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


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


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)


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


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?


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    $\begingroup$ Could you please ask a more specific question than "what's going on"? What does it mean "What's going on"? You described what's going on. There are other things that are going on, too, but it is not clear which of them you find interesting or confusing. There's surely no contradiction in the text you wrote. Some theorems hold when the epsilons are safely separated, some effects appear when they're not, and so on. $\endgroup$ Jul 11, 2013 at 12:35
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    $\begingroup$ @LubošMotl Thanks for your comment, my question was written under rush. I've tried to add more focused questions, and more details. In short I want to know if the Berry phase and the adiabatic theorem are compatible, and to which extend they are (if they are). Please tell me if what I added is still insufficient to make any sense. Thanks again. $\endgroup$
    – FraSchelle
    Jul 11, 2013 at 14:02
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    $\begingroup$ Berry phase and the adiabatic theorem are compatible. Some statements of the adiabatic theorem omit to mention the phase of the final wavefunction. Berry phase is an elaboration in that it says explicitly what that phase factor is. That's all there is to it. $\endgroup$
    – Dan Piponi
    Jul 11, 2013 at 23:54
  • $\begingroup$ @DanPiponi Thanks for your comment. Would you then say that the proximity to a degeneracy point is not a problem at all, and that the adiabatic theorem can be expanded to include the degeneracy point(s) ? If yes, would you please elaborate a bit more about that. Thanks in advance. $\endgroup$
    – FraSchelle
    Jul 12, 2013 at 7:52
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    $\begingroup$ @Oaoa: I wonder if part of the issue is the age of the references you are using? They were written before Berry's paper. They could be (implicitly) redefining the states to remove the Berry's phase, without considering the effect of closed loops in parameter space. Perhaps just a more modern reference like Nakahara would clear this up. Also proximity to degeneracy points is not an issue, if you go slowly enough, its going through degeneracy points that breaks the adiabatic theorem and hence berry's phase. $\endgroup$ Jul 12, 2013 at 14:19

2 Answers 2


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.

  • $\begingroup$ Thanks a lot for your really enlightening answer. So if I understand correctly the adiabatic theorem always presents a Berry's phase, but usually this phase factor is just 1 (or the phase $= 0 \equiv 2\pi$) and there is no effect associated with it. When there is a degeneracy point somewhere, in contrary, the phase can be $\neq 0 \equiv 2\pi$ and there are effects associated with it. The question which remains is: what somewhere means? Does the distance from the degeneracy is a clear notion? What defines the space ? Perhaps I should ask an other question about that. Thanks again. $\endgroup$
    – FraSchelle
    Jul 30, 2014 at 9:06
  • $\begingroup$ Well, I posted an other question regarding the space in which the degeneracy occurs: physics.stackexchange.com/questions/128754 $\endgroup$
    – FraSchelle
    Jul 30, 2014 at 9:35
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    $\begingroup$ Well, the adiabatic theorem only gives a Berry phase for $\textit{closed}$ loops. This is a very stringent condition meaning that you must end up where you started in the parameter space. If there is no closed loop, the phase factor is the usual "just a phase" that can be gauged away. Sorry it took me so long to get back to you. $\endgroup$
    – Xcheckr
    Aug 10, 2014 at 22:08
  • $\begingroup$ No problem for the delay, I was actually thinking I've been too enthusiast a few days ago. You're perfectly right, Berry phase only exist as a topological obstruction. Thanks again for your answer, which put me back on the right track. $\endgroup$
    – FraSchelle
    Aug 12, 2014 at 4:00
  • $\begingroup$ @Xcheckr The gif has been removed. Is it similar to this one Berry's geometric phase: a review? $\endgroup$
    – L.K.
    Apr 12, 2017 at 16:08

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.


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