Matrix elements of the commutator of the delta function with $p^2$

Question

Suppose we have potential $V(x) = \delta(x)$. I want to evaluate $\langle x' \vert [p^2, \delta(x)] \vert x'' \rangle$. Unprimed quantities are operators while primed quantities are eigenvalues/eigenstates of the position operator.

My attempt is below. I am confident in the working, up until the final step where I assert the result is zero, based on a hand-wavy pseudo-mathematical argument. My physical intuition tells me that the commutator should be non-zero, because this question arises directly from asking whether the Hamiltonian for a particle subject to a delta-potential commutes with the free particle Hamiltonian (and I would think not).

The ask is to check/correct the mathematical working (right at the end) based on this disagreement between the answer I expected (not zero) and the answer I obtained. The justification for asking this question is that I don't know how to handle this type of integral and my book (Sakurai) doesn't seem to cover it.

$$\begin{align} \langle x' \vert \delta(x) \vert x'' \rangle &= \langle x' \vert \lim_{\sigma -> 0} \frac{1}{\sqrt{2 \pi \sigma^2}}\exp\bigl(\frac{-x^2}{2\sigma^2}\bigl) \vert x'' \rangle\\ &= \lim_{\sigma -> 0}\frac{1}{\sqrt{2 \pi \sigma^2}}\exp\bigl(\frac{-x''^2}{2\sigma^2}\bigl) \langle x' \vert x'' \rangle \\ &= \lim_{\sigma -> 0}\frac{1}{\sqrt{2 \pi \sigma^2}}\exp\bigl(\frac{-x''^2}{2\sigma^2}\bigl) \delta(x'-x'') \\ &= \delta(x'') \delta(x'-x'')= \delta(x'') \delta(x'), \end{align}$$

$$\begin{align} \langle x' \vert p^2 \vert x''' \rangle = (-i\hbar)^2\frac{\partial^2}{\partial x'^2}\delta(x'-x'''), \end{align}$$

$$\begin{align} \langle x' \vert \delta(x)p^2 \vert x'' \rangle &= \int{\langle x' \vert \delta(x) \vert x'''\rangle\langle x''' \vert\ p^2 \vert x''\rangle dx'''} ,\\ \end{align}$$

$$\begin{align} \langle x' \vert p^2 \delta(x) \vert x'' \rangle &= \int{\langle x' \vert p^2 \vert x'''\rangle\langle x''' \vert\delta(x)\vert x''\rangle dx'''} ,\\ \end{align}$$

$$\begin{align} \langle x' \vert [\delta(x),p^2] \vert x'' \rangle &= \int{\langle x' \vert \delta(x) \vert x'''\rangle\langle x''' \vert\ p^2 \vert x''\rangle dx'''} \\ &-\int{\langle x''' \vert\delta(x)\vert x''\rangle \langle x' \vert p^2 \vert x'''\rangle dx'''} \\ &=\int{\delta(x''') \delta(x'-x''') (-i\hbar)^2\frac{\partial^2}{\partial x'''^2}\delta(x'''-x'') dx'''} \\ &-\int{\delta(x'') \delta(x'''-x'') (-i\hbar)^2\frac{\partial^2}{\partial x'^2}\delta(x'-x''') dx'''} \\ &=\int{\delta(x''') \delta(x'-x''') (-i\hbar)^2(-1)^2(2!)\frac{\delta(x'''-x'')}{(x'''-x'')^2} dx'''} \\ &-\int{\delta(x'') \delta(x'''-x'') (-i\hbar)^2(-1)^2(2!)\frac{\delta(x'-x''')}{(x'-x''')^2} dx'''} \\ &= 0. \end{align}$$

The (false) hand-wavy argument: intuitively, the only non-zero contribution will come from $x' = x'' = x''' = 0$ and the divergent contribution of the denominators will be the same in each integrand. The integrands will then have the same "values" and cancel (whatever this "value" really is).

For the derivatives, I have used the expression $x'^n \delta^{(n)}(x') = (-1)^n (n!) \delta(x')$, where $f^{(n)}$ denotes the nth derivative of $f$. I chose this identity because the integrals include multiple $\delta$-functions, so the usual integration rule with a single $\delta$-function multiplying an "ordinary" function did not seem applicable. Thus the appeal to intuition above.

