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In another Phys.SE question, I've proposed the next-cited proof of this statement:

the momentum matrix elements in position representation, $\langle x'|\hat{p}|x\rangle$, are all not null

I'm limiting to one particle in one dimension. This is the proof:

The last part of the proof is show that $\hat{p}$ operators contains all the off-diagonal elements; I give you one attempt, but is a little bit sloppy (and maybe the statement is false, but I don't find literature about, so I'll ask another question on this point):

  • $\hat{p}$ has at least one off-diagonal element, else it commutes with $\hat{x}$ and it isn't;
  • $\hat{p}$ is invariant under reference translations (think on it's eigenstates $\langle x | p \rangle = \varphi_p (x) = exp(\frac{ixp}{\hbar})$), so it matrix elements depends only on the "distance" $x - x'$ (and it's sign, because it has to be hermitian)
  • under scale transformations of positions $\hat{p} \rightarrow \lambda \hat{p}$ (think again on the action on its eigenstates)

so if one elements is not null also the others has to be not null, because their distance can be reduce to the one of the not-null elements with a scale transformation.

My question is: is the statement true? and is the proof correct? If not, but the statement is true, can you give me a right proof? Else, can you show me where I'm making a mistake?

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The matrix elements $\hat p$ are given by $$ <x|\hat p|x'>= -i\hbar \partial_x <x|x'> = -i\hbar\partial_x\delta(x-x')= -i\hbar\delta'(x-x') $$ As the derivative of the delta function is a distribution, its "value" at any point is not really well-defined. You must instead evaluate it on a test function. In a sense it has non-zero values only infinitely close to the diagonal.

The question then is whether we can say that $\delta'(x)$ is zero away from $x=0$. To see that in some sense it is not zero recall that $\delta(x)$ has the property that $f(x)\delta(x)=f(0)\delta(x)$, and so only needs the value of $f$ exactly at zero. On the other hand $\delta'(x)$ obeys $$ f(x)\delta'(x) = f(0)\delta'(x) - f'(0) \delta(x) $$ as is easily seen by ``integrating'' both sides of this identity against a test function. Thus $\delta'(x)$ needs information both at $x=0$ and also in an infinitesimally small neigbourhood of $x=0$ so that it can compute $f'(0)$. If we use a lattice regularization, the $\hat p$ operator will be an infinite matrix with entries on the principal diagonal and the also in the immediately neighbouring diagonals (although no really good lattice definition exists). These will become infinitesimally close to the principal diagonal as we make the lattice spacing go to zero. All furtheraway entries in the infinite matrix will be zero.

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  • $\begingroup$ I know definition and properties of distributions, but delta has a limited support , and so also it's derivative. In this sense I could say that are null outside 0 because if I integrate them against a test function null in a neighbourhood of 0 the result is null. So I could conclude that $\hat{p}$ is diagonal, but this is impossible, due to commutation rule. $\endgroup$ – Annibale Oct 2 '17 at 13:24
  • $\begingroup$ @Alessandro. Recall that $f(x)\delta(x) =f(0)\delta(x)$, so only the value at $x-0$ matters, but using test functions we can show that $f(x)\delta'(x)=f(0)\delta'(x)-f'(0)\delta(x)$. To get $f'(0)$ you need to know $f(x)$ in a neighourhood of $x=0$, so it's not really true that $\delta'(0)$ is zero away from $x=0$. I'll edit my answer to make this clear $\endgroup$ – mike stone Oct 2 '17 at 17:34

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