The time evolution of a Hamiltonian

[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

 

[DrClaude]

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 $H(t) = U(t) H(0) U^\dagger(t)$ ?

 

[etotheipi]

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

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

 

[DrClaude]

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

The prime doesn't stand for the derivative here.

 

[vanhees71]

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?

 

[vanhees71]

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.

 

[vanhees71]

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.

 

[vanhees71]

Yes, and then you have to be very careful: In the interaction picture, there is a time dependence from the operators and the explicit time dependence, i.e., you have
$$V(\hat{x}'(t),\hat{p}'(t),t)=\exp(\mathrm{i} \hat{H}_0 t) V(\hat{x}'(0),\hat{p}'(0),t) \exp(-\mathrm{i} \hat{H}_0 t),$$
i.e., $\hat{H}_0$ only provides the time evolution for the time dependence which comes in from the time dependence of the (not explicitly time dependent) operators $\hat{x}$ and $\hat{p}$ but not the explicit one (see also my other posting above).

 

[Llukis]

Perhaps I am not explaining myself properly, it must be my fault. Do not worry, I am working a solution. If someone else is interested I will post here once I have finished