Is the Heisenberg-picture commutator $[x(t),p(t)]$ between position and momentum always equal to $i\hbar$? Some misconceptions over here, For $x=$position and $p=$ momentum,
I know $[x,p]=i\hbar$ but does $[x(t),p(t)]$ still have the same relation where $t$ here represents time.
 A: $$
[\hat{x}(t),\hat{p}(t)] = [\hat{x}_0,\hat{p}_0] = i\hbar
$$
This is easily proven as follows. We have the unitary time evolution operator given by $\hat{T}(t)$. Unitary means $\hat{T}(t)\hat{T}^{\dagger}(t) = \hat{T}^{\dagger}(t)\hat{T}(t) = \hat{I}$ where $\hat{I}$ is the identity operator. 
Let $A_0$ and $B_0$ be arbitrary operators with $[A_0,B_0]=C_0$.
In the Heisenberg picture we have
\begin{align}
\hat{A}(t) &= \hat{T}^{\dagger}(t)\hat{A}_0\hat{T}(t)\\
\hat{B}(t) &= \hat{T}^{\dagger}(t)\hat{B}_0\hat{T}(t)\\
\hat{C}(t) &= \hat{T}^{\dagger}(t)\hat{C}_0\hat{T}(t)\\
\end{align}
So
\begin{align}
[\hat{A}(t),\hat{B}(t)] &= \hat{A}(t)\hat{B}(t) - \hat{B}(t)\hat{A}(t)\\
&= \hat{T}^{\dagger}(t)\hat{A}_0\hat{T}(t)\hat{T}^{\dagger}(t)\hat{B}_0\hat{T}(t) - \hat{T}^{\dagger}(t)\hat{B}_0\hat{T}(t)\hat{T}^{\dagger}(t)\hat{A}_0\hat{T}(t)\\
&= \hat{T}^{\dagger}(t)[\hat{A}_0,\hat{B}_0]\hat{T}(t) = \hat{T}^{\dagger}(t)C_0\hat{T}(t)\\
&=C(t)
\end{align}
So we see that commutation relations are preserved by the transformation into the Heisenberg picture. Actually, we see that commutation relations are preserved by any unitary transformation which is implemented by conjugating the operators by a unitary operator.
Note that unequal time commutation relations may vary. For example, if we have a Harmonic oscillator Hamiltonian
$$
\hat{H}_0 = \frac{1}{2}m\omega^2 \hat{x}^2 + \frac{1}{2m} \hat{p}^2
$$
We can derive and solve the Heisenberg equations of motion. Upon doing this we will get the usual classical equations of motion but with hats.
\begin{align}
\hat{x}(t) = \hat{x}_0 \cos(\omega t) + \hat{p}_0 \sin(\omega t)\\
\hat{p}(t) = \hat{p}_0 \cos(\omega t) - \hat{x}_0 \sin(\omega t)
\end{align}
Consider $\hat{x}(0) = \hat{x}_0$ and $\hat{x}(\frac{\pi}{2}\frac{1}{\omega}) = \hat{p}_0$.
\begin{align}
\left[\hat{x}(0),\hat{x}\left(\frac{\pi}{2}\frac{1}{\omega}\right)\right] = [\hat{x}_0,\hat{p}_0] = i\hbar
\end{align}
So we see that the commutator of $\hat{x}$ with itself at different times can be non-zero.
A: As always $\vec p = - i\hbar \vec \nabla$ this is always true in Schrödinger quantum mechanics. 
