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BEC paper question

  1. Sep 27, 2015 #1
    http://bec.science.unitn.it/infm-bec/papers/preprints/becenc.pdf

    It's in regards to expression (4)

    I don't understand why the eigenvectors and eigenvalues of this operator happen to be the single particle eigenstates and the occupation numbers as they mention just below this expression.

    I do understand that in general a hermitian operator can be written like this in terms of its eigenfunctions and eigenvalues - I just don't understand how to identify these eigenfunctions as the single particle eigenstates.

    By single particle eigenstates I understand them being the eigenstates of the Hamiltonian for a single particle.
     
  2. jcsd
  3. Sep 28, 2015 #2

    Daz

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    Think about how you construct the many-body state vector for a system of N identical particles. If the particles are all identical then you might be tempted to simply add their single-particle state vectors. But that won’t do because if the particles are identical then they must be genuinely indistinguishable. A simple addition of state vectors leaves the particles distinguishable by their generalised co-ordinates.

    There are two ways you can construct the many-body state vector: by taking a linear combination of the single-particle state vectors which is either symmetric or anti-symmetric with respect to particle exchange.

    Google “exchange interaction” to see many worked examples.

    That particular exchange-symmetry, which you are forced to impose on the many-body wavefunction when you build it up from the individual particle’s wave functions means that when you construct the density matrix in step (4) of their calculation, the interference terms vanish and the single-particle eigenstates pop back out.
     
  4. Sep 28, 2015 #3
    So basically we are talking about the wave function of the system is:

    ##\Psi(r1,r2,...) = \Psi_1 (r1) \Psi_2 (r2) ..... + \Psi_1(r2) \Psi_2 (r1) ..... + ....## all permutations

    Where ##\Psi_i## are the single particle eigenstates.

    By plugging this into expression (3) of the paper I should get (4)?
     
  5. Sep 28, 2015 #4

    Daz

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    Yes.... but.... if you’re thinking of trying it yourself, there’s a subtlety here that I’m sure you’re aware of, although I don’t suppose it would do any harm to mention it.

    You need to remember that the density matrix is fundamentally a statistical beast. It’s conventionally defined to be the ensemble average of the product of the ket and bra state vectors. You won’t be able to make the off-diagonal elements vanish without averaging over random phase in the integrands.

    The assumption behind this is that the system is in thermal equilibrium. If the system is thermal then the relative phase of the integrands is randomised by contact with the thermal reservoir. (Absolute phase is, of course, irrelevant and un-observable.) When you take the ensemble average you are effectively averaging an ensemble of random numbers which yields zero.

    (As an aside, if the system is coherently excited, it is indeed possible to obtain nonvanishing off-diagonal elements in the density matrix. The relaxation of this non-equilibrium state is called dephasing, or decoherence.)
     
  6. Sep 29, 2015 #5

    DrDu

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    The one density matrix has here little to do with statistical mechanics. Specifically, it does not involve an ensemble average.
    The one density matrix is a hermitian one particle operator and as such there exists a diagonal representation in terms of its eigenfunctions, which are usually called natural orbitals.
    Specifically, the eigenvalues and vectors fulfill
    ## n_i \phi_i(r)=\int dr' n^{(1)}(r',r) \phi_i(r') ##.
     
  7. Sep 29, 2015 #6

    Daz

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    Hi DrDu - thanks for your input. Clearly you know more about this than I do. I can see how your expression there works if the system is in a pure state but I’m struggling to see how it would work in a real system at finite temperature. I’m sure you’re right but any light you can shed on this would be greatly appreciated. The way I see it is that if you have a system consisting of a large number of interacting particles then surely the only way to describe that is statistically?
     
  8. Sep 29, 2015 #7

    DrDu

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    I suppose that for a statistical ensemble, the one density matrix is just a partial trace over the full density matrix, i.e. tracing over all but one degree of freedom: ##n^{(1)}=\int dx_2\ldots dx_N \langle x_1 x_2\ldots x_N| \rho |x_N\ldots x_2 x'_1\rangle ##.
     
  9. Sep 29, 2015 #8

    Daz

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    OK Thanks for that. I’m still struggling, to be honest, but I think I need to sleep on this and maybe I’ll have more questions later.

    To the OP (Coffee_): I think you might be well advised to follow DrDu’s guidance, rather than mine.
     
  10. Sep 29, 2015 #9

    DrDu

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    That is in deed a subtle question.
    The problem is that there is no unique one particle hamiltonian for a system of interacting electrons.
    If the system is non-interacting, then you (yes, you) can show that the electronic wavefunction is a Slater determinant formed from the eigenstates ##\phi_i## of the one-particle hamiltonian h. The natural orbitals are identical to the ##\phi_i## and the ##n_i = 1## for the occupied orbitals and 0 for the unoccupied ones.

    Edit: Ups, this holds for a system of Fermions. For a system of Bosons, as we are interested here, instead of a Slater determinant you have a permanent and the occupation numbers are not restricted to be 1. Rather, for the ground state, n_0=N.
     
  11. Sep 30, 2015 #10
    DrDu thanks for your input.

    I have tried what you mention, and even for a 2 particle state I seem to fail. Let me show you my attempt:

    http://imgur.com/LiUDLnA
     
  12. Oct 1, 2015 #11
    Hey, sorry if I'm persisting but I'm quite eager to lift this confusion/misunderstanding. Could you at least tell me vaguely where in my attempt I go wrong? I have overlooked it a few times and don't see the mistake.
     
  13. Oct 1, 2015 #12

    DrDu

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    Sorry, I couldn't read what you posted on my smartphone. If psi_1 and psi_2 are orthogonal, then the integrals vanish and it seems you are done.
     
  14. Oct 1, 2015 #13
    I feel so stupid..... That's what you get from position representation, should have seen it's just <f1 | f2 >

    Oh really how silly, but still thanks a lot.

    It would seem that this occupation level ##n_l## they speak of is just ##\sqrt(2)## is I consider the initial function to have another ##\frac{1}{\sqrt{2}}##
     
  15. Oct 2, 2015 #14

    DrDu

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    Sounds good. You should also look at the more trivial case where the ground state wavefunction is a simple product of identical functions ##\phi_1(r_1)\ldots \phi_1(r_i)\ldots\phi_1(r_N)##.
     
  16. Oct 3, 2015 #15
    Wait something must be wrong here, consider eq(4) from the paper. Now set r=r' and integrate over the volume. The left term is 1 and the right term is a sum over ##n_l##. This doesn't result in the normalization condition they mention.
     
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