7
$\begingroup$

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?

$\endgroup$
12
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.