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

  • $\begingroup$ 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

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 $t,t'$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.