# wavefunctions of individual electrons in the Sommerfeld free electron model

just started learning about quantum statistical mechanics and the Sommerfeld free electron model of metallic solids. Sommerfeld considered the electrons in the solid to behave as though they were non interacting particles in a 3D infinite square well (following from the classical Drude model). The theory states that each electron occupies a certain eigenstate (of the square well) with a corresponding wave vector .Being fermions, a maximum of 2 electrons can occupy a single eigenstate. The density of states can then be worked out and the degeneracy pressure calculated.

Now here's my question.In the beginning of my QM course I learnt that the actual wavefunction of a particle in a given potential is actually a linear combination of all possible eigenfunctions of the particle in that potential.In that case, why do we consider the individual electronic wavefunctions in the above problem to be merely eigenfunctions of the square well and not their linear combinations? What am I missing here?

## 1 Answer

The actual wavefunction of a particle can be expressed as a linear combination of all possible eigenfunctions because the eigenfunctions constitute a complete basis set. So if the eigenfunctions are 1, 2, 3, 4, 5... the general solution for the wavefunction W is W = a*1 + b*2 + c*3 + d*4 + ...

What you are missing is that a might be 1, b, c, d ... might all be 0. so that a totally allowable wavefunction is W =1. Or b might be 1 and a, c, d... all might be 0, so another totally allowable wavefunction is W =2.

In the Sommerfeld problem you are describing, the first and second electrons are in the 1st eigenstate so they each have the same wavefunction W=1. The 3rd and 4th electrons are in the 2nd eigenstate so they each have wavefunction W=2. And so on. W=1 and W=2 are special cases of the general solution, W=1 does show W as a linear combination of all eigenfunctions, its just most of the eigenfunctions are "present" with amplitude zero.