Every time 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, $$ \rho_{ab}(x-y) = \frac{1}{2} \langle0|\{\psi_a(x),\bar{\psi}_b(y)\}|0\rangle. $$ 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 $S_{Fab} = \langle0|T\, \psi_a(x) \, \bar{\psi}_b(y)|0\rangle$.)

  • Much of the algebraic complication I see is in being able to
    handle the quirky minus 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 $\langle0|\psi_a(0)|n\rangle \langle n|\bar{\psi}_b(0)|0\rangle$ and $\langle0|\bar{\psi}_b(0)|n\rangle \langle n|\psi_a(0)|0\rangle$ 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 the relation, $$ S_F(p) = \int _0^\infty d\mu ^2 \frac{\rho(\mu^2)} {p^2 - \mu^2 +i\epsilon} $$ is satisfied?

    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?


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – Student
    Nov 18, 2011 at 0:38
  • $\begingroup$ 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. $\endgroup$
    – Jon
    Nov 20, 2011 at 10:20
  • $\begingroup$ 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. $\endgroup$
    – Student
    Nov 20, 2011 at 20:30
  • $\begingroup$ I added a clarification about in the answer. $\endgroup$
    – Jon
    Nov 20, 2011 at 22:05
  • $\begingroup$ 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! $\endgroup$
    – Student
    Nov 21, 2011 at 0:59

Take the full answer it is frommy lecture to QFT. \subsection{Boson propagator} It is prefer to write the collision between to particles in term of \textit{amplitude} of \textit{probability}. The perturbative approximation of QFT assumed that the particles propagate freely except some points, when there are emission or absorption of quanta. we write the solution of motion aquations compled as pertirbative series around of free solution of motion equations of free field. The methode uses Green's function which R.Feynman gave his probabilist interpritation of implitude. Motion equation of free boson(Kein-Gordon equation) is writing:\ \begin{equation} \left( p^{2}-m^{2}\right) \varphi(p)=0 \end{equation} Where $\varphi(p)$ is a scalar function.\ Green's function $ G(p) $, in the space of momentum is: \begin{equation} (p^{2}- m^{2})G(p)=\delta^{4}(p) \end{equation} Then $ G(p)=\frac{\delta^{4}(p)}{p^{2}-m^{2}} $, $ \delta^{4} $ is the Dirac function defined as \begin{equation} \delta^{4}(p)=\delta(p_0)\delta(p_1)\delta(p_2)\delta(p_3) \end{equation} Feynman interpritation is that this operator is as amplitude of probability that the boson propagates with quadri-momentume. Propagator = $ \frac{i}{p^{2}-m^{2}} $. In same way, feynman defined an amplitude of probability that the boson whether emitted ( or absorbed) by particle 1, and/or absorbe by particle 2 of interactions.\ The amplitude are driveded for various kinds pf interactions between various particle, the square of the magnitude of each amplitude turns out be tje probability of that particular interaction (transition) occuring. These transition amplitudes each depend on the initial real particles, the final real particles, and the virtial particles that mediate the transition. It turns out that the factor in the amplitude representing the virtual particle contribution is identical to the feynman propagator $ \Delta_F $.\ \ \begin{equation} \Delta(x,y) = \int\frac{d^{4}k}{(2\pi)^{4}} \exp^{-ik(x-y)} \frac{1}{k^{2}-m^{2}+i\varepsilon} \\ \end{equation} \section{The spenor feynman Propagator} We will follow similar steps to drive feynman propagator for Dirac fields as we did for scalar feild , similar to what we did scalars, we will, heuristical, consider the operator field $ b\psi $ to creat a virtual particle at event y, and $ \psi $ to destroy that virtual particle particle at even x. The spinor propagator incorporates this two field operators. The propagator realy corresponds to a kind of probability density function in y and x. It represents the probability density of Dirac particle appearing at y and disapiaring at x. we show that in briefly:\ for the virtual spin $ 1/2 $ particle feynman propagator were\ \begin{equation} iS_{F}(x-y) = \langle 0 \vert T\lbrace \psi(x)\bar\psi(y)\rbrace \vert 0 \rangle = \langle 0 \vert \lbrace \psi(x)\bar\psi(y)\rbrace \vert 0 \rangle \end{equation} if $ t_{y} < t_{x} $ ( particle) \begin{equation} = \langle 0 \vert[\psi^{+}(x), \bar\psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation} \begin{equation} = [\psi^{+}(x), \bar\psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ \psi^{+}(x), \bar\psi^{-}(y)]_{+} \end{equation} \begin{equation} = iS_{\alpha\beta}^{+}(x - y) = \frac{1}{2(2\pi)^{3}}\int (\not p + m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{- i}{(2\pi)^{4}}\int_{c^{+}}\frac{(\not p + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation} for the virtual spin $ 1/2 $ particle feynman propagator were\ \begin{equation} iS_{F}(x-y) = \langle 0 \vert T \lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle = - \langle 0 \vert \lbrace \bar\psi(x)\psi(y)\rbrace \vert 0 \rangle \end{equation} if $ t_{y} < t_{x} $ ( antiparticle) \begin{equation} = \langle 0 \vert[\bar\psi^{+}(x), \psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation} \begin{equation} = [\bar\psi^{+}(x), \psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ \bar\psi^{+}(x), \psi^{-}(y)]_{+} \end{equation} \begin{equation} = iS^{-}(x - y) = -\frac{1}{2(2\pi)^{3}}\int (\not p - m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{ i}{(2\pi)^{4}}\int_{c^{-}}\frac{(\not p + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation} \ The two contour integrals in the last lines () and () were combined in final step to yield the single integral over real space to get the final result for \textit{\textbf{the spinor Feynman propagator }} \begin{equation} S_{F}(x - y) = \int_{-\infty}^{+\infty}\frac{ d^{4}p}{(2\pi)^{4}}\frac{(\not p + m)e^{-ip(x - y)}}{p^{2} - m^{2} + i\varepsilon} \end{equation} The momentum space form of the propagator ( its fourier transform, is \begin{equation} S_{F}(p) = \frac{\not p+ m}{p^{2} - m^{2} + i\varepsilon} = (\not p + m)\Delta_{F}(p) \end{equation} Observe that: \begin{equation} (\not p - m)(\not p + m) = \gamma^{\mu}\gamma^{\nu}p_{\mu}p_{\nu} - m^{2} = p^{2} - m^{2} \end{equation} Then we can multiply by $ (\not p + m) $ both the numerator and the denominator in eq. and rewrite $ S_{F}(p) $ in the form, \begin{equation} S_{F}(p) = \frac{i}{\not p - m} \end{equation}

  • $\begingroup$ Sorry bψ: ψ bar, and slp : p slash = \gamma^{\nu}p_{\mu} $\endgroup$
    – Benzian
    Jul 18, 2017 at 11:54
  • $\begingroup$ use \bar\psi for $\bar\psi$ and \not p for $\not p$. $\endgroup$ Jul 18, 2017 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.