When trying to understand the idea of virtual particles, the explanations for it seem to imply that, for example, electron-positron pairing happens within the fluctuating electron field, so does this mean that, when we speak of the electron field, it by itself is responsible for both electrons and positrons, and there is no separate "positron field" to speak of?


3 Answers 3


Yes, that is correct. In fact, you don't even need quantum field theory for this. The Dirac equation was the original relativistic equations describing the quantum behavior of the electron, and it already predicts the existence of the positron. You can't describe electrons in a quantum and relativistic manner without simultaneously describing positrons.

Both electrons and positrons are excitations of the electron field. Electron are associated with modes with positive frequency, while positrons are associated to negative frequencies. Hence, different field modes give rise to either electrons or positrons.

Similar ideas apply to all other particles. The antiparticles are always described by the same field, and typically correspond to a different set of modes (some particles are their own antiparticles though).

  • $\begingroup$ @QuantumWonder Yes, that is correct $\endgroup$ Commented May 19 at 21:34
  • $\begingroup$ Makes me wonder why the type of frequency (positive or negative) for a given energy excitation only matters in the case of fermions. $\endgroup$ Commented May 19 at 21:44
  • $\begingroup$ @QuantumWonder It also matters for bosons. For example, the $W^+$ boson is the antiparticle of the $W^-$ boson. The key property is that the field must be complex for the two different excitations to "decouple" and describe different particles, and most fields you know and love are real. However, fermions are usually complex fields due to the very nature of spinors, while it is easier for bosons to have a real field (such as the photon, graviton, $Z$ boson, or Higgs). $\endgroup$ Commented May 19 at 22:45
  • $\begingroup$ There are situations in which a fermionic particle is its own antiparticle. This is a possibility for neutrinos, for example. If I'm not mistaken this happens for Majorana neutrinos, but you might want to ask a particle physicist to be sure $\endgroup$ Commented May 19 at 22:45
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    $\begingroup$ @NíckolasAlves The term Majorana particle specifically means a particle that is its own antiparticle, so Majorana neutrinos would be neutrinos that are their own antiparticle by definition. It's an open question whether they actually exist or not--as I understand it the prevailing school of thought is "probably not but we haven't proven it yet". I'm pretty sure there are no known Majorana fermions, at least not below the electroweak energy scale. $\endgroup$
    – Hearth
    Commented May 21 at 2:55

Do particle & anti-particle pairs belong to the same field?

Yes. For example, in Quantum Electrodynamics the Dirac field operator is usual written in terms of annihilation operators $a_{\vec p}^s$ and $b_{\vec p}^s$ as: $$ \psi(x) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\vec p}}}\sum_s \left( a_{\vec p}^s u^s(\vec p)e^{-ip\cdot x} + b^s_{\vec p}v^s(\vec p)e^{ip\cdot x}\right)\;. $$ (See, for example, Peskin and Schroder Chapter 3.)

In this same notation the Hamiltonian operator is $$ H = \int \frac{d^3p}{(2\pi)^3}E_{\vec p}\sum_s \left( {a^s_{\vec p}}^\dagger a^s_{\vec p} + {b^s_{\vec p}}^\dagger b^s_{\vec p} \right) $$ and the momentum is $$ \vec P = \int \frac{d^3p}{(2\pi)^3}{\vec p}\sum_s \left( {a^s_{\vec p}}^\dagger a^s_{\vec p} + {b^s_{\vec p}}^\dagger b^s_{\vec p} \right) $$ and the electric charge is $$ Q = -|e|\int \frac{d^3p}{(2\pi)^3}\sum_s \left( {a^s_{\vec p}}^\dagger a^s_{\vec p} - {b^s_{\vec p}}^\dagger b^s_{\vec p} \right)\tag{1}\;. $$ (See, Id.)

Note the minus sign in Eq. (1), which shows that the Dirac field accounts for both the negative charge carriers (electrons) and the positive charge carriers (positrons). So, ${a^s_{\vec p}}^\dagger$ creates electrons, which have mass $+m$ and charge $-|e|$ and ${b^s_{\vec p}}^\dagger$ creates positrons, which have mass $+m$ and charge $+|e|$. (N.b., it is only the sign of the charge that flips for "anti-matter," not the sign of the mass.)


That's ambiguous. Fermionic particles and anti-particles are most consisely discribed by a single field with multiple components. Its however always possible to decompose these and describe them as separately propagating while interacting.

A similar decomposition happens to the standard model fermions in order to describe parity violation: the 4 component Dirac spinnor describing both fermion helicities and matter/anti matter fields get decomposed into two 2 component ones that only discribe one helicity. This way the left-handed ones can have weak interaction terms but the right-handed ones don't. (Although I suppose the same can be achieved with a matrix like interation term with vanishing diagonal entries for the right handed fermions. This doesn't seem to be a common choice.)

For the gauge fields the notion of anti-particles is even more ambiguous. Fields that dont cary conserved charge (like the photon field) dont neatly fall into the two camps.

Also keep in mind that virtual particles are a reasoning and computational device introduced by perturbation theory, not something observable/predicted by field theory. Phycisists often talk like they are real things in nature, which is at best debatable.


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