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I am seeing in a lot of papers about quantum optics the term "non-degenerate photon pair", which seems like a very important concept.

This may seem like a silly question, but I am an EE undergrad who has just started reading papers in this field and would appreciate any clarification or even advise on some textbook references.

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  • $\begingroup$ maybe @WetSavannaAnimaaka RodVance will see this. From what I understand here arxiv.org/pdf/1304.1490.pdf "We configure the device to create and manipulate two-colour (non-degenerate) or same-colour (degenerate), path-entangled or path-unentangled photon pairs. " So the non degenerate part seems to refer to the creation of two different ffrequency photons. $\endgroup$
    – anna v
    Commented Jun 13, 2016 at 4:34
  • $\begingroup$ Thank you for the reference, I will try to investigate this topic more. $\endgroup$
    – Nen
    Commented Jun 13, 2016 at 16:02

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In general, degeneracy in quantum mechanics means that there are at least two states that have indistinguishable energy. In the case of a degenerate photon pair, the two photons have will the same frequency (as E = $\hbar$ $\omega$).

This is, of course, oversimplifying things slightly. The above describes the case when the photons are each in a single mode. In reality, most photon sources will produce photons which are multimode. But we won't go there!

It's hard to elaborate further without the specific context in which "non degenerate" appears. I'd imagine that in the context of photon pairs it is in relation to quantum inference. In the case of the paper discussed in a previous answer (arxiv.org/pdf/1304.1490.pdf) this is indeed the case.

I won't talk about quantum interference here, but maybe discussing how the states are labeled in the above paper will clear up what they mean by degenerate and non-degenerate.

Degenerate photons pairs- both in the same mode- are given by states like

$\left|n_{A} n_{B}\right>$

$n_{A}$ and $n_{A}$ are simply the number of photons in the output ports A and B of their experiment. As the photons are degenerate, there is no need to give a label to show which photons have which frequency. They are all the same!

For non degenerate states, the photons have frequencies labeled by i and s. Non degenerate pairs exist in states like

$\left|n_{i} n_{s}\right>_A$ $\left|n_{i} n_{s}\right>_B$

A and B still label the output ports, but we now have to label which mode the photons are in as well (i or s). For example $\left|1_{i} 0_{s}\right>_A$ $\left|0_{i} 1_{s}\right>_B$ is a state where there is one photon of mode i in port A and one photon in mode s in port B.

I hope this helps. If you're completely new to quantum optics, then a good place to start might be "Introduction to Quantum Optics: From the Semi-classical Approach to Quantized Light" by Grynberg, Aspect and Fabre.

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  • $\begingroup$ Thank you very much to take the time to answer my question with such details! I will definitely start reading the book you recommended. $\endgroup$
    – Nen
    Commented Jun 15, 2016 at 15:20
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The photon pair (obtained in a spontaneous parametric down conversion WITHOUT post-selection) keeps the same filter angle, reducing crosstalk in quantum optic networks, therefore increasing channel manipulation.

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  • $\begingroup$ Thank you for the answer. Could you point me to some references that provides more information about photon nondegeneracy? I am reading some papers on dispersion enabled directional couplers and finding difficulties in some of their physical concepts. $\endgroup$
    – Nen
    Commented Jun 13, 2016 at 15:55

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