Phase space in quantum mechanics and Heisenberg uncertainty principle

In my book about quantum mechanics they give a derivation that for one particle an area of $h$ in $2D$ phase space contains exactly one quantum mechanical state. In my book about statistical physics they do exactly the same, but now for $6D$ phase space (now one quantum mechanical state covers a volume of $h^3$).

I'm not sure that I interpret this well. Does this mean that in this $6D$ volume of phase space (of $h^3$), the particle can only be in exactly one location with only one possible set of momentum vectors $p_x$, $p_y$, $p_z$ (this is how I interpret "one possible state")?

If this interpretation is right, I have the following problem. Both books say that this is conform the uncertainty principle of Heisenberg, but I don't see why. It seems to me that because one can measure (f.i. in $2D$ phase space) $p_x$ and $x$ so that $∆x∆p_x = h$ and because this is exactly the surface of phase space in which there is only one state, that one has now determined exactly the position and momentum of the particle (because there is only one state in this area). Because of this I guess something is wrong with my interpretation (although it seems to match with the derivation).

Answer

Your interpretation is not quite right. One sharp interpretation one can give to this "cutting" of phase space into cubes of size $h^{2N}$ (here $N$ is the dimension of the system's configuration space), is that it allows one to use classical phase space to count the number of energy eigenstates of the corresponding quantum hamiltonian. Instead of trying to describe what I mean, let's investigate this stuff through an example.

Consider, the one-dimensional simple harmonic oscillator. The hamiltonian is $$ H(q,p) = \frac{1}{2m}p^2 + \frac{1}{2}m\omega^2 q^2 $$ Let's say I want to answer the following:

Question. Given an energy $E>0$, how many states are there with energies less than $E$?

Now, we have to be careful here because the term "state" means different things in the classical and quantum cases. In the classical case, a state is a point $(q,p)$ in phase space. In the quantum case, a state is a vector in Hilbert space. We can therefore reinterpret the question as follows:

Classical version. What is the area $A(E)$ of the region of phase corresponding to all classical states $(q,p)$ with energies less than $E$?

Quantum version. How many energies eigenstates $\Omega(E)$ does the Hamiltonian posses with energies less than $E$?

The amazing thing is that provided we measure area in phase space in units of $h$, and provided we consider $h$ to be much smaller than the other scales in the problem, both of these questions will give (approximately) the same answer! Let's show this. In the classical case, the region of phase space containing all states with energies less than $E$ is the area of the interior of the ellipse defined by $$ E< H(q,p) $$ It turns out that the area of this ellipse is $$ A(E) = \frac{2\pi E}{\omega} $$ On the other hand, in the quantum case recall that the energy eigenvalues are $E_n = (n+1/2)\hbar\omega$. This means that the number of eigenstates having energy less than $E$ is found by solving $$ (\Omega(E) + \tfrac{1}{2})\hbar\omega =E $$ which, for $h$ small gives $$ \Omega(E) \sim \frac{E}{\hbar\omega} $$ Now here's where the magic happens, notice that $$ \frac{A(E)}{\Omega(E)} \sim \frac{\frac{2\pi E}{\omega}}{\frac{E}{\hbar\omega}} = 2\pi \hbar = h $$ so we have $$ \boxed{\Omega(E) \sim\frac{A(E)}{h}} $$ In words: the area of phase space, measured in units of $h$, allows us to accurately count the number of quantum states below a given energy