1
$\begingroup$

Can anyone explain why $$U(t)A_{H}(t)|\psi_{H}\rangle = A_{I}(t)|\psi_{I}(t)\rangle~?$$

This equation is on Piers Coleman's book, Introduction to Many-body physics. It is used to prove

$$ A_{H} = U^{\dagger}(t) A_{I}(t) U(t) \tag{5.99} $$

where $U(t)$ is the time evolution operator in interaction space which is Dyson time-ordered.

It may be something hidden in plaint sight, but I cannot figure it out.

$\endgroup$
5
  • $\begingroup$ $UA = UAU^{\dagger}U$ because $U$ is unitary. $\endgroup$
    – march
    Oct 3, 2023 at 19:53
  • $\begingroup$ Yes, you can use $UA=UAU^{\dagger}U$ to prove 5.99 from the first equation. But why do we have the first equation? $\endgroup$
    – QFT
    Oct 3, 2023 at 21:37
  • 1
    $\begingroup$ I think this relation only holds if you put $t_0=0$. Indeed, if you use these results, you should arrive (modulo typos) at $A_H=e^{-iH_0t_0} U^\dagger_I(t,t_0) A_I U_I(t,t_0) e^{iH_0t_0}$. Note that, as usual, $A_H$ as well as $A_I$ depend on $t$ but also on $t_0$ as a parameter (via the time-evolution operators). Eq. 5.7 in the book suggests that the author puts $t_0=0$, as far as I can see... $\endgroup$ Oct 3, 2023 at 22:48
  • $\begingroup$ @TobiasFünke Thanks. I have worked out and explained as the following. $\endgroup$
    – QFT
    Oct 4, 2023 at 21:11
  • $\begingroup$ You're right, I made a mistake regarding the $t_0$ stuff in the comment above. Eq. $(1)$ in the linked answer actually also assumes $t_0=0$ (where $t_0$ is the time where both pictures coincide), but can be generalized easily to account for arbitrary $t_0$. So the statement in the comment above is false, I think, and only correct (and the desired result) if you also put $t_0=0$ there. But starting from the properly generalized eq. $(1)$ of the linked answer, eq. $(5.99)$ of the book immediately follows for any choice of $t_0$. You should double check all of this, though. $\endgroup$ Oct 4, 2023 at 22:16

1 Answer 1

1
$\begingroup$

Thanks @TobiasFünke

  1. I have worked through the page you mentioned. It was a great answer. I learned

$U_{I}(t, t') = e^{iH_{0}t}~U_{S}(t,t')~e^{-iH_{0}t'}$

I have one comment. In your answer $t'$ is assumed to be a second observation point. The initial point of the system is at $t = 0$. But in the question, $t_{0}$ is assumed to be the initial point of the system, where you turn on your Hamiltonian.

  1. I have worked out the trick in my question.

$\langle\psi_{H}|A_{H}(t)|\psi_{H}\rangle = \langle\psi_{I}(t)|A_{I}(t)|\psi_{I}(t)\rangle$

$\langle\psi_{H}|A_{H}(t)|\psi_{H}\rangle = \langle\psi_{I}(0)|U_{I}(t)^{\dagger}A_{I}(t)|\psi_{I}(t)\rangle$

Since $|\psi_{I}(0)\rangle=|\psi_{S}(0)\rangle=|\psi_{H}\rangle$

$A_{H}(t)|\psi_{H}\rangle = U_{I}(t)^{\dagger}A_{I}(t)|\psi_{I}(t)\rangle$

$U_{I}(t)A_{H}(t)|\psi_{H}\rangle = A_{I}(t)|\psi_{I}(t)\rangle$

  1. Unfortunately for me equ (5.7) in Piers Coleman's book is questionable. He simply wrote

$|\psi_{I}(t)\rangle = U(t)|\psi_{I}(0)\rangle$

$U(t) = e^{iH_{0}t}e^{-iHt}$ (5.7)

Where he put $H$ as a constant. However the theme of this chapter and the goal of interaction picture is separate the time dependent part $V$ and the constant $H_{0}$. Thus $H$ is supposed to be time dependent, and should be something like a Dyson Time-ordered operator.

Best.

$\endgroup$
1
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Oct 4, 2023 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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