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For phonons, I understand why we can translate everything back into the first Brillouin Zone: there is a minimum wavelength defined by two lattice sites.

For electrons, which are delocalized, I see no such argument. From my original understanding, we plot the dispersion, then translate everything back into the first BZ (assuming we are using the reduced zone scheme). In reality, then, the higher bands do correspond to higher momentum states than the lower bands. Is this correct?

Then, if that is correct, I don't understand direct vs. indirect band gaps. Wouldn't all gaps be indirect in that case, just requiring the emission of a phonon with a wavenumber equal to a reciprocal lattice vector?

Thanks!

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  • $\begingroup$ Whether electrons are 'delicalized' or not is irrelevant - the solutions to the wave functions in the crystal that the electrons occupy must conform to the lattice symmetry. This enables them to be folded back in to the first zone. No phonons are needed for a vertical transition. $\endgroup$ – Jon Custer Dec 27 '15 at 22:18
  • $\begingroup$ I recently put an explanation and image on wikipedia on this topic: See en.wikipedia.org/w/… and the red+blue graphs on the right. $\endgroup$ – Steve Byrnes Dec 30 '15 at 15:10
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The picture you have in mind is the free electron gas model. Next step which is typically done is nearly free electron gas . Here the electrons are not plane waves anymore but are described by Bloch functions, which have lattice periodicity. Hence you are in the same situation as with phonons, and backfolding appears.

Simply you can understand it as following: assume you have high-k electron, i.e. ki-periodic plane wave. Put in a periodic potential, which is multiply your plane wave with cos(x/a), a is lattice constant. You will have then two waves with k=ki$\pm$$\frac{1}{a}$.

Regarding the band gaps: there are two origins for them:

1) within previous consideration: they open at the edge of the BZs, electrons cannot propagate due to Bragg reflection. These are always direct in 1D (See Comment to this question) but can be indirect in higher dimensions (due to different band curvature in different directions).

2) there are bands which originate from other orbitals. Too understand this it useful to read about tight binding-approximation (I find this more useful than wiki). Shortly each band (or sub-band-structure previously discussed) originates from certain atomic orbital. Thus while for instance s band can have peak in $\Gamma$-point d-orbital might have dip elsewhere.

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The problem is resolved when you realize that the Bloch wave functions $$...\psi_{k-G}, \psi_k, \psi_{k+g}...$$ etc are all the same thing. I won't explain why this is so as this is evident from the definition of the Bloch wave function:

$$\psi_k = \sum_G C_{k+G} e^{i(k+G)x}$$

Once you know this fact, you can see that given any $\psi_k$ outside the first Brillouin Zone, we can find a $\psi_{k'+G}$ which is equivalent to the first one, and thus, we shift the wave function back into the first Brillouin Zone.

Now we can see how energy bands come into the picture:

When we write the periodic potential as a Fourier series, and then plug in $\psi_k$ into the Schrödinger's equation, we get an infinite set of equations (practically, a finite set) of the form:

$$\big( \frac {hk^2}{2m} - E \big)C_k + \sum_G U_G C_{k+G} = 0$$

where the k runs over all $k-2G, k-G, k, k+G, k+2G$, and E is the energy of the wave function.

This set of equations can be solved, and then one attains an increasing set of values for E, which correspond to the various energy bands. Note that we could have plugged in the wave function $\psi_{k\pm n G}$, and still would have got the same energy spectrum, since these are really the same wave function.

Now answering your question about direct band gap crystals; since the same wave vector can correspond to different energies, thus we can jump from one energy band to another, without changing the wave vector $k$. Thus, this would be the case in a direct band gap material.

Also, we shouldn't really be talking about momentum so simply, because the momentum $hk$ actually corresponds to a free or nearly free particle, and isn't really the momentum for the electron in a lattice potential. The only thing that maters mathematically are the selection rules for the wave vector, which come into play only when there is a change in the wave vector.

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