# How does N00N state interferometry give a cosine dependence on $N\phi$

$$\newcommand{\ket}[1]{\left|#1\right>}$$ A Mach-zahnder interferometer with $$\ket{\alpha}$$ input in port A and $$\ket{0}$$ in port B gives an intensity difference in the two detectors after the second beam splitter as $$|\alpha|^2\cos{\phi}$$ where $$\phi$$ is the phase difference between the two interfering rays coming from the arms of the interferometer. This dependence on $$\cos{\phi}$$ is improved to $$N\cos{N\phi}$$ if the N00N state is used. Where

$$\ket{N00N} = \frac{\ket{N}_C\ket{0}_D + \ket{0}_C\ket{N}_D}{\sqrt{2}}$$

and $$\ket{N}$$ is a fock state, or a number state of light, and $$\ket{\alpha}$$ is the coherent state.

Now, as I understand, the N00N state should exist between the arms of the interferometer, that is, after the first beam splitter, if there are N photons in the arm D, there are none in arm C, and vice versa. When working with $$N=2$$, after going through a phase shift $$\phi$$ in arm D such that $$\ket{N} \xrightarrow{\phi} e^{-iN\phi}\ket{N}$$ and then through the second beam splitter, I am getting the output state

$$\ket{\psi_{out}} = \frac{i\sqrt{2}(1-e^{-2i\phi})\ket{2}_{D2}\ket{0}_{D1}}{4} - \frac{(1+e^{-2i\phi})\ket{1}_{D2}\ket{1}_{D1}}{2} - \frac{i\sqrt{2}(1-e^{-2i\phi})\ket{0}_{D2}\ket{2}_{D1}}{4}$$.

From this, I am getting $$\left = \left = 0$$. where $$I$$ for the detectors are the number operators at those detectors. Can someone please tell me where people are getting the dependence of $$\cos{2\phi}$$ from? Am I wrong conceptually somewhere? Are they measureing some other operator? The case for $$N=2$$ will be fine. Thanks.

PS: The beam splitter transformations are:-

$$\hat{a}_{A}\rightarrow\frac{\hat{a}_C-i\hat{a}_D}{\sqrt{2}}$$ , $$\hat{a}_{B}\rightarrow\frac{-i\hat{a}_C+\hat{a}_D}{\sqrt{2}}$$

and

$$\hat{a}_{C}\rightarrow\frac{\hat{a}_{D2}-i\hat{a}_{D1}}{\sqrt{2}}$$ , $$\hat{a}_{D}\rightarrow\frac{-i\hat{a}_{D2}+\hat{a}_{D1}}{\sqrt{2}}$$

where $$\hat{a}$$ is the general annihilation operator for the number state.

• I haven't read everything carefully, but from your formula you can get a $\cos(2\phi)$ by manipulating the factors $1\pm e^{-2i\phi}$: observe that $1\pm e^{-2i\phi}=e^{-i\phi}(e^{i\phi}\pm e^{-i\phi})$, and you can probably get the conclusion from here – glS Dec 11 '18 at 10:58