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Suppose that $p=-i\hbar \frac{\partial }{\partial x}$ was known, i.e. $\langle x|p=-i\hbar \frac{\partial}{\partial x}\langle x|$. Suppose the only other known condition was $[x,F(p)]=i\hbar \frac{\partial F(p)}{\partial p}$.(which could be derived from the previous one. )

Then how to derive $\langle p|x=i\hbar\frac{\partial }{\partial p}\langle p|$ directly?

I tried to complete $\langle p|$ into an operator by $|m\rangle\langle p|$ then apply the commutation relationship $[x,|z\rangle\langle p|] =-i\hbar \frac{\partial |z\rangle\langle p|}{\partial p}$. However, after expanding it, I was not able to get rid of a term $-|z\rangle\langle p|x$, which indicated something wrong.

Is there anyway to derive $\langle p|x=i\hbar\frac{\partial }{\partial p}\langle p|$ from $[x,F(p)]=i\hbar \frac{\partial F(p)}{\partial p}$?

Cosmas Zachos
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ShoutOutAndCalculate
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    What makes you think that what you are trying to prove is correct? More precisely the RHS is ill-defined. – DanielC Oct 30 '19 at 07:31
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    Can't you get from $\langle x | \hat{p} | \psi\rangle = -i\hbar \partial_x \langle x | \psi \rangle$ the form of the eigenfunctions of $\hat{p}$ as plane waves? and then what you are looking for can be just derived by integration by parts? –  Oct 30 '19 at 10:10
  • Near duplicate. – Cosmas Zachos Oct 30 '19 at 15:21

1 Answers1

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Consider $$ \langle p|\hat x |x\rangle = x \langle p |x\rangle =x \frac{1}{\sqrt{2\pi \hbar }} e^{-ipx/\hbar}=i\hbar \partial_p e^{-ipx/\hbar} \frac{1}{\sqrt{2\pi \hbar }}= i\hbar \partial_p \langle p |x\rangle , $$ and note it holds for all x.

Can you now sandwich the above between $ | p \rangle $ and $\langle x| $ and integrate over x and p to get the more memorable $$ \bbox[yellow]{\hat x= \int \!\! dp ~~|p\rangle ~i\hbar \partial_p \langle p | } ~~~~~~~? $$

Cosmas Zachos
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  • Got it. (I got really messed up by not realizing those $x$ and $p$ s were no longer operators after a long night.) But still, It troubles me why we need all those fourier relationships and integral identities, I felt like we should able to derive the expression directly from $[x,F(x)]$, where $[p, F(x)]$ was obvious. It seemed to me that the duality has already been fully characterized, so why can't we just go pass the integral, and get a purly differential solution? – ShoutOutAndCalculate Oct 30 '19 at 16:00
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    Yes, the 2nd equation says effectively that, in the momentum space representation, $\hat x $ presents as $i\hbar \partial_p$, so $[i\hbar \partial_p , F(p)]= i\hbar \partial_p F(p)$. – Cosmas Zachos Oct 30 '19 at 16:51