Adiabatic theorem and Berry phase

Question

As far as I can check, the adiabatic theorem in quantum mechanics can be proven exactly when there is no crossing between (pseudo-)time-evolved energy levels. To be a little bit more explicit, one describes a system using the Hamiltonian $H\left(s\right)$ verifying $H\left(s=0\right)=H_{0}$ and $H\left(s=1\right)=H_{1}$, with $s=\left(t_{1}-t_{0}\right)/T$, $t_{0,1}$ being the initial (final) time of the interaction switching. Then, at the time $t_{0}$, one has

$$H\left(0\right)=\sum_{i}\varepsilon_{i}\left(0\right)P_{i}\left(0\right)$$

with the $P_{i}$'s being the projectors to the eigenstates associated with the eigenvalue $\varepsilon_{i}\left(0\right)$, that we suppose known, i.e. $H_{0}$ can be exactly diagonalised. Then, the time evolution of the eigenstates is supposed to be given by

$$H\left(s\right)=\sum_{i}\varepsilon_{i}\left(s\right)P_{i}\left(s\right)$$

which is fairly good because it just requires that we are able to diagonalise the Hamiltonian at any time, what we can always do by Hermiticity criterion. The adiabatic theorem (see Messiah's book for instance)

$$\lim_{T\rightarrow\infty}U_{T}\left(s\right)P_{j}\left(0\right)=P_{j}\left(s\right)\lim_{T\rightarrow\infty}U_{T}\left(s\right)$$

with the operator $U_{T}\left(s\right)$ verifying the Schrödinger equation

$$\mathbf{i}\hslash\dfrac{\partial U_{T}}{\partial s}=TH\left(s\right)U_{T}\left(s\right)$$

can be proven exactly if $\varepsilon_{i}\left(s\right)\neq\varepsilon_{j}\left(s\right)$ at any time (see e.g. Messiah or Kato).

Now, the Berry phase is supposed to be non vanishingly small when we have a parametric curve winding close to a degeneracy, i.e. precisely when $\varepsilon_{i}\left(s\right) \approx \varepsilon_{j}\left(s\right)$. For more details, Berry defines the geometric phase as

$$\gamma_{n}\left(C\right)=-\iint_{C}d\mathbf{S}\cdot\mathbf{V}_{n}\left(\mathbf{R}\right)$$

with (I adapted the Berry's notation to mine)

$$\mathbf{V}_{n}\left(\mathbf{R}\right)=\Im\left\{ \sum_{m\neq n}\dfrac{\left\langle n\left(\mathbf{R}\right)\right|\nabla_{\mathbf{R}}H\left(\mathbf{R}\right)\left|m\left(\mathbf{R}\right)\right\rangle \times\left\langle m\left(\mathbf{R}\right)\right|\nabla_{\mathbf{R}}H\left(\mathbf{R}\right)\left|n\left(\mathbf{R}\right)\right\rangle }{\left(\varepsilon_{m}\left(\mathbf{R}\right)-\varepsilon_{n}\left(\mathbf{R}\right)\right)^{2}}\right\} $$

for a trajectory along the curve $C$ in the parameter space $\mathbf{R}\left(s\right)$. In particular, Berry defines the adiabatic evolution as following from the Hamiltonian $H\left(\mathbf{R}\left(s\right)\right)$, so a parametric evolution with respect to time $s$. These are eqs.(9) and (10) in the Berry's paper.

Later on (section 3), Berry argues that

The energy denominators in [the equation for $\mathbf{V}_{n}\left(\mathbf{R}\right)$ given above] show that if the circuit $C$ lies close to a point $\mathbf{R}^{\ast}$ in parameter space at which the state $n$ is involved in a degeneracy, then $\mathbf{V}_{n}\left(\mathbf{R}\right)$ and hence $\gamma_{n}\left(C\right)$, is dominated by the terms $m$ corresponding to the other states involved.

What annoys me is that the Berry phase argument uses explicitly the adiabatic theorem. So my question is desperately simple: what's the hell is going on there? Can we reconcile the adiabatic theorem with the Berry phase elaboration? Is the Berry phase a kind of correction (in a perturbative expansion sense) of the adiabatic theorem? Is there some criterion of proximity to the degeneracy that must be required in order to find the Berry phase?

REFERENCES:

• Berry, M. V. Quantal Phase Factors Accompanying Adiabatic Changes. Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences, 392 45–57 (1984).

• Kato, T. On the Adiabatic Theorem of Quantum Mechanics. Journal of the Physical Society of Japan, 5 435–439 (1950).

• Messiah, A. Mécanique quantique Dunod (French reprint of the 1958 edition) (1995).

Answer 1

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:

Berry Phase

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.

Answer 2

The compatibility with the adiabatic theorem has been made clear by the answer from @Xcheckr. I just would like to add some insight to the degeneracy issue.

This degeneracy, which gives rise to the acquisition of a nontrivial Berry phase, lives in the abstract parameter space. It is in general not to be traversed by the interested closed path of adiabatic parameter evolution somehow because it is a singularity (topological defect) in terms of the adiabatic construction. And in regards to the spin-$\frac{1}{2}$ example of a $3D$ parameter space, the degeracy at $B=0$ is nothing but a magnetic monopole of the Berry curvature fictitious gauge field defined upon the parameter space. The presence of such a monopole renders the $U(1)$ bundle nontrivial, hence the $\pi_2(S^2)=\mathbb{Z}$ classification.

In simple words, once you have a monopole, no wonder you obtain nonzero Berry phase by calculating the fictitious magnetic flux covered by the subtended solid angle. If you know the Weyl semimetal, which has a $\vec{k}\cdot\vec\sigma$ Hamiltonian in $3D$ near the band crossing (Weyl point), you immediately know that the aforesaid $B=0$ degeneracy corresponds directly to the $k=0$ Weyl point.