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The sum of two 3-momenta $\vec{p}_1$ and $\vec{p}_2$ can be zero if their magnitudes are equal and they are directed opposite to each other.

What about the sum of two 4-momenta $p_1^\mu$ and $p_2^\mu$?

Let's assume that the sum can be zero which implies that their individual components must add up to zero. In particular, $p_1^\mu+p_2^\mu=0$ implies for the zeroth components that $E_1+E_2=0$. Can this condition ever be satisfied other than the trivial case when $E_1=E_2=0$? Is this possible for a particle-antiparticle pair because antiparticles have negative energy?

Solidification
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2 Answers2

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On the event horizon of a static (Schwarzschild) black hole or on the ergosphere of a rotating (Kerr) black hole, a quantum fluctuation may produce a pair of particles one with positive energy and the other with negative energy. The particle with positive energy just before the horizon/ergosphere and the particle with negative energy just after. The negative energy particle is allowed by the time Killing vector switching to spacelike crossing the horizon/ergosphere.

It is a specific case, but in principle the 4-momentum addition of the two particles may give zero.

Michele Grosso
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It is clear that $p_1^\mu+p_2^\mu=0$ if and only if $$ E_1+E_2=0\qquad\text{and}\qquad \vec p_1+\vec p_2=\vec 0 $$

OP correctly argued that $\vec p_1+\vec p_2=\vec0$ is perfectly possible, and asks whether $E_1+E_2=0$ is possible as well. The answer is yes. The origin of energies is irrelevant, and therefore if $E_1+E_2=\mathcal E_0$, it suffices to redefine $E_i\to E_i-\mathcal E_0/2$ to obtain $E_1+E_2=0$.

One could argue that the choice of origin we made above is not natural. If you choose, instead, the origin of energies such that $E=m$ when the particle is at rest, then $E=\sqrt{\vec p^2+m^2}$ is positive-definite. The sum of two positive-definite functions is also positive-definite, and therefore $E_1+E_2>0$, strictly. In that case, the sum of energies cannot vanish.

Finally, if you use massless particles (which have no rest frame), then $E=|\vec p|$, which is semi-positive definite. The only way to get $E_1+E_2=0$ is that $E_i=0$, that is, that $\vec p_i=\vec 0$. But a particle with no mass, no momentum, and no energy, is not really a particle at all: it is the vacuum. Whether this qualifies is a matter of opinion.

To clarify a misconception in the OP: anti-particles have positive energy. The expression $E=\sqrt{m^2+\vec p^2}$ is valid both for particles and anti-particles.

AccidentalFourierTransform
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