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Bosons and Fermions in a rigorous QFT

  1. Feb 8, 2012 #1
    I'm wondering, is there still a sharp distinction between Bosons and Fermions in a rigorous QFT, if exsits?
    My question is motivated by the following, consider one of the equations of motion of QED:
    [tex]\partial_\nu F^{\nu \mu} = e \bar{\psi} \gamma^\mu \psi[/tex]
    In our familiar perturbative QED (Here I'm not 100% sure if I use the word "perturbative" correctly. I simply mean fields are quantized as free fields, and we introduce an interaction built from free fields operators, like an iteration method), LHS is made of Bosonic operators and RHS is made of Fermionic operators, and since the Bosonic sector and Fermionic sector are independent in the total Fock space, perturbative QED fails to satisfy this equation of motion.
    I suppose if a rigourous QED exists, this equation of motion should be satisfied, but this in turn means the fermion operator and bosonic operator must act on a Hilbert space they share together, then is there still a sharp distinction between Bosons and Fermions?
     
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  3. Feb 8, 2012 #2

    DrDu

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    One of the greater successes of axiomatic QFT was the proof that at least in some quantum field theories bosonic and fermionic behaviour results quite naturally for localized charges.
    See e.g. the book by R. Haag, Local quantum physics.
     
  4. Feb 8, 2012 #3

    tom.stoer

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    The above mentioned field equation cannot be quantized directly b/c it has to be gauge fixed. In A°=0 at least for the time-indep. constraint (the Gauss law) this equaton is implemented on the physical Hilbert space. This does not require any "free field approach".
     
  5. Feb 8, 2012 #4

    strangerep

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    The product of two spin-1/2 operators is bosonic (being a superposition of spin-0 and spin-1 in general). (You might want to review Clebsch-Gordan coefficients and associated angular momentum decomposition theory in ordinary QM if you're not already familiar with it.)

    Yes, in the sense that there's a superselection rule between them. But when you start forming products, things get more complicated.
     
  6. Feb 9, 2012 #5

    tom.stoer

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    Perhaps it makes sense to consider a simple qm example; the problem for the two-dim. harmonic oscillator as a toy model would be

    [tex]H_i = \frac{1}{2}p_i^2 + \frac{1}{2}x_i^2 = a_i^\dagger a_i + \frac{1}{2}[/tex]

    [tex]H = H_1 + H_2[/tex]

    Now we can easily solve equations like

    [tex](H - N)|N\rangle = 0[/tex]

    for some eigenvalues N, but structurally the equation "bosonic operator = fermionic operator" would be something like

    [tex](H - N) = 0[/tex]

    and this is obviously not allowed as an operator equation b/c

    [tex](H - N) = 0 \;\Rightarrow\; (H - N)|m,n\rangle = 0 \;\forall m,n \;\Rightarrow\; (m+n+1-N) = 0\;\forall m,n [/tex]

    Gauge fixing introduces some additional structures like resolving "unphysical bosonic operators" in terms of fermionic operators via Gauss law, but I still don't see how this is sufficient to resolve the problems for the remaining operator equations. Neverthelesse there should be some solution e.g. for QE´D and QCD where these problems have been treated w/o using perturbation theory.
     
  7. Feb 9, 2012 #6

    DrDu

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    Is there anybody who believes that a rigorous QED exists?
     
  8. Feb 9, 2012 #7

    tom.stoer

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    I don't believe in a rigorous QED ...

    ... but this problem seems to be trivial and there must be a solution in "standard textbook QED with canonical quantization using Fockspace".

    I think the subtlety is the regularization of the operator product on the r.h.s. which requires e.g. point splitting with gauge-field insertion (in order not to destroy gauge invariance); this would introduce a gauge field dependency on the r.h.s. whereas the fermionic contribution on the l.h.s. comes from the solution of the Gauss law constraint i.e. A° expressed in terms of fermionic charge density.

    That means that quantization, gauge fixing and regularization translates the equation

    "bosonic operator = fermionic operator"

    into something like

    "bosonic + fermionic operator = fermionic + bosonic operator"

    which has a chance to hold as operator equation in terms of standard Fock space creation and annihilation operators.

    This is of course no rigorous proof but is indispensable already for standard textbook QED.
     
