Operator Product Expansion as shown in Schwartz

  • #1
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Summary:
Need help in understanding the operator product expansion as shown in Schwartz
I have included here the screen shot of the page I am referring to.


I am unsure of how this non-local Lagrangian of Eqtn(32.68) has been constructed. Have they just integrated the interaction Lagrangian densities over two different sets of points (x & y) ?
If so, then why is there no P_L in there, why just a gamma matrix ?
And in this Eqtn(32.68) have they used the full electroweak theory ?
The paragraph above claims that they have integrated out the W boson ,then got this expression but then why have the written the W propagator in there as well ?
In the next step Eqtn(32.69) I dont get how the expression for the propagator is modified ie where does the DeAlembertian come from ?
From this it seems to me that the thing responsible for converting a non-local Lagrangian to a local one,is simply the position space delta function resulting from the momentum integral of the exponential.

From what I understand, the interaction due to the full electroweak Lagrangian should be non-local while for the lower energy 4 Fermi theory, it should be local. So would the answer have been the same had they used the 4 Fermi theory instead ?
 

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  • #2
samalkhaiat
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I am unsure of how this non-local Lagrangian of Eqtn(32.68) has been constructed. Have they just integrated the interaction Lagrangian densities over two different sets of points (x & y) ?
If so, then why is there no P_L in there, why just a gamma matrix ?
And in this Eqtn(32.68) have they used the full electroweak theory ?
The paragraph above claims that they have integrated out the W boson ,then got this expression but then why have the written the W propagator in there as well ?
He seems to invent names for himself. There are no non-local Lagrangians in QFT.
1) What he calls “non-local Lagrangian” is the scattering operator for second order processes: [tex]S^{(2)} = - \frac{1}{2} \int d^{4}x d^{4}y \ T \Big\{\mathcal{L}_{I}(x) \mathcal{L}_{I}(y) \Big\} ,[/tex] where [tex]\mathcal{L}_{I}(x) = g \ N \left[ J^{\mu}(x) W_{\mu}(x) \right] ,[/tex] is the (normal-ordered) local interaction Lagrangian.
2) By “integrating out” the [itex]W[/itex] field he means applying the Wick theorem to the product of fields in [itex]S^{(2)}[/itex] and picking the term which contains the contraction of the two [itex]W[/itex] fields: [tex]\overline{W_{\mu}(x)W_{\nu}}(y) \equiv \langle 0 |T \{ W_{\mu}(x)W_{\nu}(y)\}|0 \rangle = iD_{\mu\nu}(x - y) ,[/tex] where [itex]D_{\mu\nu}(x)[/itex] is the propagator of the (very) massive vector field [itex]W_{\mu}(x)[/itex] [tex]D_{\mu\nu}(x - y) = \frac{1}{(2\pi)^{4}} \int d^{4}p \ \frac{- \eta_{\mu\nu}}{p^{2} - M^{2}} \ e^{ip(x - y)} .[/tex] So, basically you obtain a process similar to the Moller scattering in QED: current emits a [itex]W[/itex] boson at [itex]x[/itex] which propagates to [itex]y[/itex] and get absorbed by the second current: [tex]S^{2}(2 \to 2) = - g^{2} \int d^{4}xd^{4}y \ N\left[ J^{\mu}(x)J^{\nu}(y)\right] \ D_{\mu\nu}(x - y).[/tex]
where does the DeAlembertian come from ?
This follows from the following (easy to show) identity [tex]\left( \partial^{2} + M^{2}\right) \int d^{4}p \ \frac{e^{ipx}}{p^{2} - M^{2}} = - \int d^{4}p \ e^{ipx},[/tex] by applying [itex](\partial^{2} + M^{2})^{-1}[/itex] to both sides: [tex]\int d^{4}p \ \frac{-1}{p^{2} - M^{2}} \ e^{ip(x - y)} = \frac{1}{\partial^{2}_{x} + M^{2}} \int d^{4}p \ e^{ip(x - y)} .[/tex]
From what I understand, the interaction due to the full electroweak Lagrangian should be non-local while for the lower energy 4 Fermi theory, it should be local.
All interactions in the relativistic QFT are local.
So would the answer have been the same had they used the 4 Fermi theory instead ?
No, because there is no boson in the Fermi phenomenological model. Fermi (4 fermions) interaction is a contact interaction.
 
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  • #3
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Thanks a lot,that was quite helpful.
So if I follow you,Schwartzs` first equation is just about writing the amplitude for a scattering process at second order ? And these Psis are meant to be left Weyl spinors ,I take it ?
Additionally I am a bit doubtful if I understood his distinction between local and non local. Seems like if the fields are at the same point (x) he is calling it local.
 
  • #4
samalkhaiat
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So if I follow you,Schwartzs` first equation is just about writing the amplitude for a scattering process at second order ?
Yes.
And these Psis are meant to be left Weyl spinors ,I take it ?
I don’t know what he did before or after those calculations. I believe he treated the W boson as massive photon just to demonstrate the idea. One can get away with such simplification because, had one used the full charged current interaction Lagrangian [tex]\mathcal{L}_{cc}(x) = \frac{g}{\sqrt{2}}\left(J^{+ \mu}(x)W^{+}_{\mu}(x) + J^{- \mu}(x)W^{-}_{\mu}(x)\right),[/tex] with [itex]J^{+ \mu} = \bar{\nu}_{L}\gamma^{\mu}e_{L} + \cdots [/itex], and applied the Wick theorem, one then simply gets two contractions, one for the [itex]W^{+}[/itex] and the other for [itex]W^{-}[/itex].
Seems like if the fields are at the same point (x) he is calling it local.
That is okay in the Lagrangian or the Hamiltonian but not in the perturbative expansion of the scattering operator.
 
  • #5
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A further question regarding OPE.
I have seen it quoted from multiple sources that in OPE the effective operators express the long distance effects and the Wilson coefficients the short distance effect.
Yet I cant seem to see how the derivation by Schwartz shows this long and short distance dependence anywhere.
 

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