The time evolution of a Hamiltonian

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##.
  • #1
Llukis
19
8
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 :biggrin:
 
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  • #2
What happens if, for example, ##V(t) = \alpha \sin(\omega t)##, such that ##V'(0) = 0##?
 
  • #3
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)## ?
 
  • #4
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##?
 
  • #5
etotheipi said:
Maybe I'm going crazy, but doesn't ##V'(0) = \alpha \omega##?
The prime doesn't stand for the derivative here.
 
  • #6
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 :biggrin:
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.
 
  • #7
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?
 
  • #8
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)##.
 
  • #9
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.
 
  • #10
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##.
 
  • #11
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.
 
  • #12
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).
 
  • #13
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 :smile:
 

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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