三种统计分布的相图
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
The Polylogarithm is defined as
$$ \begin{align} \mathrm{Li}_s(z) = \sum_{k = 1}^{\infty} \frac{z^k}{k^s} \end{align} $$
Condensation of free Boson
For free Bosons, when the density fixed, lower the temperature, the chemical potential will approach
$$
k_B T_C = \left(\frac{n}{\mathrm{Li}_{3/2}(1)}\right)^{2/3}\frac{1}{\frac{m}{2\pi\hbar^2}}
$$
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
$$
\begin{align}
n(T, 2\mu)
= \left(\frac{M k_B T}{2 \pi \hbar^2} \right)^{3/2}\mathrm{Li}_{3/2}(z)
\end{align}
$$
where $z=e^{\frac{2\mu}{k_B T}}$. We can get the condensate critical temperature of pairs
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,
$$
\begin{align}
n(T, \mu) =& \frac{4\pi}{(2\pi)^3}\int_0^{\infty} \mathrm{d}k
\frac{k^2}{z^{-1}e^{ \frac{1}{k_B T} \frac{\hbar^2 k^2}{2m} } + 1}\\
=& \frac{1}{\lambda_{\mathrm{d.B.}}^3} \frac{4}{\sqrt{\pi}}
\int_0^{\infty}\mathrm{d}x \frac{x^2}{z^{-1}e^{x^2} + 1}\\
=& \frac{1}{\lambda_{\mathrm{d.B.}}^3} \frac{2}{\sqrt{\pi}}
\int_0^{\infty}\mathrm{d}\varepsilon \frac{\sqrt{\varepsilon}}{z^{-1}e^{\varepsilon} + 1} \\
=& \frac{1}{\lambda_{\mathrm{d.B.}}^3} \left[-\mathrm{Li}_{3/2}(-z)\right] \\
=& \left(\frac{m k_B T}{2 \pi \hbar^2} \right)^{3/2}\left[-\mathrm{Li}_{3/2}(-z)\right]
\end{align}
$$
Maxwell-Boltzmann distribution
Reference
- Polylogarithm
- Gamma function
- Pathria, R. K., and Paul D. Beale. Statistical Mechanics. 3rd ed. Amsterdam ; Boston: Elsevier/Academic Press, 2011. Chap.7 and Appendices D, E.
- Stoof, Henk T. C., Koos B. Gubbels, and Dennis B. M. Dickerscheid. Ultracold Quantum Fields. Theoretical and Mathematical Physics. Dordrecht ; New York: Springer, 2009. Chap.4.3.