Degeneracy conduction electrons

In summary: In other words, the maximum allowed value of n is determined by the area in k-space that can be occupied by N electrons. This is similar to the classical case where the maximum allowed value of the cyclotron radius is determined by the size of the sample.
  • #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?
 
Last edited:
Physics news on Phys.org
  • #2
Zero field? What is "B"?
 
Last edited:
  • #3
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
 
  • #4
da_willem said:
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?
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 [itex]\mathbf{B}=B \mathbf{\hat{z}} [/itex], 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.
 
  • #5
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?
 
  • #6
da_willem said:
Do I understand it correctly if I make the analogy with classically the cyclotron radius of the electrons exceeding the sample size?
The way I use the classical analogy is by saying that the maximum number of cyclotron orbits I can fit into a given area (say, the area of the sample) determines the degeneracy.
 

1. What is degeneracy conduction in electrons?

Degeneracy conduction in electrons refers to a phenomenon in which multiple energy levels of conduction electrons are present at the same energy level. This is caused by the Pauli exclusion principle, which states that no two electrons can occupy the same quantum state.

2. How does degeneracy affect the behavior of conduction electrons?

Degeneracy can affect the behavior of conduction electrons in several ways. It can lead to increased conductivity, as more energy levels are available for the electrons to move through. It can also affect the thermal and electrical properties of materials, as degeneracy can influence the movement and interactions of electrons.

3. What factors can influence the degree of degeneracy in conduction electrons?

The degree of degeneracy in conduction electrons can be influenced by factors such as temperature, magnetic fields, and the density of the material. Higher temperatures and stronger magnetic fields can decrease the degree of degeneracy, while higher densities can increase it.

4. How does degeneracy conduction affect the electronic properties of materials?

Degeneracy conduction can significantly impact the electronic properties of materials. It can affect the electrical conductivity, thermal conductivity, and magnetic properties of materials. It can also influence the band structure of a material, leading to unique electronic and optical properties.

5. Can degeneracy in conduction electrons be manipulated for practical applications?

Yes, degeneracy in conduction electrons can be manipulated for various practical applications. For example, materials with high degeneracy can be used for thermoelectric devices, which convert heat into electricity. Degenerate materials can also be used in the development of electronic devices, such as transistors and diodes.

Similar threads

  • Atomic and Condensed Matter
Replies
13
Views
3K
  • Advanced Physics Homework Help
Replies
8
Views
1K
Replies
2
Views
1K
  • Quantum Physics
Replies
3
Views
714
  • Quantum Physics
Replies
16
Views
1K
Replies
1
Views
584
  • Atomic and Condensed Matter
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
727
  • Atomic and Condensed Matter
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
1K
Back
Top