2
$\begingroup$

I've been reading a textbook on QFT, and learned that we can calculate the probability amplitude that a system in state $|i\rangle$ will "collapse" into state $|f\rangle$ after some amount of time $t$, by calculating each term in the Dyson Series and adding them all together. My problem with this interpretation is that when calculating the Dyson Series for 2 incoming pions with momentum $k_1$ and $k_2$ and leaving with momentum $k_3$ and $k_4$, the first term is infinity if $k_1$ = $k_3$ and $k_2$ = $k_4$. My understanding of this infinity is that we get it from working with momentum eigenstates which cannot describe real systems. By constructing "wave-packets" of these states, this divergence goes away. That's all fine and dandy, but for the next term we have $-i\lambda V$ in the dyson series, where $\lambda$ is the coupling strength and $V$ is the volume that the particle may exist in when measured. What's a bit confusing to me is that this term will depend on the coupling constant and the volume, and both can be large or small. So how can we ensure that the squared amplitude will be less than or equal to one if the coupling constant and volume can be any number?

(Pions are defined here as particles associated with the scalar field)

$\endgroup$
1

1 Answer 1

-1
$\begingroup$

They convert to 0 asyou increase the parameters. So, substitute in larger parameter valeus.

$\endgroup$
1
  • $\begingroup$ What do you mean by "converting" to 0? Can you elaborate on this and how it solves the problem? $\endgroup$
    – user310742
    Commented Nov 18, 2021 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.