How does the density of states change with temperature?

[unscientific]

Homework Statement

Part (a): Plot fermi energy as a function of N
Part (b): Derive the density of states and find its value
Part (c): How many atoms reside at 20% of fermi energy? Estimate diameter of cloud
Part (d): For the same atoms without spin, why is the cloud much smaller? Estimate the transition temperature.

Homework Equations

The Attempt at a Solution

Part (a)

The Fermi Energy is the highest energy level occupied by the atoms at T = 0.

We know that $\epsilon_f \propto \left( \frac{N}{V}\right)^{\frac{2}{3}}$ so the graph looks like:

Part (b)

I'm not sure why there is a factor of $\omega^3$, as the density of states seem to be independent:

$$g_{(k)} dk = (2S+1) \frac{1}{8} \times \frac{4\pi k^2 dk}{(\frac{\pi}{L})^3}$$
$$g_{(k)} dk = \frac{V}{\pi^2} k^2 dk$$

Using the substitution $E = \frac{\hbar^2 k^2}{2m}$ still doesn't produce any $\omega$ What does it mean when they say $\epsilon >> \hbar \omega$?

 

[DrClaude]

Using the substitution $E = \frac{\hbar^2 k^2}{2m}$ still doesn't produce any $\omega$

That equation for the energy is valid for a free particle, not for one confined to a harmonic oscillator.

What does it mean when they say $\epsilon >> \hbar \omega$?

It means you can neglect the discreteness of the energy levels and consider them to be continuously distributed.

 

[unscientific]

That equation for the energy is valid for a free particle, not for one confined to a harmonic oscillator.It means you can neglect the discreteness of the energy levels and consider them to be continuously distributed.

Part (b)
For one confined to a harmonic oscillator, $E = (n+\frac{1}{2})\hbar \omega$. I've been thinking what the density of states would look like in n-space. Since $\epsilon>>\hbar \omega$, I can simply take $\epsilon \approx n\hbar \omega$.

Would it be something like:
$$g_{(n)} dn = (2S+1) \frac{1}{8} \frac{4 \pi n^2 dn}{1}$$
$$g_{(\epsilon)} d\epsilon= \frac{\pi}{\hbar^3} \frac{\epsilon^2}{\omega^3} d\epsilon$$

Therefore $\alpha = \frac{\pi}{\hbar^3}$.

At T = 0 K, the occupation number up till fermi energy is 1. After fermi energy, it is 0. It is a heavyside function.

$$N = \sum n_{(\epsilon)} = \int g_{(\epsilon)} d\epsilon$$
$$N = \int_0^{\epsilon_F}\frac{\pi}{\hbar^3} \frac{\epsilon^2}{\omega^3} d\epsilon$$
$$\epsilon_F = \hbar \omega\left(\frac{3N}{\pi}\right)^{\frac{1}{3}}$$

For $N = 10^6$ and $\omega = 2\pi \times 10^5$,
$$\epsilon_F = 6.52 \times 10^{-27} J = 41 \space neV$$

Checking online, the fermi energy of Lithium is 4.7 eV. I'm off by 10 orders of magnitude..

 

[DrClaude]

For $N = 10^6$ and $\omega = 2\pi \times 10^5$,
$$\epsilon_F = 6.52 \times 10^{-27} J = 41 \space neV$$

Checking online, the fermi energy of Lithium is 4.7 eV. I'm off by 10 orders of magnitude..

I guess you stumbled upon the Fermi energy for electrons in solid lithium. For a degenerate Fermi gas of atoms, that's indeed the order of magnitude. The corresponding Fermi temperature is $T_\mathrm{F} \approx 0.5\ \mathrm{mK}$, which is why advanced cooling techniques are required!

 

[unscientific]

I guess you stumbled upon the Fermi energy for electrons in solid lithium. For a degenerate Fermi gas of atoms, that's indeed the order of magnitude. The corresponding Fermi temperature is $T_\mathrm{F} \approx 0.5\ \mathrm{mK}$, which is why advanced cooling techniques are required!

Thanks a lot, that does make sense. Where does the degeneracy of $\frac{(n+1)(n+2)}{2}$ come into play here? Should it go into the density of states or the partition function?

 

[unscientific]

Any input on whether my density of states is right?

[Edit]I'm concerned about the missing degeneracy given in the question of $\frac{(n+1)(n+2)}{2}$