I've been reading up a bit on semiconductor quantum wells, and came across a selection rule for an infinite quantum well that says that "Δn = n' - n = 0", where n' is the quantum well index of an excited electron state in the conduction band, and n is the index of the valence band state where the electron came from (see, for example, p. 126 of "Optical Properties of Solids" by Mark Fox (2001)). Let's assume that I'm operating within all the basic approximations (e.g., we're using light to excite this electron, the quantum well is oriented parallel to x and y, and perpendicular to z, the dipole approximation is valid such that the interaction Hamiltonian is given by E*x, light is incident normal to the quantum well such there is no component of E oriented normal to the quantum well, screening and Coulomb interactions are ignored, etc.).(adsbygoogle = window.adsbygoogle || []).push({});

It is usually the case that a relevant conservation law must exist because of some underlying symmetry. For example, I can understand that the selection rule "Δn ∈ evens" is a consequence of parity conservation, which is in turn a consequence of inversion symmetry.

My question is this: Is there an underlying symmetry in the Hamiltonian forcing the selection rule "Δn = 0" to exist? What is it?

I suppose my best guess is that it has something to do with momentum conservation, although it's a little hard to wrap my head around this since the quantity conserved is not "k" in the usual sense, but rather "k_nz = n*pi/w", (where "n" is a positive integer and "w" is the well width), and there is no translation symmetry along the "z" direction. Maybe there's a symmetry related to continuous transformations of the well width?

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# In an infinite quantum well, why Δn=0?

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