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How to prove $$\hat{p}|x\rangle=i\hbar\frac{\partial}{\partial x}|x\rangle,$$ using $$[\hat{x},\hat{p}]=i\hbar~?$$ The question seems to be uncomplete because for any $f(x)$ $$[\hat{x},\hat{p}+f(x)]=i\hbar.$$

However, there is a seemingly right answer:

$$ \hat{Q} = e^{-i \frac{u}{\hbar} \hat{p}},[\hat{x},\hat{Q}]=u \hat{Q} $$ $$ \hat{x}\hat{Q}|x\rangle = (x+u)\hat{Q}|x\rangle $$ $$ \hat{Q}|x\rangle = |x+u\rangle $$ $$ \hat{T} = e^{i \frac{u}{\hbar} \hat{x}},[\hat{p},\hat{T}]=u \hat{T} $$ $$ \hat{p}\hat{T}|p\rangle = (p+u)\hat{T}|p\rangle $$ $$ \hat{T}|p\rangle = |p+u\rangle $$

Therefore

$$ \langle x | p \rangle = e^{i\frac{px}{\hbar}} \langle 0 |_x |0 \rangle_p $$

$$ \delta(p-p') = \langle p | p' \rangle = 2\pi \hbar |\langle 0 |_x |0 \rangle_p|^2 \delta(p-p') $$ $$ \langle 0 |_x |0 \rangle_p = \sqrt{\frac{1}{2 \pi \hbar}}$$ $$ \hat{p}|x\rangle = \int \hat{p} |p\rangle\langle p |x\rangle= \int \hat{p} |p\rangle e^{-i\frac{px}{\hbar}}\sqrt{\frac{1}{2 \pi \hbar}} = i\hbar\frac{\partial}{\partial x}|x\rangle $$ Which assumption is made here?

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marked as duplicate by Qmechanic Oct 16 '14 at 9:02

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ Hi Doris. Please don't repost a closed question in a new entry. $\endgroup$ – Qmechanic Oct 16 '14 at 9:03
  • $\begingroup$ @Qmechanic I am sorry, I edited my old question and found it was closed before, so I asked a new one, because the old one was closed and I still do not understand this question. $\endgroup$ – Doris Oct 16 '14 at 11:07