Quantum mechanics position/momentum state operator proof Is there any way to prove
$$
e^{-i\beta p}|q\rangle = |q+\beta\rangle
$$ 
just by using these identities $$
[q,\mathcal{F}(p)]=i\hbar \mathcal{F}'(p) \;\;\;\;[q,p]=i\hbar
$$
in quantum mechanics?
 A: Yes, you just have to check that $e^{-i\beta P/\hbar}|q\rangle$ is an eigenket of $Q$ (uppercase represents operators) with eigenvalue $q+\beta$. We evaluate
$$
Q \big (  e^{-i\beta P/\hbar}|q\rangle \big )
$$
by using the first identity (which is actually a consequence of the canonical commutation relation):
$$
Q e^{-i\beta P/\hbar} = e^{-i\beta P/\hbar}Q + i\hbar \left ( \frac{ -i \beta}{\hbar} e^{-i\beta P/\hbar} \right ),
$$
which upon substitution gives
$$
Q \big ( e^{-i\beta P/\hbar}|q\rangle \big ) = (q + \beta ) \big ( e^{-i\beta P/\hbar}|q\rangle \big ).
$$
This means that $e^{-i\beta P/\hbar}|q\rangle$ is proportional to an eigenket of $Q$ with eigenvalue $q+\beta$, namely $c|q+\beta\rangle$. Fortunately, since the displacement operator $e^{-i\beta P/\hbar}$ is unitary, $c$ must satisfy $|c|^2 = 1$ to preserve normalization of the position eigenkets. This means $c$ is just an arbitrary phase factor which can promptly be chosen to be unity and then we have our final result:
$$
e^{-i\beta P/\hbar}|q\rangle = |q + \beta\rangle
$$
