<!--
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$$
\def\moy#1{{\langle #1 \rangle}}
\def\abs#1{{\left| #1 \right|}}
\def\dd{\mathrm{d}}
\def\ed{\mathrm{e}}
\def\norm#1{{\parallel #1 \parallel}}
\def\ket#1{{| #1 \rangle}}
\def\bra#1{{\langle #1|}}
\def\braket#1#2{{\langle #1 | #2 \rangle}}
\def\un{\small1\normalsize\kern-.33em1}
$$
# HOM Metrology
### Swap operator
$$
S\ket{\omega_1}\otimes \ket{\omega_2} = \ket{\omega_2}\otimes \ket{\omega_1}
$$
### Coincidence probability $P_c$
$$
P_c = \frac{1}{2}\left(1-\bra{\psi}U^{\dagger}SU\ket{\psi}\right)
$$
where $U = e^{-i\kappa H}$
As $SS= \un$, $P_c$ can also be written as
$$
P_c = \frac{1}{2}\left(1-\bra{\psi}U^{\dagger}SUS S\ket{\psi}\right)
$$
and $U^{\dagger}SUS = e^{i\kappa H}e^{-i\kappa SHS}$.

Therefore
$$
P_c(\kappa) = \frac{1}{2}\left(1-\bra{\psi}e^{i\kappa H}e^{-i\kappa SHS}S\ket{\psi}\right)
$$

### Near $\kappa = 0$, we can write
$$
e^{i\kappa H}e^{-i\kappa SHS} = (\un + i\kappa H - \frac{1}{2}\kappa^2H^2)
(\un - i\kappa SHS - \frac{1}{2}\kappa^2SH^2S) + \cdots \\
= \un + i\kappa(H-SHS) -\frac{\kappa^2}{2}(H^2 + SH^2S -2 HSHS) + \cdots
$$
and thus
:::info
$$
P_c(\kappa) = \frac{1}{2}\left(1-\bra{\psi}S\ket{\psi} +i\kappa\bra{\psi}[H,S]\ket{\psi}
-\frac{\kappa^2}{2}\bra{\psi}(H^2 + SH^2S - 2HSHS)S \ket{\psi}
\right)
$$
:::
now we can consider special cases
* $S\ket{\psi} = \pm \ket{\psi}$
* $[H,S] = 0$

Considering $S\ket{\psi} = \pm \ket{\psi}$ then the linear term in $\kappa$ becomes zero, thus near $\kappa = 0$ we have

$$
P_c(\kappa) = \frac{1}{2}\left(1-\bra{\psi}S\ket{\psi} -
\frac{\kappa^2}{2}\bra{\psi}(H^2 + SH^2S - 2HSHS)S \ket{\psi}
\right)
$$
also if either $S\ket{\psi} = \pm \ket{\psi}$ or $[H,S] = 0$
then $\bra{\psi}H^2 + SH^2S - 2HSHS\ket{\psi} = \pm \bra{\psi}(H-SHS)^2\ket{\psi}$
thus
$$
P_c(\kappa) = \frac{1}{2}\left(1\pm 1 \pm
\frac{\kappa^2}{2}\bra{\psi}(H- SHS)^2 \ket{\psi}
\right)
$$

### For all $\kappa$ (not necessarily near zero) we can write

$$
\frac{d}{d\kappa} e^{i\kappa H}e^{-i\kappa SHS} = i\left(H - e^{i\kappa H} SHS e^{-i\kappa H}\right)e^{i\kappa H}e^{-i\kappa SHS}
$$

## Fisher Information
For a Bernouilly probability distribution $P_c(\kappa)$, $1-P_c(\kappa)$, the Fisher information $I$ is given by
$$
I = \frac{1}{P_c(1-P_c)}\left[\frac{dP}{d\kappa}\right]^2
$$
With $P_c(\kappa)$ given by
$$
P_c(\kappa) \simeq \frac{1}{2}\left(1-\bra{\psi}S\ket{\psi} +i\kappa\bra{\psi}[H,S]\ket{\psi}
-\frac{\kappa^2}{2}\bra{\psi}(H^2 + SH^2S - 2HSHS)S \ket{\psi}
\right)
$$
we obtain at $\kappa=0$
$$
I = \frac{\abs{\moy{[H,S]}}^2}{(1+ \moy{S})(1-\moy{S})}; \text{ if } \moy{S} \neq 0.
$$
Only in the case $\moy{S}=0$, the Fisher information is given by
$$
I = 2 \frac{d^2P_c(\kappa)}{d\kappa^2}; \text{ if } \moy{S}= 0.
$$

