I What exactly is the amplitude of an interaction?

Natchanon
Messages
31
Reaction score
3
I've been reading Griffths' intro to elementary particles and I encountered this symbol that looks similar to "M" called amplitude, which can be calculated by analyzing the Feynman diagram of an interaction. What exactly is it? When I hear amplitude I imagine waves, but not sure what this one's supposed to mean.
 
Physics news on Phys.org
This is the "probability amplitude". You multiply it by its complex conjugate in order to get a type of probability density for the interaction, similarly to the way in ordinary QM the position probability density ##P(\vec x) = |\psi(\vec x)|^2 = \psi^*(\vec x)\psi(\vec x)##.
 
  • Like
Likes Natchanon
If you have not seen a quantum mechanical amplitude, it is likely that Griffiths is too advanced for you at the moment. I would suggest backing off to a book on QM, and when you have that down, return to Griffiths,
 
Last edited:
Vanadium 50 said:
If you have not seen a quantum mechanical amplitude, it is likely that Griffiths is to advanced for you at the moment. I would suggest backing off to a book on QM, and when you have that down, return to Griffiths,
I have read his book intro to quantum mechanics and have taken an intro to QM class. I know about Schrodinger eq and how to calculate probability from it.
 
jtbell said:
This is the "probability amplitude". You multiply it by its complex conjugate in order to get a type of probability density for the interaction, similarly to the way in ordinary QM the position probability density ##P(\vec x) = |\psi(\vec x)|^2 = \psi^*(\vec x)\psi(\vec x)##.
So intergral of|M|^2 is the prob that particular interaction will occur?
 
Natchanon said:
So intergral of|M|^2 is the prob that particular interaction will occur?

Not quite. In particle physics there are two kind of processes: scattering and decays. There are two famous observables that you can calculate with QFT: cross section for the first and decay width for the second.

For both you need ## | \mathcal M | ^ 2 ##, but also some kinematics of the process.

## \mathcal M ## represents somehow the probability, but it is not as direct as in QM.
 
  • Like
Likes Natchanon
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