Why the volume of a cell in phase space should be equal to $(2\pi \hbar)^s$?

Question

We want to properly define the concept of entropy using the Boltzmann's Definition of it. But there is a big problem: the coarse graining problem (Id est: How do we count the number of microstates in the allowed region of phase space? By dividing the phase space in cells? Good idea! But how big should the cells be? )
Fortunately Quantum Mechanics comes in our help: if we consider, for example, the case of our system being an ideal gas, then we can show, apparently, that the volume of a cell in an $s$-dimentional phase space should be $(2\pi \hbar)^s$. (Reference: Landau, Theorical Physics Volume 5, page 40)
This fact should arise, I think, from considering the motion of a free particle confined in an $n$-dimentional box; the energy states of the particle, as well as its momentum, then should be quantized, we can then use this fact to state that the volume of a cell in phase space should be big enough to include into itself only one of the possibile quantized states. In this way the choice of the volume of the cell is not arbitrary, and we can derive it exactly.

However I have three big problems regarding this approach:

1. For a particle in a box energy and momentum are indeed quantized, but the position is not; phase space should be composed of position and momentum, but only one of this two is quantized, so seems impossibile to me to use quantization to determine what volume the cell should have.
2. If somehow we can solve my first problem, then what mathematical derivation tells us that the volume necessary to have only one quantum state in a cell is $(2\pi \hbar)^s$?
3. All this reasoning is based upon the hypotesis of dealing with an ideal gas, so no interaction and we can consider, for the derivation of the volume of the cell, a free particle in a box with its simple quantization of momentum; but what if we are dealing with something else? Should we change our value $(2\pi\hbar)^s$ for the volume or there is some trick that allows us to keep using it even when not dealing with an ideal gas?

Answer 1

The fact that we can associate a quantum state to a cell of phase space is a general result (not just for the particle in a box) and is physically meaningful, unlike what is suggested in the comments to your question.
What's missing in this discussion is how the quantum states are connected to the classical phase space. This gap is neatly bridged by the quasi-classical WKB approximation for solving the Schrödinger equation. You can find the discussion leading to the result in Landau Lifsitz (III, §48), as always.

I'll try to give a brief summary here but the explanation is somewhat lengthy.

The classical limit in quantum mechanics is analogous to geometric optics, the simplified description of waves (quantum states) in term of rays (particle trajectories). In geometric optics Fermat's principle states that light moves as to minimise the travel time, $T = ∫dt$, which up to a constant¹ is the wave phase.
Particles, similarly, satisfy Hamilton's principle that states the action, $S = ∫(pdx - Edt)$, is minimised by the particle true motion. Here $p$ is the momentum and $E$ is the total energy.

The analogy motivates the WKB ansatz²: $Ψ(x, t) = A(x, t) e^{iS(x, t)/ℏ}$ for the wave function. You can now plug this $Ψ$ into the (time-independent) 1D Schrödinger equation and solve for $A, S$ in the limit $ℏ \to 0$. The approximate solution is $$ ψ(x) = \frac{C_1}{\sqrt p} e^{i/ℏ ∫pdx} \qquad \text{where}\;\; p = ±\sqrt{2m(E - U(x))} \;, $$ in agreement with the classical expression for the momentum³. The result is valid everywhere except where $E = U(x)$: the turning points of the classical motion. You can still build global solutions by taking clever linear combinations, in between two turning points $a<x<b$ you get: $$ ψ_1(x) = \frac{C_1}{\sqrt p} \cos\underbrace{\left(\frac{1}{ℏ} ∫_a^x\!\! pdx - \frac{π}{4}\right)}_{φ_1}, \quad \text{or} \quad ψ_2(x) = \frac{C_2}{\sqrt p} \cos\underbrace{\left(\frac{1}{ℏ} ∫_b^x\!\! pdx + \frac{π}{4}\right)}_{φ_2} \;. $$ The two wave functions are equivalent, so must we have $|C_1| = |C_2|$ and $φ_1 + φ_2 = \frac{π}{2} + nπ$. From the latter you immediately get $$ \frac{1}{πℏ} ∫_a^b pdx = \frac{1}{2πℏ} ∮ pdx = n + \frac{1}{2} $$ The "∮" symbol means integration over the full period ($a→b→a$) and is in fact the area enclosed by the curve $E=\text{const}$ in the phase space.

The final step is showing that $n$ is not just any integer but the number of zeros of the wave function (easy) and that this is directly related to the quantum number of the state (not easy). The latter is a result of the Schrödinger equation being a Sturm-Liouville problem and is called the oscillations theorem.

For a fully separable system with $s$ degrees of freedom you get $s$ such integrals, so you can combine them to form the phase space volume: $$ ∫dpdx = (2πℏ)^sN $$ In a general system the result still holds true, but you can't use the WKB method to prove it.

Anyway, the fact that $∮pdx = 2πℏ (n + ½)$ is quite significant: before the Schrödinger equation, in the so-called old quantum theory, this formula was the rule for obtaining the energy spectrum of a system. Remarkably, it gives the exact eigenvalues for several key systems like the hydrogen atom and the quantum harmonic oscillator, even if it's an approximation.

The $2πℏ$ factor may never enter ensemble averages, true, but the idea of a phase space cell arises in many contexts and is the basis of quantities like the thermal de Broglie wavelength $Λ$ (in thermodynamics and statistical mechanics) or the quantum concentration $n_q$ (in plasma and semiconductors physics).
It's also worth mentioning that there is direct experimental evidence that supports this, as the entropy of an ideal gas does depend on $2πℏ$, can be measured and agrees very well with the theory⁴.

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¹ $dt = dl/v = kdl/ω$, since the phase velocity is $v = ω/k$.

² In geometric optics this is called the Eikonal ansatz. The ℏ here is a dimensional constant, you can think of it as an alternative to the definition $E = hν$.

³ Note that $p$ is imaginary where $E < U$, in this case the exponential becomes real.

⁴ See "Experimental tests of the Sackur-Tetrode equation" in "Thermal Physics" - Kittel, Kroemer.