Answer 1

Maybe it would be easier to just notice that $\delta(x)$ is a Hermitian operator so it can be applied either left or right: $$\langle x^\prime|\delta(x)|\psi\rangle=\delta(x^\prime)\langle x^\prime|\psi\rangle ;\qquad \langle \psi|\delta(x)|x^{\prime\prime}\rangle=\delta(x^{\prime\prime})\langle \psi|x^{\prime\prime}\rangle.$$ Then it is straightforward to manipulate \begin{align} \langle x^\prime|p^2\delta(x) |x^{\prime\prime}\rangle-\langle x^\prime| \delta(x)p^2|x^{\prime\prime}\rangle&=\delta(x^{\prime\prime})\langle x^\prime|p^2 |x^{\prime\prime}\rangle-\delta(x^\prime)\langle x^\prime| p^2|x^{\prime\prime}\rangle\\ &=[\delta(x^{\prime\prime})-\delta(x^{\prime})](-i\hbar)^2\frac{\partial^2}{\partial x^{\prime 2}}\delta(x^\prime-x^{\prime\prime}). \end{align} The term in $[]$ vanishes when $x^{\prime\prime}=x^\prime$ and you can argue that the derivative of the delta function should vanish everywhere away from $x^\prime=x^{\prime\prime}$. But to actually argue what happens at $x^\prime=x^{\prime\prime}$ is much more subtle, because we have a product of vanishing and diverging terms. You should treat this as a distribution and check how it integrates against some test function $f(x^\prime,x^{\prime\prime})$.

Let's try this integration trick with some sufficiently differentiable function $f$. We integrate by parts once with vanishing boundary terms to find \begin{align} \int_{-\infty}^\infty dx \int_{-\infty}^\infty dy f(x,y) [\delta(y)-\delta(x)]\frac{\partial^2}{\partial x^{2}}\delta(x-y)&=\int_{-\infty}^\infty dx f(x,0) \frac{\partial^2}{\partial x^{2}}\delta(x)-\int_{-\infty}^\infty dx \int_{-\infty}^\infty dy \delta(x-y)\frac{\partial^2}{\partial x^{2}}[f(x,y) \delta(x)]\\ &=\frac{\partial^2 f(x,0)}{\partial x^2}\bigg{|}_{x=0}-\int_{-\infty}^\infty dx \frac{\partial^2}{\partial x^{2}}[f(x,x) \delta(x)]\\ &=\frac{\partial^2 f(x,0)}{\partial x^2}\bigg{|}_{x=0}. \end{align} The last term went to zero either because of the fundamental theorem of calculus or because the integrals of all four terms in the expansion of the second derivative will cancel when integration by parts brings all the derivatives to $f$. We certainly expected that this product is nonzero somewhere; otherwise the operator $p^2$ and $\delta(x)$ would both be diagonal in the $x$ basis, which they are not.

Another basis-independent relation we could have used is \begin{align}[p^2,\delta(x)]&=p[p,\delta(x)]+[p,\delta(x)]p\\ &=p([p,x]\frac{\partial \delta(x)}{\partial x})+([p,x]\frac{\partial \delta(x)}{\partial x})p\\ &=-i\hbar [p\delta^\prime(x)+\delta^\prime(x)p],\end{align} using the rule for commutators of functions.

Answer 2

To see how the doubly singular distribution $\langle x' \vert [\delta(x),p^2] \vert x'' \rangle$ behaves, consider two plain functions allowing indefinite integrations by parts,
$$ \int\!\! dx' dx'' ~\langle x' \vert [\delta(x),p^2] \vert x'' \rangle f(x') g(x'')\\ =\int\!\! dx' dx'' dx'''\langle x' \vert [\delta(x),p^2] \vert x'' \rangle f(x') g(x'') \\ = -\hbar^2\int\!\! dx' dx'' dx'''~f(x') g(x'')~~~~~\times \\ \left ( {\delta(x''') \delta(x'-x''') \frac{\partial^2}{\partial x'''^2}\delta(x'''-x'') } - {\delta(x'') \delta(x'''-x'')\frac{\partial^2}{\partial x'^2}\delta(x'-x''') } \right )\\ = -\hbar^2 (f(0)g''(0)- f''(0) g(0)), $$ manifestly not vanishing. Your $\infty \cdot 0$ is indeterminate.

You should be able to also verify this by reguralizing the δ functions through your Gaussians before taking the vanishing width limit!