Suppose we have a Hamiltonian $H(\mathbf{R})$ which depends on some parameters $\mathbf{R}$. For each value of $\mathbf{R}$, the Hamiltonian will have some set of eigenvectors $\{ | \phi_{j} (\mathbf{R}) \rangle \}$.

If we change the parameters $\mathbf{R}$ around some closed loop $C$ in parameter space, we can consider the effect that this has on quantum states. Throughout we assume that the change is made adiabatically, starting and ending at the point $\mathbf{R_{0}}$ in parameter space. We will also ignore all 'dynamical' phases of the form $e^{-i E t / \hbar}$ for simplicity.

Case 1: $H(\mathbf{R})$ is non-degenerate

In this case, if we start off in the eigenstate $| \phi_{i} (\mathbf{R_{0}}) \rangle$, this traversal results in a phase factor between the initial and final states

$$ | \psi_{\mathrm{final}} \rangle = e^{i \gamma_{C}} | \phi_{i} (\mathbf{R_{0}}) \rangle, $$

with the phase $\gamma_{C}$ given by

$$ \gamma_{C} = \int_{C} \mathrm{d} \mathbf{R} \cdot \mathbf{A}(\mathbf{R}), $$

where $\mathbf{A}(\mathbf{R}) = i \langle \phi_{i} (\mathbf{R}) | \nabla_{\mathbf{R}} \phi_{i} (\mathbf{R}) \rangle$ is the Berry vector potential.

$\gamma_{C}$ is known as the Berry phase, and is often called a 'geometric phase' since it depends only on the geometry of the path $C$, not the time taken to traverse it.

Case 2: $H(\mathbf{R})$ is degenerate

If instead the Hamiltonian is degenerate, then this traversal through parameter space will result in a unitary operation on the initial state

$$ | \psi_{\mathrm{final}} \rangle = U_{C} | \phi_{i} (\mathbf{R_{0}}) \rangle, $$

with the unitary operator $U_{C}$ given by

$$ U_{C} = \hat{P} \exp\Big( - \int_{C} \mathrm{d} \mathbf{R} \cdot \mathbf{A}(\mathbf{R}) \Big), $$

where $\hat{P}$ is the path-ordering symbol, and now $\mathbf{A}(\mathbf{R})$ is a matrix with entries

$$ \mathbf{A}_{i j}(\mathbf{R}) = \langle \phi_{i} (\mathbf{R}) | \nabla_{\mathbf{R}} \phi_{j} (\mathbf{R}) \rangle. $$

(Note: this operation is unitary because $\mathbf{A}$ is anti-Hermitian.)


I am trying to understand how the Berry phase relates to the statistics of particle exchange. If I understand correctly, we can model particle exchange in the way I have set out above, by changing some parameters $\mathbf{R}$ of the Hamiltonian. For example, we can imagine $H(\mathbf{R})$ to be some trapping potential which localises the particles at positions $\mathbf{R}$. By changing this trapping potential, we can move the particles around each other.

As far as I'm aware, the phase (or more generally, the unitary operation) that results from this sort of path traversal is only dependent on the topology of the path. Yet from the exposition I've given above, I can see no reason why this phase wouldn't change if the path $C$ were deformed slightly. So my question is

Question: Under what conditions is the Berry phase (or unitary operation) only dependent on the topology of the path through parameter space? Are there any differences in these conditions between Cases 1 and 2?

Thoughts on an answer

From reading the answer to this question, it seems that the Berry phase can be thought of in relation to parallel transport of vectors around loops in the parameter space. Apparently, if the "Berry curvature" is zero except at some isolated points, then the Berry phase is only dependent on whether the loop contains one of these points. In this sense, it is only dependent on path topology.

If this answer is along the right lines, please could someone flesh it out? In particular, why should the Berry curvature be zero?


1 Answer 1


In general, the Berry's phase is not path-independent -- that's why we call it a geometric phase and not a topological phase.

