Simple Quantum Mechanics Question about The Commutator of Translation Operators Say there is $\hat{J} = \exp[-i \hat{p} l/ \hbar]$ and $\hat{U}= \exp[-i\hat{H}t/ \hbar]$, where $\hat{H}$ is time-independent. 
Can we say anything about $[\hat{J},\hat{U}]$? Is it zero? How do we show this? 
For example if $\hat{H} = \hat{p}^2 /2m + m\omega^2 \hat{x}^2/2$.
 A: It depends on the Hamiltonian.
In general in quantum mechanics, if $V$ is a unitary operator representing some symmetry, then we say that $H$ is invariant under that symmetry provided the Hamiltonian is invariant under conjugation by $V$;
\begin{align}
  V^{-1} H V = H.
\end{align}
Notice that this condition can also be written as $[H,V]=0$.  Now, if a Hamiltonian is invariant under such a symmetry, then we can multiply both sides by $-it/\hbar$, take the operator exponential of both sides, and use the identity $e^{A^{-1}BA} = A^{-1}e^BA$ to obtain
\begin{align}
  V^{-1}UV = U
\end{align}
which can be written as
\begin{align}
  [U,V]=0.
\end{align}
On the other hand, suppose that $[U,V]=0$, then expand the commutator in powers of $t$.  This gives
\begin{align}
  [I -(it/\hbar)H +\cdots, V]=0
\end{align}
which, after equating all coefficients of powers of $t$ on the left to zero implies $[H,V]=0$.  So we have shown that

The hamiltonian is invariant under a symmetry $V$, if and only if the time evolution operator $U$ commutes with $V$.

In the special case of spatial translations $T$ (which you have rather non-standardly labeled as $J$), the property $[U,T]=0$ holds if and only if $H$ is translation-invariant.
In the case of $H=P^2/2m+m\omega^2X^2$, the Hamiltonian is not translation-invariant because
\begin{align}
  T^{-1}HT = H + \frac{it}{\hbar}[P,H] + O(t^2)
\end{align}
and $[P,H]\neq 0$ for this Hamiltonian since $[P,P^2]=0$ but $[P,X^2]\neq 0$.
A: Note that in the very particular case of the quantum harmonic oscillator, it is interesting to use different representations. For instance: 
$$P \sim i(a-a^+), H \sim a^+a \tag{1}$$
You have formulae which allow you to disentangle $a$ and $a^+$ (below $\lambda$ is real): 
$$e^{\lambda(a^+-a)}=e^{\lambda a^+} e^{-\lambda a} e^{- \frac{\lambda^2}{2} }\tag{2}$$
This reduces the problem of finding a commutator $[e^{-i\alpha P}, e^{-i\beta H}]$, to finding commutators  $[e^{\alpha a}, e^{-i\beta a^+a}]$ and $[e^{-\alpha a^+}, e^{-i\beta a^+a}]$.
A: This is a good formula to remember, or at least, to think of, when you're dealing with the exponential of operators:
Baker–Campbell–Hausdorff formula
In particular, for your case, the braiding identity is useful. We see that $[J,U] \neq 0$.
