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Statistical Mixture of N States in the Hartree-method.

  1. Jan 3, 2007 #1
    In the Hartree model of interacting electrons one assumes that the wave function is in the form
    [tex] \Psi(x_1, x_2, ..., x_N,t) = \psi_1(x_1,t) \psi_2(x_2, t) ... \psi_N(x_N,t) [/tex]
    which of course is a quite crude approximation since it for example does not take into account Pauli principle.

    I have studied some recent articles in plasma physics which starts from this expression and then derives a fluid equation (for example, G. Manfredi, arxiv:quant-ph/0505004).

    From the form above it is then claimed that this can be represented by a one-particle density matrix
    [tex] \rho(x,y) = \sum_{i=1}^N p_i \psi_i^*(x) \psi_i(y) [/tex]
    where [tex] p_i [/tex] are the "occupation probabilities".

    Does any one know how to obtain this form of the density matrix?

    If one would have started with a completely anti-symmetric wave function and calculated the density matrix (by tracing over [tex] N-1 [/tex] particles) then I can guess that the result would be similar to this, but then I think that the probabilites would be [tex] 1/N [/tex].

    Does the form of the density matrix above possible to derive from that the particles are intdistinguishable or is it just some plausible arguments which gives the form above?
     
  2. jcsd
  3. Jan 3, 2007 #2
    It is iteration method, see in: P. Ring, P. Schuck: The nuclear Many-Body Problem, Springer Verlag, New York 1980. Really good book!
     
  4. Jan 3, 2007 #3
    Thx for the tips!
     
  5. Jan 3, 2007 #4

    Dr Transport

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    If memory serves me corectly, use a Slater Determinant for the wave function and it take care of the Exlusion Principle.
     
  6. Jan 3, 2007 #5
    Yes, the exclusion principle can be accounted for. It is called Hartree-Fock theory, but I don not need do consider the antisymmetry for the moment.

    The thing is that I don't understand how they get the density matrix. Usually, if you have a distribution for different states that a particle may have [tex] \{ \psi_i(x) \} [/tex] with probabilities [tex]p_i [/tex] then the density matrix is
    defined as the one above.

    However, I am dealing with many particles, but the density matrix is a one-particle density matrix. (?)
     
  7. Jan 3, 2007 #6
    1. As You probably know it is impossible to solve many-body (more than 4) with interactions. That's why HF procedure leads to one-body problem i.e. many particles (but without interactions between them) embeded into ,,avarage" potential. You can then treat particle as if it has no neighbour. HF procedure works, but in my opinion is not physical.
    2. If memory serves me corectly, old interpretation of density matrix is that it represents ensemble of particles. It is no difference between density matrix which represents one particle in fixed state and density matrix which represents many particles, all in the same state! (see: Feynmann, ,,Statistical Mechanics")
     
  8. Jan 18, 2007 #7
    Hmm, the reason to introduce the density matrix is allways to allow for a statistical distribution of states. In some cases however you may use the single or two-particle density matrix when doing calculations on a many-particle system.

    The density matrix for a particle in a fixed state [tex] \psi [/tex] is
    [tex] \rho = \left| \psi \right> \left< \psi \right| [/tex]

    and the density matrix for a many particle system where the particles are all in the same state (and hence has to be bosons) is given by
    [tex] \rho = \left| \psi \right> \left< \psi \right| \otimes \left| \psi \right> \left< \psi \right| \otimes \dots \otimes \left| \psi \right> \left< \psi \right| [/tex],
    so apparently there is a difference. However, since all particles are in the same state you can trace out all particles but one som the single particle density matrix above can be used.
     
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