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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 $[H, P] = 0$?

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  • $\begingroup$ Maybe if and only of I have a free theory, with No potential? $\endgroup$
    – Henry
    Jan 11, 2015 at 21:35
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    $\begingroup$ Do you know of the canonical commutation relation? $\endgroup$
    – Kyle Kanos
    Jan 11, 2015 at 21:36
  • $\begingroup$ @KyleKanos I just made the calculation, and I obtained [H, P] = -ih d/dx v(x) That is zero iff V = 0, that is in free theory! $\endgroup$
    – Henry
    Jan 11, 2015 at 21:40
  • $\begingroup$ What kind of answer are you looking for? One could be that $[H,P]=0$ when the system has translational invariance $\endgroup$
    – glS
    Jan 11, 2015 at 21:42
  • $\begingroup$ @glance does my above calculation holds? $\endgroup$
    – Henry
    Jan 11, 2015 at 21:43

3 Answers 3

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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}

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  • $\begingroup$ Isn’t $p^2$ the scalar operator $\mathbf{p}\cdot\mathbf{p}$? In that case, you’re actually calculating $[A\cdot B,C]$… is it the same? $\endgroup$
    – ric.san
    May 27, 2022 at 17:20
  • $\begingroup$ @ric.san: yes, you're right. That's exactly what I did to write out the commutator relation. $\endgroup$
    – user115625
    May 29, 2022 at 5:56
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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 $P$).

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    $\begingroup$ can I see the calculation please? I mean the commutation relation of the infinitesimal generators that commute with H. Also the book page would be awesome! $\endgroup$
    – Henry
    Jan 11, 2015 at 21:50
  • $\begingroup$ it just comes from expanding $e^{ipa}H(x)e^{-ipa} =H(x)$ around $a = 0$ and equating terms at each order. The sought relation comes from the first order IIRC $\endgroup$
    – Phoenix87
    Jan 11, 2015 at 21:56
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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 $[\vec{P},H]=\vec{0}$.

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    $\begingroup$ This answer is incorrect, the Hamiltonian is not the time derivative 'operator' (which is not itself an operator on Hilbert Space) and everything that follows is consequently also wrong. The answer by Phoenix87 is good. $\endgroup$
    – jacob1729
    May 30, 2019 at 9:51
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    $\begingroup$ To expand on prev comment: the Hamiltonian operator is represented by a suitable combination of operators representing contributions to energy. It is not true that $\hat{H} = i \hbar \partial /\partial t$. Rather, physical evolution over time is such that kets satisfy the Schr. equation. The fact that two operators have the same effect on a subset of all kets does not imply the operators are the same, even though in this example they will have the same effect on all physically allowed kets. $\endgroup$ May 30, 2019 at 9:53

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