I was trying to calculate commutator $\left[\hat{P}, \hat{T}(\vec{dx})\right]$ where $\hat{P}$ is momentum operator and $ \hat{T}(\vec{dx})$ is Translation operator
My approach:
I tried calculating the terms
\begin{eqnarray}
\hat{P} \hat{T}(\vec{dx}) |x\rangle \text{ and } \hat{T}(\vec{dx})\hat{P}|x\rangle
\end{eqnarray}
Now I know that $\hat{T}(\vec{dx})|x\rangle = |x + dx\rangle$ but I dont know how will $\hat{P}$ operate on state $|x\rangle$ or $|x + dx\rangle$
You know that $\langle p|x\rangle = e^{-ixp/\hbar}/\sqrt{2\pi/\hbar}$, so $$ \langle p|\hat T (dx)|x\rangle = \langle p|x+dx\rangle= \frac{e^{-ip(x+dx)/\hbar}}{\sqrt{2\pi\hbar}}\\ =\langle p|x\rangle e^{-ipdx/\hbar}= \langle p|e^{-i\hat Pdx/\hbar}|x\rangle \\ \leadsto \hat T (dx)= e^{-i\hat P dx /\hbar}, ~~\implies~~ [\hat P, \hat T]=0. $$
Heed the counterintuitive minus sign in the exponent: kets translate oppositely to functions and bras!
Geeky : If you enjoy integrating delta functions by parts, you might well consider two handy representations of this unitary operator, $$ \hat T(a) =\int \!\! dp~|p\rangle e^{-ipa/\hbar}\langle p| = \int \!\! dx~|x\rangle e^{-a \partial_x}\langle x|. $$
