Let's take an energy level density function for a particle as $\rho(E)=const {V} {\delta(E-\varepsilon)}$, $V$ is a gas volume, $\delta(E-\varepsilon)$ - delta-function, $\varepsilon$ - energy level of excited particle (it is not as usual $\rho(E)\sim E^{\frac{d-2}{2}}$). It means that particles in this exotic gas can be only on two energetic levels - ground - with the energy $E_0=0$, and excited - $E=\varepsilon$ with degeneracy proportional to $V$ which in termodynamic limit goes with the number of particles $N$ to infinity so that the particle density $n=N/V$ would remain constant. To define critical temperature $1 /{\beta_c}$ we have:

$$N_0+\int_0^\infty \frac{\rho(E)dE}{e^{\beta E}-1}=N,$$

where $N_0$ - the number of particles in ground state. So, for critical particle density $n_c$ instead 0f well-known power dependence $n_c{\sim}(1/{\beta_c})^{d/2} $ `we have: $$n_c\sim\frac {1}{e^{\beta_c\varepsilon } -1}$$

Question: Are there physical systems or theoretical models, with such a "two-level condensation"?

I would also be grateful for the rationale behind the lack of physical sense of the proposed model, if any.

P.S. I would like to emphasize that we have not two states - one ground and one excited, but one ground and infinitely many (in the thermodynamic limit) excited states with the same energy $\varepsilon$. And in each of the infinite number of excited states there can be an infinite(bosons!) number of particles Two states would be in case of $V\equiv1$

P.P.S. Of course, this model can be formulated without a transition to continuum energy, integration, delta-function, etc. Within the framework of a discrete model without reference to an ideal gas model, this looks like this: $N$ indistinguishable particles can be placed in a $G + 1$ cells, in one of the cells they have zero energy (be in the ground state), in the all other $G$ cells they have the same energy $\varepsilon$. If we introduce the particle density in the "space of cells" $n=N/G$ (the mean number of particles in a cell) and thermodynamical limit ($N, G \to\infty$, $n=const$ ) then for the critical density $n_c$ we have: $$n_c =\frac {1}{e^{\beta_c\varepsilon } -1}$$

P.P.P.S. Mathematically, the proposed model is closely related to the well-known mathematical problem of the number of weak compositions of the integer $n$, i.e. the number of odered partitions of integer $n$ into exactly $G$ parts allowing them to be zero: $${\displaystyle {n+G-1 \choose G-1}}$$ So for the canonical partition function (the number of particles N fixed) we have:

$$ \sum\limits_{n=0}^{N}\dfrac{(n+G-1)!}{n!(G-1)!}\exp({-\beta\varepsilon}n)$$

And for grand сanonical partition function:

$$\frac{1}{1-z}(\frac{1}{1-ze^{-\beta\epsilon }})^G$$

where $z=e^{\beta \mu}$ is fugacity.

IMHO this is the simplest model, demonstrating the main essence of BEC

P.P.P.P.S. Interesting detail, if we replace "quantum stat.weight" in this model: $${\displaystyle {n+G-1 \choose G-1}}$$

on "(semi)classical" one (whithout $1/n!$) : $$G^n,$$
we get a molecular zipper model introduced by C.Kittel in 1969 (https://wwwusers.ts.infn.it/~milotti/Didattica/Biophysics/handouts/Kittel_1969.pdf)


closed as unclear what you're asking by Emilio Pisanty, Kyle Kanos, ZeroTheHero, Jon Custer, David Hammen May 11 '17 at 22:49

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  • $\begingroup$ Normally, condensation is defined as the presence of a macroscopic fraction of the population in a single state out of a continuum. In a two-level system, this is always the case even at infinite temperature, so how are you defining condensation in that case? $\endgroup$ – Emilio Pisanty May 7 '17 at 11:56
  • $\begingroup$ @Emilio Pisanty: I don't understand your question: we have energy level with infinite degeneracy, so we have continuum exited states and single ground state $\endgroup$ – Aleksey Druggist May 7 '17 at 12:06
  • 1
    $\begingroup$ No, that's not what "continuum" means - you don't seem to understand the model you have postulated. The density of states $\delta(E-\varepsilon)$ describes a single state at energy $\varepsilon$. You can have a macroscopic population on your single excited state, but that still doesn't turn it into a continuum of states (in the same way that in standard BECs having a macroscopic population in the ground state does not mean you have a continuum of ground states). As currently written, the question is fundamentally ill-posed and essentially unanswerable. $\endgroup$ – Emilio Pisanty May 7 '17 at 14:23
  • $\begingroup$ @EmilioPisanty: "The density of states δ(E−ε) describes a single state at energy ε" -no, not a state but energy level. How we define the number of states? We divide phase volume of the system 0n $2{\pi}h$. Since we have single momentum(energy) so the number of states of the particle would be $(1*V)/(2{\pi}h)$ i.e. infinite in thermodynamic limit $\endgroup$ – Aleksey Druggist May 7 '17 at 14:44
  • $\begingroup$ ...And in each of the infinite number of excited states there can be an infinite(bosons!) number of particles $\endgroup$ – Aleksey Druggist May 7 '17 at 14:53