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

  • 1
    $\begingroup$ Thanks! Very helpful and concise! $\endgroup$ – fielder Jan 21 at 9:42

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