Just a simple question, I didn't manage to find out in my books...

The fugacity $z = e^{\beta \mu}$ in the case we have condensation in a bose statistics. Is it always 1 or $z \to 1$?

In the critical temperature situation, the limit case when we have condensate or not, do we have a condensate fraction of 0 that is $F = \frac{\langle n_0 \rangle}{\langle N \rangle} \to 0$?

Thanks for your answers!


1 Answer 1


I think your question is some old, but I'll answer anyway.

The condensate occurs when the fugacity $z=1$. Given the particle number for the Bose-Einstein statistics \begin{align} N(T,V,z)=\sum_{k}^\infty \langle n_k\rangle^{BE}=\sum_{k}^{\infty}\frac{1}{z^{-1}e^{\beta \epsilon_k}-1}, \end{align} we can easily obtain that \begin{align} \mu\leq 0\quad \text{or}\quad 0\leq z\leq 1\quad, \forall\epsilon_k. \end{align} Using the density of states $g(\epsilon)=\frac{2\pi V}{h^3}(2m)^{3/2}\epsilon^{1/2}$ with the dispersion relation $\epsilon_k=\frac{\hbar^2 \vec{k}^2}{2m}$ and recognizing that at large scale the energy lines will be continuous, one can obtain that \begin{align} \boxed{N(T,V,z)\simeq\frac{V}{\Lambda^3}g_{3/2}(z)+\frac{z}{1-z}\equiv N_\epsilon+N_0, }\end{align} where $N_\epsilon$ is the excited particles and $N_0$ the ground state particles which I introduced "by hand" because the approximation is very bad for low energies; $\Lambda$ is the De Broglie thermal wavelength given by \begin{align} \Lambda=\left(\frac{h^2}{2m\pi kT}\right)^{1/2}, \end{align} and $g_n(z)$ the polylogarithmic function, which can be approximated as a series using the condition $0\leq z \leq 1$ and the change of variable $x=\beta \epsilon$ \begin{align} g_n(z)=\frac{1}{\Gamma(n)}\int_{0}^{\infty}\frac{x^{n-1}dx}{z^{-1}e^x-1}\simeq\sum_{k=1}^{\infty}\frac{z^k}{k^n}. \end{align} Note that the polylogarithmic function $g_n(z)$ is the Riemann zeta function $\zeta(n)$ when $z=1$, i.e. \begin{align} g_n(1)=\sum_{k=1}^{\infty}\frac{1}{k^n}\equiv \zeta(n). \end{align} Riemann zeta function has analytical values and numerical values: \begin{align} \zeta(1)\to\infty \qquad \zeta(3/2)\approx 2.612 \qquad \zeta(2)=\frac{\pi^2}{6}\\ \zeta(5/2)\approx 1.341 \qquad \zeta(3)\approx 1.202 \qquad \zeta(7/2)\approx 1.127\\ \zeta(4)=\frac{\pi^4}{90}\qquad \zeta(6)=\frac{\pi^6}{945} \qquad \zeta(8)=\frac{\pi^8}{9450} \end{align} If $z=1$ in the boxed equation for the particle number one can see that the $N_0$ term dominates, giving a divergence for the ground state particles, i.e., all the new particles introduced in the system (without "cost", because the chemical potential is negative) will going to the ground state. If you're not convinced yet, we can show that, for $z=1$ the excited particles gives \begin{align} N_\epsilon=\frac{V}{\Lambda^3}g_{3/2}(1)\equiv \frac{V}{\Lambda^3}\zeta(3/2)\approx \frac{V}{\Lambda^3}\;2.612, \end{align} i.e. \begin{align} \frac{V}{\Lambda^3}g_{3/2}(z) \leq N_\epsilon \leq \frac{V}{\Lambda^3}\zeta(3/2)=N_\epsilon^{max}. \end{align} The excited particles reach a maximum value when $z=1$. If we add new particles they can never be excited: they will always remain in the ground state. So in $z=1$ Bose-Einstein condensate occurs.


Note that polylogarithmic zeta function can be written explicitly like \begin{align} g_n(z)=\sum_{k=1}^{\infty}\frac{z^k}{k^n}=z+\frac{z^2}{2^n}+\frac{z^3}{3^n}+\dots, \end{align} so for $z\ll 1$, we can write \begin{align} g_n(z)\approx z \qquad (z\ll 1). \end{align} This condition ensures that polylogarithmic function has a maximum value when $z=1$

  • $\begingroup$ Now z is a function of $\mu$ and $T$ so at what chemical potential and temperature z=1? $\endgroup$
    – Paul
    Dec 24, 2019 at 14:39

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