91影视

In order to show that the average particle energy of a fermion at low temperature$\mathbf{(}\mathbf{~}\mathbf{0}\mathit{K}\mathbf{)}$can be written as just$\mathbf{\left(}\frac{\mathbf{3}}{\mathbf{5}}\mathbf{\right)}{\mathit{E}}_{\mathbf{F}}$, (the Fermi energy), the equations for the total energy of a group of fermions, the average energy. And the Fermi energy will be needed.

The total energy${\mathit{U}}_{\mathbf{\text{local}}}$ofnumber of fermions of mass$\mathit{m}$is:

${\mathit{U}}_{\mathbf{\text{local}}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{10}}{\mathbf{\left(}\frac{\mathbf{3}{\mathbf{蟺}}^{\mathbf{2}}{\mathbf{鈩�}}^{\mathbf{3}}}{{\mathbf{m}}^{\mathbf{3}\mathbf{/}\mathbf{2}\mathbf{V}}}\mathbf{\right)}}^{\mathbf{2}\mathbf{/}\mathbf{3}}{\mathit{N}}^{\mathbf{5}\mathbf{/}\mathbf{3}}$ 鈥�..(1)

Here$\mathit{V}$is their total volume.

The average energy$\stackrel{\mathbf{炉}}{\mathbf{E}}$of a group of particles is their total energy$\mathit{E}$divided by the number of particles$\mathit{N}$:

$\stackrel{\mathbf{炉}}{\mathbf{E}}\mathbf{=}\frac{\mathbf{E}}{\mathbf{N}}$ 鈥�..(2)

The Fermi energy${\mathit{E}}_{\mathbf{F}}$of an$\mathit{N}$number of particles with spin$\mathit{s}$and mass$\mathbf{\text{m}}$is:

${\mathit{E}}_{\mathbf{F}}\mathbf{=}\frac{{\mathbf{蟺}}^{\mathbf{2}}{\mathbf{鈩�}}^{\mathbf{2}}}{\mathbf{m}}{\mathbf{\left[}\frac{\mathbf{3}}{\mathbf{(}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{蟺}\sqrt{\mathbf{2}}\mathbf{V}}\frac{\mathbf{N}}{\mathbf{V}}\mathbf{\right]}}^{\mathbf{2}\mathbf{/}\mathbf{3}}$ 鈥�..(3)

Here$\mathbf{\text{V}}$is their total volume.

To show that the average energy can be written in terms of the Fermi energy, it would help to get a simplerexpression for the Fermi energy first. Equation $\left(3\right)$is

used with s being$\frac{1}{2}$ (since as fermions. They're spin$\frac{1}{2}$ particles):

$\begin{array}{c}{E}_F=\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3}{(2s+1)蟺\sqrt{2}}\frac{N}{V}\right]}^{2/3}\\ =\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3}{(2\left\{\frac{1}{2}\right\}+1)蟺\sqrt{2}}\frac{N}{V}\right]}^{2/3}\\ =\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3N}{2\sqrt{2蟺V}}\right]}^{2/3}\\ =\frac{{\mathrm{鈩�}}^2}{m}{\left[\frac{3{蟺}^{3}N}{2^{3/2}蟺V}\right]}^{2/3}\\ =\frac{{\mathrm{鈩�}}^2}{2m}{\left(\frac{3{蟺}^{3}N}{V}\right)}^{2/3}\end{array}$

The average energy is found by the use of equation , with equation (1) replace the,$E$and combine the N's:

$\begin{array}{c}\stackrel{炉}{E}=\frac{E}{N}\\ =\frac{1}{N}\left[\frac{3}{10}{\left(\frac{3{蟺}^2{\mathrm{鈩�}}^3}{m^{3/2}V}\right)}^{2/3}{N}^{5/3}\right]\end{array}$

$=\frac{3}{10}{\left(\frac{3{蟺}^2{\mathrm{鈩�}}^3}{m^{3/2V}}\right)}^{2/3}{N}^{2/3}$

That can be simplified some by factoring the and$\mathrm{鈩�}$out of the parentheses, and the $\text{N}$into it:

$\begin{array}{c}\stackrel{炉}{E}=\frac{3}{10}{\left(\frac{3{蟺}^2{\mathrm{鈩�}}^3}{m^{3/2V}}\right)}^{2/3}{N}^{2/3}\\ =\frac{3{\mathrm{鈩�}}^2}{10m}{\left(\frac{3{蟺}^{2}N}{V}\right)}^{2/3}\end{array}$

And then rewrite it slightly:

$\begin{array}{c}\stackrel{炉}{E}=\frac{3{\mathrm{鈩�}}^2}{10m}{\left(\frac{3{蟺}^{2}N}{V}\right)}^{2/3}\\ =\frac{3}{5}\left[\frac{x^2}{2m}{\left(\frac{3{蟺}^{2}N}{V}\right)}^{2/3}\right]\end{array}$

The term in brackets is the same as equation (4), so it can be written in terms of that:

$\begin{array}{c}\stackrel{炉}{E}=\frac{3}{5}\left[\frac{x^2}{2m}{\left(\frac{3{蟺}^{2}N}{V}\right)}^{2/3}\right]\\ =\frac{3}{5}\left(E_F\right)\\ =\frac{3}{5}{E}_F\end{array}$