# More on the Berry phase

I know that students find their first encounter with the Berry phase quite a challenge. On the other hand, I’m guessing that the adiabatic approximation that we discuss first seems a lot more straightforward. My aim in this note is to show that that this apparent difference in difficulty is illusory. I suppose this may be good or bad news....

Recall that the idea of the adiabatic approximation was to express the state of our system in terms of the
instantaneous eigenbasis of the Hamiltonian. For the spin-1/2 system discussed in the lectures, this is

$$|\Psi(t)\rangle = c_{+}(t)|+_{t}\rangle+c_{-}(t)|-_{t}\rangle$$

One then finds that in the adiabatic approximation (when the Hamiltonian changes sufficiently slowly -- see the notes for the precise formulation) the evolution of the coefficients under the time dependent Schrödinger equation is very simple

$$c_{\pm}(t) = \exp\left(-\frac{i}{\hbar}\int_{0}^{t}E_{\pm}(t')\, dt'\right)c_{\pm}(0).$$

This is very appealing, as it seems like a modest generalisation of the evolution of stationary states. Unfortunately, it is meaningless! The problem is that, since $$|\pm_{t}\rangle$$ are defined as eigenstates, their phase is not fixed. We could just as well take some states differing by arbitrary phases at each instant in time. Now bear in mind that we are searching for a solution of the time dependent Schrödinger equation, completely determined (including its phase) by the initial state. So this arbitrariness will not do!

The solution is to fix once and for all the phase of the eigenstates for each value of the instantaneous Hamiltonian. That’s why we switch to labelling them by $$\mathbf{H}$$ rather than time. This seems rather modest, but it has dramatic consequences.

In our original derivation of the adiabatic approximation, we assumed (tacitly) that as the Hamiltonian changed by a differential amount, we only had to account for the change of the eigenstate in an orthogonal direction. If we were working with real vectors, this would be fairly clear, as the change of a normalised vector can only be perpendicular to the vector. For complex normalised vectors, we have the phase to worry about, but we ignored this subtlety on the first pass.

It turns out that it is not possible to choose the states so that the change in the states is only perpendicular to the states. The best way to convince yourself of this is to actually find the eigenstates explicitly in terms of the direction of $$\mathbf{H}$$ that defines the Hamiltonian. There are many ways to do this, corresponding to different assignments of phase, but one way is done in Eq. (2.37-2.39). Then you find

$$\langle \mathbf{H},+|(\delta|\mathbf{H},+\rangle)=i \mathbf{A}_+\cdot\delta \mathbf{H},$$

which will be nonzero. Any other assignment of the phase to the eigenstates can only lead to a gauge transformation of $$\mathbf{A}_+$$. Such a transformation can never eliminate $$\mathbf{A}_+$$ altogether. One way to see that is to find the curl of $$\mathbf{A}_+$$. It is nonzero, and can’t be changed by a gauge transformation.

In particular, the total phase change (in addition to the dynamical phase involving the energy discussed above) is the line integral of $$\mathbf{A}_+$$ around a closed loop in parameter space. It is related to the integral of the curl over a bounding surface, by Stokes’ theorem, and gives a gauge invariant phase change:
Berry’s phase.

As I say in the notes, I do recommend looking at
Berry’s paper