# How BEC being described by the single-particle density matrix?

1. Jul 21, 2014

### csky

Hello everybody,

this is my first time being here. I am a beginner learning some introductions on Bose-Einstein Condensation (BEC) on my own. Often times in the literature (say, [1], [2] (p.409) ) it comes the one-body(single-particle) density matrix, as

$$<\psi|\mathbf{\Psi(r)^\dagger\Psi(r')}|\psi>=N\int dx_2...dx_N~\psi^*(r,x_2,...,x_N)\psi(r',x_2',...,x_N')$$

I am not sure how to derive the above equation... My first step is to write $<\psi|\mathbf{\Psi(r)^\dagger\Psi(r')}|\psi>$ as

$$<\psi|\mathbf{\Psi(r)^\dagger\Psi(r')}|\psi>=\int dx_1...dx_N \int dx_1'...dx_N' \psi_t^*(x_1,...,x_N)<x_1,...,x_N|\mathbf{\Psi(r)^\dagger\Psi(r')}|x_1',...,x_N'>\psi_t(x_1,...,x_N)$$

then I am not sure how to handle $<x_1,...,x_N|\mathbf{\Psi(r)^\dagger\Psi(r')}|x_1',...,x_N'>$. Any ideas?

You probably know that e.g. $\mathbf{\Psi(x_1)\Psi(x_2)}|0>=|x_1,x_2>$ and so on for the position eigenstates of n particles in genera. Furthermore, you know the commutation properties of the Psi operators, $\{\mathbf{\Psi^+(x_1),\Psi(x_2)}\}=\delta(x_1-x_2)$. This should be sufficient to work out the matrix element.