Statistical weight for $N$ harmonic oscillators in microcanonical ensemble

Question

I would like to compute the statistical weight for the microcanonical ensemble for $N$ harmonic oscillators. To do that i use the hamiltonian of the harmonic oszillator: $$H(q,p)=\sum\limits_{i=1}^N \frac{\vec{p}^2_i}{2m}+\frac{m\omega^2}{2}\sum\limits_{i=1}^N \vec{q}^2_i $$ The statistical weight is given as: $$\Omega(N,E,V)=lim_{\bigtriangleup E->0}\frac{1}{\bigtriangleup E} \int_{E\leq H(q,p) \leq E+\bigtriangleup E}^. \! \frac{1}{N!} \, \prod\limits_{i=1}^N \frac{d^3 p_i d^3 q_i}{(2\pi h)^3}$$ At first i took a look on the same problem but with the hamiltonian of a free particle. In this case the hamiltonian is only dependent from $p_i$ so you can rewrite your limit to a radius of a sphere in momentum space with $p=(2mE)^{0,5}$ and use the equation for a n-dimensional volume of a sphere. The integration according to $q_i$ you can do immediately and you will get $V^N$. The assumption was that $0\leq q_i\leq L$ is so you get $V=L^3$. In this case the solution is clear to me. Now i struggle to find a approach for the harmonic oscillator. My first idea was to do the same thing as with the free particle and rewrite your hamiltonian so you can get your limits for q and p. But then you have an additional dependence of q in your integral and the question is which limits do you set for the q? Or you can do the same thing but at first for q instead of p and get an additional dependence of q in your integral. A another idea was to concentrate on the "shape" of the 3d-harmonic oscillator in the phase space. In 1d it is an ellipse. So you maybe should change the coordinates? I am grateful for any helpfull approaches or comments.

Answer 1

Wellcome @StefanBoltzmann, this is a paradigmatic example and the solution can be found in many introductory books. Usually, the steps to follow are:

1. Approximate the surface integral by a volume integral (much easier to solve): $\lim_{N\rightarrow\infty} \int_{E\leq H(q,p) \leq E+\bigtriangleup E}^. \! \, \prod\limits_{i=1}^N d^3 p_i d^3 q_i \sim \lim_{N\rightarrow\infty} \int_{H(q,p)\leq E}^. \! \, \prod\limits_{i=1}^N d^3 p_i d^3 q_i$
   This is just telling that in high dimensions the volume is a good approximation to the surface ($R^{3N} \sim R^{3N-1})$

2. As you said, identify the volume in phase space. To me, it becomes much more clearer if I write the integral in terms of a Heaviside function (Wikipedia).
   $\int_{H(q,p)\leq E} \, \prod\limits_{i=1}^N d^3 p_i d^3 q_i=\int \, \prod\limits_{i=1}^N d^3 p_i d^3 q_i \ \theta(E-\sum\limits_{i=1}^N \frac{\vec{p}^2_i}{2m}-\frac{m\omega^2}{2}\sum\limits_{i=1}^N \vec{q}^2_i)$
   With the change of variables $\overrightarrow{\widetilde{q}}_i=\overrightarrow{q}_i\sqrt{\frac{m\omega^2}{2}}$ , $\overrightarrow{\widetilde{p}}_i=\overrightarrow{p}_i\sqrt{\frac{1}{2m}}$ this integral reduces to the volume of a 6N dimensional sphere with radius $\sqrt{E}$.

3. The tricky part is related with the term $\frac{1}{N!}$. In the case of the free particle (the ideal gas), it was indeed needed in order to obtain an extensive entropy (Gibb's paradox). In this case, "there is no need" of such a factor, even more, if you inclue it, the entropy will not be extensive! There are several ways to understand this appearent lack of consistency. One of such ways is to think that these particles are localized and that there are exactly $N!$ ways of putting the oscillators in a lattice, so $\frac{N!}{N!}=1$. Classic texts appeal to the distinguishability (see Pathria page 65).