# Inconsistency caused by the usage of matrix elements of the momentum operator in position representation

Suppose $$p$$ is the momentum operator, we know that the matrix element of $$p$$ in the $$q$$-basis reads $$\langle q'|p|q''\rangle=-i\hbar\frac{\partial}{\partial q'}\delta(q'-q'')=i\hbar\frac{\partial}{\partial q''}\delta(q'-q'')$$.

Now consider the eigenequation $$p|p'\rangle=p'|p'\rangle$$. If we start with the LHS and insert two sets of completeness relations, we have $$p|p'\rangle=\int dq'\int dq'' |q'\rangle\langle q'|p|q''\rangle\langle q''|p'\rangle=\int dq'\int dq'' |q'\rangle\left[i\hbar\frac{\partial}{\partial q''}\delta(q'-q'')\right]\langle q''|p'\rangle$$.

We next integrate by part with respect to $$q''$$ to get

$$p|p'\rangle=\int dq' |q'\rangle\left[i\hbar\delta(q'-q'')\langle q''|p'\rangle\right]|^{q''=+\infty}_{q''=-\infty}-\int dq'\int dq'' |q'\rangle\left[\frac{\partial}{\partial q''}\langle q''|p'\rangle\right][i\hbar\delta(q'-q'')]$$ $$=\left[i\hbar|q''\rangle\langle q''|p'\rangle\right]|^{q''=+\infty}_{q''=-\infty}+\int dq' |q'\rangle\left[-i\hbar\frac{\partial}{\partial q'}\langle q'|p'\rangle\right]$$

$$=\left[i\hbar|q''\rangle\langle q''|p'\rangle\right]|^{q''=+\infty}_{q''=-\infty}+\int dq' |q'\rangle\left[p'\langle q'|p'\rangle\right]$$

$$=\left[i\hbar|q''\rangle\langle q''|p'\rangle\right]|^{q''=+\infty}_{q''=-\infty}+p'|p'\rangle$$.

Is there any mistake in the derivation that causes the appearance of the boundary term $$\left[i\hbar|q''\rangle\langle q''|p'\rangle\right]|^{q''=+\infty}_{q''=-\infty}$$?

Update: Let's look at the equation we obtained in the $$p$$-basis: $$\langle p''|p|p'\rangle=\left[i\hbar\langle p''|q''\rangle\langle q''|p'\rangle\right]|^{q''=+\infty}_{q''=-\infty}+\langle p''|p'|p'\rangle$$.

If $$p''=p'$$, we have $$\langle p'|p|p'\rangle=\left[i\hbar\langle p'|q''\rangle\langle q''|p'\rangle\right]|^{q''=+\infty}_{q''=-\infty}+\langle p'|p'|p'\rangle$$

$$=\left(\frac{i}{2\pi}\right)|^{q''=+\infty}_{q''=-\infty}+\langle p'|p'|p'\rangle=0+\langle p'|p'|p'\rangle=\langle p'|p'|p'\rangle$$. No problem.

If $$p''\neq p'$$, we have $$\langle p''|p|p'\rangle=\left[\frac{i}{2\pi}e^{i\frac{q''(p'-p'')}{\hbar}}\right]|^{q''=+\infty}_{q''=-\infty}+\langle p''|p'|p'\rangle$$.

To get a consistent result, we expect that $$\left[\frac{i}{2\pi}e^{i\frac{q''(p'-p'')}{\hbar}}\right]|^{q''=+\infty}_{q''=-\infty}=0$$. This sounds weird...

Nevertheless, I found the following argument in R. Shankar's book "Principles of Quantum Mechanics" (page 66):

The limit $$\lim_{q''\to\infty}e^{i\frac{q''(p'-p'')}{\hbar}}$$ should be defined to be "the average over a large interval":

$$\lim_{q''\to\infty}e^{i\frac{q''(p'-p'')}{\hbar}}=\lim_{Q\to\infty,\Delta\to\infty}\frac{1}{\Delta}\int^{Q+\Delta}_Qdq''e^{i\frac{q''(p'-p'')}{\hbar}}=0$$, if $$p'\neq p''$$.

If we choose to accept this, it seems that the inconsistency can be removed...

