Changing Hamiltonian with some eigenvalues constant

[hilbert2]

TL;DR

A system where the adiabatic theorem holds in some cases even for a fast change in the potential energy function.

Suppose some quantum system has a Hamiltonian with explicit time dependence $\hat{H} := \hat{H}(t)$ that comes from a changing potential energy $V(\mathbf{x},t)$. If the potential energy is changing slowly, i.e. $\frac{\partial V}{\partial t}$ is small for all $\mathbf{x}$ and $t$, then the adiabatic theorem says that an initial state $\left|\psi\right.\rangle$ at time $t_1$ will be the equivalent state in the basis of new instantaneous eigenstates of $\hat{H}$ at a later time $t_2$. This is what is called adiabatic passage.

Now, if for example the lowest three instantaneous eigenvalues of $\hat{H}(t)$, denoted $E_1 (t)$, $E_2 (t)$ and $E_3 (t)$, stay constant in time despite the Hamiltonian changing, I would suppose that an initial state superposed from $\left|\psi_1 \right.\rangle$, $\left|\psi_2 \right.\rangle$ and $\left|\psi_3 \right.\rangle$ at time $t_1$ will be the equivalent state in the new instantaneous eigenbasis at time $t_2$ even if the "passage" is made arbitrarily much faster with scaling $V(\mathbf{x},t) \rightarrow V(\mathbf{x},\lambda t)$ where $\lambda > 1$.

There seem to be some ways to simulate a faster than normal adiabatic passage even for all eigenvalues non-constant, such as in these references:

https://inspirehep.net/literature/1975583
https://www.nature.com/articles/s41467-021-27900-6

and this has some applications in quantum control theory.

But am I right or missing something here when I assume that an initial state, made of eigenstates of $\hat{H}$ that correspond to time-constant eigenvalues, can change "adiabatically" no matter how fast the change is?

 

[andresB]

Let's start from the beginning.

Let's say we have a time-dependent Hamiltonian $\hat{H}(t)$ with a discrete spectrum, and there is no level crossing for the time considered. The adiabatic theorem states that if we have an initial state $\left|\psi_{n0}\right\rangle$ such that $\hat{H}(t=0)\left|\psi_{0}\right\rangle =E_{n}(t=0)\left|\psi_{n0}\right\rangle$ then, if the time evolution is sufficiently “slow”, the initial state will evolve such that $\left|\psi_{n}(t)\right\rangle$ obeys the equation $\hat{H}(t)\left|\psi_{n}(t)\right\rangle =E_{n}(t)\left|\psi_{n}(t)\right\rangle .$

Now, what if the conditions of the adiabatic evolution are not met but we still want that our initial state $\left|\psi_{n0}\right\rangle$ evolve such that $\hat{H}(t)\left|\psi_{n}(t)\right\rangle =E_{n}(t)\left|\psi_{n}(t)\right\rangle$ is still true for all times? in that case, we can proceed as follows:

Instead of considering our original Hamiltonian $\hat{H}(t)$ we look for an alternative Hamiltonian $\hat{\widetilde{H}}(t)$ such that the non-adiabatic time evolution generated by $\hat{\widetilde{H}}(t)$ is

$$\hat{U}(t,t_{0})\left|\psi_{n0}\right\rangle =\left|\psi_{n}(t)\right\rangle $$

where $\hat{U}(t,t_{0})$ is solution to the Schrödinger equation $$\frac{d}{dt}\hat{U}(t,t_{0})=\hat{\widetilde{H}}(t)\hat{U}(t,t_{0}).$$

Notice that the time evolution is not generated by the original Hamiltonian. We have to find a new and different Hamiltonian that generates the desired path in the state space.

 

[hilbert2]

That's what they seem to do in those publications where they attempt to produce a pseudo-adiabatic change faster than how slow it usually has to be done. In the idea I wrote about, some part of the spectrum of $\hat{H}$ doesn't change in time. It seems to me that then you only have to consider the finite-dimensional subspace of the states with constant eigenvalue, if the initial state is a superposition of them. And then with regards to those states it's the same as if the Hamiltonian were time-independent in Schrödinger picture.

I'm not sure how one would experimentally put a particle in a time-variant potential that keeps some eigenvalues of $\hat{H}(t)$ constant, but that kind of $V(x,t)$ can be generated numerically by starting with some $V(x,t_0 )$ and step-by-step adding small perturbation terms (small enough for 1st order perturbation theory to be applicable) that keep some of the eigenvalues constant. That can be done by choosing a few perturbation potential functions $\tilde{V}_1 , \dots \tilde{V}_n$ and solving a linear system on each time step to find a linear combination of those $\tilde{V}_k$ that doesn't change the given eigenvalues in the 1st order approximation. It's a linear system because in 1st order the effect of several perturbation terms is additive.

 

[andresB]

Well, yes, the experimental realization can indeed be hard to implement. You have to find a new experimental set-up that mimics non-adiabatically the adiabatic evolution given by the old setup.

However, I think the theory is easy enough, take a look at the examples given in
https://sci-hub.se/https://iopscience.iop.org/article/10.1088/1751-8113/42/36/365303

 

[DrClaude]

Here is a good review of the STA (shortcut to adiabaticity) approach:
https://doi.org/10.1103/RevModPhys.91.045001

Note that the point of STA is not to find an alternative adiabatic route, but to allow intermediate changes of level so long as the final state is close enough to the one that would be arrived at by adiabatic transport.

 

[hilbert2]

I guess a possible example of a shortcut to adiabaticity would be if there's some way to adiabatically put a hydrogen atom in an electric field (Stark effect) with faster switching on of the $\vec{E}$-field when there's also a magnetic field during the process but $\vec{B} = \vec{0}$ at the beginning and end point.