# Free particle 1D phase space

I'm trying to draw the phase space for a particle moving freely between 0 and $$L$$. I guess $$H=E$$(total energy, constant)$$=\frac{p_{x}^2}{2m}$$ so $$p_{x}=\pm\sqrt{2mE}$$ for every x between 0 and L, and taking the positive sign when going from 0 to L and the negative sign when going from L to 0. If I draw the phase space for this particle, with energy between E and $$\delta E$$ I think I should get what i drew on the image. However, it seems a bit odd, is there something wrong?

• If the particle is confined to move from 0,L then that's just the infinte square well, right? – InertialObserver May 4 '19 at 17:41
• Yes it is... but i had never tried to draw its phase space... I used to solve Schrodingers equation and graph a couple of wavefunctions, that was it – Juan Pablo Arcila May 4 '19 at 17:44
• From your comment, it looks like you are discussing a quantum particle, for which there are ferocious "uncertainty-principle"-type constraints... You are probably graphing an aspirational Wigner function distribution, but then the cartoon you sketched is not accurate... See Exercise 0.2 , p 29 here. – Cosmas Zachos May 4 '19 at 18:38
• If you were really serious about this, see Belloni et al, AmJPhys 72 (2004) 1183. – Cosmas Zachos May 4 '19 at 19:27

Your situation is a particle in a box, and as such the energy is quantized to be

$$E= E_n = E_1 n^2$$.

If you plot $$E$$ vs. $$n^2$$ for $$n>0$$ you get one half of a discrete parabola. What you want is to know the number of states that lie between energies $$E$$ and $$E + dE$$. That is,

$$dn = \frac{dn}{dE} dE$$

We call $$\frac{dn}{dE}$$ the density of states. To find this we solve for $$n$$ obtaining

$$n = \sqrt{\frac{E}{E_1}}$$

Therefore, the density of states is

$$\frac{dn}{dE} = \frac{1}{2\sqrt{E_1}} \cdot \frac{1}{\sqrt{E}}$$