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).


1 Answer 1


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

  • $\begingroup$ Thank you, that made things more clear. I have still some questions though. If we take, in the case of the harmonic oscillator, an ellips in phase space the amount of quantum states is the area of the ellips divided by $h$. But I wonder how I have to interpret the case, if we don't take an ellips, but f.i. half an ellips or a square in the phase space. Because in these books, they say that (in the case of a free particle) for a random area of $dp_xdx$ in phase space (generalized coordinates are now $p_x$ and $x$), the amount of states in this area is $\frac{dp_xdx}{h}$. $\endgroup$
    – yarnamc
    May 5, 2013 at 10:18
  • $\begingroup$ So although I understand how we can speak about "an amount of states" in an ellips in phase space (in the case of the harmonic oscillator), I don't really see how we can generalize this to a random area in phase space. How can one physically interpret an amount of states in a random area? $\endgroup$
    – yarnamc
    May 5, 2013 at 10:22
  • $\begingroup$ @yarnamc I agree that it seems that the procedure I have outlined does not seem to be immediately generalizable to say arbitrary phase volume. My feeling is that when authors are making claims about such things, they are simply being imprecise. After all, there is more to quantum mechanics than classical phase space can tell us. In particular, we should only expect that classical statements agree with quantum ones in some limit (like in my response where we considered $h\to 0$). In short, I kind of think these authors are full of it :) and I also don't really see how we can generalize. $\endgroup$ May 5, 2013 at 20:14
  • $\begingroup$ I have to say however, I would be very interested if someone else on physics.SE knows something that I don't. Let's hope more people read this! $\endgroup$ May 5, 2013 at 20:16
  • $\begingroup$ I think now that the amount of states in a random area is just a mathematical concept without physical interpretation. Thanks for your answers :D $\endgroup$
    – yarnamc
    May 8, 2013 at 9:26

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