Multiplying $X$ and $P$ operators in quantum mechanics using delta functions (on the $X$ basis)

Question

Alright so I'm trying to figure out how to find the operator $XP$ in the $x$ basis, knowing that the elements of $X$ and $P$ are $x \delta(x-x')$ and $-ih \delta'(x-x')$ respectively. I know how to do it by assuming $X=x$ and $P=-ihd/dx$, but I don't want to do it that way, I want to actually calculate the elements of the new matrix. Now normally I would assume I just have to multiply the $xx'$ element of $X$ with the $x'x$ element of $P$ and then integrate over $x'$, analogous to how I would multiply two finite matrices. However I would then end up with an integral over $x'$ which contains two delta functions, $\delta(x-x')$ and $\delta'(x'-x)$. I have no idea how to integrate that. I believe there is a gap in my understanding of infinite matrices, delta functions, etc. Does anyone have any ideas?

Answer 1

How about using momentum eigenstates as the states being summed over: $$ \langle x|\hat x\hat p|x'\rangle= \int \frac{dp}{2\pi} \langle x|\hat x|p\rangle \langle{p}|\hat p|x'\rangle \\ = \int \frac{dp}{2\pi} x e^{ipx} p e^{-ip x'}\\ = x \int \frac{dp}{2\pi} e^{ipx} i\frac{\partial}{\partial x'} e^{-ip x'}\\ ix\frac{\partial}{\partial x'} \int \frac{dp}{2\pi} e^{ip(x-x')} \\ = ix\frac{\partial}{\partial x'} \delta(x-x')\\ =- ix\frac{\partial}{\partial x}\delta(x-x') $$ It's not exactly rigorous mathematics, but it makes me less nervous that multiplying two distributions!

Answer 2

The dirty secret here is that not all operators are infinite-dimensional "matrices" (the ones that are in a reasonable sense are Hilbert-Schmidt operators but $x$ and $p$ are not among them) and that $\lvert x\rangle$ and $\lvert p\rangle$ are not elements of the Hilbert space (but instead of a larger rigged Hilbert space, see user1504's excellent answer on the topic). The operators $x$ and $p$ simply can't both act on things like $\lvert x\rangle$ in a well-defined manner, because they are far outside the domain of definition for $p$.

As you noticed, in this case pretending these things are ordinary matrices and vectors lands you in the ill-defined world of products of distributions. They are not, and you cannot blindly apply naive generalizations of finite-dimensional linear algebra to them. You already know the correct way to represent $xp$, namely by taking $p$ as the derivative operator. Just do that if you're looking for rigor and sanity.

Answer 3

I'm not sure I understand the question, but Dirac's picture, in his landmark book, instructs you to use $$ \hat X = \int dx ~~|x\rangle x \langle x| ,\\ \hat P = \int dx ~~|x\rangle \frac{\hbar}{i}\partial_x \langle x| , $$ so $$ \hat X \hat P = \int dx dx' ~~|x\rangle x \langle x |x'\rangle \frac{\hbar}{i}\partial_{x'} \langle x'| = \int dx dx' ~~|x\rangle x ~ \delta ( x -x') ~\frac{\hbar}{i}\partial_{x'} \langle x'| \\ = \frac{\hbar}{i} \int dx ~|x\rangle x ~ \partial_{x} ~\langle x| , $$ which evokes your matrix multiplication vision.

The "matrix" vision is normally illustrated by "quantum mechanics around the clock, in a "clock" of N hours, expounded in Weyl's celebrated book, where you see that $\hat P$ (found by Santhanam & Tekumalla) is not diagonal, but, in a cagey/crafty large N limit devolves to the above (which is also really not diagonal).