Corrections SET

--- tags: Calculs new SET ---  $$ \def\moy#1{{\langle #1 \rangle}} \def\abs#1{{\left| #1 \right|}} \def\dd{\mathrm{d}} \def\ed{\mathrm{e}} \def\norm#1{{\parallel #1 \parallel}} \def\ket#1{{| #1 \rangle}} \def\bra#1{{\langle #1|}} $$ # Corrections SET J'ai refait les calculs, voici ce que je trouve, en précisant les définitions de certains des opérateurs en essayant d'être cohérent avec la version soumise de l'article. D'aileurs, je pense que l'Eq. 3 de l'article, qui définit l'opérateur déplacement n'est pas correcte. Si on veut déplacer $a \rightarrow a+\alpha$ telle que $D^\dagger a D = a + \alpha$ il faut que $D$ soit défini comme $D = e^{\alpha a^\dagger -h.c}$. ## Définitions ### $S$ and $U$ $$ S = \gamma \iint dk dk'L(k,k')a_s^\dagger a_i^\dagger; \quad U = e^{(\hat{S} - h.c)} $$ ### Schmidt décomposition $$ L(k,k') = \sum_n \sqrt{\lambda_n} \psi_n(k)\phi_n(k') $$ with $\psi_n$ and $\phi_n$ solutions of $$ \int_k' K_1(k,k') \psi_n(k') dk' = \lambda_n\psi_n(k); \quad \int_k' K_2(k,k') \phi_n(k') dk' = \lambda_n\phi_n(k) $$ with $K_1$ and $K_2$ given by $$ K_1(k,k') = \int du L(k,u) L^*(k',u);\quad K_2(k,k') = \int du L(u,k) L^*(u,k') $$ #### Remarks : * $K_i$ are hermitian kernels $$ K_i(k,k') =K_i^*(k',k) $$ * If $L(k',k) = e^{i\phi} L(k,k')$ then $K_1(k,k') = K_2(k,k')$ hence $\phi_n(k)=\psi_n(k)$ * $K_1(k,k') = \sum_n \lambda_n\psi_n(k)\psi_n^*(k')$ * $K_2(k,k') = \sum_n \lambda_n\phi_n(k)\phi_n^*(k')$ ### Schmidt modes $$ b^\dagger_n = \int dk \psi_n(k) a_s^\dagger(k); \quad c_n^\dagger = \int dk \phi_n(k) a_i^\dagger(k); $$ or $$ a_s^\dagger(k) = \sum_n \psi_n^*(k) b_n^\dagger; \quad a_i^\dagger(k) = \sum_n \phi_n^*(k) c_n^\dagger $$ ### U as a function of Schmidt modes $$ U = \Pi_n \exp\left[\sqrt{\lambda_n}(\gamma b_n^\dagger c_n^\dagger - h.c)\right] $$ ### without seed $$ U^{-1} a_s(k_s) U = \sum_n \psi_n(k_s) \left[ \cosh(\abs{\gamma}\sqrt{\lambda_n})b_n + e^{i\theta}\sinh(\abs{\gamma}\sqrt{\lambda_n})c_n^\dagger \right] $$ where $\gamma = \abs{\gamma}e^{i\theta}$. Hence $$ \bra{\text{vac}}U^{-1} a^\dagger_n(k_s)a_s(k_s) U\ket{\text{vac}} = \sum_n \abs{\psi_n(k_s)}^2 \sinh^2(\abs{\gamma}\sqrt{\lambda_n}) $$ ### With seed #### Displacement operator $D = D(\{\beta(k)\})$ $$ D = \exp\left(\int \beta(k) a_i^\dagger(k) dk - h.c.