A question involving position and momentum operators 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?
 A: 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.
