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If you think that this question is likely to get closed then please do not answer and only say that in the comments since this system doesn't let me delete the question once it has answers.

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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up vote 3 down vote accepted

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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