# Momentum distribution function for a particle in a 1D box

In these notes on statistical thermodynamics (pp. 62), I encountered this [topic: particle in a 1D box]:

We shall adopt the initial condition that the probability distribution function has the form $$\rho(p; q : 0) = \delta(q) B(p)$$ which initially confines all the particles in the ensemble to the center $q = 0$. The momentum distribution function $B(p)$ is evenly distributed over the allowed range: $$B(p)=\frac{1}{2(p_{max}-p_{min})}[\Theta(p-p_{max})-\Theta(p-p_{min})+\Theta(p+p_{max})-\Theta(p+p_{min})]$$

Questions:

1. How is the function $B(p)$ evenly distributed. It is not clear (one can easily plug in $p=p_{max}$ and $p=p_{min}$ and see they are not equal in general). Is $\Theta$ some well-known function in math/physics?

2. Why is such a complicated function used? Won't $B(p)=$ constant work (to create an evenly distributed function)?

• $\Theta$ is probably the Heaviside step function. Commented Jul 6, 2015 at 11:39
• It must've used this then: $\int \rho(p,q:t) dp dq=1$ Commented Jul 6, 2015 at 11:46
• 1. $\Theta$ is the Heaviside step function. 2. The expression gives $B(p)=constant$ (normalised) inside the allowed interval, zero outside.
– fqq
Commented Jul 6, 2015 at 11:54
• @fqq thanks. Could you add that as an answer showing the derivation of $B(p)$ Commented Jul 6, 2015 at 12:32

Maybe the important step is to realise that the allowed range of momenta is $$R=[-p_\text{max}:-p_\text{min}]\cup[p_\text{min}:p_\text{max}].$$

Then the first two $\Theta$'s give one if $p$ is positive and in the allowed range, whereas the last two $\Theta$'s give a contribution that's only 1 if the $p$ is in the negative allowed range. The term out front normalizes this, such that $\int B=1$.

The way to see this is to consider the distinct possibilities. E.g. looking at the first two $\Theta$'s: $$\Theta(p-p_\text{max})-\Theta(p-p_\text{min}).$$ Then if $p>p_\text{max}$, both arguments are positive and $1-1=0$, so we don't get any contribution. If, on the contrary, $p<p_\text{min}$, then both arguments are negative and both $\Theta$'s evaluate to zero straightaway. The interesting case is when $p\in(p_\text{min}:p_\text{min})$. Then only the second $\Theta$ is 1 and we get $-1$.

It appears that there is a mistake in the distribution, in that it is the negative what it should be. But you get the idea.