# Microcanonical Ensemble - probability of finding momentum

I was trying to calculate the probability of finding a particle with momentum $$p$$ in the microcanonical ensemble in a 3-dimensional box.

$$\rho_i(\vec p)= \langle\delta(\vec p - \vec p_i)\rangle= \int \mathrm d^3q_1\ldots \mathrm d^3q_N\int \mathrm d^3 p_1\ldots \mathrm d^3p_N \;\rho(\vec q_1,\ldots,\vec p_1,\ldots,\vec p_i,\ldots,\vec p_N)\delta(\vec p-\vec p_i).$$

In the microcanonical ensemble, the phase space density is constant, so our work is to evaluate the last integral over the momentum space. Does the second integral correspond to a volume of a 3 dimensional "hypersphere" or a $$3N$$ dimensional hypersphere of radius $$p$$?

• What is the system? Is this the classical approximation or do you need in quantum approximation? – cosmoscalibur Oct 13 '15 at 3:17
• It's the quantum approximation indeed. However, I suppose the difference between the classical and quantum approximation relies on the N! factor when you count the states (concerning to (in)distinguishably particles). In this particular example, is there another difference? ty. – CAIO FERNANDO Oct 13 '15 at 14:47
• But what is the particular example? This a free particle, an harmonic oscillator or what system or hamiltonian is? According to hamiltonian of the system (which is the shape of boundary of integration) the integral could be an hipersphere, an hipersllipsoid or a hipercube or any hipervolume. – cosmoscalibur Oct 14 '15 at 18:25
• Also, if this is free particles, only momentum integral appear on equation and that is a 3N dimensional hipervolume but if hamiltonian depend of position, that is a 6N dimensional hipervolume. – cosmoscalibur Oct 14 '15 at 18:30

$$\mathcal{H} = \sum\limits_{i= 0}^{3N-1} \left( \frac{p_i^2}{2m} + 2m\pi^2 \nu^2q_i^2\right)$$
$$x_i = \sqrt{2m}\pi \nu q_i \qquad \qquad x_{3N+i} = \frac{p_i}{\sqrt{2m}}$$