Particle density operator

[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)= lψ(r)l²

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

lψ(r)l² = ∫dr' ψ*(r')δ(r-r')ψ(r')

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

 

[naima]

We have <r|Ï|r> = ∫dr' ψ*(r')δ(r-r')ψ(r')
where do you see a possible identification?
|ψ(r)|² is found on the diagonal of rho not on delta.

 

[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?

 

[naima]

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