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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?

Qmechanic
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Allan Kane
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2 Answers2

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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.

Community
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Robin Ekman
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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.

Andrii Magalich
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