I am now reading some articles about Dirac fermions in condensed matter physics and the most famous example is graphene. I am now trying to understand page 5 in this article : https://arxiv.org/abs/1410.4098

In the literature it is stated that since the wavefunction around the $\bf K$ and $ \bf K'$ points are related by time-reversal symmetry, the Berry phases (or winding numbers) have opposite sign. That is, if the Berry phase is $+\pi$ around $\bf K$ then the Berry phase will be $-\pi$ around $\bf K'$.

I am now trying to prove this property. I first use the time-reversal property to relate the states around $K$ with those around $\bf K'$.

If I write the states around $\bf K$ as $\left| {{\bf K} + {\bf q}} \right\rangle$ where $|\bf q|$ $<<$ $|\bf K|$. Then the time-reversal partner with the same energy will be $\Theta \left| {{\bf K} + {\bf q}} \right\rangle \approx \left| {-{\bf K} - {\bf q}} \right\rangle$. Now, use the property that ${\bf K'} + {\bf K}$ is a reciprocal lattice vector in graphene, the state $\left| {-{\bf K} - {\bf q}} \right\rangle$ is the same as $\left| {{\bf K'} - {\bf q}} \right\rangle$.

The Berry phase at $\bf K$ is \begin{equation} \phi_{\bf K} = \oint\limits_{around\ \bf K} {\left\langle {{\bf{K}} + {\bf{q}}} \right|{\nabla _{\bf{q}}}\left| {{\bf{K}} + {\bf{q}}} \right\rangle } \cdot d\bf q \end{equation} while the Berry phase at $\bf K'$ is \begin{equation} \phi_{\bf K'} = \oint\limits_{around\ \bf K'} {\left\langle {{\bf{K'}} + {\bf{q}}} \right|{\nabla _{\bf{q}}}\left| {{\bf{K'}} + {\bf{q}}} \right\rangle } \cdot d\bf q \end{equation}

I know that it will be trivial if I explicitly write down the spinor expression of the graphene $\approx \left[ {\begin{array}{*{20}{c}} {1}\\ {e^{i\phi_{\bf q}}} \end{array}} \right]$.

But I would like to seek some more general proof for the statement that the Berry phase (or winding number) has opposite sign for time-reversal related Dirac point without the explicit expression of the wavefunction. Is it possible?

I am also confused with another point. I actually don't know how to well-define the path for the line-integral. For example, if I use counterclockwise $\bf q$-path and get the Berry phase $=\pi$, then I can reverse the $\bf q$-path to be clockwise then I will get Berry phase $= -\pi$. Therefore I can use a clockwise path for $\bf K$ while a counterclockwise path for $\bf K'$ and it will lead to same Berry phase. Does this make sense?

I am still new in this field and will be extremely grateful for any suggestions! :)


1 Answer 1


I believe your Berry phase formula misses the imaginary factor $i$. The handwaving argument would be: apply time-reversal (TR) to everything, the $i$ changes to $-i$.

As TR effects also the $k$-point you are now at the TR connected Dirac point but with the opposite sign in the Berry connection.


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