The Berry connection, following the derivation of the Berry phase for a non degenerate system, is

$\mathcal{A}_{k}(\lambda) = i \langle n|\frac{\partial}{\partial \lambda^{k}}|n\rangle$

This result is gauge-dependent, as seen making an analogy with the Aharonov-Bohm effect.

What I don't see is why this is a connection, and how it's related with the parallel transport of eigenstates along the parameter space. Also I haven't found literature that explained this didactically. Why is the Berry connection a connection?


1 Answer 1


The following figure depicts a closed path in the parameter space and shows the variation of the state's normalized eigenvectors in $n$ discrete point along the path. The last point of the path $n$ coincides with the first point $0$.

enter image description here The Berry phase is the total of the accumulated phases along the path: $$ \phi = \lim_{n \to \infty}\arg(\langle 0|1 \rangle\langle 1 |2 \rangle .,.,.,\langle n-1 | n\rangle\langle n | 0\rangle)$$ This definition makes sense, since it is gauge invariant and a local phase transformation on every point of the path $$|k' \rangle = e^{i \alpha_k}|k \rangle$$ does not change its value because: $$\langle l' |k' \rangle = e^{i \alpha_k - i\alpha_l}\langle l' |k' \rangle$$ We can choose a gauge transformations such that: $$|0' \rangle = |0 \rangle$$ (Initial condition), and $$\arg(\langle k' |k' +1\rangle) = 0 $$ The second condition makes the direction of the eigenvector at the $k+1$ -th point parallel to that of the $k$ -th point. Thus in the basis $|k' \rangle$, the vectors are parallel transformed.

Please notice that the last term in the Berry phase $\langle n | 0\rangle$ cannot be rotated to zero, because the phase of $|0'\rangle$ is fixed by the initial condition and the phase of $|n'\rangle$ is fixed by the condition $\arg(\langle n'-1 |n' \rangle)=0 $. Thus we are left with only one term in the Berry phase : $$ \phi = \arg(\langle n' | 0'\rangle)$$ It must be emphasized that the above expression is valid only in the parallel transport gauge system $\{|k' \rangle\}_{k=0, ., ., ., n}$. The value of this phase is equal to the original value due to gauge invariance.

Going to the continuum, we have: $$\arg(\langle k |k+1\rangle) \rightarrow \arg(\langle n(t) |n(t+\Delta t)\rangle) $$ Where in the continuum, we use the points $n$ in the parameter space to parametrize the eigenvectors; We have: $$|n(t+\Delta t)\rangle = |n(t)+\frac{d n}{d t} \Delta t \rangle = |n(t)\rangle + \frac{\partial |n(t)\rangle}{\partial n^i}\frac{dR^i}{dt}\Delta t$$ Thus: $$\arg(\langle n(t) |n(t+\Delta t)\rangle) =\mathrm{ Im}(\langle n(t) | \frac{\partial |n(t)\rangle}{\partial n^i})\frac{dn^i}{dt}\Delta t$$

(because the vector is normalized, we have $\arg(1+x) = \mathrm{Im}(x)$). The inner product $\langle n(t) | \frac{\partial |n(t)\rangle}{\partial n^i})$

is pure imaginary because:

$$\mathrm{ Re}(\langle n(t) | \frac{\partial |n(t)\rangle}{\partial n^i}) \rangle= \frac{1}{2} (\langle n(t) | \frac{\partial |n(t)\rangle}{\partial n^i}\rangle + \langle \frac{\partial |n(t)\rangle}{\partial n^i} | n(t)\rangle) = \frac{\partial }{2\partial n^i} \langle n(t)| n(t) \rangle = 0$$ (since $\langle n(t)| n(t) \rangle =1$).

Thus we may write the Berry phase as: $$ \phi = \oint \langle n(t) | \frac{\partial |n(t)\rangle}{\partial n^i})\frac{dn^i}{dt}d t$$

The integrand in the last expression is the Berry connection.

  • $\begingroup$ Thanks for your answer. Let me see if I got the idea (it's been a bit tricky to me): We have a $U(1)$ bundle represented by our subspace of eigenstates, so inside that $U(1)$, an eigenvector, in a loop on the parameter space, has to be parallel transported in order to see if in that subspace a loop takes that vector to itself. When it's not the case, we have a berry phase that measures the obstruction to parallel transport that vector, i.e., shows that the topology of the bundle is nontrivial... $\endgroup$
    – vbarcelo
    Nov 27, 2017 at 17:42
  • $\begingroup$ ....We use the Berry connection to do that. Okay, so, for computing the Berry phase, we integrate that connection over the cycle, because it has all the information to do that. I see now that this connection is a differential projected into $U(1)$, applying the chain rule to get the dependence in the parameter ¿Is that correct?. And this is analogous to a covariant derivative in that subspace i.e., a differential that parallel transports the eigenvectors (We are assuming that our states are normalized of course). Is that so? $\endgroup$
    – vbarcelo
    Nov 27, 2017 at 17:42
  • 1
    $\begingroup$ Yes, your explanation is correct; the Berry connection has the same gauge properties as the vector potential in electromagnetism. The only thing that needs more precision in your explanation is the bundle description: There are several bundles at play here: The eigenvectors form a line bundle as they are linear spaces of dimension $1$ (The eigenvectors at the fibers of the line bundle are not normalized). The structure group of this line bundle is $U(1)$, so there exists a principal U(1) bundle that the eigenvector line bundle is associated to. $\endgroup$ Nov 28, 2017 at 7:57
  • $\begingroup$ cont. In addition, we can map the line bundle to a universal bundle over the complex projection space, which is itself associated to a principle $U(1)$ bundle over the complex projective space (I did not talk about the latter bundles in the answer, but they are very important in deeply understanding of the Berry phase phenomena). Please, see lib-extopc.kek.jp/preprints/PDF/2000/0032/0032169.pdf $\endgroup$ Nov 28, 2017 at 7:58
  • $\begingroup$ As the state $|n\rangle$ undergoes evolution it may pick up not just a phase. The wavefunction at $|n(t_1)\rangle$ and $|n(t_2)\rangle$ might be quite different. Although for a circular evolution the start point and the end point must differ at most by a phase. $\endgroup$
    – Alex
    Mar 4, 2023 at 10:13

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