  9. Feb 9, 2012 #8
    That sounds quite a long way to go, is there a layman explanation of it?
     
  10. Feb 9, 2012 #9
    Then what's the underlying theory that makes perturbative QED plausible?
     
  11. Feb 9, 2012 #10
    Can you give a concrete example? Better a QED example since I'm not familiar with QCD.
     
  12. Feb 9, 2012 #11

    tom.stoer

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    My example is described in #7

    We must not mix two issues
    - perturbative QED
    - Fock space with creation and annihilation operators

    The latter does not imply perturbation theory
     
  13. Feb 9, 2012 #12
    Could you elaborate? I have always thought a Fock space only makes sense for perturbation theory, since I would imagine an interaction should destroy the simple direct product structure of different sectors.
     
  14. Feb 9, 2012 #13

    DrDu

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  15. Feb 9, 2012 #14

    Physics Monkey

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    A useful perspective on this question is provided by lattice gauge theory. For example, Gauss' law, which is the 0th component of the equation you wrote, is satisfied as an identity on the physical hilbert space. In other words, in the true hilbert space electric field lines can only end where charges are located. Nevertheless, there are many ways to distinguish bosons and fermions. In this model the fermions carry charge while the gauge bosons do not. There are composite operators made of fermions that are bosonic and carry charge, but there is still a fermion number that remains sensible.

    Of course, this is not to say that there is no blurring of the lines. Bosonization in one dimension is a procedure for exchanging bosons and fermions (and is even relatively rigorous). In higher dimensions one can have solitons built from gauge and bosonic matter degrees of freedom that can carry weird charges and even be fermions.
     
  16. Feb 9, 2012 #15

    tom.stoer

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    Of course going beyond perturbation theory is still difficult, but neither impossible nor excluded in principle

    Think about a Hamiltonian

    [tex]H = \frac{p^2}{2} + V(x)[/tex]

    You can rewrite this as

    [tex]H = \frac{p^2}{2} + \frac{x^2}{2} + \left[V(x) - \frac{x^2}{2}\right] = \frac{p^2}{2} + \frac{x^2}{2} + \tilde{v}(x) = a^\dagger a + \frac{1}{2} + \tilde{u}[/tex]

    Now it's up to you to solve this problem exactly or to use perturbation theory in u.

    I think it's possible to rewrite a QFT in terms of creation and annihilation operators acting on Fock spaces, but I don't see why a perturbative treatment is mandatory.
     
  17. Feb 10, 2012 #16

    DrDu

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    I don't consider this problem as too representative for the situation in QFT. The main problem in field theory is that the perturbed and free fields live in different Hilbert spaces -- a problem usually absent in ordinary QM.
    Superconductivity can be taken as a toy model on how to solve this problem:
    The BCS Hamiltonian can be diagonalized introducing new field operators by the Bogoliubov Valatin transformation. The new field operators are also fermionic but describe the interacting particles.
    They cannot be obtained perturbationally from the free electron gas.
     
  18. Feb 10, 2012 #17

    tom.stoer

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    I never said that.

    What I am saying is that one may start with Fock space operators and then use some non-perturbative techniques. Bogoljubov transformation, bosonization, ... are examples. That is more than just a solution, it's a kind of formal re-definition of the theory.

    There is no need to use perturbation theory only b/c of Fock space states, neither before nor after Bogoljubov transformation.
     
  19. Feb 10, 2012 #18
    But here if you creat the Hilbert space using creation operators, it might not be the same with the true Hilbert space, then the best you can get is a perturbation theory.
     
  20. Feb 10, 2012 #19
    I'd like the ask the question again, if a rigorous QED doesn't exist, what would be underlying theory of perturbative QED?
     
  21. Feb 11, 2012 #20

    tom.stoer

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    At least in QM tis is not true.

    Which states do I miss?

    I can construct nearly arbitrary operators from the creation and annihilation operators; look at the method of coherent states, for example.

    And please not that up to now we haven't defined the Hilbert space, neither for x and p, nor for the creation and annihilation operators! So if we need to change the entire Hilbert space for some reason we are not forced to do this by introducing creation and annihilation which are nothing else but linear combinations of x and p. There is physics behind it so far.
     
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