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 ($d\le 2$).


1 Answer 1


"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"

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 \begin{equation} T_c^{(1)}=\frac{N}{\log N}\frac{\hbar\omega}{k_b}, \end{equation} 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.

  • $\begingroup$ Note that, as mentioned in the abstract to that paper, there is no true phase transition (at finite temperature) for noninteracting atoms in 1D. The given expression is for a 'pseudo-transition,' as defined in detail in the text. $\endgroup$
    – Rococo
    Commented Jun 23, 2019 at 18:10

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