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Everytime I am faced with an analysis of the spectral function it looks like a "new" unintuitive set of jugglery with expectation values and I am unable to see a general picture of what this construction means. I am not sure that I can frame a coherent single question and hence I shall try to put down a set of questions that I have about the idea of spectral functions in QFT.

  • I guess in a Dirac field theory one defines the spectral function as follows,

$\rho_{ab}(x-y) = \frac{1}{2} <0|\{\psi_a(x),\bar{\psi}_b(y)\}|0>$

Also I see this other definition in the momentum space as,

$S_{Fab}(p) = \int _0 ^\infty d\mu^2 \frac{\rho_{ab}(\mu^2)}{p^2 - \mu ^2 + i\epsilon}$

Are these two the same things conceptually? I tried but couldn't prove an equivalence.

(I define the Feynman propagator as $S_{Fab} = <0|T\big(\psi_a(x)\bar{\psi}_b(y)\big)|0>$)

  • Much of the algebraic complication I see is in being able to handle the quirky "_" sign in the time-ordering of the fermionic fields which is not there in the definition of the Feynman propagator of the Klein-Gordon field (..which apparently is seen by all theories!..) and to see how the expectation values that one gets like $<0|\psi_a(0)|n><n|\bar{\psi}_b(0)|0>$ and $<0|\bar{\psi}_b(0)|n><n|\psi_a(0)|0>$ and how these are in anyway related to the Dirac operator $(i\gamma^\mu p_\mu +m)_{ab}$ that will come-up in the far more easily doable calculation of the spectral function for the free Dirac theory.

  • Is it true that for any QFT given its Feynman propagator $S_F(p)$ there will have to exist a positive definite function $\rho(p^2)$ such that a relation is satisfied like,

$S_F(p) = \int _0^\infty d\mu ^2 \frac{\rho(\mu^2)} {p^2 - \mu^2 +i\epsilon}$

So no matter how complicatedly interacting a theory for whatever spin it is, its Feynman propagator will always "see" the Feynman propagator for the Klein-Gordon field at some level? (..all the interaction and spin intricacy being seen by the spectral function weighting it?..)

  • One seems to say that it is always possible to split the above integral into two parts heuristically as,

$$\begin{eqnarray}S_F(p) &=& \sum (\text{free propagators for the bound states})\\ &&+ \int_\text{states} \big( (\text{Feynman propagator of the Klein-Gordon field of a certain mass})\\ && ~~~~~~~~~~~~~~\times(\text{a spectral function at that mass})\big)\end{eqnarray}$$

Is this splitting guaranteed irrespective of whether one makes the usual assumption of "adiabatic continuity" as in the LSZ formalism or in scattering theory that there is a bijection between the asymptotic states and the states of the interacting theory - which naively would have seemed to ruled out all bound states?

To put it another way - does the spectral function see the bound states irrespective of or despite the assumption of adiabatic continuity?

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You would like to state the Källen-Lehman representation through the Fermonic field commutator, that is $\{\bar\psi(x),\psi(y)\}$ but, as far as the proof goes, you can only use this in a time-ordered way. I mean, you should use $\theta(x_0-y_0)\psi(x)\bar\psi(y)-\theta(y_0-x_0)\bar\psi(y)\psi(x)$. This will grant the due appearance of the Klein-Gordon propagator in the final formula. When you will do that, the standard view, seen through states and bound states, holds true.

About adiabatic continuity, you will always get a weighted sum of free propagators with all the spectrum of the theory, free and bound states, that is in agreement with such a hypothesis. The effect of the interaction will be coded in the weights and the spectrum itself.

Finally, positivity of the spectral function can only be granted, and a proof holds, when the states behave in a proper way. This is not exactly the case for a gauge theory and some of the difficulties arising in proving the existence of a mass gap can be tracked back to a problem like this. E.g. see this book by Franco Strocchi.

Further clarification: When you insert the operator generating a translation in the bosonic field, the same is somewhat different for the spinorial case. You will get

$$U^\dagger\psi U=S\psi $$

with $S$ the one I think you studied in the proof of Lorentz invariance of the Dirac equation. Now, you are almost done. This will give for you matrix element

$$\langle 0|\psi(0)|\alpha\rangle=\sqrt{Z}u(\alpha)$$

being $\alpha$ running both on momenta and spin. You are practically done as, using the known relations $\sum_s u\bar u= \gamma\cdot p+m$ and $\sum_s v\bar v= \gamma\cdot p-m$, you will get back Källen-Lehman representation.

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The "problem" is with the minus sign that time-ordering of the fermionic fields introduces. After one introduces a complete set one is left with two terms which look like, $<0|\psi_a(x)|n><n|\bar{\psi_b(y)}|0>$ and $<0|\bar{\psi_b(y)}|n><n|\psi_a(x)|0>$. In the corresponding terms of the calculation with scalar fields these terms are equal but for the Dirac field I don't know whether these are related or not..things would be simple if one could argue that these two terms differ by exactly a negative sign -- but i don't know either way. –  user6818 Nov 18 '11 at 0:38
    
I do not know if you mean this. When you use translational invariance of the vacuum, that is you consider $\psi(x)=e^{ipx}\psi(0)e^{-ipx}$ and $e^{ipx}|0\rangle=|0\rangle$, you should also consider in this case the spin contribution with respect to the scalar field. This contribution, managed through standard formula like $\sum u\bar u=\gamma p+m$ will permit you to recover the standard contribution $iZ/\gamma p-m$ and in the end you will get the K-L representation for Fermions. –  Jon Nov 20 '11 at 10:20
    
I thought of this but am not clear as to how or where the spin contribution is going to come from. If you can show as to how the $\sum u\bar{u}$ - the free Dirac field term is going to emerge from the matrix elements I typed above. –  user6818 Nov 20 '11 at 20:30
    
I added a clarification about in the answer. –  Jon Nov 20 '11 at 22:05
    
Thanks for your efforts. I have tried doing everything you are saying. When I insert that general $U$ in that matrix element then I will get those factors of $S$ but that doesn't help me understand as to why one should get the free-field $u_\alpha$ out of it? The main query of the question! And also why do you say that just translations should affect the spinor fields any non-trivially? I would think that only when one rotates does the non-triviality of the spin come into play - translations are always a phase! –  user6818 Nov 21 '11 at 0:59
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