# 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?

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: $$e^{-i\beta P/\hbar}|q\rangle = |q + \beta\rangle$$