Let us assume a direct semiconductor with parabolic band structure around the $\Gamma$ point (which is not bad for most semiconductors) in momentum ($k$) space. Now we excite it with light:

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I encircled the start and end state in red. The shaded area resambles the photon energy with momentum $q$. So my question to this kind of situation is, is it possible to have an electron at exactly $k=0$? My concerns are, that this would mean, that the electron would have no kinetic energy when $E=\hbar^2k^2/2m_e$. That means, that the position has maximal uncertainty after Heisenberg. And that would mean that the electrons wave function would be extended into infinity. Since a semiconductor is typically smaller than infinity, this can't be possible.

  • $\begingroup$ This infinite medium causes some interpretation issues. If the electron wave function gives equal probability everywhere, and the total probability is one, then, at each point, the wave function is zero. However, this is a limiting case which does not exist. In reality, the volume will be finite, thus the wave function will be non-zero everywhere for a total probability = 1. $\endgroup$ – fffred Aug 21 '13 at 21:26
  • $\begingroup$ Putting the details of the question aside, I think the source of the confusion is mixing up physical and crystal momenta. Just because a state has $\mathbf{k}=0$ does not mean it has zero kinetic energy; that goes for all bands except the lowest one. In the extended-zone scheme any two states in nearest bands with same $\mathbf{k}$ are separated in physical momenta by $\hbar\mathbf{G}$ where $\mathbf{G}$ is the reciprocal lattice vector. In other words, for a vertical transition $\hbar\mathbf{k}_f = \hbar\mathbf{k}_i + \hbar\mathbf{G}$ in the extended-zone picture. $\endgroup$ – NanoPhys Aug 23 '13 at 11:44
  • $\begingroup$ (continued) Sorry, there's a typo above. I meant to write $\mathbf{p}_f = \mathbf{p}_i + \hbar\mathbf{G}$. For a vertical transition $\mathbf{k}_f = \mathbf{k}_i$. That was the whole point I was trying to make! The typo could have been seriously misleading. $\endgroup$ – NanoPhys Aug 23 '13 at 11:52

Your expression for electron energy is missing a constant.

The wavefunction of the nearly free electron model satisfies Bloch's theorem: $$\psi_\boldsymbol{k}(\boldsymbol{r}) = e^{i \boldsymbol{k \cdot r}}u_\boldsymbol{k}(\boldsymbol{r}) $$
where $u$ has the periodicity of the lattice. Near $\boldsymbol{k}=0$, the electron energy $E$ is approximately: $$E = \epsilon_0 + \frac{\hbar^2 k^2}{2m_e} $$

where $\epsilon_0$ is the enegy of $u_0$.

These are all bound states (within the crystal), with lower energies than a free electron outside the crystal.

Reference: Kittel, "Introduction to Solid State Physics", Chapter 10 in the 4th ed.


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