Fermi Gas

1. Mar 17, 2007

cepheid

Staff Emeritus
Hi,

I have a question about the discussion of the free-electron (Fermi) gas in my solid-state physics notes. In the free electron model, you basically have particles in a box, and the state of any particle is described by four quantum numbers, nx, ny, nz, and ms, the spin magnetic quantum number. Furthermore, the wavefunction of a particle is given by:

$$\psi(\mathbf{r}) = e^{i\mathbf{k} \cdot \mathbf{r}}$$

where k is defined as follows:

$$k_x = \frac{2 \pi}{l_x}n_x$$

et cetera. I have assume that the box has dimensions

V = lxlylz

Here is the step I am having trouble understanding:

The number of states associated with an element $d^3k = dk_xdk_ydk_z$ in k-space is then given by

$$2dn_xdn_ydn_z = \left(\frac{V}{8 \pi^3} \right) 2dk_xdk_ydk_z$$

Although this follows if you sort of consider each k component as a function of each corresponding n component, it doesn't make a lot of sense

Questions:

1) $n_x$, $n_y$, and $n_z$ are each $\in \mathbb{Z}$, so why are k and n suddenly being treated as continous variables?

2) Basically, other treatments I have seen divide k-space into a discrete set of blocks or unit cells, each of which is associated with a point (kx,ky,kz). The number of states in each block is then just...2. So what does the statement in bold (the number of states "associated with an element") even mean?

2. Mar 17, 2007

marcusl

The important point is that the free electron states approach a continuum. The dimensions of your box are huge compared to atomic spacings, the number of states is huge, the spacing between successive k's is negligible, etc. That's why you can treat the quantities as continuous.

Given that, then dk_x/dx from your equation is
$$dk_x = \frac{2 \pi}{l_x}dn_x$$

Multiplying the expressions for x,y and z gives the result quoted.

3. Mar 17, 2007

cepheid

Staff Emeritus
Thanks for your reply. Sure, I inferred that they were making a continuum approximation, and your justification for that *sort of* makes sense (see below why not completely). I even understand the math:

$$dn_xdn_ydn_z = \frac{l_xl_yl_z}{(2 \pi)^3} dk_xdk_ydk_z = \frac{V}{8 \pi^3} dk_xdk_ydk_z$$

What I don't understand is why they multiply both sides by 2 and then call this "the number of states associated with a volume element in k-space."

Another related question I have (related because it stems from the same lack of understanding) is as follows. If we assume still that there are two states for every point in the (now continuous) k-space, then suddenly there are an uncountable and infinite number of states in any finite volume! Clearly this is wrong. There are supposed to be a countable and finite number of states in any volume in k-space. They get around this problem by stating (as above) that the number of states at each point is no longer 2...it is infinitesimal. I don't understand the justification for this or what they are doing at all. This doesn't match with the physical picture of what is going on.

Last edited: Mar 17, 2007
4. Mar 17, 2007

PRB147

2--Spin degeneracy

5. Mar 17, 2007

cepheid

Staff Emeritus
Hi,

Yes, I realize that there are two spin states for every energy eigenstate defined by given values of the three principle quantum numbers. I said that numerous times, and I even stated in my OP that the fourth quantum number was $m_s = \pm \frac{1}{2}$. If this is what you mean by spin degeneracy, then unfortunately, this statement does not actually address what I was asking in question 2 of post #1. Thanks for your response though.

I would really like it if somebody could address the issue I identified in post #3, as that is really the main point of confusion. Thanks.

Last edited: Mar 17, 2007
6. Mar 17, 2007

marcusl

What PRB said about the factor of 2. As for the rest, the states look enough like a continuum to allow the differentiation operations, but they are still countable, and have a finite density (number per volume of k space).