However, one situation when the Berry curvature is forced to be zero is when you have time-reversal symmetry. For example, consider the simple case when the wavefunction $|\psi_{\textbf{R}}\rangle$ can be taken to be purely real for all values of $\textbf{R}$. Then clearly the Berry vector potential $\textbf{A} = i \langle \psi_{\textbf{R}} | \nabla_{\textbf{R}} | \psi_{\textbf{R}} \rangle$ is purely imaginary. On the other other hand, $\textbf{A}$ is also supposed to be purely real (which can be verified using the orthogonality condition $\langle \psi_{\textbf{R}}| \psi_{\textbf{R}} \rangle = 1$), so the only possibility is that $\textbf{A} = 0$ and hence the curvature $\nabla \times \textbf{A} = 0$

(Side note, if $\textbf{A} = 0$ why doesn't this imply that the Berry phase is zero? The answer is that even in the presence of time-reversal symmetry, you might not be able to find a single gauge that makes the wavefunction real for all values of $\textbf{R}$. But you should be able to do it in the vicinity of any given value of $\textbf{R}$. Since the curvature $\nabla \times \textbf{A}$ is gauge-invariant, the above argument implies that it is zero everywhere, although we might not be able to find a gauge such that $\textbf{A} = 0$ everywhere.)

Now, particle exchange statistics are still supposed to be well-defined and topologically invariant even in the absence of time-reversal symmetry (e.g. in a magnetic field). But you cannot simply identify these exchange statistics with a single Berry's phase in this case. Instead, you have to consider a difference of the phase accumulated for several different processes such that the local, non-topological contributions cancel out. This is discussed (though not strictly in terms of Berry's phase) in https://arxiv.org/abs/cond-mat/0302460

  • $\begingroup$ Thanks for your answer, and for the link to the paper -- I'll have a read of it when I have time. I'm not sure I understand your second paragraph though. First, isn't the Berry vector potential always real because the gradient operator $\nabla_{\mathbf{R}}$ is anti-Hermitian? Second, how does that argument depend on time-reversal symmetry? $\endgroup$
    – anon1802
    Jan 30, 2018 at 21:47
  • $\begingroup$ @OliverLunt I don't think it makes sense to call a gradient operator acting in parameter space (not Hilbert space) anti-Hermitian. But anyway I did indeed state that "$\textbf{A}$ is supposed to be purely real". But if the wavefunction is real then $\textbf{A}$ is also purely imaginary, which implies that $\textbf{A} = 0$. In the simplest case, time-reversal symmetry is implemented by complex conjugation on the wavefunction, so saying the wavefunction is real is the same as saying you have time-reversal symmetry. $\endgroup$ Jan 30, 2018 at 21:49
  • $\begingroup$ Yes, I think you're right about not calling $\nabla_{\mathbf{R}}$ anti-Hermitian -- the minus sign comes from the orthonormality condition, so $\nabla_{\mathbf{R}} ( \langle \phi_{i} (\mathbf{R}) | \phi_{j} (\mathbf{R})\rangle ) = \nabla_{\mathbf{R}} \delta_{ij} = 0$. With that, I think I understand your second paragraph now, so +1. $\endgroup$
    – anon1802
    Jan 30, 2018 at 23:46
  • $\begingroup$ Regarding your final paragraph, I had a read of the paper you linked, but found it a bit hard to make the connection with Berry phases. In a search for answers, I found this paper by Berry (the Berry). Here he connects particle exchange statistics to a topological phase factor "associated with non-contractible circuits in the doubly connected (and nonorientable) configuration space of relative positions with identified antipodes." What I'm wondering is how this connects with what you said in your final paragraph? $\endgroup$
    – anon1802
    Jan 30, 2018 at 23:51
  • $\begingroup$ @OliverLunt I don't have much of a sense of what the motivation for the construction in the Berry paper. They are introducing additional unphysical degrees of freedom, which is quite strange. The artificiality of their construction is demonstrated from the fact that they purport to prove the spin statistics theorem, yet this theorem does not hold in general for nonrelativistic particles. $\endgroup$ Jan 31, 2018 at 4:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.