1
$\begingroup$

Quantum Mechanics Volume One page 188 by Claude Cohen Tannoudji.

In $q$ and $p$ state vectov formalism.

$QS(\lambda) |q\rangle=(q+\lambda)S(\lambda)|q\rangle$, where $S(\lambda)=e^{-i\lambda P/\hbar}$. Thus when only consider $p$ and $q$, one may effectively think $S(\lambda)|q\rangle=|\lambda+q\rangle$.

However, what if there is a third set of eigenvectors say $l$, does $S(\lambda)|p\rangle$ still holds? Meaning if there is a third set of eigenvectors say $l$, can one still regard there is no difference between $S(\lambda)|q\rangle$ and $|\lambda +q\rangle$?

(consider $[S(\lambda),L]$)

$\endgroup$
  • 2
    $\begingroup$ "a third set of eigenvectors, say l". Eigenvectors of what? If of an operator of Q and P, all goes through. If of an operator commuting with Q and P, all goes through too. What is troubling you? $\endgroup$ – Cosmas Zachos Sep 14 '18 at 21:33
0
$\begingroup$

The properity closely related to the fact that $|q>$ in heliber space is considered as a complete set of basis(Use of 'complete' as in 'complete set of states' or 'complete basis'), that is $S(\lambda)|q>$ as a states with eigenvalue of $\lambda +q$, there is "nothing more to say" about the vector $S(\lambda)|q>$ as it has been fully expressed in the $|q>$ space.

However, this is only valied in the space vector formalism, and once the language translated into wave foramilism or expression of finite dimesion of matrix, thre results may not hold.

More mathematical proofs are still welcome.

$\endgroup$

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.