Editing Fermi gas (section)

=== Thermodynamic limit ===
In the [[thermodynamic limit]], the total number of particles ''N'' are so large that the quantum number ''n'' may be treated as a continuous variable. In this case, the overall number density profile in the box is indeed uniform.

The number of [[quantum state]]s in the range <math>n_1 < n < n_1 + dn</math> is:
<math display="block">D_n(n_1)\, dn = 2 \, dn\,.</math>

[[Without loss of generality]], the zero-point energy is chosen to be zero, with the following result:

<math display="block">E_n = \frac{\hbar^2 \pi^2}{2 m L^2} n^2 \implies dE = \frac{\hbar^2 \pi^2}{m L^2} n \, dn
= \frac{\hbar \pi}{L}\sqrt{\frac{2E}{m}} dn \,.</math>

Therefore, in the range:
<math display="block">E_1=\frac{\hbar^2 \pi^2}{2 m L^2} n^2_1< E < E_1 + dE\,,</math>
the number of quantum states is:
<math display="block">D_n(n_1) \, dn = 2\frac{dE}{dE/dn} = \frac{2}{\frac{\hbar^2 \pi^2}{m L^2}n} \, dE
\equiv D(E_1) \, dE\,.</math>

Here, the [[Density of states|degree of degeneracy]] is:

<math display="block">D(E)=\frac{2}{dE/dn}
=\frac{2L}{\hbar \pi}\sqrt{\frac{m}{2E}}
\,.</math>

And the [[density of states]] is:

<math display="block">g(E)\equiv \frac{1}{L}D(E)=\frac{2}{\hbar \pi}\sqrt{\frac{m}{2E}}\,.</math>

In modern literature,<ref name=":0" /> the above <math>D(E)</math> is sometimes also called the "density of states". However, <math>g(E)</math> differs from <math>D(E)</math> by a factor of the system's volume (which is <math>L</math> in this 1D case).

Based on the following formula:

<math display="block">\int^{E_\mathrm{F}}_{0} D(E) \, dE = N \,,</math>

the Fermi energy in the thermodynamic limit can be calculated to be:

<math display="block">E_{\mathrm{F}}^{(\text{1D})}=\frac{\hbar^2 \pi^2}{2 m L^2} \left(\frac{N}{2}\right)^2\,.</math>