A gas consisting in identical bosons, say, atoms with integer spin, obeys at sufficiently low temperatures the Bose-Einstein statistics. Let's assume for simplicity that the only relevant quantum number is the energy level. The main feature of such a gas is that more than one atom may occupy the same energy level.
Now, the concept of quantum non-degeneracy means that given a set Ê = {Ê(1), Ê(2), . . ., Ê(N)} of operators, for a single value Ek = {E(1)k, E(2)k, . . ., E(N)k} of this set there exists a single state a quantum system (in this case the set Ê of operators is said to be complete).
In our case, only the energy value distinguishes between states. My question is whether there is any connection between the concept of quantum non-degeneracy, and a non-degenerate quantum gas. To my understanding these are different concepts, i.e. what I saw in literature is that a degenerate Bose gas is a B-E condensate.
If at most one single atom occupies an energy level, the statistics won't be Bose-Einstein, it would be Fermi-Dirac. Still, I don't feel completely sure on the terminology of degenerate and non-degenerate for quantum gases, and this is why I ask the above question.
Cf. http://web.mit.edu/8.333/www/lectures/lec24.pdf and related sections preceding it. In a non-degenerate gas the occupation number is given by the usual expression for the Bose gas-and it's not zero or one, but finite. The statements about non-degeneracy aren't correct; degeneracy for quantum gases isn't used in the same way as for energy levels of one-particle Hamiltonians, that's all.
That's terminology. Non-degenerate means close to Boltzmann, when one can neglect plus minus 1 in the distributions.
Metal is an example of a degenerate electron gas while a semiconductor is usually non-degenerate. A degenerate semiconductor behaves like a metal.
Sofia @
What you will most often find in the literature (because it is the commonly most useful quantity) is the mean occupation number,
= 1/[exp((Ei - mu)/(kBT) -1]
where mu is the chemical potential, which regulates the total boson density of a actual type.
Note added: You perhaps want to know the probability of finding Ni bosons, resulting in the mean number above. It is
P(Ni) = C * exp(-x Ni), Ni=0,1,2,...,
where x = (Ei - mu) and C is a normalization constant. The same expression is valid for fermions except in this case Ni = 0, 1 only. Finally, the distribution for Boltzmann particles is
P(Ni) = C * exp(-x Ni)/Ni!, Ni=0,1,2,...
I am not very happy with words, because they invite to misinterpretations and misunderstandings and endless empty discussions. I think what you and I (and everyone else of knowledge) mean by quantum degeneracy is the rather trivial fact that sometimes several quantum levels may have the same energy. A degenerate quantum gas is a horrible designation, which should have been banished from the language of physics. Because it is used both for Fermi and Bose gases, in ways which I find impossible to rationalize as a common feature.
This is well demonstrated by Igor on this thread, where he explain its use in Fermi gases, although you asked about Bose gases. To the extend there is something common, it is the concept of zero-temperature quantum gases (which do not need any additional descriptions).
Dear Kåre,
I know all these things. I was confused about the terminology "degeneracy". It is an ambiguous term, because a degenerate quantum number means that different (mutually orthogonal) quantum states display that quantum number. On the other hand, the concept of degenerate gas is connected with the condensate phase.
But, maybe you would consider my last question: it is about the chemical potential. It is at the site
https://www.researchgate.net/post/What_is_the_role_of_the_chemical_potential_m_in_the_Bose-Einstein_statistics
Maybe you can help about that. It's a long time since I learnt thermodynamics and quantum statistics, and meanwhile I didn't have need of them. So, my knowledge in them gathered a lot of dust.
With thanks,
Sofia
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