Equation of motion for expectation value of a quantum particle in a momentum eigenstate: $$\frac{d}{dt} \langle x \rangle = \frac{1}{i h} \langle [x,H] \rangle$$

and since it's in a momentum eigenstate, $$\frac{1}{i h} \langle [x,H] \rangle = \frac{1}{i h} \langle p \vert [x,H] \vert p \rangle$$

Expanding this,

$$\frac{1}{i h} \langle p \vert (x \frac{p^2}{2m} + xV(x)) - (\frac{p^2}{2m}x + V(x)x)\vert p \rangle = \frac{1}{i h} (\frac{p^2}{2m} - \frac{p^2}{2m}) \langle p \vert x \vert p \rangle + \frac{1}{i h} \langle p \vert[x,V(x)] \vert p \rangle = 0$$

since $V$ is $V(x)$. But $\frac{d}{dt}\langle x \rangle = \frac{p}{m}$. Where is my mistake?

  • $\begingroup$ Note that $\hat p$ is an operator, you cannot use it like a variable $\endgroup$
    – Courage
    Jun 5 '16 at 18:08
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    $\begingroup$ Hint: what is $\langle p | x | p \rangle$? $\endgroup$ Jun 5 '16 at 18:14
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/14116/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 5 '16 at 18:24

Well, you know that the eigenfunction of $\hat p$ is $\exp(ipx)$, so let's try to find what the expectation value of $x$ is, to begin with: $$\langle p | x | p\rangle = \int \exp(-ipx) x \exp(ipx) \, dx = \int x\, dx$$ and this integral doesn't exist. Then it shouldn't come as a surprise that trying to take the time derivative gives nonsense. The underlying reason is that $\exp(ipx)$ isn't normalizable. This answer to a similar question gives more details on how to resolve this.

  • $\begingroup$ thank you! this hit the nail on the head. <x'|p|x''>=ih delta(x'-x'')/(x'-x'') which clears up my confusion (and is consistent with the form of the momentum operator in the position basis since delta(x)/x= - d delta(x)/dx $\endgroup$
    – Allan Kane
    Jun 6 '16 at 16:07

Update: Comments by Robin pointed out my confusion.

Consider the following:

$$ [x, p^2] = x p p - p p x = x p p - p x p + p x p - p p x = [x,p] p + p [x,p] = 2 i h p $$

If you plug this in the initial expression, you will get exactly what you would expect from this observable: $p/m$.

But formally correct manipulations that you performed provide a different answer, showing that something is not right. For the conclusion, please see Robin's answer.

Original mistaken answer:

You cannot take the momentum term outside the average - momentum p is an operator that does not commute with x.

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    $\begingroup$ -1: Since the state is an eigenstate of $\hat p$, you can replace $\hat p$ with the eigenvalue $p$ (acting to the left on the bra and to the right on the ket). $\endgroup$ Jun 5 '16 at 18:22
  • $\begingroup$ No, I think you can't. Following your logic, the following is true: $ \langle x p \rangle = \langle x \rangle p$ and hence $ \langle [x,p] \rangle = 0 $. $\endgroup$ Jun 5 '16 at 18:29
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    $\begingroup$ You need to make sure that all the quantities involved actually exist, see here physics.stackexchange.com/q/14116/2451 $\endgroup$ Jun 5 '16 at 18:30
  • $\begingroup$ I think you are right $\endgroup$ Jun 5 '16 at 18:45
  • $\begingroup$ But I can get the correct answer if I use commutation relation of x and p, this is confusing $\endgroup$ Jun 5 '16 at 18:48

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