Consider the following: $$[ \hat H, \hat x]=\left[-\frac{\hbar^2 \hat p^2}{2m}+V,\hat x\right]\ne0 \text{ in general}$$ But $$[ \hat H, \hat x]=\left[i\hbar \frac{\partial }{\partial t},\hat x\right]=0 $$ Why are these not in contradiction?


marked as duplicate by Quantum spaghettification, Qmechanic Nov 20 '15 at 20:11

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    $\begingroup$ Because the time derivative is not an operator on the space of states, so your second equation is meaningless, see e.g. this question $\endgroup$ – ACuriousMind Nov 20 '15 at 17:42
  • $\begingroup$ @ACuriousMind are you basically saying that states are a function of position only and the time part just describes the evolution of that position, as such to have an operator that takes a time derivative would be operating on something that we should consider to be 'fixed'? $\endgroup$ – Quantum spaghettification Nov 20 '15 at 17:45
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    $\begingroup$ @Joseph In addition to ACuriousMind's link you might find this helpful too, in particular Qmechanic's answer as to why $\hat H\neq i\hbar \partial _t$. physics.stackexchange.com/a/17479 $\endgroup$ – AngusTheMan Nov 20 '15 at 18:22
  • $\begingroup$ @ACuriousMind Thanks for your links, these pretty much answer my question and I think this one should be closed as a duplicate of either of them. I have added another question on a similar topic, using the ideas in these links, physics.stackexchange.com/q/219669. $\endgroup$ – Quantum spaghettification Nov 20 '15 at 19:21
  • $\begingroup$ @AngusTheMan see my comment above. $\endgroup$ – Quantum spaghettification Nov 20 '15 at 19:22