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Quantum Mechanics: Expression for electron density

  1. Mar 5, 2008 #1
    I'm a little confused about the expression for the electron density.

    For example in Marder's Condensed Matter Physics, he writes this

    [tex] \rho(\vec{r}) = N \int \Psi^{*} \delta( \vec{r} - \vec{r}_1) \Psi d\vec{r}_1 \ldots d\vec{r}_N \delta [/tex]


    what's going on?
     
    Last edited: Mar 5, 2008
  2. jcsd
  3. Mar 5, 2008 #2
    What are you confused about? If you're just getting started in QM you might want to pick up Griffiths book.
     
  4. Mar 5, 2008 #3

    olgranpappy

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    Homework Helper

    He's taking the many-electron wavefunction, squaring it, and integrating over all but one of the coordinates.
    I.e.,
    [tex]
    \rho(\vec r) \equiv N \int d^3r_2 d^3r_3 d^3r_4\ldots d^3r_N |\Psi(\vec r,\vec r_2,\vec r_3,\vec r_4,\ldots,\vec r_N)|^2
    [/tex]
    it desn't matter which one is singled out because the wavefunction is either symmetric or antisymmetric in all it's coordinates and thus the squared wavefunction is symmetric in it's coordinates. multiplication by N is because we want rho normalized such that
    [tex]\int d^3r \rho = N[/tex]
    rather than normalized to 1.
     
  5. Mar 5, 2008 #4
    Ah thank you! So am I right to say that this integral

    [tex]
    \int d^3r_2 d^3r_3 d^3r_4\ldots d^3r_N |\Psi(\vec r,\vec r_2,\vec r_3,\vec r_4,\ldots,\vec r_N)|^2
    [/tex]

    is just the probability of finding an electron in any volume element?
     
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