# The time evolution of a Hamiltonian

• I
Llukis
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

Last edited:

Mentor
What happens if, for example, ##V(t) = \alpha \sin(\omega t)##, such that ##V'(0) = 0##?

Llukis
What happens if, for example, ##V(t) = \alpha \sin(\omega t)##, such that ##V'(0) = 0##?
Yes, so this counterexample is enough to reject my assumption, I guess.
So, then, when can we write a general time-dependent Hamiltonian as ##H(t) = U(t) H(0) U^\dagger(t)## ?

What happens if, for example, ##V(t) = \alpha \sin(\omega t)##, such that ##V'(0) = 0##?

Maybe I'm going crazy, but doesn't ##V'(0) = \alpha \omega##?

Mentor
Maybe I'm going crazy, but doesn't ##V'(0) = \alpha \omega##?
The prime doesn't stand for the derivative here.

Gold Member
2022 Award
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
Be careful. That's a trap you get into easily. It's important to distinguish implicit and explicit time dependence. In this case you have
$$\hat{V}'=V(\hat{x}',t),$$
where the ##t## dependence is only the explicit time-dependence, but in the interaction picture also ##\hat{x}## (and ##\hat{p}##) depend on time. By definition they "move with ##\hat{H}_0##", i.e., the equations of motion for the operators read
$$\mathrm{d}_t \hat{x}'=\frac{1}{\mathrm{i} \hbar} [\hat{x}',\hat{H}_0].$$
Note that ##\hat{H}_0'=\hat{H}_0##. The solution of this equation of motion is
$$\hat{x}'(t)=\exp(\mathrm{i} \hat{H}_0 t/\hbar) \hat{x}'(0) \exp(-\mathrm{i} \hat{H}_0 t/\hbar).$$
But this implies that
$$V(\hat{x}'(t),t)=V[\exp(\mathrm{i} \hat{H}_0 t/\hbar) \hat{x}'(0) \exp(-\mathrm{i} \hat{H}_0 t/\hbar),t]=\exp(\mathrm{i} \hat{H}_0 t/\hbar) V(\hat{x}'(0),t)\exp(-\mathrm{i} \hat{H}_0 t/\hbar) .$$
This implies that you have to take into account the explicit time dependence in addition to the one coming in through the intrinsic time dependence of ##\hat{x}'(t)##, and that's why the equation of motion for ##V## reads
$$\mathrm{d}_t V(\hat{x}',t)=\frac{1}{\mathrm{i} \hbar}[V(\hat{x}',t),\hat{H}_0]+\partial_t V(\hat{x}',t),$$
where ##\partial_t## refers only to the explicit time dependence.

Llukis
Be careful. That's a trap you get into easily. It's important to distinguish implicit and explicit time dependence. In this case you have
$$\hat{V}'=V(\hat{x}',t),$$
where the ##t## dependence is only the explicit time-dependence, but in the interaction picture also ##\hat{x}## (and ##\hat{p}##) depend on time. By definition they "move with ##\hat{H}_0##", i.e., the equations of motion for the operators read
$$\mathrm{d}_t \hat{x}'=\frac{1}{\mathrm{i} \hbar} [\hat{x}',\hat{H}_0].$$
Note that ##\hat{H}_0'=\hat{H}_0##. The solution of this equation of motion is
$$\hat{x}'(t)=\exp(\mathrm{i} \hat{H}_0 t/\hbar) \hat{x}'(0) \exp(-\mathrm{i} \hat{H}_0 t/\hbar).$$
But this implies that
$$V(\hat{x}'(t),t)=V[\exp(\mathrm{i} \hat{H}_0 t/\hbar) \hat{x}'(0) \exp(-\mathrm{i} \hat{H}_0 t/\hbar),t]=\exp(\mathrm{i} \hat{H}_0 t/\hbar) V(\hat{x}'(0),t)\exp(-\mathrm{i} \hat{H}_0 t/\hbar) .$$
This implies that you have to take into account the explicit time dependence in addition to the one coming in through the intrinsic time dependence of ##\hat{x}'(t)##, and that's why the equation of motion for ##V## reads
$$\mathrm{d}_t V(\hat{x}',t)=\frac{1}{\mathrm{i} \hbar}[V(\hat{x}',t),\hat{H}_0]+\partial_t V(\hat{x}',t),$$
where ##\partial_t## refers only to the explicit time dependence.

Yes, you are right. I supposed that my time-dependent term ##V(t)## does not depend on ##x##, though.
When is this expression ##V^\prime (t) = U_I(t) V^\prime(0) U^\dagger_I(t)## valid?

Gold Member
2022 Award
I don't understand what ##V## then should be. Is it just that you switch on (and maybe off) a constant potential, i.e., ##V(t)## indepencent of ##\hat{x}##? Then the time-dependence is entirely explicit and ##\hat{U}_I## commutes with ##V## at all times, i.e., you simply have ##V'(t)=V(t)##.

Llukis
I don't understand what ##V## then should be. Is it just that you switch on (and maybe off) a constant potential, i.e., ##V(t)## indepencent of ##\hat{x}##? Then the time-dependence is entirely explicit and ##\hat{U}_I## commutes with ##V## at all times, i.e., you simply have ##V'(t)=V(t)##.

Well, ##V^\prime(t) = V(t)## is valid if ##V(t)## commutes with ##H_0## at all times, which is not the case, generally.

Gold Member
2022 Award
If ##V## doesn't depend on any operators but is simply a constant (i.e., ##\propto \hat{1}##) at all times it commutes with ##H_0##.

Llukis
If ##V## doesn't depend on any operators but is simply a constant (i.e., ##\propto \hat{1}##) at all times it commutes with ##H_0##.

But, in my case, ##V(t)## is a general time-dependent operator.

$$V(\hat{x}'(t),\hat{p}'(t),t)=\exp(\mathrm{i} \hat{H}_0 t) V(\hat{x}'(0),\hat{p}'(0),t) \exp(-\mathrm{i} \hat{H}_0 t),$$