1
$\begingroup$

The optical theorem links the imaginary part of the forward scattering amplitude to the total decay width of a particle: $\mathrm{Im}\,M_{i\to i} = m\Gamma_{tot}$. Here $\Gamma_{tot} = \frac{1}{2m} \sum_x |M_{i\to x}|^2$. My QFT lecturer told us that the Breit-Wigner propagator $\frac{i}{p^2-m^2+im\Gamma_{tot}}$ follows from that. How?

I tried to use the LSZ formula to replace the forward amplitude to the propagator. However I don't even know if the LSZ formula is applicable here (due to the complex mass). If one can use it I find something like (for a scalar particle) $iM_{i\to i} \sim (p^2 - m^2)^2 \cdot \frac{i}{p^2-m^2+i\varepsilon}$ which should simply vanish for on-shell particles, whether or not $m$ has an imaginary part.

So how does the Breit-Wigner propagator follows from the optical theorem? In particular why is the $\Gamma$ in the Breit-Wigner propagator the same as in the optical theorem?

$\endgroup$
  • $\begingroup$ WP useful? $\endgroup$ – Cosmas Zachos Feb 5 at 20:53
  • $\begingroup$ Sorry, but how does anything on this Wikipedia page answer this question? :D $\endgroup$ – toaster Feb 6 at 21:43
  • $\begingroup$ It sends you to L Brown's friendly derivation to the generic propagator of an unstable particle, and whence the relativistic Breit-Wigner formula. I did not stick to the somewhat off-mainstream twists and turns of your proposed path to it. $\endgroup$ – Cosmas Zachos Feb 6 at 21:46
  • $\begingroup$ This just explains that a complex pole of propagator is related to a exponential decay of the propagator. This does not show that the $\Gamma$ in the propagator is actually the total decay width as calculated in scattering theory: $\Gamma_{tot} = \frac{1}{2m} \sum_x |M_{i \to x}|^2$. $\endgroup$ – toaster Feb 7 at 12:24
  • $\begingroup$ I took you to Lowell Brown's (6.314-6.3.23), but I can't make you drink... Perhaps if you took that primary section into consideration in restructuring your question you might have better luck. $\endgroup$ – Cosmas Zachos Feb 7 at 16:34

Your Answer

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

Browse other questions tagged or ask your own question.