# Does the Breit Wigner formula apply to intermediate virtual particles?

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?

• Possibly related: physics.stackexchange.com/q/4349/2451 – Qmechanic May 28 '12 at 19:21
• Rest mass of a virtual photon is non-zero. – Siyuan Ren 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

The formula in Nogueira's answer tells the story.

For example, in positron electron annihilation, they form a photon which might eventually decay into another two particles.

In the formula we see a pole in the complex plane. Crossections are real numbers and calculating the square of the amplitude will give real numbers in the formula, which is the Breit-Wigner formula.

Can we calculate the resonant cross section for this process with the Breit Wigner Formula as well?

A virtual particle, even though it is described by this pole, is not represented by a real function, so cannot be interpreted as a cross section , it is part of the integration that will give a cross section. Within the integral it is off mass shell for the particle it is describing, i.e. the virtual photon in the question. Even the specific resonance plotted within the integral is like a mountain in the complex plain. It is only the integration which can relate the pole with physical crossections, and virtual particles disappear .

A virtual particle, even though unphysical in energy and momentum carries the other quantum numbers identifying particles, as spin , charge etc, that is why it is useful in drawing Feynman diagrams that are a shorthand for the integrations that give cross sections. One should not confuse it with a real particle because its manifestation is in the unphysical complex plane. It is after the integrations are carried through that the effect of the pole in the propagator can be compared with data/cross-sections.

$E_0$ 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.

• "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 --- ex-moderator kitten Jan 25 '13 at 20:08

The Breit Wigner Formula in the cross-section reveals a pole in the scattering amplitude living in the upper half-plane of complex values of energy if you do a analytic continuation.

The state of the system have this form: (We are free to choose the $\pm$)

$$\psi^{(\pm)}_g(t)=\int d\alpha\, e^{-iE_\alpha t} g^{(\pm)}(\alpha) \phi_\alpha + \int d\alpha\,\int d\beta\, \frac{e^{-iE_\alpha t} g^{(\pm)}(\alpha)T_{\beta \alpha}}{E_\alpha -E_\beta \pm i\epsilon} \phi_\beta$$

Where $g^{(\pm)}(\alpha)$ is a suitable superposition related to the choice $\pm$.

The Breit Wigner Formula in the cross-section implies that $T_{\beta \alpha}$ have a pole in $E_\alpha= E_r - i\Gamma/2$ that yields a term in the amplitude that behaves like $exp(-iE_r)exp(-\Gamma t / 2)$. So the probability for this state been measure in the $E_r$ decays exponentially with characteristic time $1/\Gamma$ (the life-time of a unstable particle). For times close enough to zero, only the states very close to the $E_r$ contribute with the integral, so we can think that the state is approximately the $E_r$, but only approximately. This state is approximately stationary for small times (compared to $1/\Gamma$).

The virtual particle is only a picture resulting from insertion of projection operators into the $T_{\beta \alpha}$ in a pertubative expansion. We may found some poles in the pertubative terms but they are integrate internally in $T_{\beta \alpha}$ and don't create a pole for $E_\alpha$. But of course that this terms create such poles for $E_\alpha$, but is not because the poles associated to a virtual particle.