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1. What is being asked?

A 3‑D time‑reversal‑invariant topological insulator (TI) is characterized by a (Z_{2}) bulk invariant

[ \nu=\frac{1}{2\pi}\Bigl[\;\underbrace{\oint_{\partial (A+B)}!{\bf A}\cdot d{\bf k}}{\text{line integral}} \;-\;\underbrace{\iint{A+B}!! \bigl(\nabla_{\bf k}\times{\bf A}\bigr)\,d^{2}k}_{\text{Berry‑curvature flux}}\Bigr]\;\;{\rm mod}\;2 . \tag{1} ]

The student wonders why the surface states of a TI acquire a Berry phase of (\pi) when their momentum winds once around the Dirac point, and how this (\pi) shows up in the expression (1).
In other words we must show that, for a non‑trivial TI ((\nu=1)),

[ \boxed{\;\oint_{\cal C}{\bf A}!\cdot d{\bf k}= \pi\;({\rm mod}\;2\pi)\;} \tag{2} ]

where ({\cal C}) is any closed loop that encloses the surface Dirac point (or, equivalently, the boundary of the half‑Brillouin‑zone used in (1)).

We shall do this purely from the Berry connection/curvature and the time‑reversal symmetry constraint, without invoking charge polarization or transport arguments.

2. Detailed derivation

2.1 Berry connection, curvature and the (Z_{2}) formula

For a (single) occupied Bloch band (n)

[ \boxed{{\bf A}{n}({\bf k}) = i\langle u{n}({\bf k})|\nabla_{\bf k}u_{n}({\bf k})\rangle}, \qquad \boxed{\Omega_{n}({\bf k}) = \bigl(\nabla_{\bf k}\times{\bf A}_{n}({\bf k})\bigr)_z } . ]

In a time‑reversal‑invariant (TRI) crystal the anti‑unitary operator (\Theta) satisfies

[ \Theta^{2}=-1\qquad (\text{spin‑}\tfrac12\;\text{electrons}) . ]

Acting on a Bloch state we can write

[ |u_{n}(-{\bf k})\rangle = \sum_{m} w_{nm}({\bf k})\,\Theta |u_{m}({\bf k})\rangle ,\qquad w({\bf k})\in U(N_{\rm occ}) . \tag{3} ]

From (3) one obtains the TR constraint on the Berry connection

[ {\bf A}(-{\bf k}) = -\,{\bf A}({\bf k}) + i\,w^{\dagger}({\bf k})\nabla_{\bf k} w({\bf k}) . \tag{4} ]

The second term is a pure gauge. Consequently the Berry curvature is odd:

[ \boxed{\Omega(-{\bf k}) = -\Omega({\bf k})}\ . \tag{5} ]

Hence the total Chern number of the whole Brillouin zone (BZ) vanishes, (\displaystyle \int_{\rm BZ}\Omega\,d^{2}k =0).

2.2 Why we integrate only over half the BZ

Because (\Omega) is odd, the integral over a half BZ, denoted (A+B) in Fig. 1 of Fu’s thesis, need not be zero. The two halves are related by TR, but the gauge choice on their common boundary is constrained by (4). This is precisely what makes the quantity (1) gauge‑invariant and integer‑valued (mod 2).

Define the Wilson loop (Berry phase) of the occupied subspace along the closed contour (\partial(A+B)),

[ \gamma \equiv \oint_{\partial(A+B)} {\bf A}!\cdot d{\bf k}\; . \tag{6} ]

Using Stokes’ theorem on a simply‑connected region would give (\gamma = \int_{A+B}\Omega). However the region (A+B) is not a closed surface in the periodic BZ: its opposite edges are identified only after a time‑reversal operation. Because of (4) the line integral on the two opposite edges does not cancel, and the difference

[ \gamma - \int_{A+B}\Omega = 2\pi\,C_{A+B} \tag{7} ]

is an even multiple of (2\pi). Therefore the expression in (1) reduces to

[ \nu = \frac{\gamma}{\pi}\;\;{\rm mod}\;2 . \tag{8} ]

Thus the (Z_{2}) invariant is nothing but the Berry phase (6) measured in units of (\pi).
If (\nu=1) (non‑trivial TI) we must have (\gamma = \pi\;({\rm mod}\;2\pi)); if (\nu=0) then (\gamma =0\;({\rm mod}\;2\pi)).

2.3 Explicit calculation for the surface Dirac cone

The low‑energy surface Hamiltonian of a strong TI is the massless Dirac model

[ H({\bf k}) = v\bigl(k_{x}\sigma_{y} - k_{y}\sigma_{x}\bigr) , \qquad {\bf k}=(k_{x},k_{y}) . \tag{9} ]

Its (conduction‑band) eigenstate can be written as

[ |u_{+}({\bf k})\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\[2pt] e^{i\theta_{\bf k}} \end{pmatrix}, \qquad \theta_{\bf k}= \arg(k_{x}+ik_{y}) . \tag{10} ]

Compute the Berry connection:

[ \begin{aligned} {\bf A}({\bf k}) &= i\langle u_{+}|\nabla_{\bf k}u_{+}\rangle = i\frac{1}{2}\bigl(0,\,e^{-i\theta}\partial_{\bf k}e^{i\theta}\bigr)
&= -\frac{1}{2}\nabla_{\bf k}\theta_{\bf k} . \end{aligned} \tag{11} ]

Take a circular path ({\cal C}) of radius (R) that encloses the Dirac point once, parametrized by (\theta\in[0,2\pi]). The line integral is

[ \begin{aligned} \gamma &\equiv \oint_{\cal C}{\bf A}!\cdot d{\bf k} = -\frac12\oint_{\cal C}\nabla_{\bf k}\theta_{\bf k}\cdot d{\bf k} = -\frac12\Delta\theta_{\bf k}
&= -\frac12\bigl(\theta(2\pi)-\theta(0)\bigr) = -\frac12(2\pi)= -\pi . \end{aligned} \tag{12} ]

Modulo (2\pi) this is (\boxed{\gamma = \pi}). The same result follows for the valence band (the sign flips, but the phase is still (\pi) modulo (2\pi)).

Thus a single surface Dirac cone carries a Berry phase of (\pi).

2.4 Linking the surface Berry phase to the bulk (Z_{2}) invariant

From (8) we have

[ \nu = \frac{\gamma}{\pi}\;{\rm mod}\;2 . \tag{13} ]

For a strong TI the bulk invariant is (\nu=1). Consequently the surface Wilson loop must satisfy (\gamma=\pi) (mod (2\pi)), exactly what we have just obtained from the explicit Dirac Hamiltonian.

Conversely, if the surface were topologically trivial ((\nu=0)), the Berry phase around any closed loop would be (0) (or an even multiple of (\pi)); a Dirac point could be gapped out without breaking time‑reversal symmetry.

Hence the (\pi) Berry phase of the surface state is a direct manifestation of the bulk (Z_{2}) invariant encoded in the half‑BZ integral (1).

3. Final answer

The Berry phase (\pi) of a topological‑insulator surface state follows from the bulk (Z_{2}) invariant. The invariant can be written as

[ \nu=\frac{1}{2\pi}\Bigl[\oint_{\partial (A+B)}{\bf A}\cdot d

Original question: Where does the Berry phase of $\pi$ come from in a topological insulator? on Physics Stack Exchange, licensed CC BY-SA.