Another way to see that $$\left[e^{iq''(p'-p'')/\hbar}\right]|^{q''=+\infty}_{q''=-\infty}=\left[e^{iq''(k'-k'')}\right]|^{q''=+\infty}_{q''=-\infty}=0$$: $$\left[e^{iq''(k'-k'')}\right]|^{q''=+\infty}_{q''=-\infty}=\int^\infty_{-\infty}d[q''(k'-k'')]e^{iq''(k'-k'')}=(k'-k'')\int^\infty_{-\infty}dq''e^{iq''(k'-k'')}=2\pi(k'-k'')\delta(k'-k'')=0$$.

• This derivation is currently pretty difficult to read because of having to keep track of all of the primed variables, and a lack of distinction between the momentum operator and a momentum eigenvalue. If you replace $q'$ with $x$ and $q''$ with $y$, replace $p$ with $\hat{p}$ and $p'$ with $p$, then it might be significantly easier to parse. Commented Sep 30, 2020 at 15:46
• In any case, at a glance this looks a bit like you've rediscovered the difficulties with assuming a self-adjoint momentum operator on the whole Hilbert space: physics.stackexchange.com/questions/143055/… Commented Sep 30, 2020 at 16:06
• No consolation, but, dotting with $\langle \psi |$ would produce $\psi^*(q'')$ normally taken to vanish at infinity. The rapidly varying exponential itself has no good limit except at p'=0. Commented Sep 30, 2020 at 20:14
• Thanks for the comment, Cosmas. What if we dot with $\langle p''|$ from the left? If $p''=p'$, then $[i\hbar\langle p'|q''\rangle\langle q''|p'\rangle]|^{q''=+\infty}_{q''=-\infty}=\frac{i}{2\pi}|^{q''=+\infty}_{q''=-\infty}=0$. This is fine. But if $p''\neq p'$, we still cannot bypass the limit $\frac{i}{2\pi}\exp(iq''(p'-p''))|^{q''=+\infty}_{q''=-\infty}$... Commented Oct 1, 2020 at 1:53

If we start with the LHS and insert two sets of completeness relations, we have $$p|p'\rangle=\int dq'\int dq'' |q'\rangle\langle q'|p|q''\rangle\langle q''|p'\rangle=\int dq'\int dq'' |q'\rangle\left[i\hbar\frac{\partial}{\partial q''}\delta(q'-q'')\right]\langle q''|p'\rangle$$

We already know that $$\langle q''|p'\rangle = e^{ip'q''/\hbar}/\sqrt{2\pi\hbar}$$, so we can make this somewhat more clear:

$$\hat p|p'\rangle = \frac{1}{\sqrt{2\pi\hbar}}\int dq' \int dq'' |q'\rangle\left[ i\hbar\frac{d}{dq''}\delta(q'-q'')\right]e^{ip'q''}$$

Integrating by parts yields

$$\frac{1}{\sqrt{2\pi\hbar}}\int dq' |q'\rangle \lim_{a\rightarrow-\infty}\lim_{b\rightarrow \infty} \left[i\hbar \delta(q'-q'')e^{ip'q''/\hbar}\right]_{q''=a}^{q''=b} - \frac{1}{\sqrt{2\pi\hbar}}\int dq' |q'\rangle(- p') e^{ip'q'/\hbar}$$

The first term is zero. The reason is that those limits are taken at fixed $$q'$$, and then the result is integrated over all $$q'$$. For any fixed $$q'$$, those limits evaluate to zero, and the integral of zero is zero. As expected,

$$\hat p|p'\rangle =p'\left( \frac{1}{\sqrt{2\pi\hbar}} \int dq' |q'\rangle e^{-ip'q'/\hbar}\right)= p'|p'\rangle$$

If you don't like that justification, note that the very definition of the (distributional) derivative of the delta function is the distribution $$\delta'$$ defined by

$$\int_{-\infty}^\infty \delta'(x-y) f(x) dx = -f'(y)$$

The heuristic "integration by parts" argument is ultimately a smokescreen; the apparently troublesome term is never actually there in the first place.

• Thanks for the detailed answer, especially for introducing me the "distributional derivative", which really helps here. Commented Oct 1, 2020 at 14:10