From 2b3a58d326c03410e84d91e09e7e6f20bc375276 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Arturo=20Mena=20L=C3=B3pez?= Date: Tue, 24 Oct 2023 18:14:29 +0200 Subject: [PATCH] Update 2023-09-13-the-hong-ou-mandel-effect.md --- _posts/2023-09-13-the-hong-ou-mandel-effect.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/_posts/2023-09-13-the-hong-ou-mandel-effect.md b/_posts/2023-09-13-the-hong-ou-mandel-effect.md index b547010d9e1b..775955aa8410 100644 --- a/_posts/2023-09-13-the-hong-ou-mandel-effect.md +++ b/_posts/2023-09-13-the-hong-ou-mandel-effect.md @@ -147,7 +147,7 @@ where we used that our chosen $B$ matrix satisfies $B=B^{-1}$. Therefore, if we $$ \begin{array}{ll} -\vert\Psi\rangle_{in} \!\!\!\! & = \vert 0\rangle_1 \vert 0\rangle_2 \\[1ex] +\vert\Psi\rangle_{in} \!\!\!\! & = \vert 1\rangle_1 \vert 1\rangle_2 \\[1ex] &= a^\dagger_1 a^\dagger_2 \vert 0\rangle_1 \vert 0\rangle_2 \\[2ex] \vert\Psi\rangle_{out} \!\!\!\! &= \frac{1}{2}(a^\dagger_3 + a^\dagger_4)(a^\dagger_3 - a^\dagger_4) \vert 0\rangle_3 \vert 0\rangle_4\\[1ex] &= \frac{1}{2}(a^{\dagger 2}_3 + a^\dagger_4 a^\dagger_3 - a^\dagger_3a^\dagger_4 - a^{\dagger 2}_4) \vert 0\rangle_3 \vert 0\rangle_4\\[1ex]