# Minimal phase space volume $h^3$ out of Heisenberg's principle

So the explanation in my text book is this: we have a phase space volume $\Delta x\Delta y\Delta z\Delta p_x\Delta p_y\Delta p_z$. Now they state that Heisenberg's uncertainty principle states $\Delta p_x\Delta x\geq h$ so our minimal phase space volume is $h^3$.

However, shouldn't Heisenberg's uncertainty principle be $\Delta p_x\Delta x \geq \frac{h}{4\pi}$?

If you take a particle in a box (3D infinite square well), then the density of momentum states per unit volume is given by $$g(p)\ dp = g_s\ \frac{4\pi p^2}{h^3}\ dp,$$ where $g_s$ is a spin degeneracy factor (e.g. $g_s=2$ for electrons, protons etc. because they can have 2 spin states per momentum state). This means that the volume of phase space occupied by one momentum state must be $h^3$.
The treatment you give in your question is a rough attempt to justify this result in terms of the (usually) more familiar Heisenberg uncertainty principle. Such a treatment cannot be exact since for a particle in a 1D infinite potential well, the product of uncertainty in momentum and position is actually given by $$\Delta p \Delta x = \frac{h}{4\pi} \left(\frac{n^2 \pi^2}{3} - 2\right)^{1/2}$$ (see this link), where $n \geq 1$ is the principle quantum number and thus the product of momentum and position uncertainty is always bigger than $h/4\pi$ and would then need to be averaged in some way over all the momentum states.