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When I look at the circle of the Dirac cone around the Dirac point of, let's say, $Bi_2Se_3$, then the electron winds around and it is true that it goes from momentum $-k$ and spin-up to $+k$ and spin-down. Now how can I use this fact to show that the Berry phase of $\pi$ arises?

When the electron completes one circle it did go from $k$ to $k$ and from spin-up to spin-up. So basically nothing changed...?

Picture that shows the Dirac cone in the Brillouin zone and the spin on the left hand side:

enter image description here

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It's important to note that the Berry phase is the phase picked up by the wave function after traversing a loop in the parameter space. The best way to figure out the Berry phase, at least in this case, is to compute the Bloch wave function and see how it changes after one complete loop in $\mathbf{k}$-space. The Bloch Hamiltonian for the surface of a topological insulator is given by $$H_{{\rm surf}}(\mathbf{k})=v_{F}\left(\sigma^{x}k_{y}-\sigma^{y}k_{x}\right)$$ where the $\sigma_x$ and $\sigma_y$ are the Pauli matrices in the spin space. If you diagonalize this Hamiltonian you will (evidently) get the Dirac cone; in other words, you will calculate the eigenvalues and eigenvectors. Since this Bloch Hamiltonian is a 2 $\times$ 2 matrix, the Bloch eigenstates can in general be written in the spinor form $$\psi_{\pm}(\mathbf{k}) \propto \frac{1}{\sqrt{2}}\left(\begin{array}{c} e^{i\theta_{\mathbf{k}}/2}\\ \pm e^{-i\theta_{\mathbf{k}}/2} \end{array}\right)$$ where $$\tan(\theta_{\mathbf{k}})=\frac{k_{y}}{k_{x}}$$ It’s easy to see that when you go around a circle in the Brillouin zone, i.e. after setting $ \theta_{\mathbf{k}}\rightarrow\theta_{\mathbf{k}}+2\pi$, you pick up an extra phase of $\pi$. This happens because there are $\theta_{\mathbf{k}}/2$ terms in both spinor components. You could think of it, in a really handwavy way, as the $\theta_{\mathbf{k}}$ in the exponential being split between the two spinor components. On the other hand, if you had the Schrödinger equation, you will have simple scalar wave functions instead of spinors. Another (much fancier) way of putting this is: the Bloch Hamiltonian will be diagonal in spin space in the absence of spin-momentum locking.

One side note: even graphene has a $\pi$ Berry phase; although the origin of it is very different. For all practical purposes people write the Bloch Hamiltonian of graphene to be diagonal in the spin space; that’s because graphene has very poor spin-orbit coupling. However, Bloch Hamiltonian has off-diagonal terms in the sublattice space. As a result, we can still write down the Bloch wave functions as spinors and get a $\pi$ Berry phase.

So to answer your question: yes, we can sort of guess the existence of $\pi$ Berry phase by looking at the winding of spin around the Dirac cone. The above math can validate our guess. We can say: since the spin rotates by $2\pi$ the spinor must pick up a $\pi$. Perhaps people don’t compute it explicitly (as I did above) because of their intuition from graphene.

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+1. I really like your answers, always detailed, clear, accurate and correct. – Heidar Jul 15 '13 at 14:22
@Heidar: Thanks. – NanoPhys Jul 15 '13 at 16:03

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