Does the +/- 1 term in bosonic and fermionic statistics matter

[KFC]

I am reading an articles introducing the Nobel Price on Bose-Einstein condensates from where I have further reading on Bosonic and Fermionic statistics on some texts. I know one of the mathematical difference is the +/- 1 term in the denominator of the distribution function as below

$f_{BE} = \dfrac{1}{A\exp(E/k_BT) - 1}$

$f_{FD} = \dfrac{1}{B\exp((E-E_f)/k_BT) + 1}$

for the Fermi-Dirac distribution, again the term "+1" arise from the fact of indistinguishable particles. But since there is a Fermi energy when the total energy lower than the Fermi energy (so $E-E_f<0$), at extreme low temperature #\exp((E-E_f)/k_BT) \to 0# so the probably becomes 1. So the "+1" becomes important.

However, the similar reasoning applied on the Bose-Einstein statistics is misleading me. In some book, it is said that the "-1" term for BE distribution arise from the fact that the particles are indistinguishable so to increase the likelihood of multiple occupation of one energy state. Well, if we subtract a positive number from the denominator of a distribution, you get bigger output for $f_{BE}$, I think that's why they said "increasing the likelihood"? But in the case when T approaches extreme low temperature so $\exp(E/k_BT)$ become a really big number, can we ignore that "-1" so to conclude that #f_{BE} \to 0#. But it seems opposite to what I learn from the article where it said at very low temperature, all particles occupying the same state so it is almost 100% to see all particles in ground state. Am I missing or misunderstanding something here?

 

[Simon Bridge]

...on some texts... In some book,...

... which texts, which book?

Well, if we subtract a positive number from the denominator of a distribution, you get bigger output for $f_{BE}$, I think that's why they said "increasing the likelihood"?

That is correct. The important difference is that you can only get one fermions per state while you can have any number of bosons. This means that the number of bosons in a given state may be arbitrarily high.

But in the case when T approaches extreme low temperature so $\exp(E/k_BT)$ become a really big number, can we ignore that "-1" so to conclude that #f_{BE} \to 0#. But it seems opposite to what I learn from the article where it said at very low temperature, all particles occupying the same state so it is almost 100% to see all particles in ground state. Am I missing or misunderstanding something here?

We can only ignore things that make a small difference. This is called "making an approximation". As you point out, that "+/-1" in the denominator makes a big difference at low temperatures, so we cannot ignore it.

The maths predicts that, at close to absolute zero, all bosons are in the lowest possible state.
For an example of a super-cold collection of bosons all in the same state that has been produced in the lab see: "liquid helium".

 

[mfb]

It depends on which limit you consider. For a fixed energy E>0, in the limit of T->0 the occupancy goes to zero - because all the particles are then at even lower energies. At a fixed temperature, for very low energies the Bose-Einstein statistics grows without limit, while the Fermi-Dirac statistic stays bound.