# Band gaps: are they at the centre or at the edge of the Brillouin zone?

Band gaps open at the edges of the Brillouin zone (BZ), since that is where the Bragg scattering occurs.

I am slightly confused, as most band structures show a gap opening at $\Gamma$ which is at the centre of the BZ. What am I missing?

• You're not missing anything. What you said is correct. I have no idea what that quotation could be talking about. Need more context. Maybe it's talking about something very specific, i.e. a specific model of a specific system, rather than band gaps in general. – Steve Byrnes Mar 20 '14 at 13:30
• @SteveB Maybe I'm missing something. The quoted sentence looks good to me. Take a look at the "Details" in my answer, and see if you can clarify our differences. – garyp Mar 20 '14 at 15:08
• This gives the answer to your question: thiscondensedlife.wordpress.com/2018/02/11/… – Xcheckr Feb 13 '18 at 15:58

They are shown at the $\Gamma$ point in special diagrams called the reduced zone scheme in which a band will be shown folded back on itself. This way of showing the band structure is convenient for a few reasons, one of which is that it saves space on the page. If you look at that band gap at $\Gamma$ and follow the lower band down to lower energies, you will find that at some other symmetry point the band will change direction and then continue down to the zero-energy state at the $\Gamma$ point.

There's a another scheme, the extended zone scheme, in which the band gap will appear more explicitly to be at a zone boundary. These diagrams take up space, and it's also difficult to tell which states share a single wave vector. In the reduced zone scheme, all states that line up vertically have the same wave vector.

Both schemes show the same information.

Details: The $\Gamma$ point has wave vector $k=0$. The edge of the 1st BZ occurs when $k=\pi/a$ where $a$ is the lattice period. The edge of the second BZ occurs at $k=2\pi /a$. But that means that the wave length of that state is equal to the lattice spacing. So for example if a wave peak occurs at a lattice site, every wave peak will occur at a lattice site. The wavefunction has the same value at every single lattice site. This is exactly the same thing that happens at when $k=0$ and the wavelength is infinite.

That is to say that the wave vector $k=2\pi / a$ is identical to the wave vector $k=0$. The second zone boundary occurs at $k=2\pi/a$ which is the same as saying that the second zone boundary occurs at $k=0$. The second zone boundary does indeed occur at the $\Gamma$ point. This seemingly peculiar state of affairs is a consequence of the mechanics of a periodic potential, such as we have in a perfect crystal.

This periodic unravelling of wave vectors is what's exploited in the reduced zone scheme.

Consider the illustration showing a nearly free electron in a 1-d periodic potential. Two views of the same dispersion relation: first in the extended zone scheme, then in the reduced zone scheme. The second zone is folded back onto the first, so that the 2-3 zone boundary occurs at $k=2\pi / a$ in the extended zone scheme, and $k=0$ in the reduced zone scheme. But in a periodic potential, $k=2\pi /a$ and $k=0$ are the same wave vector. Both diagrams contain exactly the same information.

Figure credit: Introduction to Condensed Matter Physics, Vol I, Feng and Jin, World Scientific 2005

• Second BZ isn't just an upscaled version of first BZ, so its edge can't be at $2k_1$ in general. – Ruslan Mar 20 '14 at 15:14
• If I have a transition from the center of the first BZ, to the boundary between the second and third BZ, I would NOT describe this transition as a "band gap ... at the edge of the BZ". – Steve Byrnes Mar 20 '14 at 15:23
• @SteveB Neither would I. But at the boundary between the second and third BZs there are two states. They would be degenerate in the absence of the potential. With the potential, they split and a gap opens up. I'd call that a band gap at $\Gamma$. – garyp Mar 20 '14 at 16:18
• @Ruslan For simplicity I'm thinking of a one-dimensional periodic potential. In that case the 2nd BZ can be folded back onto the first BZ (reduced zone scheme). The boundary between BZ 1 and BZ 2 occurs at $k=\pi/a$. The boundary between BZ 2 and BZ 3 ocuurs at $k=2\pi/a = 0$ – garyp Mar 20 '14 at 16:24
• @Steve and gary (in fact, mostly gary) in the absence of a potential, the unfolded E-k chart would, presumably, resemble a parabola, right? I'm struggling to see, how there are two states at a zone boundary (2nd to 3rd), unless you take the gamma-point to be the other state. – LLlAMnYP May 18 '18 at 20:08