Let's consider a 1D ring made up of $M_s$ ideal point-like particles.

The simplest possible ring has $M_s=3$ particles. Hence the relevant first Brillouin zone (BZ) is the set of wavenumbers $k=\{-1,0,1\}$. Is it right?

Now let's turn to a ring made up of $M_s=4$ particles. The first BZ should include one element more, that means: $k=\{-1,0,1,2\}$. Is it right? What's the difference between including $k=-2$ or including $k=+2$? Do they correspond to the same Bloch function ?


1 Answer 1


I suppose by 1D-ring you mean a 1D chain with periodic boundary conditions. In this case for $M_s=3$, all possible $k$-values are given by $k \in \{-\frac{2\pi}{L}, 0, \frac{2\pi}{L}\}$, where $L$ is the circumference of the ring, because $e^{ikx} \overset{!}{=} e^{ik(x+L)}$ (PBC).

The brillouin zone boundary in this case is at $k=\frac{\pi}{a} = \frac{3\pi}{L}$, because the spacing between the atoms is $a=\frac{L}{3}$.

For $M_s=4$, the $k$-values are $k \in \{-\frac{2\pi}{L}, 0,\frac{2\pi}{L}, \frac{4\pi}{L} \}$ or, as you correctly assumed, $k \in \{ -\frac{4\pi}{L}, -\frac{2\pi}{L}, 0,\frac{2\pi}{L}, \}$. This comes about because as you can see, the outermost $k$-value lies on the brillouin zone boundary, which is now at $k= \frac{4\pi}{L}$. Whenever the $k$-vector is exactly on this boundary, this corresponds to a sign change of the bloch factor from atom to atom, which you can easily verify. Choosing the positive or negative value does nothing to change this fact (also easily verified). So indeed, they correspond to the same bloch wave.

It is also worth noting that $k$-vectors can be shifted by reciprocal lattice vectors to get equivilant $k$-vectors, which is basically the reason for why it is almost always sufficient to look at the first brillouin zone. In the latter case, $-\frac{4\pi}{L}$ and $\frac{4\pi}{L}$ are separated exactly by such a reciprocal lattice vector, from which one can also see that they are equivilant.


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