# Why very strong fields are required for a photon to split?

Photon splitting does not occur in free space as energy and momentum cannot be conserved in any Lorentz frame. But it does occur in the presence of a strong field. Consider the example of a Magnetar. The typical field of a Magnetar is around $10^{10}$ teslas. X-ray photons have been found to split into two in these magnetars.

I understand that the field has momentum which can take part in the process, but why cant the process occur when a weak field is present? What is the scale of the magnetic field at which photon splitting becomes possible.

• the problem involving the splitting of photons was studied in more details by dozens of groups, please refer to: slac.stanford.edu/econf/C980518/proc/PDF/chistyakov.pdf cds.cern.ch/record/300629/files/9604028.pdf Here you can find more details and citations for further study of the subject. The underlying theory is still being developed ... Pavel Jan 4, 2015 at 10:32
• Doesn't this have to do with photon-photon interactions and the total energy of the photons needs to $\geq$ 1.022 MeV? I think the high magnetic fields means high photon energies, but I am not an expert in quantum so I would wait for someone else to clarify. Jan 5, 2015 at 20:30

I believe it would be incorrect to term this as "photon-splitting". What is really happening is that a photon is being annihilated and subsequently two new photons are created by means of a non-linear optical process. Such a process that is routinely harnessed in quantum optics laboratories is spontaneous parametric downconversion (SPDC).

The reason why this effect is not seen in weak fields is simply because non-linear effects become significant only when the field strengths are large. This becomes clear if you consider the expansion of the electric polarization $\vec{P}$ in the various orders of the electric field $\vec{E}$. In the weak-field limit, only the first order term linear in the electric field survives.

In this limit, the Hamiltonian $H\propto \vec{D}\cdot \vec{E}$ (where $\vec{D}=\epsilon_{0}\vec{E}+\vec{P}$) of the process when expressed in terms of the mode annihilation and creation operators essentially contains only terms of the form,

$H\propto \sum_{j,k} A_{jk}\hat{a}^{\dagger}_{j}\hat{a}_{k}$

Such a Hamiltonian commutes with the total photon number operator $\sum_{k}\hat{a}_{k}^{\dagger}\hat{a}_{k}$ and hence the photon number cannot change in the process. This constraint of photon number conservation is the hallmark of any linear-optical process; and as a result a photon cannot annihilate and produce two new photons.

However, in the strong field limit the higher order contributions to $\vec{P}$ become significant and as a result the Hamiltonian now contains terms of the form $\hat{a}_{i}\hat{b}_{j}^{\dagger}\hat{c}_{k}^{\dagger}$ which effectively imply the annihilation of one photon and creation of two others. It is precisely these processes that result in the phenomena that you describe.

In conclusion, I note that if the Hamiltonian was time-independent then the total energy of the two new photons would be equal to the energy of the original photon. Other symmetries like translational and rotational invariance if present, will lead to momentum and angular momentum conservation (such constraints are referred to as "phase-matching conditions") too. Nevertheless, it is clear that the process by itself is not anything as simple as "splitting", which is anyway quantum-mechanically meaningless.