I am trying to understand the explanation of Bose-Einstein condensation for non-interacting bosons given in Piers Coleman's "Introduction to Many-Body Physics", pg. 85-86. Coleman first writes that the density of a gas of bosons at finite temperature is given by $$\rho = \int \frac{d^3 k}{(2\pi)^3} (e^{\beta (E_k - \mu)}-1)^{-1}, \tag{1}\label{1}$$ where we have $E_k = \hbar^2 k^2/(2m)$. He then points out that the chemical potential $\mu$ must be negative, presumably since this avoids a divergent integral yielding an infinite density, and that as the temperature is lowered ($\beta$ increases), $\mu$ must increase (become less negative) for a fixed particle density $\rho$. This is clear from a simple inspection of the integral---$\mu(\beta)$ increases monotonically (from a negative value) with $\beta$. As one increases $\beta$, at some point $\mu(\beta) = 0$. This is called the Bose-Einstein condensation temperature, and it can be computed in terms of $\rho$ by setting $\mu = 0$ in the integral and solving for $\beta$.

So far so good. Now I quote the part I don't understand:

Below this temperature, the number of bosons in the $k = 0$ state becomes macroscopic, i.e. $$n_{\epsilon = 0} = \frac{1}{e^{-\beta \mu} - 1} = N_0(T) \tag{2}\label{2}$$ becomes a finite fraction of the total particle number. Since $N_0(T)$ is macroscopic, it follows that $$\frac{\mu}{k_B T} = -\frac{1}{N_0(T)} \tag{3}\label{3}$$ is infinitesimally close to zero. For this reason, we must be careful to split off the $k = 0$ contribution to the particle density, writing $$N = N_0(T) + \sum_{k \neq 0} n_k \tag{4}\label{4}$$ and then taking the thermodynamic limit of the second term. For the density, this gives $$\rho = \frac{N}{V} = \rho_0(T) + \int \frac{d^3 k}{(2\pi)^3} \frac{1}{e^{\beta (E_k)} -1}. \tag{5}\label{5}$$ The second term is proportional to $T^3$. Since the first term vanishes at $T = T_0$, it follows that, below the Bose-Einstein condensation temperature, the density of bosons in the condensate is given by $$\rho_0(T) = \rho \left[1 - \left(\frac{T}{T_0}\right)^{3/2}\right].\tag{6}\label{6}$$

(a) I understand the use of the Bose-Einstein distribution to obtain the particle number in the $k = 0$ state. But at the condensation temperature $T_0$, $\mu = 0$, and this quantity becomes infinite. I know somehow this is going to be the point, but I don't understand precisely how: as we cross $T = T_0$ evidently something in our theory has broken down (for example, at $T = T_0$, $\mu = 0$, and we obtain an infinite particle density \eqref{1}). What actually limits the particle density from becoming infinite (what assumption in our theory breaks down here?)? And what does it mean to talk about what happens "below" $T_0$ within the equations we have written down? For example, within these equations, as $T$ goes below $T_0$, $\mu$ cannot cross zero and become positive, otherwise we get divergences. So in \eqref{2}, is $\mu$ positive or negative?

(b) How can one clean up the math leading to \eqref{5}, or at least give some reasonable explanation? I cannot make sense of this derivation. Can one take some integral, split it into an $\epsilon$-neighborhood of the origin and the remainder, evaluate, and take the limit $\epsilon \to 0^+$ to observe a finite contribution from the origin and the second term as written?

(c) In the last sentence, why does it say $\rho_0(T)$ vanishes at $T = T_0$? Isn't it exactly the opposite ($\rho_0(T_0) = \infty$, by dividing \eqref{2} by $V$ and sending $\mu \to 0^-$)?


1 Answer 1


In general, I think that your issue is in identifying how the different quantities scale with $N$ in the thermodynamic limit.

  1. At finite $N$, there is no phase transition. In fact $\mu$ is a decreasing function in $T$, and always satisfies $\mu<0$. The subtlety is that it scales differently in $N$ below $T_c$ in the thermodynamic limit. Below $T_c$, the finite quantity is rather $N\mu$. In the thermodynamic limit:
  • $T_c\leq T<+\infty$: $\mu$ is finite negative and increases from $-\infty$ to $0$ as $T$ decreases. Accordingly, $N\mu=-\infty$ in this range.
  • $0<T\leq T_c$: $\mu$ is constant equal to $0$. However, according to (3) and (6), $N\mu$ increases from $-\infty$ to $0$ as $T$ increases.

