# Bose-Einstein condensation in 1D harmonic oscillator and its density of states

I have troubles understanding how (and whether) Bose-Einstein condensation works in 1-D harmonic oscillator. From my calculation it seems that in limit of infinite number of particles, almost all of them are in the ground state regardless of temperature.

I have found quite a few interesting articles about this system, both from mathematical and physical perspective (in the latter case people are even realizing it by constructing cigar-shaped optical traps), but I have not found anywhere answer satisfying me.

Here is my reasoning:

In canonical ensemble (i.e. number of particles is fixed), the thermal state is $\rho=\frac{\exp(-H/T)}{\text{Tr}[\exp(-H/T)]}$, so that eigenstate to energy $E$ has population proportional to $\exp(-E/T)$. The fact that underlying system is of bosonic nature should not - in my opinion, I might be wrong here - affect the form (as matrix exponent/Gibbs state) of thermal state anyhow.

The Hamiltonian of harmonic oscillator is $H=\sum\limits_{k=0}^{\infty} k a_k^\dagger a_k$, where $a_k$ acts on $k$-th excitation subspace and naturally $H$ can be viewed as sum of $(\text{energy of excitation})\times(\text{number of excitations})$. I have set the energy spacing $\hbar\omega$ to $1$, and shifted the ground state to $0$ for convenience. The eigenstates of Hamiltonian are simple Fock states, e.g. $H|0,2,1,0,3,7,\ldots\rangle=(2\times 1+1\times 2+3\times4+7\times5)|0,2,1,0,3,7,\ldots\rangle=51|0,2,1,0,3,7,\ldots\rangle$.

From now on, the number of particles is fixed and denoted by $N$

In the thermal state, population of some energy level $E$ is its degeneracy times $\exp(-E/T)$ (times normalization constant). In the case of harmonic oscillator (with $\hbar\omega=1$ and null ground state energy) in order to determine the degeneracy of each energy a bijection between Fock states of given energy $E$ and integer partitions of $E$ can be drawn: each integer partition with length at most $N$, e.g. $$51=1+1+2+4+4+4+5+5+5+5+5+5+5$$ corresponds to a Fock state: $k_i$ repetitions of integer $i$ resemble $k_i$ of $i$-th excitation. The partition above can be thus interpreted as Fock state $|0,2,1,0,3,7,\ldots\rangle$. If the length of integer partition is not equal to the number of particles, we just populate the ground state - or add zeroes to the partition. Therefore, number of integer partitions of $E$ with length at most $N$ is the degeneracy of energy $E$ in the system of $N$ bosons in harmonic trap. Also, the maximum length of all integer partitions of $E$ is of course $E$, as $E=1+1+\ldots+1$.

Let us set the number of particles $N=10^6$ and temperature $T=60$ (which are, if I remember correctly, about the right parameters for real-life optical traps). Quick calculation shows that the most populated state is around energy $E=2000$, but all states in this energy range have almost all particles in ground state (e.g. if 1% of particles are not in the ground state, the energy is at least $0.01*10^6=10^4$).

In the limit of infinite number of particles, this happens regardless of temperature: almost all (i.e. except for finitely many) particles in the most probable states are in the ground state.

Is this result correct? I am confused, since Bose gas in a box has completely different behavior: depending on the dimensionality, there exists a well-defined condensation temperature ($d\ge 3$) or it is impossible ($d\le 2$).

Condensation of ideal Bose gases in harmonic traps in 1, 2 and 3 dimensions is discussed in Mullin, WJ, "Bose-Einstein condensation in a harmonic potential" (1997). JOURNAL OF LOW TEMPERATURE PHYSICS. 55. https://scholarworks.umass.edu/physics_faculty_pubs/55. In Eq. (43) Mullin provides the expression for the condensation temperature in 1D $$T_c^{(1)}=\frac{N}{\log N}\frac{\hbar\omega}{k_b},$$ which grows with $N$. So your observation for the ideal Bose gas in a 1D harmonic potential is correct. Note that he credits D. F. Goble and L. E. H. Trainor, Can. J. Phys. 44, 27 (1966); Phys. Rev. 157, 167 (1967), for this result, but those references are behind paywalls.