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Kostya
Kostya
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But, if we get the correlation functions, then, can't we just employ the LSZ formula to find the scattering amplitudes, and so we need to calculate fewer diagrams?

I think that you are missing the perturbation theory in your picture. -- If you know the correlation functions, then you get the amplitudes by LSZ. But in order to get the correlation function you need to calculate some diagrams -- usually representing terms in some perturbation theory.

I put a photon "wiggle" over an external electron line.
Can I simply multiply the original diagram with my regulated electron propagator?

I thought that is exactly what LSZ is about -- the scattering amplitude is a correlation function times a bunch of $\frac{1}{\sqrt{Z}} (p^2-m^2)$factors: $$\langle f|i\rangle \sim \frac{p^2-m^2_1}{\sqrt{Z_1}} \cdot \frac{q^2-m^2_2}{\sqrt{Z_2}} \cdot \dots \Gamma(p,q,\dots)$$ Which on the language of Feynman diagrams means that you need to "amputate" everything that hangs on the external lines of the correlation function $\Gamma$: $$\Gamma(p,q,\dots) = \frac{Z_1}{p^2-m^2_1}\cdot\frac{Z_2}{p^2-m^2_2}\cdot\Gamma_{amp}(p,q,\dots)$$$$\Gamma(p,q,\dots) = \frac{Z_1}{p^2-m^2_1}\cdot\frac{Z_2}{p^2-m^2_2}\dots\Gamma_{amp}(p,q,\dots)$$ Or, finally: $$\langle f|i\rangle \sim \sqrt{Z_1}\cdot\sqrt{Z_2}\cdot\Gamma_{amp}(p,q,\dots)$$$$\langle f|i\rangle \sim \sqrt{Z_1}\cdot\sqrt{Z_2}\dots\Gamma_{amp}(p,q,\dots)$$ I hope that is what you are asking...

But, if we get the correlation functions, then, can't we just employ the LSZ formula to find the scattering amplitudes, and so we need to calculate fewer diagrams?

I think that you are missing the perturbation theory in your picture. -- If you know the correlation functions, then you get the amplitudes by LSZ. But in order to get the correlation function you need to calculate some diagrams -- usually representing terms in some perturbation theory.

I put a photon "wiggle" over an external electron line.
Can I simply multiply the original diagram with my regulated electron propagator?

I thought that is exactly what LSZ is about -- the scattering amplitude is a correlation function times a bunch of $\frac{1}{\sqrt{Z}} (p^2-m^2)$factors: $$\langle f|i\rangle \sim \frac{p^2-m^2_1}{\sqrt{Z_1}} \cdot \frac{q^2-m^2_2}{\sqrt{Z_2}} \cdot \dots \Gamma(p,q,\dots)$$ Which on the language of Feynman diagrams means that you need to "amputate" everything that hangs on the external lines of the correlation function $\Gamma$: $$\Gamma(p,q,\dots) = \frac{Z_1}{p^2-m^2_1}\cdot\frac{Z_2}{p^2-m^2_2}\cdot\Gamma_{amp}(p,q,\dots)$$ Or, finally: $$\langle f|i\rangle \sim \sqrt{Z_1}\cdot\sqrt{Z_2}\cdot\Gamma_{amp}(p,q,\dots)$$ I hope that is what you are asking...

But, if we get the correlation functions, then, can't we just employ the LSZ formula to find the scattering amplitudes, and so we need to calculate fewer diagrams?

I think that you are missing the perturbation theory in your picture. -- If you know the correlation functions, then you get the amplitudes by LSZ. But in order to get the correlation function you need to calculate some diagrams -- usually representing terms in some perturbation theory.

I put a photon "wiggle" over an external electron line.
Can I simply multiply the original diagram with my regulated electron propagator?

I thought that is exactly what LSZ is about -- the scattering amplitude is a correlation function times a bunch of $\frac{1}{\sqrt{Z}} (p^2-m^2)$factors: $$\langle f|i\rangle \sim \frac{p^2-m^2_1}{\sqrt{Z_1}} \cdot \frac{q^2-m^2_2}{\sqrt{Z_2}} \cdot \dots \Gamma(p,q,\dots)$$ Which on the language of Feynman diagrams means that you need to "amputate" everything that hangs on the external lines of the correlation function $\Gamma$: $$\Gamma(p,q,\dots) = \frac{Z_1}{p^2-m^2_1}\cdot\frac{Z_2}{p^2-m^2_2}\dots\Gamma_{amp}(p,q,\dots)$$ Or, finally: $$\langle f|i\rangle \sim \sqrt{Z_1}\cdot\sqrt{Z_2}\dots\Gamma_{amp}(p,q,\dots)$$ I hope that is what you are asking...

Source Link
Kostya
Kostya
  • 20.4k
  • 6
  • 71
  • 99

But, if we get the correlation functions, then, can't we just employ the LSZ formula to find the scattering amplitudes, and so we need to calculate fewer diagrams?

I think that you are missing the perturbation theory in your picture. -- If you know the correlation functions, then you get the amplitudes by LSZ. But in order to get the correlation function you need to calculate some diagrams -- usually representing terms in some perturbation theory.

I put a photon "wiggle" over an external electron line.
Can I simply multiply the original diagram with my regulated electron propagator?

I thought that is exactly what LSZ is about -- the scattering amplitude is a correlation function times a bunch of $\frac{1}{\sqrt{Z}} (p^2-m^2)$factors: $$\langle f|i\rangle \sim \frac{p^2-m^2_1}{\sqrt{Z_1}} \cdot \frac{q^2-m^2_2}{\sqrt{Z_2}} \cdot \dots \Gamma(p,q,\dots)$$ Which on the language of Feynman diagrams means that you need to "amputate" everything that hangs on the external lines of the correlation function $\Gamma$: $$\Gamma(p,q,\dots) = \frac{Z_1}{p^2-m^2_1}\cdot\frac{Z_2}{p^2-m^2_2}\cdot\Gamma_{amp}(p,q,\dots)$$ Or, finally: $$\langle f|i\rangle \sim \sqrt{Z_1}\cdot\sqrt{Z_2}\cdot\Gamma_{amp}(p,q,\dots)$$ I hope that is what you are asking...