First principle wave mechanical formulations reveal that a quantum particle in any potential trap has non-zero energy in its ground state. This implies that no atom of trapped dilute boson gas in a harmonic potential trap used to achieve its BEC state can have zero energy/momentum. However, people in the field claim that a macroscopically large number of atoms (say about 60% just for an example) in the lowest temperature BEC state achieved by them have zero momentum. Can any body kindly help me understand the reality? This is a state of great confusion.
To simplify your response to this question, let me put my understanding of the reality for your kind perusal. The BEC in trapped dilute gas occurs in a single quantum state of non-zero energy (or corresponding non-zero momentum). In what follows if you can indicate whether I am right or wrong will be sufficient for me.
If wrong, I would greatly appreciate your reason.
The situation for an ideal Bose gas in a harmonic trap is fairly straightforward. At zero temperature, all of the atoms occupy the lowest energy harmonic oscillator state. In this state, the particles have a distribution of momenta determined by the Fourier transform of the ground state wave function. The situation for a uniform system is essentially the same, but here the lowest energy state is the zero momentum state. All of the particles reside in this state at zero temperature. The common element in both situations is that one is talking about the lowest quantum state, not the state with zero momentum.
The situation with interactions is more complicated. Essentially what happens is that interactions deplete the occupation of the lowest quantum state. In the case of liquid helium, the occupation of the zero momentum state is about 10%. For a dilute harmonically trapped Bose gas, the occupation of the lowest quantum state is in excess of 90% (depending on the strength of the interaction). The subtlety here is what one means by the 'lowest quantum state'. The rigorous answer to this question is that it is the eigenfunction of the single-particle density matrix corresponding to the largest eigenvalue which is interpreted as the number of particles in this state. When this number is macroscopically large, in the sense that it is a finite fraction of the total number of particles N, we say that the system is Bose-condensed. When people talk about the 'occupancy of the lowest quantum state' they are really referring to this fraction.
Dr Zaremba, hearty thanks for your response to the question. To help me further for having better understanding of the p=0 condensate, I request your valuable comments and clarification on the following points.
In what follows, all particles of an ideal Bose gas in a harmonic trap at T=0 occupy the ground state of a single particle trapped in the said trap. Evidently, they all have identically equal non-zero energy in their ground state or in what we call as BEC state at T=0. The state obviously has zero entropy.
It is well known that the ground state of a particle in a harmonic trap is not the eigen state of the momentum operator of the particle indicating that particle momentum in this state is not a good quantum number. It is also evident that the expectation value of the momentum operator in every state of a particle in a harmonic trap is zero. Nevertheless, it is interesting, that people determine the momentum distribution of particles in this state by obtaining the Fourier transform (FT) of the ground state wave function just to prove that the state has p=0 condensate. In what follows from the underlined statement, I wonder if this momentum distribution is relevant for unravelling the physical behaviour of the system. Further since the said ground state wave function is neither a plane wave of single momentum, nor a superposition of only two plane waves of equal and opposite momentum, its FT is expected to conclude that particles are distributed on states of different momenta. It sounds OK in the frame work of mathematics but not in that of physics since particles having different momenta constitute a state of non-zero entropy which can not be reconciled because the state is a zero entropy state, as concluded in the first paragraph.
The ground state wave function of a particle trapped in a harmonic trap of finite size has significant amplitude only inside the trap, particles are expected to have momentum p > h/2L (with L being the size of the trap); in other words even non-interacting bosons in the said trap are not expected to have zero momentum and hence p=0 condensate. It is a different thing that people assume that h/2L is as good as zero since it is ~10^{-2) times smaller than h/2d (where d is inter-particles separation). In summary, particles of even non-interacting boson gas in harmonic trap do not have p=0 condensate; it is a condensate in the ground state of momentum p = h/2L and this state is not an eigen state of momentum operator; rather it is a state of the superposition of two plane waves of equal and opposite momenta (p, -p).
The same is true for uniform system of non-interacting bosons, since, rigorously, speaking the ground state wave function is a standing wave (not a plane wave) where momentum ceases to be a good quantum number of the state and its lowest value is h/2L, -not ZERO. I wish to emphasize this reality because it helps in understanding the truth that superfluidity has no relation with the so called p=0 condensate as emphasized by Landau, and the BEC in a uniform system of even non-interacting bosons is not a p=0 condensate.
Following the FT of the ground state of a single particle in harmonic trap and taking h/2L ~ 0, it appears that even in case of an ideal Bose gas, p=0 condensate at no temperature is 100%; this is different from the case of uniform system where it reaches 100% value at T=0.
It is obvious that these problems become more pronounced at all T in case of systems of interacting bosons