The previous Lecture exhibited formal aspects of quantum theory. Consider now the application of that theory to statistic mechanics. We have previously written the ensemble average of a variable as
where the sum passes over all states of the system, and where is the value of A in the state j. To give this expression a quantum mechanical form, recall that to make a quantum observation we replace all observables A with their corresponding operators . If one measures A for a system in state , unless is an eigenstate of one gets different values for A in different measurements. If a large series of replicas of a given system are examined, the quantum average value of A is . Applying these principles to eq. 11.1 replaces E and A with their operator forms, so for a quantum system
the sum on j passing over all states of the system.
Equation 11.2 is significant because it embodies a double average, one average being quantum mechanical and the other statistical mechanical:
1) The quantum averages are indicated by the terms and . The first term gives the expectation value of in the state , while the second term gives the expectation value of in the same state. If the states were eigenstates of and , the quantum average would have a particularly simple form, namely eq. 11.1. If the were not eigenstates of and , taking the expectation values would require expanding the in appropriate eigenstates, as shown in eqs. 10-31 and 10-32.
2) The statistico-mechanical average is indicated by the sums on j, which reflect averages over all states of the system, the various terms being given their canonical statistical weights.
Throughout the treatment of quantum systems, all ensemble averages will have this double nature. In each case, there will be a thermal average over different states of the system. Within each term of the thermal average, there will be a quantum average to obtain the expectation values of and within that state. The sequence of taking the averages is significant, the quantum average being taken separately for each state, the various quantum averages then being combined to yield a thermal average. This order of taking the two averages is consistent with the thus-far established image of an ensemble average. To average within the classical canonical ensemble, the function A is evaluated separately in each element of the ensemble, a thermally weighted average of A from different states of the ensemble then being taken. In the quantum canonical ensemble, evaluating in each element of the ensemble requires computing its expectation value. After the expectation value has been evaluated separately for each element of the quantum ensemble, the thermally weighted statistico-mechanical average of is taken, as in the non-quantum case.
In Lectures 8 and 9, statistical mechanics was applied to quantized systems: the rigid rotor, the quantized-spin in an external field, and the harmonic oscillator. The partition function was written as a sum over the energy eigenstates of the system. Energies computed quantum-mechanically, and a discrete form of eq. 11.1, then gave values for Q, <E>, and other quantities.
A first question is whether or not we were justified in using the list of energy eigenstates of the system as a correct list of all states of the system. There are variables which do not commute with . If we had written the sum of eq. 11.1 while using some other set of eigenstates, instead of using energy eigenstates, would we have obtained the same value for Q? If we get different values for Q when we use different sets of basis vectors, how d0 we know which eigenstates we should use to take an average over the canonical ensemble? Fortunately, it turns out that the choice of eigenstates is irrelevant. The terms and can be viewed as matrix elements of and . The sums over states of eq. 11.1 are traces of matrices. The trace of a matrix is invariant to the choice of basis vectors. While, e. g. , changes when the basis vectors are changed, the sums in eq. 11.1 are independent of the choice of basis vectors. Any set of basis vectors which is a complete orthogonal set of states is equally acceptable for evaluating Q or <A>, because eq. 11.2 gives the same value for regardless of the choice of basis vector.
The independence of ensemble averages and partition functions from the choice of basis vectors has practical applications in statistical mechanics. To obtain the thermodynamic properties of a quantum system, one only needs to evaluate its partition function
Q is easily evaluated if the are energy eigenstates, because in that case and
Suppose one has a system for which a set of basis vectors (a ``complete set of states'') have been found, but those basis states are not the energy eigenstates. Q can be computed from eq. 11.3 while using the basis vectors as the states of the system. The individual matrix elements will be more tedious to evaluate, because involve the quantum average of over the mixed (with respect to energy eigenstates) state . However, use of the rather than the energy eigenstates will not affect the sum Q of the diagonal matrix elements of .
As a demonstration of the independence of Q and the choice of basis vectors, consider the isolated spin system of Lecture 8. For a single spin the energy was for , the partition function being written
This Q may be rewritten as a double quantum and thermal average, using the notation of the previous Lecture. Eq. 11.5 is
for and . Eqs. 11.5 and 11.6 are written in terms of energy eigenstates. An alternative to energy eigenstates are the helicity eigenstates
in terms of which the partition function is
It may be confirmed by direct computation that Q from eq. 11.9 agrees with Q in eq. 11.5.
A technical issue arises in the quantum evaluation of ensemble averages. There is no physical guarantee that A and commute, in which case and need not be equal. This issue arises already in conventional quantum mechanics, in the evaluation of such averages as . The orthodox resolution for calculating the expectation value of non-commuting variables is to evaluate the symmetrized product . [When I last investigted the question, no one had proposed an experiment to test the orthodox resolution.]