Feynman's approach to precision QED

Spinnor
Gold Member
Messages
2,227
Reaction score
419
Feynman's approach to precision QED calculations, see his book "Quantum Electrodynamics", was shown(?) to be equivalent to the method of quantum fields. Both methods get the right the get the right answers to some problems.

What quantum problems does Feynman's method (there is a name for that I guess) not work or work as well when compared to the method of Quantum Fields?

Thanks for any help!

The book, much of which can be read (edit, some of which can be read),

https://books.google.com/books?id=xt-Vvhloo8YC&printsec=frontcover&dq=feynman+quantum+electrodynamics&hl=en&sa=X&ved=0ahUKEwiDgseIoZ3KAhULGj4KHQBHCwgQ6AEILTAD#v=onepage&q=feynman quantum electrodynamics&f=false
 
Physics news on Phys.org
I don't know the book, but I guess you mean perturbation theory. It works well if the coupling strength is weak. That is true in QED and for the weak interaction, and it works reasonably well for QCD at high energies. It fails for low-energetic QCD processes.

Perturbation theory uses a series that is not convergent - if you would keep calculating higher and higher orders, at some point the results would become less precise again. We are far away from that limit, however, so this does not play a role in today's calculations.
 
Thanks mfb, need time to refine my question, the dentist calls.
 
If I understand what Spinnor is asking, there is no difference. They are equivalent. There are many important instances in which QED differs from classical electromagnetic fields (e.g. quantum tunneling which is a purely quantum phenomena not found in classical electromagnetism), but QED is a quantum field theory.
 
Spinnor said:
Feynman's approach to precision QED calculations, see his book "Quantum Electrodynamics", was shown(?) to be equivalent to the method of quantum fields. Both methods get the right the get the right answers to some problems.

What quantum problems does Feynman's method (there is a name for that I guess) not work or work as well when compared to the method of Quantum Fields?

Thanks for any help!

The book, much of which can be read (edit, some of which can be read),

https://books.google.com/books?id=xt-Vvhloo8YC&printsec=frontcover&dq=feynman+quantum+electrodynamics&hl=en&sa=X&ved=0ahUKEwiDgseIoZ3KAhULGj4KHQBHCwgQ6AEILTAD#v=onepage&q=feynman quantum electrodynamics&f=false
Are you referring to "Old-Fashioned Perturbation Theory"? (In this approach, Feynman diagrams are different from the usual approach taught in schools nowadays, several OFPT may correspond a single "model" Feynman diagram). It is completely equivalent to the modern approach.
 
  • Like
Likes Spinnor and vanhees71
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top