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I am currently learning about Bose-Einstein condensation (BEC). I understand that the ground state is rapidly populated when the temperature goes below the critical temperature. This macroscopic occupation of the ground state is the BEC.
However, most texts I have read immediately assert that all excess particles occupy the ground state. Is there a reason why the BEC cannot occupy the excited states?
The expected boson occupation for a state with energy $\epsilon_j$ is given by
$$ \langle n_j \rangle = \frac{1}{e^{\beta (\epsilon_j - \mu)}-1} = \frac{1}{z^{-1}e^{\beta\epsilon_j}-1} = \frac{z}{e^{\beta \epsilon_j}-z} $$
Here $\beta = \frac{1}{kT}$ and I have defined the fugacity $z=e^{\beta\mu}$ where $\mu$ is the chemical potential. Without changing the physics we can add an arbitrary offset to the energies so that the ground state energy, $\epsilon_0 = 0$. We see that for $\langle n_j \rangle$ to be positive it is necessary that $0<z<1$.
To understand BEC we must understand the behaviour of $z$ as a function of temperature. For concreteness I assume a fixed atom number $N$ and a 3D isotropic harmonic oscillator with frequency $\omega_0$. The energy levels are then space by $\hbar \omega$. It is then reasonable to define a dimensionless temperature $\tilde{T} = \frac{kT}{\hbar \omega_0}$. Below I plot $z$ as a function of $\tilde{T}$ for a non-interacting bosonic gas in a 3D harmonic oscillator potential for various atom numbers $N$.
This plot shows fugacity $z$ versus temperature $\tilde{T}$ for atom number $N = \{10^1, 10^2, 10^3, 10^4\}$ from left to right. We see that as $T$ decreases $z$ increases linearly towards $1$ until $T_c$ at which point $z$ saturates to 1. As $T$ is lowered below $T_c$ $z$ still increases but now more slowly since it has saturated. The transition from linear growth to saturate becomes sharper and more "phase transition-y" as atom number $N$ is increased.
Let us now consider the ground state population.
$$ \langle n_0 \rangle = \frac{z}{1-z} $$
We see that as $z \rightarrow 1$ that $\langle n_0 \rangle$ will become very large.
Now consider the first excited state population.
$$ \langle n_1 \rangle = \frac{z}{e^{\frac{\epsilon_1}{kT}}-z} $$
I note that for experimental BECs the quantity $\frac{\epsilon}{kT} \ll 1$*. I will now consider two limits of this function, the $T>T_c$ and $T<T_c$ limits.
For $T>T_c$ we have that $z<1$ so we can approximate $e^{\beta \epsilon_1} \approx 1$ and write
\begin{align} \langle n_1 \rangle \approx \frac{z}{1-z} \end{align}
As temperature decreases towards $T_c$ this function increases since $z$ increases towards $1$.
For $T<T_c$ we have $z\approx 1$ and we can no longer approximate $e^{\beta \epsilon_1} \approx 1$ so we have
$$ \langle n_1 \rangle \approx \frac{1}{e^{\beta \epsilon_1}-1} \approx \frac{1}{1+\beta \epsilon_1-1} \approx \frac{kT}{\epsilon_1} $$
We see that this function decreases as $T$ is decreased.
Thus we see that the excited state population $\langle n_1 \rangle$ decreases as $T$ is either increased or decreased away from $T_c$. Thus, the excited state population has a maximum at $T=T_c$. The question then of whether the excited state can be macroscopically occupied is a question of how large the excited state population is at the critical temperature. For a 3D Harmonic oscillator we have
$$ kT_c \approx \hbar \omega N^{\frac{1}{3}} $$
So the fraction of atoms in the first excited states at the transition is
$$ \frac{n_1}{N} \approx \frac{N^{\frac{1}{3}}}{N} = N^{-\frac{2}{3}} $$
So we see that the excited state fraction decreases as the total atom number $N$ increases and we move deeper and deeper into the thermodynamic limit. Below I plot the excited state fraction as a function of $\frac{T}{T_c}$ for $N = \{10^2, 10^3, 10^4, 10^5\}$ atoms.
See W. Ketterle and N.J. van Druten, Phys. Rev. A 54, 656 at W. Ketterle and N.J. van Druten, Phys. Rev. A 54, 656 for a more thorough discussion of finite atom number effects in Bose-Einstein condensation.
*This is a critically important point about Bose-Einstein condensation. The energy corresponding to the critical temperature is MUCH larger than the energy corresponding to the first excited state. There is a trivial single particle effect which is that if you decrease the temperature so much that $kT \ll \epsilon_i$ then of course you expect to find most particles in the ground state. This would be true even of a classical gas of distinguishable particles. I emphatically point out that this is not the physics of BEC.
