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

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?

• Could you write out the integral you're having trouble with in your question? Jun 5, 2020 at 18:28
• Well, I could try but it definitely seems wrong to me. I'm on a mobile so writing it in detail seems a bit hard now, I simply tried to integrate over x' the product of the elements Xxx' and Px'x as I described them above. Jun 5, 2020 at 18:31

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!

• I thought about that but it's kinda what I wanted to avoid. I wanted to understand how I could multiply two infinite matrices in some sort of rigorous manner. Also where did that 2pi denominator come from? Jun 5, 2020 at 18:53
• The $2\pi$ is just my normalization of $|p\rangle$: $\langle p|p'\rangle= 2\pi \delta(p-p')$. There's no way to define $\hat x$ so that both it and $\hat p$ turn into infinite discrete matrices. $\hat x$ is undefined in a periodic system, and with rigid walls $\hat p$ is not self adjoint. Jun 5, 2020 at 18:57
• Oooh alright thanks, I missed that. Jun 5, 2020 at 18:58
• @AndreasC that integral is analogous to matrix multiplication. It's almost exactly the same as writing out $A = B \cdot C$ as $A_{ij} = \sum_k B_{ik}C_{kj}$. Jun 5, 2020 at 20:26
• Which integral? Jun 5, 2020 at 21:11

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.

• Oh wow, have I been lied to? Lots of textbooks make it sound like you can do that... Huh. Do you know any books or material that I can read on this subject? I want to understand the mathematics behind QM a bit better and more rigorously because I feel like I don't have a great grasp and a lot of things seem really hand wavy or misleading... That's the whole reason I tried to do what I tried to do here, I thought, hey, the textbooks pretend it's all infinite dimensional matrices etc, can't I just multiply their elements? Jun 5, 2020 at 19:14
• Oh lol I just noticed the post you linked to answered a similar question... Jun 5, 2020 at 19:15
• Still, I'd appreciate some sort of neat little book that explores the math about QM and what you can and can't do. Jun 5, 2020 at 19:16
• @AndreasC It's not available as a "neat little book", it's basically the entire mathematical field of functional analysis. A classic for the physicist interested in rigor is the multi-volume series (!) by Simon and Reed. Jun 5, 2020 at 19:27
• Jesus Christ I just checked it lol. It's actually huge. Yeesh. Idk where to start now... Jun 5, 2020 at 21:15

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).

• Huh... That reminds me of something I actually did earlier on thinking it was just bullshit... Interesting to see it isn't but I was mostly concerned about why what I described in my original post didn't work. Still, your post is definitely helpful. Ευχαριστώ ;) Jun 5, 2020 at 21:21
• Link to S&T is here. I can't do it myself, but I have watched RPF use "bullshit" to deadly effect. Jun 5, 2020 at 21:29
• What is RPF? Also, this is pretty interesting, I'll look into it. Jun 5, 2020 at 21:45
• Ooooh alright. Well, yeah, unsurprising. He got results though. My bullshit didn't map exactly to what you mentioned but it was kinda similar in concept. To be completely honest I don't remember exactly what I did... I did it a few days ago when that question first sprang into my mind and for a while I thought "yeah, that makes sense". Then today I thought "hold on" and started looking further into it. I remember I still used delta functions somehow but I integrated over what I called x and y, and somehow it gave the correct result for the few operators I tried it on... Jun 5, 2020 at 21:59