It seems that you want a rather general answer. I don't know if you can describe the modifications of the different BZ's generally in plain maths, since the topology of the BZ may change when you strain the lattice (different number of corners and edges) and I am not aware of any simple way to describe the BZ in general. If you can work with the reciprocal lattice vectors corresponding to your Bravais lattice vectors things are way easier:

The reciprocal lattice vectors are in terms of the Bravais lattice vectors, R:

B = 2pi*(R^{-1})^T

the transpose of the inverse of the matrix of lattice vectors (column vectors, since you want the strain to operate from the left). So in your notation, introducing S=(1+strain tensor) and ignoring the factor 2pi (which will disappear in the last step anyway):

B(new) = [(S*R(old))^{-1}]^T = (rules for inverse and transpose of products) =

= (S^{-1})^T * ((R(old))^{-1})^T

= (S^{-1})^T * B(old)

Hope this helps and remember to always check other people's maths.

Hi,

In strained quantum wells the main change is on the valance band. The strain lifts the degeneracy and you can see clearly the heavy-hole (hh) and light-hole (lh) states at the top if the valance band. This in certain cases change the bandgap and the type of strain (compressive or tensile) changes the hh (or) the lh to be on the top of VB. Accordingly different type of strained QWs are useful for different types of devices.

Hi,

I missed another point in your question, for uniform strain (hydrostatic) there is (possibly) symmetric lift of the degeneracy, but I think biaxial strain modify the band structure differently in different directions [along and ]. As the valance band maxima lies at the center of the BZ visualization is little easy I think, so the hh and lh states would be aniostropically lifted along different directions of the BZ depending on the strain unisotropy (~ the strain tensor).

Please someone correct my mistakes wherever appropriate!!!

Dear Dhamodaran Santhanagopalan:

Thank you very much for your answer. However, I was not looking for the strain effect on the band structure (E-k). My interest is the BZ. Let me clarify. Say for example, the BZ of quantum well is hexagonal with 6 equal side. My question is, how this hexagon will be modified under strain. This is important when you want to see the band structure along different direction of BZ and when you want to integrate some function over the first BZ. Thanks.

OMG, okay let me think about it.

Let me assume the lattice be fcc with BZ along be hexagonal in face with 6 equal sides. Now I assume that the lattice in all three directions be compressed (uniformly) to half of the initial value (important point is the typical strain in reality are not so strong to compress the lattice to half its initial value). Then all the symmetry points in the first BZ would move two times in k-space w.r.t. the initial position. So the hexagonal face would still be hexagonal but the sides (in k-space) would be two times larger than the initial one.

If the strain is not uniform then the structure changes and accordingly the hexagonal face wont be having equal sides (in fact it may not be even hexagonal).

Happy New Year!