What is the relation between adiabatic elimination and adiabatic theorem? Does adiabatic elimination come from adiabatic theorem?

# [Physics] the relation between adiabatic elimination and adiabatic theorem

quantum mechanics

#### Related Solutions

The adiabatic theorem is required to derive the Berry phase equation in quantum mechanics. Therefore the adiabatic theorem and the Berry phase must be compatible with one another. (Though geometric derivations are possible, they usually don't employ quantum mechanics. And while illuminating what is going on mathematically, they obscure what is going on physically.)

The question of degeneracy points is a little more subtle, but let me make one thing clear: **if one crosses a degeneracy point, the adiabatic theorem is no longer valid** and one cannot use the Berry phase equation that you have written in the question (the denominator will become zero at the degeneracy point).

Now, let us take the spin in magnetic field example as an illustration of the Berry phase. Suppose we have a spin-1/2 particle in a magnetic field. The spin will align itself to the magnetic field and be in the low energy state $E=E_-$. Now, we decide to adiabatically change the direction of the magnetic field, keeping the magnitude fixed. Adiabatically means that the probability of the spin-1/2 particle transitioning to the $E=E_+$ state is vanishingly small, i.e. $\hbar/\Delta t<<E_+-E_-$. Suppose now that the magnetic field traces out the loop below, starting and ending at the red point:

In this case, one will pick up a Berry phase equal to:

\begin{equation} \textbf{Berry Phase}=\gamma = -\frac{1}{2}\Omega \end{equation}

where $\Omega$ is the solid angle subtended. This formula is proven in Griffiths QM section 10.2. However, it is not that important to understand the overall picture.

I chose this example because there a couple things to note that make it relevant to your question:

1) The adiabatic theorem is **critical** in this problem for defining the Berry phase. Since the Berry phase depends on the solid angle, any transition to the $E=E_+$ state would have destroyed the meaning of tracing out the solid angle.

2) The degeneracy point lies at the center of the sphere where $B=0$, where $B$ is the magnetic field. Though the spin may traverse any loop on the sphere, it cannot go through this degeneracy point for the Berry phase to have any meaning. This degeneracy point is ultimately responsible for the acquisition of the Berry phase, however. We must in some sense "go around the degeneracy point without going through it" for one to obtain a Berry phase.

## Best Answer

Yes, the adiabatic elimination is a particular application of the adiabatic theorem in a more complicated context.

The adiabatic theorem says that if the Hamiltonian is changing as a function of time very slowly and there is a gap in the spectrum around the eigenvalue $E$, the eigenstate with this eigenvalue will continuously evolve into the "same" eigenstate of the gradually changing Hamiltonian, without getting a contribution from other eigenstates.

Adiabatic elimination is a "practical" application in systems with at least three states. The first-to-second transition may be driven by an AC field and the second-to-first decay may be eliminated by the effect described in the adiabatic theorem. The time-dependent Hamiltonian in this case involves the gradual increase of the Stark shift.