# Are the two interference patterns at the output of an interferometer quantum entangled?

When an interferometer creates an interference pattern a second interference pattern is always created. The two patterns are opposites of each other. Knowing one pattern lets me accurately predict the other pattern 100% of the time. I have found this to be true in Michelson, Mach Zehnder, Sagnac, Fabry and even double slit interferometers. I understand that conservation of energy is the reason that is given for this. My question: Is quantum entanglement the underlying process? Are the two interference patterns quantum entangled?

• You have the same effect when you send classical light on an interferometer. No entanglement there. Mar 23 '20 at 20:27

Yes (see edit below though).

The two interference patterns are quantum entangled, meaning they are described by a single and common state vector.

If a single photon hits a beam-splitter, it can be either reflected to path $$a$$ or transmitted to path $$b$$: However, you cannot describe the two paths as (classically) independent and unrelated. The beam splitter has connected the two paths, it has entangled them, for what occurs on one of them (e.g. a measurement) also affects what occurs on the other one. No matter the distance, even if they are separated by light years.

So you should write the full state as $$|\Psi\rangle = 1/\sqrt{2} \cdot \left ( \,\,\, |1\rangle_a \otimes |0\rangle_b + |0\rangle_a \otimes |1\rangle_b \,\,\,\right ) ,$$

where $$|1\rangle$$ and $$|0\rangle$$ mean having one or zero photons.
Paths $$a$$ and $$b$$ are entangled because you cannot spit this state vector in $$( ... )_a \otimes (...)_b$$, which would allow you to treat them separately.

EDIT

Actually, reading your question more carefully, I think I have chosen a very niche example. In my case then yes, it's entanglement.

The opposite nature of the two patterns is because of the $$\pi$$ phase shift that light that is reflected receives wrt to the one that is transmitted.