# Particle density operator

1. Dec 20, 2013

### aaaa202

There is something I do not understand. One way to define the current density operator is through the particle density operator Ï(r). From the fundamental interpretation of the wavefunction we have:

Ï(r)= lÏˆ(r)l2

But my book takes this a step further by rewriting the equality above:

lÏˆ(r)l2 = âˆ«dr' Ïˆ*(r')Î´(r-r')Ïˆ(r')

And thus identifies the particle density operator as the delta function above. How does this make sense in any way?

2. Dec 21, 2013

### naima

We have <r|Ï|r> = âˆ«dr' Ïˆ*(r')Î´(r-r')Ïˆ(r')
where do you see a possible identification?
|Ïˆ(r)|Â² is found on the diagonal of rho not on delta.

3. Dec 21, 2013

### K^2

In general, we write expectation of operator A as $\displaystyle \langle \psi |A|\psi \rangle = \int \psi^*(r') A \psi(r') dr'$. Substituting $A = \delta(r-r')$ gives you the correct expression for expectation of $\rho(r)$. So what's the problem?

4. Dec 21, 2013

### naima

Yes but
<r|Ïˆ><Ïˆ|r'> = <r|Ï|r'> is not identified to Î´(r-r') like aaa202 said.
it is not here an average value but an element of a density matrix.