From 71129b9e3baa075dba4f009009bdd395ba74f0f1 Mon Sep 17 00:00:00 2001 From: Matias Bundgaard-Nielsen <56378038+mabuni1998@users.noreply.github.com> Date: Thu, 9 Jan 2025 20:20:59 +0100 Subject: [PATCH] small docs fixes --- docs/src/beamsplitter.md | 2 +- docs/src/multiplewaveguides_v2.md | 2 ++ docs/src/tutorial.md | 4 ++-- 3 files changed, 5 insertions(+), 3 deletions(-) diff --git a/docs/src/beamsplitter.md b/docs/src/beamsplitter.md index 3d337e4..866df66 100644 --- a/docs/src/beamsplitter.md +++ b/docs/src/beamsplitter.md @@ -123,7 +123,7 @@ nothing #hide ``` ![hom_plot](hom.svg) -We could also have created this plot by performing the beamsplitter operation by hand and instead initializing the state directly in this state. The initial state we consider is a single photon in each waveguide: $$|\psi \rangle = \int \int dt_1 dt_2 \xi_1(t_1) \xi_2(t_2) w_1^\dagger(t_1) w_2^\dagger(t_2) \ket{\emptyset}$$, where $$\xi_1(t_1)$$ and $$\xi_2(t_2)$$ denote the wavefunction of the photon in waveguide 1 and 2, respectively. Notice that there is no factor of $$1/\sqrt(2)$$ in front of the initial state as the two photons occupy different waveguides. If they initially occupied the same waveguide, we would need a factor of $$1/\sqrt(2)$$ for the state to be normalized. Performing the beamsplitter operation $w_1(t) \rightarrow 1/\sqrt(2) ( w_1(t) - i w_2(t))$ and $w_2(t) \rightarrow 1/\sqrt(2) ( - i w_1(t) + w_2(t))$, we arrive at the transformed state: +We could also have created this plot by performing the beamsplitter operation by hand and instead initializing the state directly in this state. The initial state we consider is a single photon in each waveguide: $$|\psi \rangle = \int \int dt_1 dt_2 \xi_1(t_1) \xi_2(t_2) w_1^\dagger(t_1) w_2^\dagger(t_2) \ket{\emptyset}$$, where $$\xi_1(t_1)$$ and $$\xi_2(t_2)$$ denote the wavefunction of the photon in waveguide 1 and 2, respectively. Notice that there is no factor of $$1/\sqrt{2}$$ in front of the initial state as the two photons occupy different waveguides. If they initially occupied the same waveguide, we would need a factor of $$1/\sqrt{2}$$ for the state to be normalized. Performing the beamsplitter operation $w_1(t) \rightarrow 1/\sqrt{2} ( w_1(t) - i w_2(t))$ and $w_2(t) \rightarrow 1/\sqrt{2} ( - i w_1(t) + w_2(t))$, we arrive at the transformed state: $$\begin{align*} diff --git a/docs/src/multiplewaveguides_v2.md b/docs/src/multiplewaveguides_v2.md index f2ba1d6..3dffa54 100644 --- a/docs/src/multiplewaveguides_v2.md +++ b/docs/src/multiplewaveguides_v2.md @@ -76,6 +76,8 @@ We can even describe a simultaneous excitation in both waveguide states like $\k nothing #hide ``` +Note the absence of $$1/\sqrt{2}$$ when defining a state with one photon in each waveguide. + ## [Scattering on two-level system](@id lodahl) As an example of a system with multiple waveguide modes, we consider a two-level system coupled with two directional waveguide modes; a left and a right propagating mode. A sketch of the system can be seen below here: diff --git a/docs/src/tutorial.md b/docs/src/tutorial.md index 0aa2d90..5f6638e 100644 --- a/docs/src/tutorial.md +++ b/docs/src/tutorial.md @@ -150,7 +150,7 @@ H_twophoton = im*sqrt(γ/dt)*( ad ⊗ w_twophoton - a ⊗ wd_twophoton ) nothing #hide ``` -If we want an initial two-photon state, we instead use the function [`twophoton`](@ref) to create a two-photon state $\frac{1}{\sqrt{2}}\left[W^\dagger(\xi)\right]^2|0\rangle = \frac{1}{\sqrt{2}} \int_{t_0}^{t_{end}} d t^{\prime} \int_{t_0}^{t_{end}} d t \ \xi^{(2)}(t,t') w^\dagger(t) w^\dagger\left(t^{\prime}\right)|0\rangle $ (see [Theoretical Background](@ref theory) for details). In the following, we define the two-photon wavefunction $\xi^{(2)}(t,t') = \xi^{(1)}(t)\xi^{(1)}(t')$ which is thus a product state of two single-photons. +If we want an initial two-photon state, we instead use the function [`twophoton`](@ref) to create a two-photon state $\frac{1}{\sqrt{2}}\left[W^\dagger(\xi)\right]^2|0\rangle = \frac{1}{\sqrt{2}} \int_{t_0}^{t_{end}} d t^{\prime} \int_{t_0}^{t_{end}} d t \ \xi^{(2)}(t,t') w^\dagger(t) w^\dagger\left(t^{\prime}\right)|0\rangle $ (see [Theory](@ref theory) for details). In the following, we define the two-photon wavefunction $\xi^{(2)}(t,t') = \xi^{(1)}(t)\xi^{(1)}(t')$ which is thus a product state of two single-photons. ```@example tutorial ξ2(t1,t2,σ,t0) = ξ(t1,σ,t0)*ξ(t2,σ,t0) @@ -160,7 +160,7 @@ If we want an initial two-photon state, we instead use the function [`twophoton` nothing #hide ``` -Notice the structure of `ξ2(t1,t2,σ,t0)`, it now has two time-arguments and the remaining arguments are parameters. If we wanted to allow for two different widths of the single-photon states in the product state, we could have also defined: `ξ2(t1,t2,σ1,σ2,t0) = ξ(t1,σ1,t0)*ξ(t2,σ2,t0)`. Another important detail is the normalization. [`twophoton`](@ref) only creates $$\int_{t_0}^{t_{end}} d t^{\prime} \int_{t_0}^{t_{end}} d t \ \xi^{(2)}(t,t') w^\dagger(t) w^\dagger\left(t^{\prime}\right)|0\rangle $$ and we thus need the factor of $$1/\sqrt(2)$$ for the state to be normalized. +Notice the structure of `ξ2(t1,t2,σ,t0)`, it now has two time-arguments and the remaining arguments are parameters. If we wanted to allow for two different widths of the single-photon states in the product state, we could have also defined: `ξ2(t1,t2,σ1,σ2,t0) = ξ(t1,σ1,t0)*ξ(t2,σ2,t0)`. Another important detail is the normalization. [`twophoton`](@ref) only creates $$\int_{t_0}^{t_{end}} d t^{\prime} \int_{t_0}^{t_{end}} d t \ \xi^{(2)}(t,t') w^\dagger(t) w^\dagger \left( t^{\prime} \right) | 0 \rangle$$ and we thus need the factor of $$1/\sqrt{2}$$ for the state to be normalized. In the following, we consider the more simple case of equivalent photons. We solve the two-photon scattering in the following.