Why is Hadronic Vacuum Polarization a Matrix Element of Current Products?

ndung200790
Messages
519
Reaction score
0
Please teach me this:
In chapter 18.4 Peskin&Schoeder(QFT) they consider the annihilation of electron and positron to hadron.Ignoring the mass of the electron,we have:

σ(e^{+}e^{-})=(1/2s)ImM(e^{+}e^{-}→e^{+}e^{-}).

We have:

iM=(-ie)^{2}u^{-}(k)\gamma_{\mu}v(k_{+}(-i/s)(i\Pi^{\mu\nu}_{h}(q))(-i/s)v^{-}(k_{+}\gamma_{\nu}u(k).

I do not understand why they can write:

i∏^{\mu\nu}_{h}(q)=-e^{2}\intd^{4}xe^{iqx}<T{J^{\mu}(x)J^{\nu}(0)>.

Where J^{\mu} is the electromagnetic current of quarks.
Thank you very much for your kind helping.
 
Physics news on Phys.org
Now,I think that it is because the fluctuation of vacum(that is excited by virtue photon) making a loop that is product of two currents(not more complex including many hadronics).Is that correct?
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top