# States created by translation operator

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

• "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? – Cosmas Zachos Sep 14 '18 at 21:33

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.