# Volume of state in phase space free particle

I have to how a quantum state of a free particle between 0 and a occupies an area of $$h$$ in the phase space.

What I did was to calculate $$\Delta x \Delta p$$ and show that it was of order $$h$$, but I don't know if that means the state occupies an area $$h$$, I tought it just meant that I cannot localize the system in the phase space with more accuracy that of $$\Delta x \Delta p$$... and actually is not $$h$$, but a constant dependent of the energy level times $$\frac{h}{2}$$. I'm trying to find a proof of the fact that "every state occupies an area $$h$$ in the phase space but i haven't found any.

• Correct, the state of the particle can't localize in phase space in an area smaller than that calculated by the Heisenberg uncertainty. But I don't understand the first paragraph. – Cinaed Simson May 8 '19 at 8:07
• Related: physics.stackexchange.com/q/64212/2451 and links therein. – Qmechanic May 8 '19 at 11:09

This statement can be formulated a little more precisely as:

The dimension of the Hilbert space of a quantum system is asymptotically equal to the volume of the phase space in units of Plank's constant (to the power of the configuration space dimension).

(For quantum systems in one space dimension the volume means the area as in the question).

(In the above asymptotically refers to the limit where both quantities are large, for example in the semiclassical approximation $$\hbar \rightarrow 0$$).

So, the assertion that a state occupies a cell of volume $$h^n$$ is a consequence of the above.

There is a little difficulty in demonstrating this statement for quantum systems studied in beginner courses of quantum mechanics because elementary systems such as a free particle, particle in a box, harmonic oscillator, etc. , both quantities: the phase space volume and the Hilbert space dimension are infinite and it is hard to compare infinite quantities.

However, if we place a constraint in the phase space that has a counterpart in the Hilbert space, we can get finite values for both quantities.

One example that we can do this comparison is in counting the number of states of energy less than $$E = \hbar n$$ in a Harmonic oscillator. The phase space region corresponding to this condition is: $$\frac{1}{2}(p^2+q^2)<\hbar n$$ This is a disc of radius $$\sqrt{2\hbar n}$$ and its area is: $$A = 2\pi \hbar n$$ In the Hilbert space, we know that there is one state for every integer, thus the number of states with energy $$<\hbar n$$ (ignoring the $$\frac{1}{2} \hbar$$ correction), is: $$N = n$$ Thus, we can see that: $$N \approx \frac{A}{2\pi \hbar} = \frac{A}{h}$$

There are quantum systems, in which both number of states and volume are finite without the need of putting constraints. For example, the phase space of a qubit (which is a quantum system with two states) is the two-dimensional sphere.

It is worthwhile to mention that this correspondence plays an important role in understanding physical phenomena such as insulators or superconductors, where in both cases there are no states below some energy gap where the system can evolve into.

On the other side, the above principle was used by Witten to compute the volume of a rather complicated space of major interest in mathematics by counting the number of states of the corresponding Hilbert space.