Counting number of configurations particles in harmonic trap For a collection of $N$ particles in a harmonic trap, the hamiltonian:
\begin{equation}
H\left(\vec{p},\vec{x}\right) = \sum_{i=1}^N \left(\frac{\vec{p_i}^2}{2m} + \frac{1}{2}k\vec{x_i}^2\right)
\end{equation}
Assume the number of configurations is given by
\begin{equation}                                                         
       \Omega\left[U,N\right] = \frac{1}{N!}\int                                 
       \frac{\text{d}^Dp_1\text{d}^Dx_1...\text{d}^Dp_N\text{d}^Dx_N}       
       {\left(2\pi \hbar\right)^{ND}}\delta \left(H-U\right) \Delta U       \end{equation}
How can I show that the integral above computes to:
\begin{equation}
\Omega[U,N]= \frac{1}{N!}\frac{1}{\Gamma[ND]}\frac{U^{ND-1}\Delta U}{\left(\hbar\omega\right)^{ND}}
\end{equation}
I don't know how to even get started. Can someone point me into the right direction?
 A: On a conceptual basis I think it all boils down to what do we mean when writing an integral like the one shown by OP. Hence, including a Dirac distribution. You already understood that we want to "measure" the number of states with a given total energy $E$. Where every possible state is a point in phase space. The Hamiltonian is a function on this phase space. As usual in a lot of systems we identify the Hamiltonian as the Noether charge resulting from time-translation symmetry, which we call energy. So the states with constant energy $E$ are the ones satisfying
$$
H(\vec p_1,\dots,\vec p_N,\vec x_1,\dots,\vec x_N) = E\ .\qquad (1)
$$
If $H$ is a "well-behaved" function on phase space, equation $(1)$ implicitly defines a sub-manifold of our phase space (which in this case we could think of as $\mathbb R ^{6N}$). So the Hamiltonian suggested by OP reads
$$
H(\vec p_1,\dots,\vec p_N,\vec x_1,\dots,\vec x_N) = \sum_{i=1}^N \frac{\vec p_i^2}{2m} + \frac k 2\vec x_i^2\ .
$$
The question is now how does the sub-manifold, which is the set of all points
which satisfy eq. $(1)$, "look like".
This is probably answered by a simple example. Let us look at the points $(x, y) \in \mathbb R^2$ which satisfy
$$
x^2 + y^2 = R^2,
$$
which is just a circle of radius $R$. And the equivalent integral one would solve would be given by
$$
S = \int_{\mathbb R ^2} \delta(x^2 + y^2 - R^2)\ \text{d}x\text{d}y,
$$
which one would solve using polar coordinates because all there is to integrate is the "volume" of a circle with radius $R$. I think from here you should be able to perform your calculations in analogy.
