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From Shankar, $$[P,H]=0\rightarrow [P,U(t)]=0$$ where $P$ is the momentum operator, $H$ is the Hamiltonian, and $U(t)$ is the propagator to the Hamiltonian.

My first question is why does this follow? Shankar says that the propagator is a function of the Hamiltonian, but from what I understand it is constructed from eigenvectors, and not from the Hamiltonian itself.

My second question is why would the commutator relation automatically hold for a function of $H$? Is this only for relations where $$[\Omega,H]=0$$ so for a function $\Lambda$ of $H$, by taking the Taylor expansion $$[\Omega,\Lambda]=[\Omega,\lambda_0+\lambda H+\frac{\lambda}{2!}H^2+\cdots]=0$$ where $\lambda$ is a c number related to $\Lambda$. If say $[\Omega,H]=\text{const}$, would that immediately imply $[\Omega,\Lambda]=\text{const}$?

My third question is for time dependent Hamiltonians, $[P,H(t)]=0$ must hold for all time in order for $[P,U(t)]=0$ to also hold. Is this because if it didn't commute then the commutator would be a time dependent function, $[P,H(t)]=f(t)$, so the propagator would no longer commute with $P$ (i.e. $[P,U(t)]=f(t)$)?

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  • $\begingroup$ For t-independent H, you have $U(1)= \exp (-itH/\hbar)$, of course. In (2), commutation with H suffices for commutation with its powers. In (3), indeed commuting with H(t) for all t suffices for commutation with U(t) to vanish, as this is now a time-ordered exponential. $\endgroup$ Jul 19 at 19:01
  • $\begingroup$ @CosmasZachos so for (2) I get that the Taylor series commutes, I guess my question is that whether that is the reason why the propagator commutes, by expanding $\exp(-itH/\hbar)$ to see how if $P$ and $H$ commutes $U(t)$ will also commute. However if the commutator equals a constant for $[P,H]$ then it will be a different constant for the commutator of $[P,U(t)]$? Could you also elaborate what it means as a time-ordered exponential? $\endgroup$ Jul 19 at 20:19
  • $\begingroup$ @CosmasZachos is my understanding for (1) and (2) correct though? $\endgroup$ Jul 19 at 21:17
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I suspect you need to go back and refresh your evaluations of commutators, like the Hadamard lemma and basic identities, $$ e^A B e^{-A}= B + [A,B]+[A,[A,B]]/2!+ ... \\ [B,A^2]=\{ A, [B,A]\}, $$ etc. Now apply them to your items. You've muffed several points above, so I don't wish to rebuke them.

  1. For a time-independent H, note $U(t)= \int \!\!dE~ |E\rangle e^{-itE/\hbar}\langle E|=\exp (-it H/\hbar)~~~\leadsto$ $$ U^{-1}PU-P=0, ~~\implies U^{-1} [P,U]=0 ~~\implies [P,U]=0. $$

  2. See above. In general, the zero is useful, since you may prove recursively that
    $$[\Omega, H]=0~~\implies [\Omega, H^2]=0 \\ ~~\implies [\Omega, H^3]=0~...~\implies [\Omega, H^n]=0, $$ so also any function f(H). Convince yourself this fails for a constant non vanishing commutator. That is why people use the Hadamard lemma in Lie groups. Just don't go there.

  3. Most books work out the time-ordered exponential for $[H(t),H(t')]\neq 0$, $$ U(t)= \prod_0^t e^{a(t') \, dt'} \equiv \lim_{N \rightarrow \infty} \left( e^{a(t_N) \, \Delta t} e^{a(t_{N-1}) \, \Delta t} \cdots e^{a(t_1) \, \Delta t} e^{a(t_0) \, \Delta t} \right), $$ where the time moments {$t_ , …, t_N$} are defined as ti ≡ i Δt for i=0, …, N , and Δt ≡t/N , and $a(t)\equiv -iH(t)/\hbar$. Might look at (22) here.

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The other answer is good, but I would like to give another point of view, by using the definition of the evolution operator as the solution of an operator-valued ODE.

With a time dependent Hamiltonian $H(t)$, the propagator $U(t,t')$ is defined by : $$i\hbar\partial_t U(t,t') = H(t)\quad \text{and} \quad U(t,t') = \mathbb I$$

Then, if $A$ is an operator with $[A,H(t)] = 0$ for all $t$, we have : $$\partial_t[A,U(t,t')] = \left[A,\partial_t U(t,t')\right] = \frac{1}{i\hbar}[A,H(t)] = 0$$ together with $[A,U(t',t')]= [A,\mathbb I] = 0$, we get $[A,U(t,t') ]=0$ for all $t,t'$.

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