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
from which we see that
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.