2
$\begingroup$

I'm confused, on a book I recently read, it talked about renormalization, the creation of renormalization is because of the infinity problem.

Here's is the problem, what is the infinity problem that needs to be eliminated in quantum mechanics? What is the cause?

$\endgroup$
1
$\begingroup$

It is all about the perturbative expansion for the interaction amplitudes. These are expressed in Feynman diagrams of increasing order.

in calculating loop Feynman diagrams in QED one finds integrals which diverge Such apparent pathologies are dealt with by the process of renormalization. This procedure can be viewed in several ways. From one perspective it is a formal manipulation, part of the definition of the quantum field theory, which allows one to calculate finite, testable expectation values and scattering amplitudes. From a more physical perspective, one starts by noting that the key divergences come from the high energy limits of the momentum integration.

So concretely it is going to large momenta that loop feynman diagrams diverge. The paper describes the procedure of avoiding the problem, which is the whole project of the renormalization procedure in this case.

$\endgroup$
  • 1
    $\begingroup$ omg what are u talking about?? lol $\endgroup$ – Nicole. C Feb 9 '17 at 23:07
  • $\begingroup$ @Nicole Cheng you wrote over here physics.stackexchange.com/a/310705/1486 "the waves stacks themselves up into a larger wave with much more amplitude" ... these are calculations for adding probability amplitudes. A divergence means that the stack adds up to infinity. We assume that in reality that doesn't happen, and renormalization is a way of representing unknown physical effects that keep the sum finite. $\endgroup$ – Mitchell Porter Feb 10 '17 at 4:55
  • 1
    $\begingroup$ Dang how you know my post??? lol $\endgroup$ – Nicole. C Feb 10 '17 at 5:04
  • $\begingroup$ "how you know my post" ? by clicking on the name and clicking on activities all is revealed $\endgroup$ – anna v Feb 10 '17 at 5:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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