Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Breit Wigner Formula describes the cross section for interactions that proceed dominantly via a intermediate particle (O*) A+B → O* → C + D:

$$σ = \frac{2\Pi}{k^{2}}\frac{Γ_{i}Γ_{f}}{(E-E_{o})^{2} + (Γ/2)^{2}}$$

A short question: Does the formula apply to situations when the intermediate particle is actually virtual?

For example, in positron electron annihilation, they form a photon which might eventually decay into another two particles. Can we calculate the resonant cross section for this process with the Breit Wigner Formula as well? If it is possible, what should we put in for $E_0$, which is supposed to be the rest mass of the intermediate particle?

share|improve this question
Possibly related: physics.stackexchange.com/q/4349/2451 –  Qmechanic May 28 '12 at 19:21
Rest mass of a virtual photon is non-zero. –  C.R. May 29 '12 at 3:45
BW (Lorentian) shape can describe both coherent (virtual) and incoherent (real) contributions to a resonance width –  Slaviks Feb 24 '13 at 21:17

1 Answer 1

$E_0$ is the resonance energy, what energy state exists in the compound system that you're populating in the compound system, O* in the question. It's not appropriate where there is no resonance, no excited state in the compound system.

Note, a single photon cannot decay into two particles, or vice versa, without interacting with the field of something else. For an isolated system, you can have a COM system for the two particles so total momentum = 0, but for a photon if the momentum = 0 ($p=h \nu /c$) then the photon energy = 0 ($E=h \nu$) and it doesn't exist.

share|improve this answer
"single photon cannot decay into two particles" Exchange of a single virtual photon is the tree level approximation to a great many interactions, and being virtual it is not constrained to have zero mass. –  dmckee Jan 25 '13 at 20:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.