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?


1 Answer 1


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.

  • $\begingroup$ Thank you. That was a very nice explanation. How can we show that "The energy corresponding to the critical temperature is MUCH larger than the energy corresponding to the first excited state" ? $\endgroup$
    – AdShil00
    Commented Nov 4, 2022 at 5:27
  • $\begingroup$ @AdSushil00 For a 3D harmonic oscillator we have $kT_c \approx \hbar \omega N^{1/3} \gg \hbar \omega$ if $N^{1/3} \gg 1$. $\endgroup$
    – Jagerber48
    Commented Nov 9, 2022 at 6:05
  • $\begingroup$ Thank you. What about the line that says ⟨𝑛1⟩ decreases as 𝑇 is either increased or decreased away from 𝑇𝑐. How does <n1> decrease if we increase T ? The expression for <n1> doesn't seem to suggest that. Also, Why is critical temperature determined by comparing occupancy of excited states and total particle, when instead by definition there has to be macroscopic occupation of ground state and not excited state ? $\endgroup$
    – AdShil00
    Commented Nov 10, 2022 at 5:29
  • $\begingroup$ Can I know how you plotted Fugacity Vs Temperature ? When I use the expected occupancy expression, it gives a plot that is different from yours. $\endgroup$
    – AdShil00
    Commented Nov 16, 2022 at 5:42
  • $\begingroup$ @AdShil00 To answer your first question: For $T>T_c$ we have $\langle n_1\rangle = z/(1-z)$. as $T$ increases $z$ decreases which means $z/(1-z)$ also decreases. To answer your second question: critical temperature is not determined by looking at the occupancy of the excited state. It is determined by looking at the occupancy of the ground state in the large particle number limit. In this answer I don't demonstrate how $T_c$ is calculated, I focus instead on the topic of the question which is occupancy of the excited state. (just to show that is technically non-zero but small). $\endgroup$
    – Jagerber48
    Commented Nov 17, 2022 at 5:55

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