# Density of states for fermions and bosons

1. Jan 8, 2014

### hokhani

To take into account the density of states for an ideal gas, we first calculate it ignoring the spin. Then to take into account the spin for a system of electrons we put the number 2 for two spin directions. Why don't we do such this for a boson gas? For example if we have a gas of spin 1 particles, why don't we put the number 3 to consider three spin directions?

Last edited: Jan 8, 2014
2. Jan 8, 2014

### samalkhaiat

The answer is Pauli exclusion principle. Spin 1 particles(bosons) do not care about it, i.e., when you interchange two bosons their wavefunction stays the same(symmertic). Fermions, on the other hand, respect the principle, i.e., have antisymmetric wavefunction, so you need that factor of two to account for all possible states.

3. Jan 8, 2014

### hokhani

Thanks, but I don't agree. We are indeed calculating the number of states accessible to a particle.Therefore for a particle with spin 1, each single level turns into three different states.

4. Jan 8, 2014

### samalkhaiat

Yes, I completely misunderstood your question! In general, if you have g states with the same energy E, then the number of non-interacting, identical particles with energy e is given by
$$n_{ i } = g_{ i } D (e_{ i } , T )$$
where $D$ is Bose or Fermi distribution function.