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This is just a conceptual question I guess. The annihilation of a particle with a finite mass and its anti-particle cannot lead to the emission of only one photon, and this is due to the conservation of energy and linear momentum.

However, how could this be shown in a mathematical way? Could this be done, perhaps, with a consideration of the four-vector momentum of the two particles?

Qmechanic
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DarthPlagueis
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  • "Could this be done, perhaps, with a consideration of the four-vector momentum of the two particles?" Er ... yes? It's easiest in the center of momentum frame. As in really, really, why-didn't-I-think-of-that easy. – dmckee --- ex-moderator kitten Oct 31 '15 at 21:40
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    The opposite: http://physics.stackexchange.com/q/13513/2451 , http://physics.stackexchange.com/q/50106/2451 , http://physics.stackexchange.com/q/145560/2451 and links therein. – Qmechanic Nov 01 '15 at 01:03

2 Answers2

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As dmckee says in a comment, the proof is ridiculously simple. Suppose we work in the centre of momentum frame so the total momentum is zero. The particle comes in with some momentum $p$ and the antiparticle comes in with the opposite momentum $-p$, and the two annihilate.

Suppose the annihilation produced a single photon. The momentum of a photon is:

$$ p = \frac{h}{\lambda} $$

but the problem is that the momentum of a photon is always $h/\lambda$. Unlike a massive particle a photon has no rest frame i.e. no frame in which it's momentum is zero. So creation of a single photon means the momentum would change from zero to $h/\lambda$ and momentum wouldn't be conserved.

For momentum to be conserved we have to create a minimum of two photons moving in opposite directions i.e. with momenta $h/\lambda$ and $-h/\lambda$.

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John Rennie
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Suppose $a\overline{a}\rightarrow\gamma$ is possible for a particle $a$ with a definite nonzero mass, $p_a^2=m^2>0$ ("mostly-minus" metric, $c=1$). Conservation of momentum implies

$p_\gamma=p_a+p_{\overline{a}}\implies p_\gamma^2=m_\gamma^2=0=p_a^2+p_{\overline{a}}^2+2p_a p_{\overline{a}}=2m^2+2 p_a \cdot p_\overline{a}$

However, the scalar product on RHS is bounded from below: We have (boldface quantities are three-vectors)

$p_a \cdot p_\overline{a}=E_a E_\overline{a} - \mathbf{p_a}\cdot\mathbf{p_\overline{a}}=\sqrt{m^2+\mathbf{p_a}^2}\sqrt{m^2+\mathbf{p_\overline{a}}^2}-\mathbf{p_a}\cdot\mathbf{p_\overline{a}}$

but since $m^2>0$ by assumption

$\geq\sqrt{\mathbf{p_a}^2}\sqrt{\mathbf{p_\overline{a}}^2}-\mathbf{p_a}\cdot\mathbf{p_\overline{a}}=\|\mathbf{p_a}\|\|\mathbf{p_\overline{a}}\|(1-\cos(\angle(\mathbf{p_a},\mathbf{p_\overline{a}}))\geq\|\mathbf{p_a}\|\|\mathbf{p_\overline{a}}\|(1-1)=0$.

Hence, $0\geq2m^2>0$, which is a contradiction.

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