# Why $U(t)A_{H}(t)|\psi_{H}\rangle = A_{I}(t)|\psi_{I}(t)\rangle$?

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.

• $UA = UAU^{\dagger}U$ because $U$ is unitary. Oct 3, 2023 at 19:53
• Yes, you can use $UA=UAU^{\dagger}U$ to prove 5.99 from the first equation. But why do we have the first equation?
– QFT
Oct 3, 2023 at 21:37
• 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... Oct 3, 2023 at 22:48
• @TobiasFünke Thanks. I have worked out and explained as the following.
– QFT
Oct 4, 2023 at 21:11
• 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. Oct 4, 2023 at 22:16

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.