Under what conditions the time evolving operator $U_t=\exp \left( -\frac{iHt}{\hbar}\right)$ and the translator operator $T_a=\exp \left( -\frac{ipa}{\hbar}\right)$ commute, and so the order of applying do not mind. This doubt appeared to me because if we work with the following notation to my point of view the results must be the same... but this only will happen if they commute, $$[U_t,T_a]=0$$

\begin{align} &\text{Time evolution and after translation} \\ &\langle x | \psi\rangle \quad \rightarrow \quad \langle x |U_t | \psi\rangle=\langle x | \psi(t)\rangle \quad \rightarrow \quad \langle x |T_a | \psi\rangle=\langle x -a| \psi(t)\rangle=\psi(x-a,t)\\ \\ &\text{Translation and after time evolution} \\ &\langle x | \psi\rangle \quad \rightarrow \quad \langle x |T_a | \psi\rangle=\langle x-a | \psi\rangle \quad \rightarrow \quad \langle x-a |U_t | \psi\rangle=\langle x -a| \psi(t)\rangle=\psi(x-a,t) \end{align}

Why this doesn't work, or what I'm missing? Because with this reasoning the result must be the same when applying before or after the operators.


1 Answer 1


First of all, you are evaluating the product $T_a U_t$ in both cases here, since you allow $T_a$ to act "to the left" and $U_t$ "to the right" in both cases. You are assuming they commute when you do this.

Second, you are assuming very specific forms for your bra and ket here! It is perhaps not surprising that more general choices (not position/energy eigenvectors) will give different answers.

  • $\begingroup$ Ok, so in general when we have this case $\langle\alpha |AB|\beta\rangle$ we have 2 options to no break any rule 1) Or apply all operators following the order and to the same 'direction' or 2) Apply the operators AB, as a new complete operator $\endgroup$
    – Euler
    Nov 5, 2021 at 11:39

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