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In interferometry with coherent light, the final output is differenced detectors. That is,

$\left<N\right> = \left<N_1\right> - \left<N_2\right>$

where $N_i$ is the number operator of that mode.

However, in interferometry with NOON states, the Wikipedia article states

consider the observable

A = | N , 0 ⟩ ⟨ 0 , N | + | 0 , N ⟩ ⟨ N , 0 | 

The expectation value of A {\displaystyle A} A for a system in a NOON state switches between +1 and −1 when the phase changes from 0 to $π/N$.

My confusion is with the expression and physical meaning of the operator A. If the NOON state is

$\left|\psi\right> = \frac{\left|N\right>_a\left|0\right>_b+e^{iN\phi}\left|0\right>_a\left|N\right>_b}{\sqrt{2}}$

as taken in the Wikipedia article. Then, because I'm confused about the notation of $A$, there are two possible expressions of $A$ for me. $A_1$ and $A_2$:

$A_1 = \left|N\right>_a\left|0\right>_b\left<N\right|_a\left<0\right|_b + \left|0\right>_a\left|N\right>_b\left<0\right|_a\left<N\right|_b$,

which I think would be the sum of the number operators on both the output arms of the Interferometer. However, this operators doesn't give the correct result of

$\left<A\right> = \cos 2\phi$

which Wikipedia alludes to in its change of phase from 1 to -1 statement. On the other hand, the expression

$A_2 = \left|N\right>_a\left|0\right>_b\left<0\right|_a\left<N\right|_b + \left|0\right>_a\left|N\right>_b\left<N\right|_a\left<0\right|_b$

does give the correct answer. However, I don't understand the physical realization of this operator. And if this is the measurement being performed, then this paper refers to it as

If one now carries out a simple measurement scheme in the N-photon detecting analyzer

So, what is this measurement being carried out? Which one is the correct representation, $A_1$ or $A_2$, of $A$ as written by Wikipedia? And what is the physical meaning of this operator?

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    $\begingroup$ Yes $A_2$ is the correct interpretation. Indeed it is essentially the same notation $\endgroup$
    – lcv
    Commented Apr 8, 2019 at 21:46
  • $\begingroup$ Do you know what is being measured then? $\endgroup$
    – lAPPYc
    Commented Apr 8, 2019 at 21:55
  • $\begingroup$ In QM every hermitian operator can in principle be measured. Whether it can be measured is another kettle of fish. If all you want is it's expectation value it seems that measuring the number of particles in a and b is more the enough. $\endgroup$
    – lcv
    Commented Apr 8, 2019 at 22:45

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