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Sum of two slater determinants

  1. Feb 23, 2009 #1
    a slater determinant gives an asymmetric wave function for fermions

    is the inverse right?

    i.e., can the wave function of some fermions always be written in the form a slater determinant?

    To make things concret, can the sum of two slater determinants be put into the form of a new slater determinant?
  2. jcsd
  3. Feb 23, 2009 #2
    i guess if all the orbits in the two slater determinants are orthogonal to each other, then the sum of the two can be put into a new slater determinant.
  4. Feb 24, 2009 #3


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    The product, yes. The sum, no, because it's not necessarily antisymmetric any more to begin with.
    I'm not sure why you'd want to form a sum of two different wavefunctions though.
  5. Feb 24, 2009 #4
    If it is expressed in product form of orbitals, a general anti-symmetric wave function could be written as a linear combination of slater determinants. This will normally mess up the total spin of the particles, but you get anti-symmetry which is good. For N particles you have 2^N spinor (anti-symmetric) eigen states to take a linear combination of.
  6. Feb 24, 2009 #5
    Hmm sorry, I meant that a single slater determinant does not necessary preserve the total spin eigenstate, but that you can obtain such if you take a proper linear combination of the slater determinants (alpha, beta spinor functions). Generally you don't need Slater determinants to obtain anti-symmetry, but it could be easier. Also, there exist cases where you could find exact solutions of the two-electron wave function (harmonic oscillator) which is not of the slater product form, but F(r)*G(R) (center of mass R, and relative coordinate r). Clearly this cannot be decomposed into slater functions. It is also common to use extended sums of Slater orbitals in order to reduce the correlation problem.
  7. Feb 27, 2009 #6
    Yes, thank you for your good example.

    It is a shame that i do not understand you well
  8. Feb 27, 2009 #7
    If the sum of two different slater determinant can be made into the form of a single slater determinant, it is a desirable thing, isn't it?
  9. Feb 27, 2009 #8


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    What I meant was that I'm not quite sure what kind of system/situation you'd have to need/want to form the sum of two Slater-determinant wavefunctions. It'd mean the two Hamiltonians are independent. So why merge them?
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