The path integral formulation of quantum statistical mechanics is particularly useful for situations in which particle spin statistics can be largely ignored. In the quantum ideal gases, we have a situation in which the spin statistics determine all of the interesting behavior! The fully quantum treatment of the ideal gas will be the subject of the next several lectures.
The spin degree of freedom of a particle is a purely quantum mechanical
aspect (with no classical analog). In quantum mechanics, spin is
analogous to an angular momentum. It is described by a Hermitian
operator 
, where the components satisfy
angular momentum type commutation relations:
The spin operators for a spin-s particle are represented by
matrices (which define different
representations of the group SU(2)). For example,
for a spin-1/2 particle, such as an electron,
the three spin operators are
which can be shown to satisfy the above commutation relations. Since
the three components of spin to do not commute, we choose, by convention,
to work in a basis in which 
is diagonal. Thus, there will
be (2s+1) eigenvalues given by 
. In the example
of the spin-1/2 particle, we see that the allowed spin eigenvalues
(denoted m)
are m=-hbar/2 and 
. The corresponding eigenstates are just
which are denoted the ``spin-up'' and ``spin-down'' states, respectively.
Note that the operator 
is also diagonal
so that the spin-up and spin-down eigenstates of are also
eigenstate of 
, both having the eigenvalue 
.
Thus, given a Hamiltonian H for a system, if H is independent
of spin, then the eigenstates of H must also be eigenstates of
and since all three can be simultaneously diagonalized.
What happens in quantum mechanics when we have systems of identical particles
of a given type of spin? Consider the simple example of a system of
two identical spin-1/2 particles. Suppose we perform a measurement which
is able to determine that one of the particles has an eigenvalue
of 
and the other 
such that 
.
Is the state vector of the total system just after this measurement
where, in the first state, particles 1 and 2 have eigenvalues
and , respectively, and, in the second state,
it is the reverse of this? The answer is that neither state is the
correct state vector since the measurement is not able to
assign the particular spin states of each particle. In fact,
the two state 
and 
are
not physically equivalent states. Two states 
and
can only be physicall equivalent if there is
a complex number 
such that
and there is no such number connecting
and . However, it is possible to construct
a new state vector 
such that
is physically equivalent to
. Let
If we require that
then
from which we see that
or
from which 
and 
. This gives us two
possible physical states of the system
which are symmetric and antisymmetric, respectively, with respect to an exchange of the particle spin eigenvalues. The analog in ordinary one-dimensional quantum mechanics would be the case of two identical particles moving along the x axis. If a measurement performed on the system determined that a particle was at position x=a and the other was at x=b, then the state of the system after the measurement would be one of the two following possibilities:
The standard postulates of quantum mechanics now need to supplemented by an additional postulate that allows us to determine which of the two possible physical states a system will assume. The new postulate states the following: In nature, particles are of two possible types - those that are always found in symmetric (S) states and those that are always found in antisymmetric (A) states. The former of these are known as bosons and the latter are known as fermions. Moreover, fermions possess only half-integer spin, s=1/2,3/2,5/2,..., while bosons possess only integer spin, s=0,1,2,....
Suppose a system is composed of N identical
fermions or bosons with coordinate labels 
and spin labels 
. Let us define,
for each particle, a combined lable 
.
Then, for a given permutation P(1),...,P(N) of the particle
indices 1,..,N, the wave function will be totally
symmetric if the particles are bosons:
For fermions, as a result of the Pauli exclusion principle, the wave function is antisymmetric with respect to an exchange of any two particles in the systems. Therefore, in creating the given permutation, the wave function will pick up a factor of -1 for each exchange of two particles that is performed:
where 
is the total number of exchanges of two particles
required in order to achieve the permutation P(1),...,P(N).
An N-particle bosonic or fermionic state can be created from
a state 
which is not properly symmetrized
but which, nevertheless, is an eigenfunction of the Hamiltonian
Noting that there will be N! possible permutations
of the N particle labels in an N-particle state, the
bosonic state 
is created from
according to
where 
creates 1 of the N! possible permutations
of the indices. The fermionic state is created from
where 
is the number of exchanges needed to create
permutation .
This simple difference in the symmetry of the wavefunction leads to stark contrasts in the properties of fermonic and bosonic systems. With these quantum mechanical rules in mind, let us work out what these properties are.