Oscillator Semiclassical Phase Space Probability versus a Quantum Mechanical Approach

 The Maxwell-Boltzmann probability factor exp(-energy/T) / normalization may be applied to a quantum oscillator using energy levels given by hbar w (n+.5). This leads to the derivative in 1/T of  a geometric series and yields an average energy: Eave = hbar w/2 + hbar w / (-1+exp(hbar w/T)).

    If one were to consider a semiclassical approach using p (momentum and x), energy would be described by pp/2 + xx/2 where for simplicity m=1 and k=1. This would appear in the MB factor with a temperature T and this expression would be expected to hold for high T or w/T small i.e. the quantum phonon jumps would be hardly noticeable compared to the average energy T in this limit. 

    If one tries to write the classical energy, for w/T small, in the form of  quantum energy, one has hbar w n in this limit, where n is the number of phonons (hbarw/2 is small in this limit) so: zz=  {pp/2+kk/2}/w is the number of phonons which should be multiplied by hbar w/T. One may then ask: What function has the limit hbar w/T? A solution is:  {1-exp(-hbarw/T)}. 

   Thus the semiclassical normalized phase space probability:  P(p,x) = {1-exp(-hbarw/T)}. Exp{ - zz {1-exp(-hbarw/T)} } yields the small hbar w/T limit  quantum average energy. As a result, the average energy in this limit should match for both the classical phase space distribution and the quantum average energy expression. This classical distribution happens to be identical to the Husimi distribution.

   One may note (as is done in (1)), that {1-exp(-hbarw/T)} is the probability for the quantum oscillator particle to be in the ground state. In the limit of hbar w/T small, this probability is hbar w/T which is a measure of the effect of the quantum phonon leaps as well. Fisher’s information, as noted in (1), is equivalent to this ground state probability, but it is also the average energy which is proportional to the temperature type value, in this case {1-exp(-hbarw/T)}. This follows from the same kind of calculation done for a gas with particles with energy pp/2m, although instead of a factor of .5T one T because there is a contribution from pp/2 and xx/2.