You therefore never cross $\mu=0$. This is consistent with (2) as since $N_0$ becomes macroscopic, in the thermodynamic limit, $N_0=\infty$. Only the intensive quantity: $$ \rho_0 = \frac{N_0}{V} $$ is finite for $T\leq T_c$, which is consistent with the fact that $N\mu$ has a finite limit.

  1. The issue is that the Thomas-Fermi approximation breaks down. A unified way would be to use the discretised orbitals, in accordance to the first quantised particle in a box. In $D$ dimensions, using $\hbar=m=1$, $L^D=N$ (extensive volume): $$ \begin{align} \rho N &= \sum_{k\in\mathbb Z^D}\frac{1}{e^{\beta (E_k-\mu)}-1} \\ E_k &= \pi\frac{2\pi}{N^{2/D}}\sum_{i=1}^D k_i^2 \end{align} $$ You can check that the Thomas-Fermi approximation breaks down, and the term $k=0$ is not negligible anymore. Alternatively, you can also directly relate it to the usual formula with the polylogarithm $\text{Li}_{D/2}$ by resumming the expression in terms of number of particles: $$ \rho N = \sum_{n=1}^\infty z^n\theta\left(\frac{2\pi\beta n}{N^{2/D}}\right)^D \\ z = e^{\beta\mu} $$ with $\theta$ being related to the Jacobi theta function: $$ \theta(t) = \sum_{k\in\mathbb Z}e^{-\pi n^2t} $$ You can check that you recover your expression in the thermodynamic limit $N\to\infty$ using: $$ \theta(t\to+\infty) = 1 \\ \theta(t^{-1}) = \sqrt t\theta(t) \\ \theta(t\to0^+) \sim t^{-1/2} $$ so by replacing each term by its limit in the sum: $$ z^n\theta\left(\frac{2\pi\beta n}{N^{2/D}}\right)^D \sim N\frac{z^n}{(2\pi\beta n)^{D/2}} \\ \rho = \frac{1}{(2\pi\beta)^{D/2}}Li_{D/2}(z) $$ which is valid for $\beta<\beta_c=\frac{\zeta(D/2)}{2\pi\rho^{2/D}}$. Intuitively, this is not valid anymore because you also have the tail: $$ z^n\theta\left(\frac{2\pi\beta n}{N^{2/D}}\right)^D \sim z^n $$ which will give the $\frac{1}{1-z}$. This contribution is negligible in the normal phase $\beta<\beta_c$ since $z$ has a finite limit and the polylogarithm contribution suffices. However, in the BEC phase, it is also important as $z\to 1$, so you get this additional term.

  2. No, $\rho_0$ vanishes at $T=T_c$. You need to be careful with extensive and intensive quantities. $n_0$ diverges at the critical temperature in the thermodynamic limit. However, rescaling it by the volume, $\rho_0$ is intensive and has a finite value below $T_c$. You should view $\rho_0$ as the order parameter:

  • $T_c\leq T<+\infty$: $\rho_0=0$, the normal phase
  • $0<T\leq T_c$: $\rho>0$, the BEC phase. It increases according to (6) as $T$ decreases.

Hope this helps.

  • $\begingroup$ I'm a bit confused by this answer, starting with (1). Don't you mean that at $T=T_c$, $\mu = 0$? And that $\mu$ increases monotonically with $T$, and varies from $-\infty$ to $0$? For the $T < T_c$ analysis, I simply do not understand which equation is my starting point. If $\mu = 0$ for $T < T_c$, then Eq. (2) is nonsensical; this equation is said to hold for $T < T_c$, so plugging in $\mu=0$ gives $\infty$. So I don't understand what comes after. Do I start with Eq. (1) plugging in $\mu = 0$? Then what? $\endgroup$
    – gilgamesh
    Jan 4 at 1:14
  • $\begingroup$ getting $+\infty$ from (2) is normal in the thermodynamic limit, $N_0$ is macroscopic. $\endgroup$
    – LPZ
    Jan 4 at 9:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.