- #1
da_willem
- 599
- 1
I'm supposed to show that the degeneracy of the energy levels of conduction electrons at fixed [itex]k_z[/tex] in zero magnetic field is given by
[tex] \frac{2L_x L_y}{\pi \hbar ^2} m \mu _B B[/tex]
Where the energy levels of the electrons are of the form (approximation):
[tex] E_{n,n_z} = E_n(k_z)= \frac{\hbar ^2 k_z ^2}{2m} + (n+\frac{1}{2})2\mu _B B [/tex]
where n is a nonnegative integer and [itex]k_z=2\pi n_z /L_z[/tex] with [itex]n_z[/itex] an integer (positive, negative or 0). The volume under consideration is[itex]V=L_x L_y L_z[/itex]
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?
[tex] \frac{2L_x L_y}{\pi \hbar ^2} m \mu _B B[/tex]
Where the energy levels of the electrons are of the form (approximation):
[tex] E_{n,n_z} = E_n(k_z)= \frac{\hbar ^2 k_z ^2}{2m} + (n+\frac{1}{2})2\mu _B B [/tex]
where n is a nonnegative integer and [itex]k_z=2\pi n_z /L_z[/tex] with [itex]n_z[/itex] an integer (positive, negative or 0). The volume under consideration is[itex]V=L_x L_y L_z[/itex]
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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