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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.

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    $\begingroup$ 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. $\endgroup$
    – Ghoster
    Commented Nov 8, 2022 at 21:23
  • $\begingroup$ 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? $\endgroup$
    – Physics101
    Commented Nov 8, 2022 at 21:30
  • $\begingroup$ 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. $\endgroup$
    – Ghoster
    Commented Nov 8, 2022 at 21:40

2 Answers 2

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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.

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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.

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