Why can't a particle interaction $A + B \rightarrow C$ conserve energy and momentum at the same time?

I learned that in the Feynman diagram for electron photon scattering, the intermediate particle represented by the middle solid line must be a virtual particle because it violates energy-momentum conservation. Two particles becoming one cannot conserve energy and momentum at the same time.

How can it be shown that this is true?

• Perhaps you can work through an attempt to see if you can match energy and momentum and then show your attempt in the question. – flippiefanus Feb 15 at 12:13

The QED vertex ($$e^- + e^+ \to \gamma$$) is kinematically not allowed because of masslessness of the photon. Since you're always free to choose your own Lorentz frame, in a center of momentum (COM) frame, the total momentum in the initial state is $$0$$. But the final state, photon, always travels with the speed of light. There exists no Lorentz frame that you can boost to, so that the photon momentum can be put to $$0$$. So the violation of conservation of momentum precludes this process from occurring in nature.

Conservation of energy is also violated, for example, in $$e^- \to e^- + \gamma$$. In COM frame, the initial state has energy of a stationary electron. The final state has an electron and a photon. This state will have an energy always larger than that of a stationary electron.

You can have $$A+B\to C$$. But only if the mass is right.

Energy and momentum conservation says that $$\vec p_C=\vec p_A+\vec p_B$$ and $$E_C=E_A+E_B$$.

Then if $$C$$ is real, not virtual, its mass is given by $$m_C^2c^4=E_C^2-p_C^2 c^2 = m_A^2c^4+m_B^2c^4+2 E_A E_B - 2 \vec p_A .\vec p_B c^2$$.

In your example, $$m_C=m_B=m_e$$ and $$m_A=0$$ so you would have to have $$E_AE_B-\vec p_A.\vec p_B c^2=0$$, which is impossible to arrange, even if $$\vec p_A.\vec p_B=p_Ap_B$$, as $$E_B>p_B c$$.

But it is possible and happens, for example with $$e^+e^- \to \psi$$, or with $$\pi^+ p \to \Delta^{++}$$, for the right energies/momenta.