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Antiparticles · Quantum Mechanics | Math Infinitum

Intuition

The negative energies would not go away, and an electron could apparently fall into them for ever, radiating as it went. Dirac’s answer in 1930 was bold: all the negative-energy states are already filled, and the Pauli principle stops electrons from falling in. A photon of more than $2mc^{2}$ can lift an electron out of this sea, leaving a hole that behaves as a particle of the same mass and opposite charge — the positron, found by Anderson in 1932. The sea itself was later dropped: in quantum field theory the negative-frequency solutions describe antiparticles of positive energy directly, and the picture works for bosons too, which have no Pauli principle. What survives is solid: every particle has an antiparticle; pairs are created when enough energy is available, at least $2mc^{2}$ ; and a particle meeting its antiparticle can annihilate, an electron and a positron at rest into two photons of 511 keV.

A car park full to the last space has no room for a new car, but if a car drives out, the empty space moves along the rows as cars shuffle into it. A hole in a filled sea behaves like a particle in its own right.

Dirac’s picture: electron states above $+mc^{2}$ , empty, and below $-mc^{2}$ , all filled, with no states in the gap of $2mc^{2}$ between. A photon of at least lifts an electron out of the sea, arrow, leaving a hole that moves like a positron.

Antiparticles

Every particle has an antiparticle of the same mass and opposite charge. Pairs are created when the energy allows — by a photon, only near a nucleus — and annihilate on meeting:

\begin{aligned}\gamma&\to e^{-}e^{+}\quad\text{needs}\ E_{\gamma}\ge2mc^{2}\\ e^{-}e^{+}&\to\gamma\gamma\end{aligned}

Properties

In Dirac’s sea picture, holes in the filled negative-energy states behave as positrons; the picture needs the Pauli principle and fails for bosons.
In quantum field theory the negative-frequency solutions describe antiparticles with positive energy; no sea is needed (the next chapter).
A photon alone cannot turn into a pair: energy and momentum cannot both be conserved, so a nucleus must take up momentum.
An electron and a positron at rest annihilate into at least two photons, back to back, of $mc^{2}=511$ keV each.
Charge conjugation maps each solution of the Dirac equation in a field to a solution for the opposite charge (stated).

Annihilation at rest gives two photons of mc^{2}

\begin{aligned}&e^{-}e^{+}\ \text{at rest}\ \Longrightarrow\ \text{not}\ \gamma,\ \text{but}\ \gamma\gamma\\ &\text{with}\ E_{\gamma}=mc^{2}\ \text{each}\end{aligned}

Before annihilation the pair has energy $2mc^{2}$ and no momentum. One photon cannot carry energy without momentum, since for light $E=pc$ . Two photons can: back to back, their momenta cancel, and sharing the energy equally, each has $mc^{2}$ .

Proof steps

E=2mc^{2},\qquad \mathbf{p}=0

The pair at rest.

E_{\gamma}=|\mathbf{p}_{\gamma}|c\ \Longrightarrow\ \mathbf{p}_{\gamma}=0\ \text{forces}\ E_{\gamma}=0\neq2mc^{2}

A single photon would have to carry no momentum and so no energy.

\mathbf{p}_{1}+\mathbf{p}_{2}=0\ \Longrightarrow\ |\mathbf{p}_{1}|=|\mathbf{p}_{2}|,\ \ E_{1}=E_{2}

Back to back with equal momenta, hence equal energies.

\begin{aligned}&E_{1}+E_{2}=2mc^{2}\\ &\Longrightarrow\ E_{1}=E_{2}=mc^{2}=511\ \text{keV}\end{aligned}

Energy conservation.

Applications

Practice

Pair Creation

Making a particle–antiparticle pair takes at least their two rest energies, $2mc^{2}$ .

E\ge2mc^{2}

Try it

What is the least energy a photon must have to create an electron–positron pair near a heavy nucleus, with $mc^{2}=0.511$ MeV, in MeV? Give three decimal places.