# Degeneracy conduction electrons

I'm supposed to show that the degeneracy of the energy levels of conduction electrons at fixed $k_z[/tex] in zero magnetic field is given by $$\frac{2L_x L_y}{\pi \hbar ^2} m \mu _B B$$ Where the energy levels of the electrons are of the form (approximation): $$E_{n,n_z} = E_n(k_z)= \frac{\hbar ^2 k_z ^2}{2m} + (n+\frac{1}{2})2\mu _B B$$ where n is a nonnegative integer and [itex]k_z=2\pi n_z /L_z[/tex] with [itex]n_z$ an integer (positive, negative or 0). The volume under consideration is$V=L_x L_y L_z$

Here's what I think, the degeneracy is in the quantum number n, which represents the angular momentum quantum number. So the degeneracy at zero magnetic field is equal to the maximum number of n. But what restricts n???

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Gokul43201
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Zero field? What is "B"?

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da_willem said:
Here's what I think, the degeneracy is in the quantum number n, which represents the angular momentum quantum number. So the degeneracy at zero magnetic field is equal to the maximum number of n. But what restricts n???
I'm not sure what you mean, but I know for one that 'n' is usually not the quantum number you use for angular momentum, so I don't know if you've made a mistake, or whether it's just an unfortunate choice of label. We usually use 'n' for the principal quantum number, 'l' for orbital angular momentum quantum number and 'm' for magnetic quantum number.

Now, assuming you mean 'm', I can tell you that:

-l <= m <= l
and 0 <= l <= n-1

Gokul43201
Staff Emeritus
Gold Member
da_willem said:
I'm supposed to show that the degeneracy of the energy levels of conduction electrons at fixed $k_z[/tex] in zero magnetic field is given by $$\frac{2L_x L_y}{\pi \hbar ^2} m \mu _B B$$ Where the energy levels of the electrons are of the form (approximation): $$E_{n,n_z} = E_n(k_z)= \frac{\hbar ^2 k_z ^2}{2m} + (n+\frac{1}{2})2\mu _B B$$ where n is a nonnegative integer and [itex]k_z=2\pi n_z /L_z[/tex] with [itex]n_z$ an integer (positive, negative or 0). The volume under consideration is$V=L_x L_y L_z$

Here's what I think, the degeneracy is in the quantum number n, which represents the angular momentum quantum number. So the degeneracy at zero magnetic field is equal to the maximum number of n. But what restricts n???
That is the important question. The hint is in the nature of the spectrum - looks like a harmonic oscillator spectrum, doesn't it? That's because it is!

Write down the Hamiltonian for an electron in a field $\mathbf{B}=B \mathbf{\hat{z}}$, and solve the TISE in the Landau gauge (make the vector potential A = (0,xB,0), say). The motion along the z-direction is unaffected by the field. So, if you factor out the z-component, you will be left with the equation for a displaced harmonic oscillator. If you then apply periodic boundary conditions, and require that the displacement term not exceed the relevant dimension of the sample (i.e, the the wavefunctions not be centered outside the sample), that will give you the required upper limit on n.

Note:
(i) these degenerate states are what are known as Landau levels,
(ii) in the zero-field limit, the degenracy vanishes.

Alas, the class on this excersise has yet passed. But it remains an intricate problem to further investigate. The problem was solved in class by investigating the surface in 2D (x and y directions) k-space occupied by N electrons and to find the maximum energy associated with this k-radius. Equating this to the second term in the expression for the energy levels the result follows by considering n/N.

I like your way better though Gokul! That way you can see where the upper limit of n physically comes from, thanks. Do I understand it correctly if I make the analogy with classically the cyclotron radius of the electrons exceeding the sample size?

Gokul43201
Staff Emeritus