# The time evolution of a Hamiltonian

• I
• Llukis
In summary, the conversation discussed the unitary time evolution of a Hamiltonian and the use of the interaction picture to solve the Schrödinger equation. The question was raised about whether the equation ##V^\prime (t) = U_I(t) V^\prime(0) U^\dagger_I(t)## is true, and it was determined that it is not always valid due to the distinction between implicit and explicit time dependence. The conversation also touched on the example of ##V(t) = \alpha \sin(\omega t)## with ##V'(0) = 0## and the need to take into account both types of time dependence in the equation of motion for ##V##.
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:
What happens if, for example, ##V(t) = \alpha \sin(\omega t)##, such that ##V'(0) = 0##?

DrClaude said:
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)## ?

DrClaude said:
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##?

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

Llukis said:
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.

vanhees71 said:
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?

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)##.

vanhees71 said:
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.

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##.

vanhees71 said:
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.

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).

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

## 1. What is the Hamiltonian in quantum mechanics?

The Hamiltonian in quantum mechanics is a mathematical operator that represents the total energy of a quantum system. It is used to describe the time evolution of a system and is crucial in understanding the behavior of particles on a quantum level.

## 2. How does the Hamiltonian affect the time evolution of a system?

The Hamiltonian determines the rate at which a quantum system evolves over time. It governs the changes in the state of a system, such as the position and momentum of particles, as time progresses.

## 3. What is the significance of the time evolution of a Hamiltonian?

The time evolution of a Hamiltonian is significant because it allows us to predict the future state of a quantum system. By solving the equations of motion described by the Hamiltonian, we can understand how a system will behave over time.

## 4. Can the Hamiltonian change over time?

Yes, the Hamiltonian can change over time, especially in systems where external forces or interactions are present. In such cases, the Hamiltonian is time-dependent and must be taken into account when studying the time evolution of the system.

## 5. How is the Hamiltonian related to the Schrödinger equation?

The Schrödinger equation is a fundamental equation in quantum mechanics that describes the time evolution of a system. The Hamiltonian is a key component of this equation, as it represents the total energy of the system. Solving the Schrödinger equation with the Hamiltonian allows us to determine the state of a system at any point in time.

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