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I apologize for my horrible math skills. But how exactly is one supposed to get ensemble averages in the 6N dimensional phase space?

Suppose I have two particles in 1D. The phase space element is then often written as $d\Gamma = dx_1dx_2dp_1dp_2$.

For a physical variable $A(q,p)$ (such as $xp$, $x^2$), do we calculate the 'ensemble' average this way?

$$<A(x,p)>(t) = \int^\infty_{-\infty} \int^\infty_{-\infty}\int^\infty_{-\infty} \int^\infty_{-\infty}A(x,p)\rho(x,x,p,p,t)dxdxdpdp$$ where please notice I am omitting all indexes and have just generic $x$ and $p$.

Am I doing it wrong? I am confused and not sure whether I should be somehow keeping the indexes. And also, should I be dividing by $2$ because the density is normalized to $ N$?

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Let's work in 1D with two particles. The phase space element is $d\Gamma = dx_1dx_2dp_1dp_2$.

The probability density in the phase space of two particles is a function $\rho(x_1,x_2,p_1,p_2)$. When multiplied by $d\Gamma$, it will be the probability of finding position of particle 1 close to $x_1$, position of particle 2 close to $x_2$, momentum of particle $1$ close to $p_1$ and similarly for $p_2$. One can see from the description that all the four arguments matter for their name/meaning as well as for their values. And this is important not only when dealing with probabilities, but also for evaluating averages.

When one is dealing with a physical observable $A(q,p)$ (such as $xp$, $x^2$), what exactly do we mean?

$$<A(x,p)> = \int^\infty_{-\infty} \int^\infty_{-\infty}\int^\infty_{-\infty} \int^\infty_{-\infty}A(x,p)\rho(x,x,p,p)dxdxdpdp$$

is mathematically meaningless.

First of all we have to understand what is the observable we want to average.

We may have observables related to a specific particle or collective observable.

Let's see an example for each case:

  1. $A_1 = x_1$. $\left< A_1 \right>$ represents the average value of the coordinate of particle 1: $$<A_1> = \int\int\int\int x_1\rho(x_1,x_2,p_1,p_2) d\Gamma$$ where the integral is extended over the relevant part of the phase space (it depends on the ensemble and on the physical system.
  2. $A_2 = \sum_{i=1}^2\frac{p_i^2}{2m}$. $\left< A_2 \right>$ represents the average value of the kinetic energy of the system of two particles: $$<A_2> = \int\int\int\int \left(\frac{p_1^2+p_2^2}{2m}\right)\rho(x_1,x_2,p_1,p_2) d\Gamma$$.

Notice, that, if $\rho$ is properly normalized, no additional factors are required.

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