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Particle density operator

  1. Dec 20, 2013 #1
    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?
  2. jcsd
  3. Dec 21, 2013 #2


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    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.
  4. Dec 21, 2013 #3


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    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?
  5. Dec 21, 2013 #4


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    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.
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