Fermi Energy of Non-Interacting Identical Li Atoms in a 3D Harmonic Potential Well

[bon]

Homework Statement

The degeneracy of the nth level above the ground state for a three dimensional harmonic oscillator is (n+1)(n+2)/2 where n takes values n=0,1,2,...

A gas of N non-interacting identical lithium atoms (mass 6amu) each having spin1/2 is confined in a 3d harmonic potential well with vibrational angular freq w. Explain what the fermi energy is and sketch a graph of the fermi energy at absolute zero of temperature in units of h/2pi w as a function of the number of particles N for N in the range 1 to 50.

Homework Equations

The Attempt at a Solution

No idea where to even begin really. Any help would be great thanks!

 

[bon]

i assume the fermi energy graph will be like a step function...but how do i include the fact that they are spin 1/2? do i need to? I can see that because of the degeneracy it will be a step funciton..

also what is the energy in each level. in the ground state is it just hbar w ? in the next state 2 hbar w etc..?

 

[fzero]

i assume the fermi energy graph will be like a step function...but how do i include the fact that they are spin 1/2? do i need to? I can see that because of the degeneracy it will be a step funciton..

How many spin 1/2 particles can occupy a single energy eigenstate?

also what is the energy in each level. in the ground state is it just hbar w ? in the next state 2 hbar w etc..?

For a 3d HO, the energies are

E_n = \hbar\omega \left(n+\frac{3}{2}\right).

To determine the Fermi energy, you will need to determine the number of states with energy less than E_F (with quantum number n_F). This gives the total number of particles N in the system. Generally you'll assume that N is large enough that you can approximate sums with integrals when counting states.

 

[bon]

hmm ok so ground state we have n=0, only 1 atom can fit. therefore fermi energy is 3/2 (in units of hbar w)

n=1 degeneracy is 3, in each state 2 atoms can fit (1 spin up, 1 spin down), therefore, 6 atoms can fit into n=1 state? so until N=7 fermi energy = 5/2 hbar w.

then for N=8 it becomes 7/2? etc...

is this right?Also the next part of the question asks me to show that the density of states for the motion of a particle confined in such a well is

g(E) = A E^2 / w^3 where A is a constant which it asks me to obtain a value for and hence derive an approximate value for Ef when N = 10^6 and w = 2pi x 10^5.

But the density of states will include a volume V and I am not given this...how am i supposed to work out the Ef without this?

Thanks!

 

[bon]

sorry that should be n=0 degeneracy is 2 therefore 2 can fit. therefore from N=3, ef = 5/2 then from N=9 becomes 7/2.. etc right?

 

[fzero]

hmm ok so ground state we have n=0, only 1 atom can fit. therefore fermi energy is 3/2 (in units of hbar w)

n=1 degeneracy is 3, in each state 2 atoms can fit (1 spin up, 1 spin down), therefore, 6 atoms can fit into n=1 state? so until N=7 fermi energy = 5/2 hbar w.

then for N=8 it becomes 7/2? etc...

is this right?

If you can get 2 atoms into each n=1 state, why can you only get one into the ground state?

Also the next part of the question asks me to show that the density of states for the motion of a particle confined in such a well is

g(E) = A E^2 / w^3 where A is a constant which it asks me to obtain a value for and hence derive an approximate value for Ef when N = 10^6 and w = 2pi x 10^5.

But the density of states will include a volume V and I am not given this...how am i supposed to work out the Ef without this?

Thanks!

You might want to explain a bit more. The density of states comes from considering volume elements in the space of quantum numbers \vec{n}, not real space.

 

[bon]

Thanks okay.

Well for the density of states..i thought it would be:

g(k) dk = (1/8 4pi k^2 dk) / (pi/L)^3 x 2

i.e. vol in k space occupied by all the states divided by the volume per state and all of this times two becomes of the spin degen. But then the L^3 on the bottom = V, the volume of the gas...?

 

[fzero]

Thanks okay.

Well for the density of states..i thought it would be:

g(k) dk = (1/8 4pi k^2 dk) / (pi/L)^3 x 2

i.e. vol in k space occupied by all the states divided by the volume per state and all of this times two becomes of the spin degen. But then the L^3 on the bottom = V, the volume of the gas...?

I'm not sure what k is, here you have quantum numbers \vec{n} and there's one state per unit cell in \vec{n} space.