What does the energy of a photon convert to with interference?

[Kurt Mueller]

TL;DR Summary

Thinking about the interferometer used in LIGO, when this interference occurs, what annihilation or conversion of energy takes place?

I was just reading about the LIGO experiment wherein an interferometer was used to detect gravity waves. This interferometer uses opposed light waves, detecting if there is a shift in their wavelength due to stretching/squeezing of a gravity wave passing through the lasers. (I hope I'm saying that right)

But what happens to the photons during this process of interference? What is the energy converted to?

Edit: Of course, I'm assuming there even is a conversion in the first place.

 

[PeterDonis]

what happens to the photons during this process of interference? What is the energy converted to?

No conversion of energy happens. The interference doesn't affect energy.

 

[Dale]

Summary:: Thinking about the interferometer used in LIGO, when this interference occurs, what annihilation or conversion of energy takes place?

Interference doesn’t change the amount of energy, so there is no conversion.

 

[Kurt Mueller]

Thank you. I was thinking along the lines of the photon being its own antiparticle and matter/antimatter annihilations.

 

[Dale]

Thank you. I was thinking along the lines of the photon being its own antiparticle and matter/antimatter annihilations.

That can happen, but not in a setup like LIGO or any other normal interferometer. It requires exceptionally luminous beams of very high energy light (gamma rays). It isn’t an interference process.

 

[vanhees71]

What are you referring to here? All I can think of is Delbrück scattering, i.e., $\gamma+\gamma \rightarrow \gamma+\gamma$ which indeed has been observed by the ATLAS collaboration at CERN in ultraperipheral lead-lead collisions, which is not precisely elastic scattering of two free photons but very close to it:

https://doi.org/10.1103/PhysRevLett.123.052001 (open access!)
https://physics.aps.org/articles/v12/s87

Within the Standard Model (or rather QED as one of its parts) it's a pure quantum effect. The lowest order perturbation theory is of order $\alpha^4$ (the box diagrams with four external photon lines).

 

[Dale]

What are you referring to here? All I can think of is Delbrück scattering

Yes, I was talking about Delbruck scattering $\gamma + \gamma \rightarrow \gamma + \gamma$ which has the incoming photons annihilate to a virtual $e^+ + e^-$ pair which then annihilates back to photons, or pair production $\gamma + \gamma \rightarrow e^+ + e^-$ which has the incoming photons annihilate to a real $e^+ + e^-$ pair.

it's a pure quantum effect.

Absolutely

 

[vanhees71]

The great thing is that it's not divergent as suggested from naive "power counting". So this box diagram's superficial degree of divergence is 0, so it could be divergent, but in fact the sum of all the box diagrams (which differ only in the topology connecting the external lines) is finite thanks to a Ward-Takahashi identity.