Time-dependence in the Hamiltonian

[Llukis]

TL;DR

How to tackle time-dependent Hamiltonians

Last week I was discussing with some colleagues how to handle time-dependent Hamiltonians. Concerning this, I would like to ask two questions. Here I go.

First question​

As far as I know, for a time-dependent Hamiltonian $H(t)$ I can find the instantaneous eigenstates from the following equation
$$H(t) |n(t)\rangle = \epsilon (t) |n(t)\rangle \: ,$$
where $\epsilon(t)$ is the eigenvalue at each instant of time. Thus, for each time $t$, the states $|n(t)\rangle$ form a basis and, therefore, I can write a general state of the system as
$$|\psi(t)\rangle = \sum_n c_n (t) |n(t)\rangle \: .$$
Is this statement correct? This means that the basis and the corresponding coefficients vary with time continuously.
If this is correct, can I go a step further and choose, for convenience, my basis at $t=0$? In other words,
$$|\psi(t)\rangle = \sum_n a_n (t) |n(t=0)\rangle \: ,$$
where the basis is static and the coefficients are different from the previous ones.

Second question​

Imagine that I can write my Hamiltonian as the sum of two terms (both depending on time)
$$H(t) = H_0(t) + H_1(t) \: ,$$
and that I am interested in working in the interaction picture where $|\psi^\prime (t) \rangle = U^\dagger_0(t) |\psi(t)\rangle$, with
$$U_0(t) = \mathcal{T} e^{-\frac{i}{\hbar}\int H_0(t^\prime)dt^\prime} \: .$$
Is this possible? Does the interaction picture remain valid when $H_0$ is a function of time? Perhaps it is not convenient to work in the interaction picture in this case, but I want to know if it is still well defined.

Thanks in advance for reading my post!

 

[vanhees71]

First question: You can of course choose this basis, but it won't help you much to solve your problem.

Second question: That's the correct procedure to switch from the Schrödinger to an interaction picture, where the eigenvectors of observable time-evolve with $\hat{H}_0$, i.e., the unitary operator $\hat{U}_0^{\dagger}$, while the state vectors time-evolve with $\hat{H}_1$ with the corresponding unitary operator
$$\hat{U}_1(t)=\mathcal{T} \exp \left [-\frac{\mathrm{i}}{\hbar} \int_0^t \mathrm{d} t' \hat{H}_0(t') \right].$$

 

[Llukis]

Thanks for taking your time to answer my questions!

First question: You can of course choose this basis, but it won't help you much to solve your problem.

So, I could write the state $|\psi (t) \rangle$ in the changing basis $|n(t)\rangle$ or choose a static basis at any given time, for instance at $t=0$. I wanted to be sure about that, thank you.
By the way, how would you face the problem of a time-dependent Hamiltonian? I would like to have some insights.

Regarding your answer to the second question, I know the procedure is correct, but I would like to know if even for a time-dependent $H_0(t)$ the interaction picture is still valid; I mean if it works in the same way that it does for a time-independent $H_0$.

Thank you so much.

 

[nucl34rgg]

The only time dependence in the basis vectors will arise as a phase so it is simplest to consider the basis the same as in the initial time and then just multiply by the time dependent phase.

 

[vanhees71]

Thanks for taking your time to answer my questions!
So, I could write the state $|\psi (t) \rangle$ in the changing basis $|n(t)\rangle$ or choose a static basis at any given time, for instance at $t=0$. I wanted to be sure about that, thank you.
By the way, how would you face the problem of a time-dependent Hamiltonian? I would like to have some insights.

Regarding your answer to the second question, I know the procedure is correct, but I would like to know if even for a time-dependent $H_0(t)$ the interaction picture is still valid; I mean if it works in the same way that it does for a time-independent $H_0$.

Thank you so much.

Formally it works the same as with a time-independent $\hat{H}_0$, but it will be quite more complicated.

 

[vanhees71]

The only time dependence in the basis vectors will arise as a phase so it is simplest to consider the basis the same as in the initial time and then just multiply by the time dependent phase.

For a Hamiltonian with some general explicit time-dependence it's unfortunately not a simple phase factor. That makes such problems so complicated.

 

[Llukis]

Sorry for my delay, I was a little bit busy so far.
Thanks for your answers. Let me add another question, just to be sure. Once I have found the instantaneous eigenstates of a $H(t)$
$$H(t) |n(t)\rangle = \epsilon_n (t) |n(t)\rangle \: , $$
can I write the spectral decomposition of the Hamiltonian? This is to say
$$H(t) = \sum_n \epsilon_n (t) |n(t)\rangle \langle n(t) | \: ,$$
at each instant $t$.
Another point I would like to clear up is if applying the time-evolution operator on a instantaneou eigenstate gives
$$\mathcal{U}(t) |n(t)\rangle = \exp \bigg( -\frac{i}{\hbar} \int_0^t \epsilon_n (t^\prime) dt^\prime \bigg) |n(t)\rangle \: . $$
Thank you all!