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Bosons and Fermions in a rigorous QFT 
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#1
Feb812, 08:27 AM

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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? 


#2
Feb812, 10:54 AM

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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. 


#3
Feb812, 04:53 PM

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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 timeindep. constraint (the Gauss law) this equaton is implemented on the physical Hilbert space. This does not require any "free field approach".



#4
Feb812, 08:34 PM

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Bosons and Fermions in a rigorous QFT



#5
Feb912, 01:24 AM

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Perhaps it makes sense to consider a simple qm example; the problem for the twodim. 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+1N) = 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. 


#6
Feb912, 02:18 AM

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#7
Feb912, 02:34 AM

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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 gaugefield 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. 


#8
Feb912, 08:28 AM

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#9
Feb912, 08:29 AM

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#10
Feb912, 08:32 AM

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#11
Feb912, 08:56 AM

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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 


#12
Feb912, 09:19 AM

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#13
Feb912, 10:42 AM

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See the article by Roberts: http://kolxo3.tiera.ru/M_Mathematics..._.pdf#page=274 


#14
Feb912, 11:11 AM

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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. 


#15
Feb912, 02:23 PM

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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. 


#16
Feb1012, 02:21 AM

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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. 


#17
Feb1012, 04:00 AM

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What I am saying is that one may start with Fock space operators and then use some nonperturbative techniques. Bogoljubov transformation, bosonization, ... are examples. That is more than just a solution, it's a kind of formal redefinition of the theory. There is no need to use perturbation theory only b/c of Fock space states, neither before nor after Bogoljubov transformation. 


#18
Feb1012, 07:43 PM

P: 679




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