三种统计分布的相图

2023-11-03

-专业笔记

Bose-Einstein distribution

Free Boson

For free Bosons, the particle density can be written as the sum of the Bose-Einstein distribution for every energy level with momentum $\mathbf{k}$, $$ n(T, \mu) = \frac{1}{V}\sum_{\mathbf{k}} \frac{1}{e^{ \frac{1}{k_B T}\left( \frac{\hbar^2 k^2}{2m} - \mu\right)} - 1} = \frac{1}{V}\sum_{\mathbf{k}} \frac{1}{z^{-1}e^{ \frac{1}{k_B T} \frac{\hbar^2 k^2}{2m}} - 1} $$ where fugacity $z = e^{\frac{\mu}{k_BT}}$. Using $$ \frac{1}{V} \sum_{\mathbf{k}}\to \frac{1}{(2\pi)^3}\int \mathrm{d}^3k $$ convert a sum over momentum states into an integral, then, $$ \begin{align} n(T, \mu) =& \frac{4\pi}{(2\pi)^3}\int_0^{\infty} \mathrm{d}k \frac{k^2}{z^{-1}e^{ \frac{1}{k_B T} \frac{\hbar^2 k^2}{2m} } - 1}\\ =& \frac{1}{\lambda_{\mathrm{d.B.}}^3} \frac{4}{\sqrt{\pi}} \int_0^{\infty}\mathrm{d}x \frac{x^2}{z^{-1}e^{x^2} - 1}\\ =& \frac{1}{\lambda_{\mathrm{d.B.}}^3} \frac{2}{\sqrt{\pi}} \int_0^{\infty}\mathrm{d}\varepsilon \frac{\sqrt{\varepsilon}}{z^{-1}e^{\varepsilon} - 1} \\ =& \frac{1}{\lambda_{\mathrm{d.B.}}^3} \mathrm{Li}_{3/2}(z) \\ =& \left(\frac{m k_B T}{2 \pi \hbar^2} \right)^{3/2}\mathrm{Li}_{3/2}(z) \end{align} $$ where we have defined the thermal de Broglie wavelength $\lambda_{\mathrm{d.B.}} = \sqrt{\frac{2\pi \hbar^2}{m k_B T}}$ by setting

$$ \frac{\hbar^2}{m}\frac{1}{\lambda_{\mathrm{d.B.}}^2} = \frac{k_B T}{2\pi}. $$

The Polylogarithm is defined as

$$ \begin{align} \mathrm{Li}_s(z) = \sum_{k = 1}^{\infty} \frac{z^k}{k^s} \end{align} $$

free-bose

free-bose.py

Condensation of free Boson

For free Bosons, when the density fixed, lower the temperature, the chemical potential will approach 0. Therefore the Bosons will condense to the zero momentum states. Therefore, we can find the condensate critical temperature $T_C$ by setting $\mu\to 0$, $$ k_B T_C = \left(\frac{n}{\mathrm{Li}_{3/2}(1)}\right)^{2/3}\frac{1}{\frac{m}{2\pi\hbar^2}} $$ where $\mathrm{Li}_{3/2}(1) = \zeta(3/2)\approx2.612$.

Condensation of Fermi pairs

The two components Fermion system on the strong BEC side, every two Fermions form a bound Boson pairs. Thus we can consider this system as a free Bosons. We label the density of pairs as n, also every component has a density n. Label the mass of a Fermion as m, so the mass of a pair is $M=2m$, and the chemical potential $2\mu$. We define the Fermi momentum as $k_F^3 = 6\pi^2 n$, and Fermi energy as $E_F = \frac{\hbar^2 k_F^2}{2m}$. Then the density of pairs can be written as $$ \begin{align} n(T, 2\mu) = \left(\frac{M k_B T}{2 \pi \hbar^2} \right)^{3/2}\mathrm{Li}_{3/2}(z) \end{align} $$ where $z=e^{\frac{2\mu}{k_B T}}$. We can get the condensate critical temperature of pairs

$$ k_B T_C = \left(\frac{n}{\mathrm{Li}_{3/2}(1)}\right)^{2/3}\frac{1}{\frac{M}{2\pi\hbar^2}} = \frac{2\pi}{\left[6\pi^2\mathrm{Li}_{3/2}(1)\right]^{2/3}} E_F \approx 0.218 E_F $$

More generally dispersion cases

$$ \int_0^{\infty}\mathrm{d}\varepsilon \frac{x^{\nu}}{z^{-1} e^{\varepsilon} - 1} = \Gamma(1+\nu)\mathrm{Li}_{1+\nu}(z) $$

where $\nu = 1/2$ for the 3D free Boson. And the gamma function

$$ \Gamma(z)=\int _0^{\infty}\mathrm{d}t\cdot e^{-t} t^{z-1} $$

which $\Gamma(3/2) = \sqrt{\pi}/2$ for the 3D free Boson.

For Fermion case

$$ \int_0^{\infty}\mathrm{d}\varepsilon \frac{x^{\nu}}{z^{-1} e^{\varepsilon} + 1} = -\Gamma(1+\nu)\mathrm{Li}_{1+\nu}(-z) $$

Fermi-Dirac distribution

For free Bosons, the particle density can be written as the sum of the Bose-Einstein distribution for every energy level with momentum $\mathbf{k}$, $$ n(T, \mu) = \frac{1}{V}\sum_{\mathbf{k}}

\frac{1}{z^{-1}e^{ \frac{1}{k_B T} \frac{\hbar^2 k^2}{2m}} + 1} $$ where fugacity $z = e^{\frac{\mu}{k_BT}}$. Convert a sum over momentum states into an integral, then, $$ \begin{align} n(T, \mu) =& \frac{4\pi}{(2\pi)^3}\int_0^{\infty} \mathrm{d}k \frac{k^2}{z^{-1}e^{ \frac{1}{k_B T} \frac{\hbar^2 k^2}{2m} } + 1}\\ =& \frac{1}{\lambda_{\mathrm{d.B.}}^3} \frac{4}{\sqrt{\pi}} \int_0^{\infty}\mathrm{d}x \frac{x^2}{z^{-1}e^{x^2} + 1}\\ =& \frac{1}{\lambda_{\mathrm{d.B.}}^3} \frac{2}{\sqrt{\pi}} \int_0^{\infty}\mathrm{d}\varepsilon \frac{\sqrt{\varepsilon}}{z^{-1}e^{\varepsilon} + 1} \\ =& \frac{1}{\lambda_{\mathrm{d.B.}}^3} \left[-\mathrm{Li}_{3/2}(-z)\right] \\ =& \left(\frac{m k_B T}{2 \pi \hbar^2} \right)^{3/2}\left[-\mathrm{Li}_{3/2}(-z)\right] \end{align} $$

free-fermi

free-fermi.py

Maxwell-Boltzmann distribution

Reference

#physics #chemical potential