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The Berry's or Zak's phase is given as

\begin{align*} \gamma & =\oint_\mathrm{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_\mathrm{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}

which can take any value between $[0,2\pi]$.

Now we want to learn the implications of the symmetries on this Berry's phase first assume that our system has only inversion symmetry such that the hamiltonian obeys

$$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$$

then we have \begin{align*} U_{p}H(\mathbf{k})U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(-\mathbf{k})U_{p}U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(\mathbf{k})U_{p}U(-\mathbf{k}) & =E(-\mathbf{k})U_{p}U(-\mathbf{k}) \end{align*}

such that $E(k)=E(-k)$ and $U_{p}U(-\mathbf{k})=e^{i\phi(\mathbf{k})}U(\mathbf{k})$ where $U_{k}$ is the vector representation of $|u(k)\rangle$. Now the Berry connection is

\begin{align*} \mathcal{A}(\mathbf{k}) & =i(U_{p}U(-\mathbf{k})e^{-i\phi(\mathbf{k})})^{\dagger}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U_{P}U(-\mathbf{k})\\ & =i(U(-\mathbf{k}))^{\dagger}e^{i\phi(\mathbf{k})}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U(-\mathbf{k})\\ & =-\mathcal{A}(-\mathbf{k})+\nabla_{\mathbf{k}}\phi(\mathbf{k}) \end{align*}

and the Berry's phase is for a 1d $\mathrm{BZ}$ for simplicity

\begin{align} \gamma & =\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\\ & =-\intop_{-\pi}^{\pi}dk\,[\mathcal{A}(-k)+\nabla_{k}\phi(k)]\\ & =\left.\phi(k)\right|_{-\pi}^{\pi}-\underbrace{\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)}_{\gamma}\label{eq:-18}\\ & =2\pi n-\gamma\label{eq:-17} \end{align}

thus we have \begin{align*} \gamma=\pi n \end{align*}

since $\gamma$ is well defined only up to $\mbox{mod }2\pi$ it can be $0$ or $\pi$ while obeying the last condition. My suspicion arises from the last step. The Berry phase should be gauge independent, but here it looks like gauge term determines its value.

Emilio Pisanty
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physshyp
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1 Answers1

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Your derivation is perfectly correct. Please see Hatsugai (page 16) who performs essentially the same calculation as you did.

I'll explain here why nevertheless the Zak phase is gauge invariant, therefore a legitimate observable.

The phase $\phi(\mathbf{k})$ is not a gauge transformation. It is the additional phase added to the state vectors as a consequence of the application of the physical operator of time reversal operator (which I'll denote by $\mathcal{T}$ for clarity). In a time reversal symmetric system as you wrote:

$$\mathcal{T} | u(\mathbf{k}) \rangle = e^{i\phi(\mathbf{k})} | u(\mathbf{-k}) \rangle $$

The function $ e^{i\phi(\mathbf{k})} $ can be any true function on the brillouin zone which has the topology of $S^1$. Thus we must have :

$$ e^{i\phi(\pi)} = e^{i\phi(-\pi)} $$ Thus $$ \phi(\mathbf{-\pi}) = \phi(\mathbf{\pi}) + 2 \pi n$$ This transformation has a winding number $n$ $$\frac{1}{2 \pi}\int e^{-i\phi(\mathbf{k})} d e^{i\phi(\mathbf{k})} = n$$

In contrast as a gauge transformation, $ e^{i\phi(\mathbf{k})}$ is a large gauge transformation. In quantum mechanics large gauge transformations correspond to physically different states. This issue was discussed in physics stack exchange in the past, please see the following questions and answers: (1), and (2).

The resons for that is that quantum theories in which states related by a large gauge transformation are distinct are completely consistent. There is no need in quatum mechanics to identify between these states, in contrast to the small gauge transformations which lead to constraints that we should get rid of in order to quantize. In reality, large gauge transformation describe superselection sectors, which explain a wide variety of phenomena. For example in the Aharonov-Bohm effect, the superselection sectors correspond to different (nonquantized) values of the flux; and for a particle moving on a torus with magnetic flux the superselection sectors correspond to different quantized fluxes. In both cases the wave functions are the same except for a phase factor, nevertheless they represent different systems.

Therefore, for a gauge transformation we should select only functions , $ e^{i\phi(\mathbf{k})}$ having a zero winding number, which makes the Zak phase gauge invariant.

David Bar Moshe
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  • Excuse me for one query: $\mathcal{T} | u(\mathbf{k}) \rangle = e^{i\phi(\mathbf{k})} | u(\mathbf{-k}) \rangle$ .How do we get this? I seem to be lost, I saw the reference also. Forgive me for my ignorance. – L.K. Nov 29 '17 at 14:17
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    @L.K. We know that the time conjugation operation reverses the momentum, therefore maps state indexed by the lattice momentum $\mathbf{k}$ to a state indexed by $-\mathbf{k}$. But since a multiplicative phase does not change the state, the most general form of the time reversed state can be of the form $e^{i\phi({\mathbf{k}})} |u(-\mathbf{k})\rangle$. The exact form of $\phi{(\mathbf{k})}$ depends on the Hamiltonian, whose specific form may necessitate such a phase in order to be time reversal invariant. – David Bar Moshe Nov 29 '17 at 15:07
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    Cont. The point is that if the winding number of this phase is not zero, it cannot be removed by a gauge transformation. – David Bar Moshe Nov 29 '17 at 15:09
  • Many thanks!! If I understand correctly. The $| u(\mathbf{k}) \rangle$ states are time reversal upto a phase factor, of the form written above(iff $| u(\mathbf{k}) \rangle$ is time reversed one)? – L.K. Nov 29 '17 at 15:49
  • @L.K. If I understood you correctly; I meant that the time reversed version of $|u(\mathbf{k})\rangle$ is $e^{i\phi{(\mathbf{k}})} |u(-\mathbf{k})\rangle$, (the states are not equal), and indeed the time reversal operation acts on the states by reversing the momentum sign and adding a phase factor. – David Bar Moshe Nov 29 '17 at 16:03
  • I am sorry to bother you so much. But there should be phase due to TR is because of the reason that it will matter the path we take to go back to the time reversed state. It is will be unity if we come back to initial state $e^{i\phi(\pi)} = e^{i\phi(-\pi)}$. – L.K. Nov 29 '17 at 17:08
  • @L.K. It's no bother; I am afraid that I misled you in my previous comment. In our case we indeed have: $|u(\mathbf{k})\rangle = \mathcal{T} |u(\mathbf{k})\rangle = e^{i\phi{\mathbf{k}}} |u(-\mathbf{k})\rangle $, because we are talking about a time reversal invariant system. The Zak-Berry phase in the question is cyclic, i.e. we always start from a given state say $|u(\mathbf{k})\rangle$ and we finish at the same state. – David Bar Moshe Nov 30 '17 at 09:02
  • cont. The computation in the question was performed with the original state first, then with the time reversed state, and consequence is obtained by equating the Zak phases of the initial state and of the time reversed state. – David Bar Moshe Nov 30 '17 at 09:02
  • Let us continue this discussion in chat. – L.K. Nov 30 '17 at 09:40