So , everywhere I have looked for pair production it is stated that it cannnot happen in vacuum . Most proofs I have seen for that , state that the conservation of energy and momentum cannot be true at the same time without a body with which the photon will interact in the first place . In most of these proofs the momentum of the electron and the positron is considered the same value p . Why that ?

My explanation why pair production cannot happen in vacuum is based in the existence of a COM frame . We consider that a COM exists for every isolated system , so one will exist for the pair production in vacuum . But if that's true the spatial momentum of the photon will be zero , which is impossible . So if we consider that the photon first interacts with another object (for example with a nucleus that initially has zero velocity ) then we can find a COM . Is this explanation ok ?

  • $\begingroup$ Your explanation is OK, it is a different way of using the conservations laws. The same momentum assumes the com system. $\endgroup$
    – anna v
    Commented Sep 17, 2020 at 17:27

4 Answers 4


In the center of mass frame, the momentum of the created electron positron pair (the magnitude of the combined momentum four vector) is 0.

This cannot be satisfied with a single photon.

There has to be either two photons (where the magnitude of the combined momentum four vector is 0), or as you say in your case, a nucleus and a single photon.

Given two photons of sufficient energy to yield at least the rest mass of an electron-positron pair, one finds that QED predicts a non-zero amplitude for the process γγ→e+e− to happen. That is all the theory tells us. No "fluctuation", no "virtual particles", nothing. Just a cold, hard, quantitative prediction of how likely such an event is.

How does gamma-gamma pair production really work?

So in the case as you say, when there is a nucleus nearby (or anything in the environment where the photon can steal momentum from), the nucleus is able to receive a recoil from the photon, and thus satisfy momentum conservation laws. In reality, the photon ceases to exist as photon, its momentum is transferred to the nucleus, and its energy is transformed to create an electron positron pair.

The flaw goes away if the photon can steal momentum from its environment. The inside of an atom has a strong electric field, which is made of (in quantum electrodynamics) "virtual" photons. Your real photon can exchange momentum with the atom by scattering from one of these virtual photons; that's where the pair creation happens.

Why is a nearby nucleus required for Pair Creation?

So your explanation is correct, and a nucleus is one example of how this process can satisfy momentum conservation laws.


Any two massive particles always have a center of momentum frame, where the total momentum is zero. So for simplicity, you may as well analyze the pair production in that frame, which implies the two particles have four-momentum $(\sqrt{m_e^2+p^2},\vec{p})$ and $(\sqrt{m_e^2+p^2},-\vec{p})$. Then by conservation of four-momentum, the original photon must have four-momentum $(2\sqrt{m_e^2+p^2},0)$ and thus mass $2\sqrt{m_e^2+p^2}$. Since photons are massless, this is a contradiction.

Your explanation is equivalent. You need to analyze the four-momenta to determine if the energy is actually equivalent for pair production, however. The energy required is always greater than $2m_ec^2$ because some of the energy goes into recoil of the other object.


Whatever is the reference frame, one needs the energy of at least $2mc^2$ to create a pair. Moreover, the momentum conservation requires that the momenta of the electron and the positron sum up to zero.


Maybe we should clarify here something subtle that can cause confusion among readers by other information sources.

You are right, spontaneous pair production of stable particles (i.e. an electron-positron pair as long they are not annihilating with other particles) is not possible in the vacuum without the presence of light and matter and their interaction. Of course there is an exception and two energetic colliding photons can produce but very rarely, a pair of stable fermions. It is called the Breit–Wheeler process and was actually observed at 2021 at LHC. So pair production is possible also by only two light beams interacting without the presence of any normal matter (i.e. without interaction of photons with fermions needed).

Another question however could ask something different that is, if the vacuum can spontaneously produce stable fermion pairs? The answer from existing theory and experiments is no.

However what is usually a point of confusion is that the vacuum zero point energy (ZPE) is in a constant state of vacuum quantum fluctuations (ZPF) thus in a state on continuous spontaneous vacuum virtual pair creation. The words here that make the difference in the meaning are "virtual" and the word "creation".

Meaning, that vacuum cannot produce out of the blue stable fermion particle pairs without the presence and interaction of light and matter inside it.

However, virtual fermion pairs are created by the vacuum and immediately after annihilated (i.e. destroyed) all the time in the vacuum, a phenomenon also known in the literature as quantum foam as a sub-sequence of the Heisenberg Uncertainty Principle (HUB) of QM and vice versa (i.e. we could also theorize that HUB is due to quantum vacuum fluctuations).

The word "virtual" emphasizes that these pairs are not the same as their stable counterparts and occur only for an instance of time before they are destroyed so small within the HUP limit that they cannot be directly observable but QFT and QM and also indirect observations of the Cosmological Constant expansion of the Universe and also in big lab experiments (i.e. Fermilab g-2 muon anomalous magnetic moment) predict their existence.

So, final answer:

Spontaneous pair productionn by the vacuum ---> No

Spontaneous virtual pair creation by the vacuum ---> Yes


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