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.

$$[\hat{x}(t),\hat{p}(t)] = [\hat{x}_0,\hat{p}_0] = i\hbar$$

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.

As always $$\vec p = - i\hbar \vec \nabla$$ this is always true in Schrödinger quantum mechanics.

$$[x,p] =i \hbar$$

Inherently mean only for equal time. That's why these type of commutation relations sometimes ,for better care, also called as equal-time commutation relation. If we calculate at unequal time, this relation will not hold instead in that case $$x$$ and $$p$$ will commute.

Physically this means that you can not simulteneously know the position and momentum of a particle. More and more accurate knowledge of x means less and less knowledge of momentum or vice a versa. But if you want to know $$x$$ at one time and $$p$$ at other different time you can know at any degree of accuracy.

• "If we calculate at unequal time, this relation will not hold instead in that case x and p will commute." <--This is not true. They COULD commute but generally do NOT. – Jahan Claes Oct 2 '18 at 4:11
• @Jahan please check, I'm totally sure about it. or if you still believe what you commented above, please give the source. – Aman pawar Oct 2 '18 at 4:44
• So you are saying $[x(t),p(t')]=\delta_{t,t'}i\hbar$? – BioPhysicist Oct 3 '18 at 2:54
• Jahan's observation here is completely right. There is an explicit example in the other answer which you can use to check. – Emilio Pisanty Oct 3 '18 at 6:39