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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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1 Answer 1

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.

Now let's answer your question:

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.

So that when the filling of electrons doesn't modify the band structure, you will always see the fermi surface perpendicular to the zone boundary if time reversal symmetry is respected.

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This merely proves that time reversal symmetry renders the BZ boundaries as isoenergetic surfaces, which doesn't answer the question. –  huotuichang Sep 30 at 14:23

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