Can you tell me which one? $\endgroup$ – daniel Jun 19 '18 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.