# Single atom states with energy 0, probability and occupation number fermions and bosons

Two mutually non-interacting atoms are trapped in a double-well potential in equilibrium at a temperature $$T$$, such that an atom can only occupy two possible single- atom quantum states, $$Ψa(x)$$ and $$Ψb(x)$$, each corresponding to the atom occupying one of the two wells of the trap.

(b) Assume both single-atom states have energy 0. Determine the probabilities pj(n) of finding n atoms in well j, as well as the average occupation number nj of each well (j = a, b), if the two atoms are:

i. bosonic atoms of the same species;

ii. fermionic atoms of the same species;

iii. of different species.

Ok, so since both states have energy 0, that means the probabilities of finding the bosons or fermions in either a or b is 1/Z, right ?

For bosons I got three possible microstates: 2 bosons occupy state A, 2 bosons occupy state B, 1 boson occupies state A, 1 boson occupies state B. So that means Z = 3 and the probability for n atoms to populate state A is 1/3, the same as for state B. Am I correct ? And for fermions I have only 1 possible micro state, right ?