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 $f(x)$ $$[\hat{x},\hat{p}+f(x)]=i\hbar.$$ But that's my homework. Can someone add some appropriate assumptions and prove the question?
Here 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?
