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Electron density expression

  1. May 18, 2014 #1
    Suppose we have a system with M nuclei and N electrons


    the electron density is defined as:


    [itex]\rho(\mathbf{r}) = \sum\limits_{\sigma = \uparrow, \downarrow} \sum\limits_{i}^N \psi^{*}_{i}(\mathbf{r\sigma})\psi_{i}(\mathbf{r\sigma})
    [/itex]

    i would like to know if it is valid to express it in the following way

    [itex]\rho(\mathbf{r}) = 2 \sum\limits_{i}^N \psi^{*}_{i}(\mathbf{r})\psi_{i}(\mathbf{r}) = 2 \sum\limits_{i}^N |\psi_{i}(\mathbf{r})|^{2}
    [/itex]
     
  2. jcsd
  3. May 18, 2014 #2
    What happens if N is odd?
     
  4. May 18, 2014 #3
    Im not seeing any difference between even or odd N.

    here is what i think

    the two factor is a consequence of the Pauli Principle, a single electron wave function can be occupied by two electrons (spin up and spin down) . And because the electron density is dependent of the 3 spatial variables and not the spin, i can right psi (r) and not psi (rσ) so the last expression is valid in my opinion.
     
  5. May 19, 2014 #4
    I still think it looks wrong. Try a trivial example with 3 electrons.
     
  6. May 20, 2014 #5
    To me it looks like all he did was to sum up/down electrons for a spin-free system.

    Looks valid if you are interested in "pure" charge density and want to forget about spin.

    Nothing wrong with it.
     
  7. May 21, 2014 #6
    3 electrons


    psi*(r1, down)psi(r1,down) + psi*(r1, up)psi(r1, up) +
    psi*(r2, down)psi(r2, down) +psi*(r2, up)psi(r2, up) +
    psi*(r3, down)psi(r3, down) +psi*(r3, up)psi(r3,up)


    2|psi(r1)|^2 + 2|psi(r2)|^2 + 2|psi(r3)|^2 = 2 sum (from i =1 to 3) |psi(ri)|² consistent with the last expression
     
  8. May 21, 2014 #7

    ChrisVer

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    1. yes it is valid... otherwise you can think of it as the spinor degrees of freedom...
    eg several times you can find that the number density is given as [itex]n= A g \int f(p) d^{3}p[/itex]
    that g is giving you the dof of the particles you care about- so if you have particles which can take values +1/2 and -1/2, then g=2.
    For a dirac s-1/2 particle, the g=4...if you had a massless spin 1 particle, then g=2 as well... if you had massive spin 1, then g=3 etc...
    Otherwise you can think of it as you did for the case of N=3

    2. N shouldn't be relevant. One reason is that you have the same wavefunction product (you have [itex]\psi_{i}^{*} \psi_{i}[/itex] which you sum its spins over, so N doesn't affect it- equivalently speaking you can interchange the summations)
     
  9. May 23, 2014 #8
    Ok! now it is much more clear.

    Just more one question

    When it is valid to approximate the density as a product of individual functions?
     
    Last edited: May 23, 2014
  10. May 23, 2014 #9

    ChrisVer

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    I don't understand the question....
    do you mean to write something like [itex]\psi_{i} \psi_{j}, i \ne j[/itex]?
     
  11. May 25, 2014 #10
    Forget my last question. What i would like to know is

    we have the general equation for electronic density:

    [itex]\rho(\mathbf{r})=N \sum_{{s}_{1}}\cdots \sum_{{s}_{N}} \int \mathrm{d}\mathbf{r}_2 \cdots \int \mathrm{d}\mathbf{r}_N[/itex]


    what are the approximations to simplify the expression to have


    [itex]\rho(\mathbf{r}) = 2 \sum\limits_{i}^N \psi^{*}_{i}(\mathbf{r})\psi_{i}(\mathbf{r}) = 2 \sum\limits_{i}^N |\psi_{i}(\mathbf{r})|^{2}[/itex]


    even with the Hartree product approximation, i can't see how to pass from the first equation to tthe last one.
     
  12. May 25, 2014 #11

    ChrisVer

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    I still don't get the question...
    You don't have infinite numbers of particles, you particles are of number N...
    In that case, the number density is given by summing the [itex]\psi_{i}[/itex]
    You don't make any integral into summation...
    But for the integral you've given, your sum appears by the operator of [itex]\rho[/itex], not by some assumption.


    Things become clearer if you work for example with momenta. Instead of the number of particle i, to work with momenta [itex]k_{i}[/itex] (go to momentum space).

    In that case,a general wf is:

    [itex] \psi_{k_{i}} (x) = \frac{1}{sqrt{V}} e^{i k^{0}_{i} x^{0} - i \vec{k}_{i} \vec{x}}[/itex]

    For finite volume, the momenta are discrete/quantized, because the wf needs to satisfy boundary conditions on the limits...

    So the number density is:

    [itex] \rho (p)= \sum_{s} \sum_{k_{i}} \delta ^{3} (\vec{p}-\vec{k}_{i}) | \psi_{k_{i}} (x)|^{2}= \frac{2}{V} \sum_{k_{i}} \delta ^{3} (\vec{p}-\vec{k}_{i}) [/itex]
    Now if you want to send the volume to infinity, then your discrete momenta values will come closer and closer to each other, going from a discrete to almost a continuous spectrum.
    In that case you can change the sum to integral (Riemann type sum).
     
    Last edited: May 25, 2014
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