# Do position operators commute at different time?

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 $$\hat{q}(t)=x$$, $$\hat{q}(t')=x'$$.
$$[\hat{q}(t),\hat{q}(t')]=xx'f-x'xf=xx'f-xx'f=0$$

• 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.
– ACat
May 17, 2019 at 7:09
• 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: $[AB,C]=A[B,C]+[A,C]B$.
– ACat
May 17, 2019 at 7:13

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 $$[X(t),X(t')]=0$$.

• 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?
– ACat
May 17, 2019 at 23:45
• 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. May 18, 2019 at 5:46
• 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. May 18, 2019 at 6:16