Fermi's Golden Rule and the S-matrix

[tomdodd4598]

TL;DR

Why can the perturbation matrix element be replaced with the transition part of the S-matrix?

Hey there,

This question was asked elsewhere, but I wasn't really satisfied with the answer.

When I learned about Fermi's golden rule, ${ \Gamma }_{ if }=2\pi { \left| \left< { f }|{ \delta V }|{ i } \right> \right| }^{ 2 }\rho \left( { E }_{ f } \right)$, it was derived from first order perturbation theory in the context of quantum mechanics.

In the context of QFT, the perturbation was replaced by the transition part $\hat { T }$ of the S-matrix, $\hat { S } ≔\hat { I } +i\hat { T }$. However, $\hat { T }$ is not necessarily given only up to first order, so why can we just make this replacement in general?

 

[A. Neumaier]

Well, Fermi's golden rule is valid only to first order, hence has second order errors. The error made by the replacement is of the same order, hence the result - though different numerically - is accurate to the same order. and therefore (without further analysis) about equally reliable.

 

[tomdodd4598]

The error made by the replacement is of the same order, hence the result - though different numerically - is accurate to the same order. and therefore (without further analysis) about equally reliable.

So is there not any good reason to expect that higher-order corrections to amplitudes will give us better predictions of decay rates and scattering cross-sections?

 

[A. Neumaier]

So is there not any good reason to expect that higher-order corrections to amplitudes will give us better predictions of decay rates and scattering cross-sections?

You get better predictions if you also use a higher order version of Fermi's golden rule. In general, if you combine different approximations the final accuracy will be more or less that of the worst link.

 

[tomdodd4598]

You get better predictions if you also use a higher order version of Fermi's golden rule. In general, if you combine different approximations the final accuracy will be more or less that of the worst link.

So in practice, is that what we do? I've read various QFT textbooks which calculate higher order diagrams, but then don't mention using higher order forms of the golden rule.

 

[tomdodd4598]

I've come a bit full-circle after a while and am still confused on this. If "the final accuracy will be more or less that of the worst link", and we're only using the first-order Fermi golden rule, then how can any higher-order matrix amplitude be reliable?