# Commutator relations for Propagators

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. $$[P,U(t)]=f(t)$$)?

• 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. Jul 19 at 19:01
• @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? Jul 19 at 20:19
• @CosmasZachos is my understanding for (1) and (2) correct though? Jul 19 at 21:17

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.

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 $$t,t'$$.