# Why is the imaginary part of the Breit-Wigner propagator given by the total decay width?

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?

• WP useful? – Cosmas Zachos Feb 5 '19 at 20:53
• Sorry, but how does anything on this Wikipedia page answer this question? :D – toaster Feb 6 '19 at 21:43
• 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. – Cosmas Zachos Feb 6 '19 at 21:46
• 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: $\Gamma_{tot} = \frac{1}{2m} \sum_x |M_{i \to x}|^2$. – toaster Feb 7 '19 at 12:24
• 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. – Cosmas Zachos Feb 7 '19 at 16:34