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?
$\begingroup$
$\endgroup$
1
-
2$\begingroup$ Closely related. Basically yes, if you appreciate how the identities lead to the coordinate representation of momentum and Lagrange's shift operator. $\endgroup$ – Cosmas Zachos Jan 21 '20 at 0:26
Add a comment
|
$\begingroup$
$\endgroup$
1
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
