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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 $\mu \to 0^-$)?

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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.

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  • $\begingroup$ 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

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