# Why does the Fermi Surface cross the Brillouin zone boundary at right angles?

I'm not sure why the fermi surface crosses the Brillouin zone boundary at right angles. I understand that this is normally the case, but not necessarily always.

I'm aware that the fermi surface is a constant energy surface up to the filling point. The Brillouin zone is in reciprocal space.

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This answer nothing to the OP's question, please don't vote up anymore; and anyone who know the answer to this question please share.

This is mainly due to the time reversal symmetry.

Consider the Bloch equation:

$$[-\frac{\hbar^2}{2m}\nabla^2+U(r)]\psi_{nk}=\epsilon_{nk}\psi_{nk}$$

Recall that $\psi_{nk}=e^{ik\cdot r}u_{nk}$, then we have:

$$[-\frac{\hbar^2}{2m}(-i\nabla+k)^2+U(r)]u_{nk}=\epsilon_{nk}u_{nk}$$

Now we want to prove $\epsilon_{nk}=\epsilon_{n-k}$, take the complex conjugate of the above equation and change $k\to-k$:

$$[-\frac{\hbar^2}{2m}(-i\nabla+k)^2+U(r)]u_{n-k}^*=\epsilon_{n-k}u_{n-k}^*$$

We can see that $\epsilon_{n-k}$ is the same set of eigenvalues as $\epsilon_{nk}$ of the same Hamiltonian $H_k$. Thus they must be equal.

Consider one particular band $n_0$, its zone boundary are $-K/2$ and $K/2$.
$\epsilon_{n_0,K/2+\Delta k}=\epsilon_{n_0,-K/2+\Delta > k}=\epsilon_{n_0,K/2-\Delta k}$
Let's $\Delta k$ tends to infinity small, the above equation just means that the first derivative of energy band near the zone boundary is zero.