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I had seen a general case that $\hat{q}(t)$ and $\hat{q}(t')$ doesn't commute at different time $t$ and $t'$, where $\hat{q}(t)$ and $\hat{q}(t')$ are Operators in Heisenberg's view. I tried to prove it. But in my proof they will commute. Could anyone tell me, where am I wrong?

My attempt: Let consider displacement along x direction $\hat{q}(t)=x$, $\hat{q}(t')=x'$.
$$[\hat{q}(t),\hat{q}(t')]=xx'f-x'xf=xx'f-xx'f=0$$

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    $\begingroup$ When you write $\hat{q}(t)=x$, $\hat{q}(t')=x'$, you are assuming that the position operators at different times share the same eigenkets (can be simultaneously diagonalized to write them in the same position-basis) which assumes that they commute. $\endgroup$
    – ACat
    May 17, 2019 at 7:09
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    $\begingroup$ You can rather approach the problem head-on and write the different-time position operators as $e^{i\hat{H}t_2}\hat{X}(0)e^{-i\hat{H}t_2}$ and $e^{i\hat{H}t_1}\hat{X}(0)e^{-i\hat{H}t_1}$ and take a deep breath and calculate the commutator which is a bit boring/tedious but not hard. Just notice that all the Hamiltonians pass right-through each other but they do not commute with the position operator $\hat{X}(0)$. If you are not familiar with such calculations, I would just point out a useful identity for commutators: $[AB,C]=A[B,C]+[A,C]B$. $\endgroup$
    – ACat
    May 17, 2019 at 7:13

1 Answer 1

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It depends on the structure of the Hamiltonian operator since $X(t)$ is intepreted as the Heisenberg evolution of $X$. That is $X(t) = e^{iHt/\hbar}Xe^{-iHt/\hbar}$.

Here are two limit cases where computations can be explicitly performed:

  1. $$H = \frac{P^2}{2m}$$ Here $X(t) = X + t\frac{P}{m}$. It is easy to obtain that $$[X(t),X(t')]= \frac{t-t'}{m}[P,X] = \frac{-i\hbar(t-t')}{m}I \neq 0$$ An explicit formula can be obtained also with $$H = \frac{P^2}{2m} + \frac{\omega^2}{2}X^2$$ proving again that $$[X(t),X(t')]\neq 0\:.$$ In general, this is the result for Hamiltonians of the form $$H = \frac{P^2}{2m} + U(X)\:.$$

  2. $$H = aP$$ Here $X(t) = X+ ta I$. In this case, evidently $[X(t),X(t')]=0$.

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  • $\begingroup$ I was wondering as to where the Hamiltonians of the form $H=aP$ appear. I presume only in theories with one spatial dimension as they would otherwise violate rotational invariance, right? $\endgroup$
    – ACat
    May 17, 2019 at 23:45
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    $\begingroup$ Indeed, I was inspired from some elementary 1D model for Dirac Hamiltonian (one particle). But I was more interested in the mathematical aspects than physical ones. $\endgroup$ May 18, 2019 at 5:46
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    $\begingroup$ Thanks for that(@Dvij Mankad and @Valter Moretti). I am satisfied with the first part of your answer. I was not able to figure out the second part. But I am working on it. $\endgroup$
    – walber97
    May 18, 2019 at 6:16

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