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

[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

&lt;\psi|\mathbf{\Psi(r)^\dagger\Psi(r&#039;)}|\psi&gt;=N\int dx_2...dx_N~\psi^*(r,x_2,...,x_N)\psi(r&#039;,x_2&#039;,...,x_N&#039;)<br />

I am not sure how to derive the above equation... My first step is to write &lt;\psi|\mathbf{\Psi(r)^\dagger\Psi(r&#039;)}|\psi&gt; as

<br /> &lt;\psi|\mathbf{\Psi(r)^\dagger\Psi(r&#039;)}|\psi&gt;=\int dx_1...dx_N \int dx_1&#039;...dx_N&#039; \psi_t^*(x_1,...,x_N)&lt;x_1,...,x_N|\mathbf{\Psi(r)^\dagger\Psi(r&#039;)}|x_1&#039;,...,x_N&#039;&gt;\psi_t(x_1,...,x_N)<br />

then I am not sure how to handle &lt;x_1,...,x_N|\mathbf{\Psi(r)^\dagger\Psi(r&#039;)}|x_1&#039;,...,x_N&#039;&gt;. Any ideas?

thanks in advance for help and comments,
C.H.

 

[DrDu]

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.