0
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ Could you write out the integral you're having trouble with in your question? $\endgroup$
    – Philip
    Jun 5, 2020 at 18:28
  • $\begingroup$ 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. $\endgroup$
    – Andreas C
    Jun 5, 2020 at 18:31

3 Answers 3

3
$\begingroup$

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!

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$
    – Andreas C
    Jun 5, 2020 at 18:53
  • 2
    $\begingroup$ 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. $\endgroup$
    – mike stone
    Jun 5, 2020 at 18:57
  • $\begingroup$ Oooh alright thanks, I missed that. $\endgroup$
    – Andreas C
    Jun 5, 2020 at 18:58
  • 1
    $\begingroup$ @AndreasC that integral is analogous to matrix multiplication. It's almost exactly the same as writing out $A = B \cdot C$ as $A_{ij} = \sum_k B_{ik}C_{kj}$. $\endgroup$
    – DanielSank
    Jun 5, 2020 at 20:26
  • $\begingroup$ Which integral? $\endgroup$
    – Andreas C
    Jun 5, 2020 at 21:11
3
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$
    – Andreas C
    Jun 5, 2020 at 19:14
  • $\begingroup$ Oh lol I just noticed the post you linked to answered a similar question... $\endgroup$
    – Andreas C
    Jun 5, 2020 at 19:15
  • $\begingroup$ Still, I'd appreciate some sort of neat little book that explores the math about QM and what you can and can't do. $\endgroup$
    – Andreas C
    Jun 5, 2020 at 19:16
  • 1
    $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    Jun 5, 2020 at 19:27
  • $\begingroup$ Jesus Christ I just checked it lol. It's actually huge. Yeesh. Idk where to start now... $\endgroup$
    – Andreas C
    Jun 5, 2020 at 21:15
1
$\begingroup$

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).

$\endgroup$
4
  • $\begingroup$ 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 ;) $\endgroup$
    – Andreas C
    Jun 5, 2020 at 21:21
  • 1
    $\begingroup$ Link to S&T is here. I can't do it myself, but I have watched RPF use "bullshit" to deadly effect. $\endgroup$ Jun 5, 2020 at 21:29
  • $\begingroup$ What is RPF? Also, this is pretty interesting, I'll look into it. $\endgroup$
    – Andreas C
    Jun 5, 2020 at 21:45
  • $\begingroup$ 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... $\endgroup$
    – Andreas C
    Jun 5, 2020 at 21:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.