## Minimum of $P_c(\kappa)$ :
The position $\kappa_m$ of the minimum of $P_c(\kappa)$ given by
$$
P_c(\kappa) \simeq \frac{1}{2}\left(1-\bra{\psi}S\ket{\psi} +i\kappa\bra{\psi}[H,S]\ket{\psi}
-\frac{\kappa^2}{2}\bra{\psi}(H^2 + SH^2S - 2HSHS)S \ket{\psi}
\right)
$$
is
$$
\kappa_m \simeq \frac{i\moy{[H,S]}}{\moy{H^2S + SH^2 - 2HSHS}}
$$

## Remark about $\moy{[H,S]}$ for a non-factorizable JSA in $\omega_{\pm}$ variables

First,
$$\moy{[H,S]} = -2i\text{Im}(\moy{HS})$$
let's write the state $\ket{\psi}$ as
$$
\ket{\psi} = \int d\omega_1\int d\omega_2 J(\omega_1,\omega_2)a_1^\dagger(\omega_1)a_2^\dagger(\omega_2) \ket{\text{vac}}
$$
which can be written as
$$
\ket{\psi} = \int d\omega_+\int d\omega_- K(\omega_+,\omega_-)
a_1^\dagger(\frac{\omega_+ + \omega_-}{2})
a_2^\dagger(\frac{\omega_+ - \omega_-}{2})
\ket{\text{vac}}
$$
where $\omega_{\pm} = \omega_1\pm\omega_2$, and $K(\omega_+,\omega_-) = J(\frac{\omega_+ + \omega_-}{2},\frac{\omega_+ - \omega_-}{2})$
Now consider that $H=\hat{\omega}\otimes\un$, then after some simple calculations we obtain
$$
\bra{\psi}HS\ket{\psi} = \int d\omega_+
\int d\omega_- \omega_+
K^*(\omega_+,\omega_-)K(\omega_+,-\omega_-) -
\int d\omega_+ \int d\omega_- \omega_-
K^*(\omega_+,\omega_-)K(\omega_+,-\omega_-)
$$
The first term is real and the second term is pure imaginary. This can be see by performing the change of variable $\omega_- \rightarrow -\omega_-$ in the integrals. This cange of variable transform the first term in its conplex congugate and the second term in the opposite of its complex conjugate.

Therefore,
$$
\moy{[H,S]} = -2i\text{Im}(\moy{HS}) = -2i\int d\omega_+ \int d\omega_- \omega_-
K^*(\omega_+,\omega_-)K(\omega_+,-\omega_-)
$$

Writting $K(\omega_+,\omega_-)$ as
$$
K(\omega_+,\omega_-) = A(\omega_+,\omega_-) e^{i\phi(\omega_+,\omega_-)}
$$
where $A(\omega_+,\omega_-) = \abs{K(\omega_+,\omega_-)}$
we have
$$
\moy{[H,S]} =-2i\int d\omega_+ \int d\omega_- \omega_-
A(\omega_+,\omega_-)A(\omega_+,-\omega_-)
e^{i[\phi(\omega_+,\omega_-) - \phi(\omega_+,-\omega_-) ]}
$$
which is in fact
$$
\moy{[H,S]} =-2i\int d\omega_+ \int d\omega_- \omega_-
A(\omega_+,\omega_-)A(\omega_+,-\omega_-)
\sin[\phi(\omega_+,\omega_-) - \phi(\omega_+,-\omega_-)]
$$
because we know that the integral is pure imaginary.
hence,
$$
\phi(\omega_+,\omega_-) - \phi(\omega_+,-\omega_-) = k\pi (k\in \mathbb{Z}) \Rightarrow \moy{[H,S]} =0
$$
For instance, if $\phi(\omega_+,-\omega_-)= \phi(\omega_+)$ does not depend upon $\omega_-$, then $\moy{[H,S]} =0$.

## Remark on the phase of the JSA
Let $J(\omega_1, \omega_2)$ the JSA. The JTA $S(t_1,t_2)$ si defined as the Fourrier transform:
$$
S(t_1,t_2) = \iint d\omega_1 d\omega_2 J(\omega_1, \omega_2) e^{-i\omega_1 t_1}e^{-i\omega_2 t_2}
$$
Performing the change of variable $\omega_{\pm} = \omega_1 \pm \omega_2$ we obtain
$$
S(t_1,t_2) = \iint d\omega_+ d\omega_- K(\omega_+,\omega_-) e^{-i\omega_+ t_+/2}e^{-i\omega_-t_-/2}
$$