Relationship between thermal (de Broglie) wavelength and Planck's constant In statistical mechanics, the term $h^{3N}$ is introduced as a measurement unit for the phase space of a system of $N$ particles in $3$ dimensions, and is usually multiplied by the Boltzmann factor $\frac{1}{N!}$ in the multiplicative constant of various characteristic quantities (for example the partition function).
In the study of the quantum statistical mechanics of systems composed by $N$ identical and indistinguishable particles, during the calculations aimed to obtain the partition function of the system from the density matrix, I encountered a similar factor, in which the Planck constant is replaced by $\lambda$, the thermal wavelength, in
$$
\frac{1}{N!\cdot{}\lambda^{3N}}
$$
As $h$ is the measurement unit in the first example, is it possible to describe the thermal wavelength in a similar way? If yes, what is its physical meaning?
Thank you in advance.
 A: I suppose you could think of it that way,
at least in the development of classical statistical mechanics.
The factor $h$ acts as a unit of action,
for each coordinate-momentum pair,
which is needed to make the classical canonical partition function dimensionless:
$$
Q = \frac{1}{N!h^{3N}} \int d^{3N}\mathbf{r} \int d^{3N}\mathbf{p} \, 
\exp\left[-\frac{K(\mathbf{p}^{3N})+\mathcal{V}(\mathbf{r}^{3N})}{k_BT}\right]
$$
where $K$ is the kinetic energy and $\mathcal{V}$ the potential energy.
It is sometimes said to define a quantum "granularity" or "discretization" in phase space,
allowing us to translate the quantum sum over states into a classical integral.
That sounds like a bit of hand-waving,
but it can be justified by examining simple systems such as the harmonic oscillator and the particle in a box,
and invoking a correspondence principle,
and I guess this is the direction from which you are approaching this question.
If we actually do the integrals over momenta,
which we can since 
$$
K = \sum_{i=1}^N\sum_{\alpha=x,y,z} \frac{p_{i\alpha}^2}{2m}
$$
this formula becomes
$$
Q = \frac{1}{N!\lambda^{3N}} \int d^{3N}\mathbf{r} \, 
\exp\left[-\frac{\mathcal{V}(\mathbf{r}^{3N})}{k_BT}\right]
$$
with the usual definition of the thermal de Broglie wavelength,
$$
\lambda = \frac{h}{p_{\mathrm{th}}} = \frac{h}{\sqrt{2\pi mk_BT}}
$$
Now $\lambda$ sets a quantum mechanical length scale.
It is the wavelength of a "typical" particle at the given temperature.
If the system is actually an ideal gas, $\mathcal{V}=0$,
the configurational integral can be done trivially,
and we get the famous result
$$
Q_{id} = \frac{V^N}{N!\lambda^{3N}} \sim \left(\frac{V/N}{\lambda^{3}}\right)^N
$$
where I have (very roughly) set $N!\sim N^N$ (the worse-than-Stirling approximation).
As the Wikipedia page says,
physically the ratio of volume-per-particle $V/N$ to $\lambda^3$
characterizes the balance between quantum and classical behaviour.
So in a similar way to $h$, which relates to 
discretization of positions and momenta together,
you could think of $\lambda$ as characterizing the discretization of positional coordinates on a quantum level, at the given temperature,
when we do the integration over positions (whether $\mathcal{V}$ is zero or not).
A high value of $Q$ (compared to $1$)
means that there are very many accessible states at 
the temperature of interest,
which indeed is consistent with a small separation between energies of states
(compared to $k_BT$) and hence the validity of the classical approximation.
