Commutation of Hamiltonian with momentum In which case does the Hamiltonian $H$ commutes with the momentum $P$?
Can anybody help me? With an example? (No particular or strange Hamiltonians and no particular momenta are involved).
How can I prove that $[H, P] = 0$?
 A: When the Hamiltonian is invariant under translations. To see this, recall that $P$ is the infinitesimal generator of translations. As shown by, e.g. Dirac in Lectures on Quantum Mechanics, any infinitesimal generator of a symmetry commutes with the Hamiltonian, which itself is the generator of time-translations, i.e. of the dynamics.
Typical examples of an Hamiltonian that commutes with $P$ is the free particle, or more generally any admissible function of $P$ alone. The QHO is an example where such a commutation doesn't hold, as the harmonic potential clearly breaks the symmetry under translation (and of course a function of the positions $Q$ might fail to commute with $P$).
A: Here's a quick proof:
\begin{equation}
[\hat H, \hat p] = [\hat T+ V, \hat p] \\
=[\dfrac{\hat p^2}{2m}+ V, \hat p] \\
=[\dfrac{\hat p^2}{2m}, \hat p] + [V, \hat p] \\
\end{equation}
Note that here, in general, the potential is a function of x, i.e. $V(x)$. Using the property of commutators that:
$$[AB, C] =A[B,C]+[A,C]B$$
Also, using the result that for any function $f$:
$$[f, \hat p]=i \hbar \dfrac{\partial f}{\partial x}$$
We get:
\begin{equation}
[\hat H, \hat p] = \dfrac{1}{2m}(\hat p[\hat p,\hat p]+[\hat p,\hat p]\hat p)+i \hbar \dfrac{\partial V}{\partial x}
\end{equation}
Operator commutes with itself! so $[\hat p, \hat p]=0$:
\begin{equation}
[\hat H, \hat p] = i \hbar \dfrac{\partial V}{\partial x}
\end{equation}
If $\dfrac{\partial V}{\partial x}=0$, i.e. $V$ has no explicit dependence upon $x$, then:
\begin{equation}
[\hat H, \hat p] = 0
\end{equation}
A: The Hamiltonian operator for a quantum mechanical system is represented by the imaginary unit times the partial time derivative. The momentum is proportional to the gradient. When you derive a system with respect to two independent variables (which is what the partial derivative does, it ignores your position as a function of time), it doesn't matter which you derive it in respect to first.
Hence the time derivative and the gradient commute.
Since the coefficients of proportionality are constant scalars, they also commute with the two derivatives, making it all cancel out and give zero.
I do not know how to make equations here so this is the best I can give you unless this works:
$$\begin{align}[P_j,H]\psi &=-i\hbar\frac{\partial}{\partial x_j}i\hbar\frac{\partial}{\partial t}\psi-i\hbar\frac{\partial}{\partial t}(-i\hbar)\frac{\partial}{\partial x_j}\psi\\ &=\hbar^2\left(\frac{\partial}{\partial x_j}\frac{\partial}{\partial t}-\frac{\partial}{\partial t}\frac{\partial}{\partial x_j}\right)\psi\\ &=\hbar^2\left[\frac{\partial}{\partial x_j},\frac{\partial}{\partial t}\right]\psi\end{align}$$
but $t$ and $x_j$ (the $j^\text{th}$ spatial variable where $x_1$ is the $x$-coordinate, $x_2=y,\ x_3=z$) are treated independently by the partial derivatives (if it was fully derived $\frac{d}{dt}$ then you'd turn some spatial coordinates into speed and such) meaning that $\left[\frac{\partial}{\partial x_j},\frac{\partial}{\partial t}\right]=0$. Hence $[P_j,H]=0$ and $[\vec{P},H]=\vec{0}$.
