# Doubt on the derivation of grand canonical probability density function

I'm following the derivation proposed in Statistical Mechanics by Huang.

He's considering two systems, the one of interest labeled 1 and a heat/particle reservoir, labeled 2. $$Q_N$$ is the partition function of the whole system (1+2).

I can follow up to $$Q_N(V,T)=\sum_{N_1=0}^N\int_{V_1} dq_1\int dp_1\frac{1}{h^{3N_1}N_1!}Q_{N_2}(V_2,T)e^{-\beta H(q_1,p_1,N_1)}$$ but then he says that

The relative probability $$\rho(q_1,p_1,N_1)$$ that there are $$N_1$$ particles in $$V_1$$ with coordinates $$\{q_1,p_1\}$$ is proportional to the summand of $$\int dp_1dq_1\sum_{N_1}$$.

And so $$\rho(q_1,p_1,N_1)=\frac{1}{h^{3N_1}N_1!}\frac{Q_{N_2}(V_2,T)}{Q_N(V,T)}e^{-\beta H(q_1,p_1,N_1)}.$$

While I find this derivation quite intuitive I am not fully convinced. Just because this function is normalized (which is the only thing I can personally infer from this proof) doesn't mean that is the right PDF. I hope that someone can give me a solid motivation on why this is the case indeed.

• Hi pp.ch.te. Linking to private clouds, dropbox, etc, is for various reasons not acceptable on SE, cf. this meta post. Oct 6, 2019 at 10:30

I agree that Huang is not being very clear here. However, what he is doing is pretty simple, since he is merely using the definition of relative probability, or, in more mathematical terms, he is calculating a marginal probability density.

Simple example with two variables: you have the joint probability density $\rho_{X,Y}(x,y)$ of the two random variables $X,Y$, and you want to get the density $\rho_X(x)$ for the random variable $X$. The recipe is simple:

$$\rho_X(x) = \int_y \rho_{X,Y}(x,y) dy \label{0}\tag{0}$$

Now, your case. The joint probability density is

$$\rho(p,q,N)=\frac 1 {h^{3N} N!} \frac{e^{-\beta H(p,q,N)}}{Q_N(V,T)} \tag{1}\label{1}$$

where

$$H(p,q,N)=H_1(p_1,q_1,N_1)+H_2(p_2,q_2,N_2) \tag{2}\label{2}$$

The marginal probability density for system 1 is obtained by integrating over the variables relative to system 2 like in Eq.\ref{0}. However, since we don't care which particles are in system 1 and which in system 2 (the particles are identical), we also have to multiply by all the possible systems with $N_1$ particles in $V_1$ and $N_2=N-N_1$ particles in $V_2$. We therefore obtain

$$\rho_1(p_1,q_1,N_1)=\frac{N!}{N_1!N_2!} \int dq_2 dp_2 \rho(p,q,N) \tag{3}\label{3}$$

Applying \ref{3} with $\rho(p,q,N)$ defined in \ref{1} and \ref{2}, you shall obtain the desired result.