Phase space density of $N$ harmonic oscillators

For one classical harmonic oscillator with Hamiltonian

$$H = \frac{p^2}{2m}+\frac{m\omega^2}{2}x^2$$

the density of states can be calculated as by calculating the number of states with Energy smaller than $$E$$:

$$\Gamma(E) = \frac{\text{area of ellipse}}{h} = \frac{E}{\hbar \omega}$$

and then by carrying out the derivative $$\frac{d\Gamma}{dE}$$ one obtains: $$g(E) = \frac{d\Gamma}{dE} = \frac{1}{\hbar \omega}$$ as the density of states.

Now I wonder if it is possible to generalize this approach to a set of N harmonic oscillators, then with Hamiltonian:

$$H = \sum_{i=1}^{N} \frac{p_i^2}{2m}+\frac{m\omega^2}{2}x_i^2$$

I can not "visualize" and understand how to calculate the area of this "ellipse" in the $$2N$$ dimensional phase space.

• You can scale things so that the bounding surface becomes a hypersphere. The volume of an n-ball is well-known and is probably somewhere in your textbook. Commented Nov 8, 2022 at 21:23
• I see that it ist possible to obtain a hypersphere for the free particle in 3d, as the hamiltonian then only depends on one variable. Is it then in the case of two variables in the hamiltonian I get two hyperspheres one for the x and one for the p variables? Commented Nov 8, 2022 at 21:30
• You get a 6D hypersphere for one particle. The six axes in its phase space are $x,y,z,p_x,p_y,p_z$. By scaling you can make $H$ look like the sum of the squares of these coordinates. Scaling was the first hint in the problem you previously posted. Commented Nov 8, 2022 at 21:40

The Hamiltonian of $$N$$ uncoupled oscillators, $$E = \sum_{i=1}^N \left(\frac{p_i^2}{2m}+\frac{k x_i^2}{2}\right),$$ defines a $$2N$$-dimensional hyper-ellipsoid with semi axes $$a_i = \sqrt{2mE}$$, $$b_i=\sqrt{2E/k}$$, $$i=1,\cdots N$$.

The volume of an ellipsoid in $$n$$ dimensions is (see Reference) $$\begin{gather} V_n = (c_1\cdots c_n) \frac{2\pi^{n/2}}{\Gamma(n/2+1)} \end{gather}$$ where $$c_i$$ are the $$n$$ semi axes and $$\Gamma(\cdots)$$ is the gamma function (not to be confused with the common symbol for phase space).

The energy contour has $$n=2N$$ semiaxes of which $$N$$ are equal to $$\sqrt{2mE}$$ and $$N$$ are equal to $$\sqrt{2E/k}$$. Using $$\Gamma(n/2+1) = \Gamma(N+1) = N!$$, the number of microstates within the hyper-ellipsoid is $$\begin{gather} \Omega(E) = \frac{V_{2N}}{h^{2N} N!} = \left(\frac{2\pi}{h^2} \sqrt{\frac{m}{k}}\right)^N \frac{E^N}{(N!)^2} . \end{gather}$$ This is the total number of microstates whose energy is less than or equal to $$E$$. The microcanonical partition function is its derivative with respect to $$E$$: $$\tag{1} \omega(E) = \frac{\partial\Omega}{\partial E} = \frac{N}{(N!)^2}\left(\frac{2\pi}{h^2} \sqrt{\frac{m}{k}}\right)^N E^{N-1} .$$

Canonical Partition Function As a bonus, let's calculate the canonical partition function, which is simpler: $$Q(\beta,N) = \int \omega(E,N) e^{-\beta E} dE = Q(\beta,N) = \frac{1}{N!}\left(\frac{2\pi}{\beta h^2} \sqrt{\frac{m}{k}}\right)^N$$ This we can write it in the equivalent form $$\tag{2} \boxed{ Q(\beta,N) = \frac{Q_1^N}{N!} }$$ where $$Q_1$$ is the canonical partition function of a single oscillator: $$Q_1 = \frac{2\pi}{\beta h^2} \sqrt{\frac{m}{k}}$$ Equation 2 states that the partition function of $$N$$ oscillators is the product of $$N$$ independent partition functions corrected for indistinguishability. This is the expected result and serves as a check of the correctness of the derivation.

For one classical harmonic oscillator with Hamiltonian $$H = \frac{p^2}{2m}+\frac{m\omega^2}{2}x^2$$

The density of states can be calculated as by calculating the number of states with Energy smaller than $$E$$... and then by carrying out the derivative $$\frac{d\Gamma}{dE}$$ one obtains: $$g(E) = \frac{d\Gamma}{dE} = \frac{1}{\hbar \omega}$$ as the density of states.

Now I wonder if it is possible to generalize this approach to a set of N harmonic oscillators, then with Hamiltonian:

Yes, it is possible.

A good first step is understanding how one arrives at $$\Gamma(E) = \frac{E}{\hbar \omega}$$ for a single oscillator in one-dimension, in a way that can be generalized.

The result for the single oscillator follows from the usual method of counting states in phase space: $$\Gamma_1 = \int \frac{dxdp}{(2\pi\hbar)}\;,$$ where the sub-script $$1$$ on $$\Gamma_1$$ means I am considering just one oscillator (in one dimension).

If you change variables to $$u=x\sqrt{m\omega^2/2}$$ and $$v=p/\sqrt{2m}$$, the integral for the number of states becomes: $$\Gamma_1 = \int \frac{du dv}{(2\pi\hbar)}|\frac{2}{\omega}|$$

In the $$(u,v)$$ coordinates the condition that the energy is less than $$E$$ is simply $$u^2 + v^2 < E$$, which means we can write an expression for $$\Gamma_1(E)$$ as $$\Gamma_1(E) = 2\pi \int^\sqrt{E}_0 \frac{dr r}{(2\pi\hbar)}|\frac{2}{\omega}| = \frac{E}{\hbar \omega}$$

Generalizing to $$N$$ oscillators in one dimension, we have: $$\Gamma_N = \frac{1}{N!}\int \frac{dx_1\ldots dp_N}{(2\pi \hbar)^N}\;.$$

Making a similar substitution for each of the $$N$$ oscillators as done above with $$u$$ and $$v$$ gives: $$\Gamma_N(E) = \frac{1}{N!}\int \frac{du_1\ldots dv_N}{(2\pi \hbar)^N} |\frac{2}{\omega}|^N = \frac{1}{N!}|\frac{2}{\omega}|^N \frac{S_{2N}}{(2\pi \hbar)^N}\int_0^\sqrt{E} dr r^{2N-1}\tag{1} \propto \frac{E^N}{(\hbar \omega)^N}\;,$$ such that the density of states is $$g(E) \propto \frac{E^{N-1}}{(\hbar \omega)^N}\;.$$

Note, if you want to work out the exact expression for $$g(E)$$ instead of just the proportionality above, you can use the fact that $$S_{2N}$$ in Eq. (1) above is the surface area of a sphere in $$2N$$ dimensions ($$2\pi^N/(N-1)!$$).

The above result is for $$N$$ oscillators in one spatial dimension. It also holds for one oscillator in $$N$$ spatial dimensions. The generalization to $$A$$ oscillators in $$B$$ spatial dimensions as well as the generalization to the case where the frequencies are not all the same is straightforward.