- #1

Llukis

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- TL;DR Summary
- How to write the time evolution of a Hamiltonian

Dear everybody,

Let me ask a question regarding the unitary time evolution of a given Hamiltonian.

Let's start by considering a Hamiltonian of the form ##H(t) = H_0 + V(t)##. Then, I move to the interaction picture where the Schrödinger equation is written as

$$ i\hbar \frac{d}{dt} |\psi^\prime(t)\rangle = V^\prime (t) |\psi^\prime(t)\rangle \: , $$

where ##|\psi^\prime(t)\rangle = \exp \big( i H_0 t/\hbar \big)|\psi(t)\rangle = U^\dagger_0(t) |\psi(t)\rangle## and ## V^\prime (t) = U_0^\dagger (t) V(t) U_0(t)##. The solution of this equation in the interaction picture is of the form

$$|\psi^\prime(t)\rangle = U_I (t)|\psi^\prime(0)\rangle \: ,$$

with the unitary evolution operator defined as

$$U_I(t) = \mathcal{T} \exp \bigg( -\frac{i}{\hbar}\int_0^t V^\prime (t^\prime) dt^\prime\bigg) \: ,$$

where I have introduced the time ordering operator ##\mathcal{T}##. So far, so good. My question is if then I can declare that the following is true:

$$V^\prime (t) = U_I(t) V^\prime(0) U^\dagger_I(t) \: .$$

Perhaps, it is a simple question but I would like to clear up my doubt.

Thanks to all of you in advance

Let me ask a question regarding the unitary time evolution of a given Hamiltonian.

Let's start by considering a Hamiltonian of the form ##H(t) = H_0 + V(t)##. Then, I move to the interaction picture where the Schrödinger equation is written as

$$ i\hbar \frac{d}{dt} |\psi^\prime(t)\rangle = V^\prime (t) |\psi^\prime(t)\rangle \: , $$

where ##|\psi^\prime(t)\rangle = \exp \big( i H_0 t/\hbar \big)|\psi(t)\rangle = U^\dagger_0(t) |\psi(t)\rangle## and ## V^\prime (t) = U_0^\dagger (t) V(t) U_0(t)##. The solution of this equation in the interaction picture is of the form

$$|\psi^\prime(t)\rangle = U_I (t)|\psi^\prime(0)\rangle \: ,$$

with the unitary evolution operator defined as

$$U_I(t) = \mathcal{T} \exp \bigg( -\frac{i}{\hbar}\int_0^t V^\prime (t^\prime) dt^\prime\bigg) \: ,$$

where I have introduced the time ordering operator ##\mathcal{T}##. So far, so good. My question is if then I can declare that the following is true:

$$V^\prime (t) = U_I(t) V^\prime(0) U^\dagger_I(t) \: .$$

Perhaps, it is a simple question but I would like to clear up my doubt.

Thanks to all of you in advance

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