\right) = \exp(\abs{\beta} s^\dagger -h.c) $$ where we have define $s^\dagger$ as : $$ s^\dagger = \int \frac{\beta(k)}{\abs{\beta}}a_i^\dagger(k) dk \text{ with } \abs{\beta}^2 = \int \abs{\beta(k)}^2 dk $$ #### $s^\dagger$ as a function of Shmidt modes : $$ s^\dagger = \sum_n \beta_n c_n^\dagger, \text{ with } \beta_n = \int \phi_n^*(k) \frac{\beta(k)}{\abs{\beta}} dk $$ Thus $$ D = \Pi_n \exp\left[ \abs{\beta}(\beta_nc_n^\dagger -h.c.) \right] $$ The displacement of $a_i(k)$ si given by $$ D^\dagger a_i(k) D = a_i(k) + \beta(k) = \sum_n \phi_n(k)(c_n + \abs{\beta}\beta_n) $$ In other words: $$ D^\dagger c_n D = c_n + \abs{\beta}\beta_n $$ #### Displacement of $U^\dagger a_s U$ $$ D^\dagger U^\dagger a_s(k_s) U D = \sum_n \psi_n(k_s) \left[ \cosh(\abs{\gamma}\sqrt{\lambda_n})b_n + e^{i\theta}(c_n^\dagger + \abs{\beta}\beta_n^*) \right] $$ which gives $$ \bra{\text{vac}} D\dagger U^\dagger a_s^\dagger(k_s) a_s(k_s) U D\ket{\text{vac}} = \sum_n\abs{\psi_n(k_s)}^2 \sinh^2(\abs{\gamma}\sqrt{\lambda_n}) + \\ \abs{\beta}^2\abs{\sum_n \psi_n(k_s)\beta_n^*\sinh(\abs{\gamma}\sqrt{\lambda_n})}^2 $$ in the limit of $\beta \gg 1$ and using $\beta_n = \int \phi_n^*(k) \frac{\beta(k)}{\abs{\beta}} dk$ we can write $$ I(k_s) = \abs{\beta}^2 \left|\sum_n \psi_n(k_s)\beta_n^*\sinh(\abs{\gamma}\sqrt{\lambda_n})\right|^2 $$ as $$ I(k_s) = \ \abs{\int dk \sum_n \psi_n(k_s)\phi_n(k) \beta^*(k)\sinh(\abs{\gamma}\sqrt{\lambda_n})}^2 $$ #### Limit as $\gamma \ll 1$ $$ I(k_s) = \abs{\gamma}^2 \abs{\int dk \sum_n \sqrt{\lambda_n}\psi_n(k_s)\phi_n(k_s)\beta^*(k)} ^2 $$ We recognize $L(k_s,k) = \sum_n \sqrt{\lambda_n}\psi_n(k_s)\phi_n(k_s)$, therefore $$ I(k_s) = \abs{\gamma}^2 \abs{\int L(k_s,k) \beta^*(k)dk }^2 $$ --- ## Interferometric signal $$ I = \int dk \moy{a_s^\dagger(k)a_i(k) + c.c} $$ where the expectation si taken over $UD\ket{\text{vac}}$. Using $$ D^\dagger U^\dagger a^\dagger_s(k)^\dagger UD = \sum_n \psi_n^*(k)\left[\cosh(\abs{\gamma\sqrt{\lambda_n}})b_n^\dagger +e^{-i\theta}\sinh(\abs{\gamma}\sqrt{\lambda_n})(c_n + \abs{\beta}\beta_n)\right] $$ and $$ D^\dagger U^\dagger a_i(k)^\dagger UD = \sum_n \phi_n(k)\left[\cosh(\abs{\gamma\sqrt{\lambda_n}})(c_n + \abs{\beta}\beta_n) +e^{i\theta}\sinh(\abs{\gamma}\sqrt{\lambda_n}) b_n^\dagger\right] $$ the only non zero term that remains after taking the expectation over $\ket{\text{vac}}$ is $$ I = 2\abs{\beta}^2\text{Re}\left\{e^{-i\theta}\sum_{nn'}\int \psi_n^*(k)\phi_{n'}(k) dk \beta_n\beta_{n'}\sinh(\abs{\gamma}\sqrt{\lambda_n})\cosh(\abs{\gamma\sqrt{\lambda_{n'}}})\right\} $$ ## In the limit $\abs{\gamma}\sqrt{\lambda_n} \ll 1$ $\sinh(\abs{\gamma}\sqrt{\lambda_n}) \simeq \abs{\gamma}\sqrt{\lambda_n}$ and $\cosh(\abs{\gamma\sqrt{\lambda_{n'}}})\simeq 1$. And using $\beta_n = \int \phi_n^*(k) \frac{\beta(k)}{\abs{\beta}} dk$ we obtain $$ I \simeq 2\abs{\gamma}\text{Re}\left\{e^{-i\theta}\iiint dk dk' dk'' \sum_n \sqrt{\lambda_n}\psi_n^*(k)\phi_n^*(k')\beta(k')\sum_n\phi_{n'}^*(k'')\phi_{n'}(k)\beta(k'') \right\} $$ We reconize $\sum_n \sqrt{\lambda_n}\psi_n^*(k)\phi_n^*(k') = L^*(k,k')$ and using $\sum_n\phi_{n'}^*(k'')\phi_{n'}(k)=\delta(k-k'')$ we finaly obtain $$ I \simeq 2\abs{\gamma}\text{Re}\left\{e^{-i\theta} \iint dk dk' L^*(k,k') \beta(k)\beta(k') \right\} $$ which can also be written as $$ I \simeq 2\abs{\gamma}\text{Re}\left\{e^{-i\theta} \iint dk dk' L(k,k') \beta^*(k)\beta^*(k') \right\} $$ which is almost the result of Perola, but with $\beta^*$ instead of $\beta$. --- # Remarques SET Je reviens sur $$ I(k_s) = \abs{\beta}^2 \left|\sum_n \psi_n(k_s)\beta_n^*\sinh(\abs{\gamma}\sqrt{\lambda_n})\right|^2 $$ $$ I(k_s) = \abs{\gamma}^2 \left|\int dk \sum_{\lambda_n\neq 0} \sqrt{\lambda_n} \psi_n(k_s)\phi_n(k)\beta^*(k)\frac{\sinh(\abs{\gamma}\sqrt{\lambda_n})} {\abs{\gamma}\sqrt{\lambda_n}} \right|^2 $$ $$ I(k_s) = \abs{\gamma}^2 \left| J_0(k_s) + J_1(k_s) \right|^2 $$ avec $$ J_0(k_s) = \int L(k_s,k)\beta^*(k) dk $$ et $$ J_1(k_s) = \int dk \sum_{\lambda_n\neq 0} \sqrt{\lambda_n} \psi_n(k_s)\phi_n(k)\beta^*(k) \left[\frac{\sinh(\abs{\gamma}\sqrt{\lambda_n})} {\abs{\gamma}\sqrt{\lambda_n}} - 1\right] $$ On peut imaginer plusieurs cas limites : ## Cas de grande intrication et faible $\abs{\gamma}$ Supposons maintenant que $\lambda_n = \lambda$ pour $n=0,1,\cdots N-1$ et $\lambda_n =0$ si $n\geq N$ Comme $\sum_n \lambda_n = N$ alors $\lambda=\frac{1}{N}$. $J_1(k_s)$ devient $$ J_1(k_s) = \left[\frac{\sinh(\abs{\gamma}\sqrt{\lambda})} {\abs{\gamma}\sqrt{\lambda}} - 1\right] \int dk\sum_{\lambda_n\neq 0}\sqrt{\lambda} \psi_n(k_s)\phi_n(k)\beta^*(k) $$ et donc $$ J_1(k_s) = \left[\frac{\sinh(\abs{\gamma}\sqrt{\lambda})} {\abs{\gamma}\sqrt{\lambda}} - 1\right] \int L(k_s,k)\beta^*(k) dk $$ Dans la limite où $N \gg \abs{\gamma}^2$, on obtient $$ J_1(k_s) = \frac{\abs{\gamma}^2}{N} \int L(k_s,k)\beta^*(k) dk $$ On rappelle que $\abs{\gamma}^2$ représente le nombre de paire de photons générées par impoulsion de pompe et $N$ est ici le nombre de Schmidt qui caractérise l'intrication. Pour un $\abs{\gamma}$ fixé, plus $L$ est non séprable et plus $J_1$ pourra être négligé. ## $\abs{\gamma}$ très grand Supposons qu'un des $\lambda_n$, que l'on note $\lambda$ dommine un peu les autres, et que $\abs{\gamma}\gg 1$ alors on pourra faire l'approximation suivante $$ \frac{\sinh(\abs{\gamma}\sqrt{\lambda_n})} {\abs{\gamma}\sqrt{\lambda_n}} \simeq \frac{e^{\abs{\gamma}\sqrt{\lambda}}}{2\abs{\gamma}\sqrt{\lambda}} $$ où $\lambda = \text{max}[\lambda_n$] on pourra donc écrire que $$ J_1(k_s) \simeq \left[\frac{e^{\abs{\gamma}\sqrt{\lambda}}}{2\abs{\gamma}\sqrt{\lambda}} -1\right] \sqrt{\lambda} \int \psi_\lambda(k_s)\phi_\lambda(k)\beta^*(k) dk $$