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Assume the Hamiltonian of a quantum particle to be independent of time and be of the form $H=\frac{1}{2m}(\hat{p}^{2}_{x}+\hat{p}^{2}_{y})+V(x,y)$. Define a new operator $\hat{p}=\hat{p}_{x}-\hat{p}_{y}$ and such that it commutes with the $H$. But then this implies that $\hat{p}$ is independent of time. In other words, $\hat{p}(t)=\hat{p}(0)$ for all $t\in [0,a]$. But we can find $\hat{p}(0)$ from the initial value of the problem or from the initial conditions of experimental setup in the lab. Therefore, $\hat{p}_{x}(t)-\hat{p}_{y}(t)=\hat{p}(0)I$ for $t>0$. Now, the commutator $[\hat{x}(t), \hat{p}_{x}(t)]=[\hat{x}(t),\hat{p}(0)I+\hat{p}_{y}(t)]=0$ for $t>0$. But how can this be true?

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    $\begingroup$ Why is $[\hat{x},\hat{p}(0)]=0$? $\endgroup$
    – Philip
    Jun 28 '20 at 22:59
  • $\begingroup$ since $\hat{p}(0)$ is just a constant and we are considering the commutator of $\hat{x}(t)$ for times $t>0$. $\endgroup$
    – user11937
    Jun 28 '20 at 23:10
  • $\begingroup$ Another question: are you imposing any condition on $V(x,y)$ so that $\hat{p}$ commutes with $\hat{H}$? It seems to me that the only way that could happen is if $V(x,y) \propto (x + y)$. But I could be wrong... $\endgroup$
    – Philip
    Jun 28 '20 at 23:35
  • $\begingroup$ the example I had in mind was exactly what you said i.e. $V(x,y)$ being proportional to $(x+y)$. $\endgroup$
    – user11937
    Jun 28 '20 at 23:38
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For clarification, we are working in the Heisenberg Picture, where the operators $\hat{O}$ follow the Heisenberg Equation of Motion:

$$i\hbar \frac{\text{d}\hat{O}}{\text{d}t} = [\hat{O},\hat{H}].$$

Your problem is that you seem to assume that $[\hat{x}(t),\hat{p}(0)]=0$, which is not true. The fact that $\hat{p}(0)$ is a constant in time does not alter the fact that it is still an operator $\hat{p}(0) = \hat{p}_x(0) - \hat{p}_y(0)$, which may not commute with $x(t)$. Now, you might be tempted to think that for $t>0$ the commutator $[\hat{x}(t),\hat{p}_x(0)] = 0$, but this not true either, since $\hat{x}(t)$ depends on $\hat{x}(0)$.

This can easily be seen if you actually solve the problem. I'm going to assume that $V(x,y) = \alpha (x + y)$, so that $[\hat{p},\hat{H}]=0$. If you solve Heisenberg's Equations of Motion for this Hamiltonian, you can show that

\begin{equation*} \begin{aligned} \hat{p}_x(t) &= \hat{p}_x(0) - \alpha t \mathbb{I}\\ \hat{p}_y(t) &= \hat{p}_y(0) - \alpha t \mathbb{I} \\ &\\ \hat{x}(t) &= \hat{x}(0) + 2 \hat{p}_x(0)t - \alpha t^2\mathbb{I} \\ \hat{y}(t) &= \hat{y}(0) + 2 \hat{p}_y(0)t - \alpha t^2\mathbb{I} \\ \end{aligned} \end{equation*}

Clearly, $\hat{p}_x(t)- \hat{p}_y(t) = \hat{p}(0)$ is independent of time, as you have shown. However, $$[x(t),p_x(t)] = [x(0),p_x(0)] = i\hbar,$$

as you would expect.

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  • $\begingroup$ thanks for answering @Philip I know that we should get the answer of the usual $I\hbar$ but in defense of the problem what if the system is prepared in such a way that $\hat{p}(0)=CI$ for some constant $C$ and $I$ is the identity operator. This, corresponds to a particle whose momentum in the x component differs by a constant from the momentum in the y component. We could even prepare as system where $\hat{p}(0)=0$. What about then? $\endgroup$
    – user11937
    Jun 29 '20 at 0:45
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    $\begingroup$ I don't think it's possible to do what you're proposing: creating a particle in a certain state doesn't change the operator. I find it's easier to understand this in the Schrodinger picture: the operators remain the same irrespective of the state in which the particle is prepared. Now, the same operator in the Heisenberg picture is simply: $$\hat{A}_H(t) = U^\dagger(t) \hat{A}_S U(t),$$ i.e. it just changes differently with time, so I don't see how you could write $\hat{p} = C \mathbb{I}$... $\endgroup$
    – Philip
    Jun 29 '20 at 1:35
  • $\begingroup$ suppose you have an source of electrons that lets out electrons with $\hat{p}_{x}=\hat{p}_{y}+CI=\hat{p}_{y}+C\sum_{p} |p><p|$. This is a valid operator acting on the Hilbert space in question. $\endgroup$
    – user11937
    Jun 29 '20 at 1:50
  • $\begingroup$ Again, I don't think this is possible: you again speak of a source of electrons that somehow changes the operator $\hat{p}_x$. Look at it this way: $\hat{p}_x(0)$ is just the momentum operator in the $x$ direction in the Schrodinger picture, and similarly for $\hat{p}_y(0)$. Now, the operator $\hat{p}(0)$ in the Schrodinger picture is just $$\hat{p}(0) \equiv \frac{\hbar}{i} \left( \frac{\partial}{\partial x} - \frac{\partial}{\partial y} \right).$$ This doesn't change whether the particle is in a state of definite momentum, or in any arbitrary state! $\endgroup$
    – Philip
    Jun 29 '20 at 2:00
  • $\begingroup$ I think it matters, because as I pointed out $\hat{p}(t)$ is independent of time and hence is equal to $\hat{p}(0)$ in other words it is a stationary state. Now, if a particle is prepared at time t=0, such that the difference of the momenta in the corresponding direction is constant then it will stay constant for the duration of some time until the system gets disturbed. Also, during that time, the operator $\hat{p}_{x}(t)$ and $\hat{p}_{y}(t)$ will have eigenvalues that differ by a constant momentum C and as a result the commutation relations don't agree with the usual ones. that's all. $\endgroup$
    – user11937
    Jun 29 '20 at 2:23

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