Graduate How Does Quantum Theory Provide a Measure for Microstates in Phase Space?

[Kashmir]

*Pathria, Statistical mechanics*"The microstate of a given classical system, at any time, may be defined by specifying the instantaneous positions and momenta of all the particles constituting the system. Thus. If $N$ is the number of particles in the system, the definition of a microstate requires the specification of $3 N$ position cuordinates and $3 N$ momentum coordinates. Geometrically, the set of coordinates $(q, p)$ may be regarded as a point in a $6N$ dimensional phase space"

( $\omega$ be the "volume" in the phase space ) the author says

"... we need to discover a fundamental volume $\omega_0$ that could be regarded as "equivalent to one microstate"Since for one microstate we have one point in phase space then how does there exist a volume in phase space that corresponds to one microstate?

 

[anuttarasammyak]

In classical physics point of no volume represents a state. But in quantum physics cell of volume h represents a states.

 

[Kashmir]

This was an excerpt from classical statistical mechanics, the author was discussing the micro canonical ensemble.

 

[anuttarasammyak]

Even classical statistical physics requires cells to describe the distribution of states, size of which are not determined. Later quantum physics provide its volume of h^3 per particle.

 

[Kashmir]

Even classical statistical physics requires cells to describe the distribution of states, size of which are not determined.

How is that?
Isn't a point in phase space a representation of microstate?

So where does the concept of cell come in?

 

[Kashmir]

This is the full excerpt
"The next thing we look for is the establishment of a connection between the mechanics of the microcanonical ensemble and the thermodynamics of the member systems. To do this, we observe that there exists a direct correspondence between the various microstates of the given system and the various locations in the phase space. The volume $\omega$ (of the allowed region of the phase space) is, therefore, a direct measure of the multiplicity $\Gamma$ of the microstates accessible to the system. To establish a numerical correspondence between $\omega$
and $\Gamma$, we need to discover a fundamental volume $\omega_{0}$ that could be regarded as "equivalent to one microstate." Once this is done, we may say that, asymptotically,
$\Gamma=\omega / \omega_{0}$"

(
$\Gamma$ is the number of microstates calculated quantum mechanically)

 

[anuttarasammyak]

\Gamma## is the number of microstates calculated quantum mechanically)

As $\Gamma$ comes from quantum mechanics, the author seems to unite classical statistical physics and quantum mechanics. In mathematics finite volume contains infinite points but finite cells, the latter of which Physics use to represent finite number of possible states.

 

[vanhees71]

That's indeed the fundamental obstacle of classical statistical physics. In classical physics there is no natural measure for phase-space volumes. This is only provided by quantum theory, where Planck's quantum of action is a natural measure for action and $\mathrm{d}^{3N} x \mathrm{d}^{3N} p /(2 \pi \hbar)^{3N}$ turns out to be the number of microstates.

Within classical physics all you can say is that due to Liouville's theorem the number of microstates must be proportional to the phase-space volume, and you can get pretty far with it as the works of Maxwell, Gibbs, Boltzmann et al in classical statistical physics show, but a really consistent formulation is only possible with quantum theory.