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Second quantization

  1. Nov 12, 2013 #1
    This is just a simple question. Im reading about the second quantization but every text I read it starts with something like: suppose we have a set of single particle basis states {la1>,la2>,...,lan>}, which are used to label the wavefunction etc.
    I just need to make sure I understand what these single particle basis states really are (they are described on wiki: http://en.wikipedia.org/wiki/Second_quantization): [Broken]
    My idea is that they are the complete set of combined eigenstates for the system. I.e. for the hydrogen atom they are the functions labeled by l,m,n (the spherical harmonics etc.) times a spinor. Is that correct?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 12, 2013 #2


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    Second quantization is a way to treat systems of interacting identical particles.

    The single particle states are the (basis) states of isolated single particle that are brought together to form the system. For example, a set of single particle states could be the energy eigenstates of a single particle.

    The single particle states are used to build basis states that span the Hilbert space of the many-particle system.

    Here are some examples http://www.phys.ens.fr/~mora/lecture-second-quanti.pdf
    Last edited: Nov 12, 2013
  4. Nov 13, 2013 #3


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    Yes, the (anti-)symmetrized products of all the single particle states span the Hilbert space (Fock space) for all many particle states .
  5. Nov 15, 2013 #4
    Okay Im not sure I understand it completely, maybe I focus too much on the single particle states. But overall I don't even see why the many particle basis induced by the slater determinants/permanents is a nice basis to work in. I mean consider a two-level system with eigenstates β12 with two bosons.
    In general the wavefunction for each particle is some linear combination of the two states above, i.e.:
    ψ1(r1)=a β1(r1) +cβ2(r1)
    ψ2(r2)=c β1(r2) +dβ2(r2)
    Imposing the symmetrization requirement for bosons:
    ψ(r1,r2) = ψ1(r12(r2) + ψ1(r22(r1)
    Which is a nasty expression, and I particularly don't see how this is nicely represented in the basis of symmetrized eigenstates.
  6. Nov 15, 2013 #5


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    I think it is nasty, but second quantization rewrites it in an equivalent but less nasty way. In the second quantized language, the (anti-)symmetrization is correctly handles automatically by the (anti-)commutation relations of the second quantized operators. The way I learnt second quantization, I think it was really tedious to show that second quantization is really equivalent to these nasty expressions (one of those things I did once just to be able to say I believe it, but never again). Perhaps there is a better way?

    http://www.mit.edu/~levitov/8514/ has notes on second quantization (the postscript file works, but the pdf doesn't) in which he says something like that.

    I think http://books.google.com/books?id=pe-v8zfxE68C&source=gbs_navlinks_s has the details in an appendix.
    Last edited: Nov 15, 2013
  7. Nov 15, 2013 #6
    hmm okay. I guess I need to take some time to understand the idea of second quantization. My book motivates it by saying that "From the slater determinant form of basisstates it is clear that in each term only the occupied single particle states play a role. It must somehow be simpler to formulate a representation where one just counts how many particleres there are in each orbital. This is achieved with the occupation number representation."
    I must admit that I don't fully understand what the author means here. What does he mean by a representation where you count how many particles there are in each orbital. Given the quantum indistinguishability does this even make sense to say? No particle is in a definite orbital since its wavefunction is connected to the other states through the wave-function symmetrization requirement. ...So maybe I'm misunderstanding this part.
  8. Nov 15, 2013 #7


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    In the example you gave, from the single particle basis states we can get this set of multiparticle basis states (I left out the normalization):

    ψ11 = β1(r11(r2) + β1(r11(r2)

    ψ12 = β1(r12(r2) + β2(r11(r2) = β2(r11(r2) + β1(r12(r2)

    ψ22 = β2(r11(r2) + β2(r11(r2)

    Because of the symmetrization, we can characterize each multiparticle basis state completely by saying:
    ψ11 has 2 particles in state 1 = |2,0>
    ψ12 has 1 particle in state 1 and 1 particle in state 2 = |1,1>
    ψ22 has 2 particles in state 2 = |0,2>

    Then we use |2,0>, |1,1>, |0,2> as basis states in the second quantized language.
  9. Nov 18, 2013 #8
    My book is many body quantum theory in CMP:
    http://www.phys.lsu.edu/~jarrell/CO...hysics Henrik Bruus and Karsten Flensberg.pdf
    On page 12 it writes how we can identify the first and second quantized states (1.49). But for me the equation stated doesn't make a lot of sense because of the notation. Why the weird indices? Why not just [itex]\nu[/itex]1,..., [itex]\nu[/itex]N rather than [itex]\nu[/itex]n1,..., [itex]\nu[/itex]nN. What good does the n do?
  10. Nov 18, 2013 #9


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    Because the authors get paid by the subscript? :rolleyes:

    No, it's because the available states have already been numbered, ν1, ν2, etc. But here we are constructing an N-particle state vector, using N creation operators. Each creation operator corresponds to some one of the available states. BUT, maybe two particles go in the same state, so two of the operators correspond to the same state. Or maybe some of the available states are unoccupied and must be skipped over. Therefore the ν's in this state vector are not simply ν1, ν2, ... They have to be numbered differently: νn1, νn2, ...
  11. Nov 18, 2013 #10
    makes sense.. Stupid of me not to see that.
    Well now we are at it ... :) on page 7 in the definition of the slater basis. Where is spin in all this? Because the determinant is a permutation in the spatial coordinates but wouldn't you also want something which for fermions for instance incorporated the possible spin states, which I would think are:
    upup downup updown downdown
    Though in Griffiths the updown and downup are only counted as one distinct state for some reason (why is that?)
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