# Semi-conductor band-gap and deformation potential

Submitting a semi-conductor to stress leads to a deformation in the energy-bands, roughly described by:$$H_{ij} = {\cal{D}}_{ij}^{\alpha\beta}\;\epsilon_{\alpha\beta}$$ $\epsilon$ being the strain (linked to the stress by Hooke's law), $H$ the perturbation Hamiltonian to the Hamiltonian describing a stress-free semiconductor, $i,j$ being indexing the energy level of the previous "free" Hamiltonian. Still, I lack intuition regarding the apparition of this term, it seems that compression enlarges the band-gap whereas dilation tightens it. Why do we have this behaviour? I have tried thinking on electrostatic arguments, the potential decreasing as $r^{-1}$ we do have an increase of energies in $r \rightarrow \alpha r$ for $\alpha < 1$ or also seeing the dilatation as a renormalization group transformation, basically going to a coarser grain (although there probably other pertinent length scales in an atomic lattice (spreading of the electronic orbitals, ...) which would make this argument wobbly). Cutting to the point, what is your hand-waved way of seeing it? Books on the subject of deformation potential don't really seem to offer intuition on it, more numerical values for specific materials.

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Isn't that the only term you can write down which is linear in the strain? The assumption would be that higher order terms $\sim\epsilon^2$ are negligible unless for some reason $\mathcal{D}^{\alpha\beta}_{ij}$ vanished. Just a guess. – Michael Brown Feb 18 '13 at 14:48

Well, possibly the simplest argument (I have no idea if this is correct) is that compressing the material reduces the configuration space available for the electrons, and increasing confinement means increased energy (differences) in quantum mechanics. Consider just the old particle-in-a-box problem, the energy levels scale as $E_n \sim \frac{n^2}{L^2}$, where $L$ is the size of the box, so a smaller box means bigger spacings between energy levels.