APhO 2021, Теория — Задача 2. Ray tracing and generation of entangled light

Theory Q2: Ray tracing and generation of entangled light · 10 т.

Условие

Useful formula:

A⃗×(B⃗×C⃗)=B⃗(A⃗⋅C⃗)−C⃗(A⃗⋅B⃗)\vec{A}\times\left(\vec{B}\times\vec{C}\right)=\vec{B}\left(\vec{A}\cdot\vec{C}\right)-\vec{C}\left(\vec{A}\cdot\vec{B}\right)

Introduction

Let E⃗\vec{E} represent the electric field, H⃗\vec{H} the magnetic field, D⃗\vec{D} the electric displacement, and B⃗\vec{B} the magnetic induction. We have D⃗=ϵ0E⃗+P⃗\vec{D}=\epsilon_0\vec{E}+\vec{P}, with P⃗\vec{P} being the polarization of the medium and ϵ0\epsilon_0 being the permittivity of free space. Only nonmagnetic dielectric media are considered in this problem, hence B⃗=μ0H⃗\vec{B}=\mu_0\vec{H}, with μ0\mu_0 being the permeability of free space. The energy density and energy flow density associated with the electromagnetic field are given by uem=12(E⃗⋅D⃗+H⃗⋅B⃗)u_{em}=\frac{1}{2}(\vec{E}\cdot\vec{D}+\vec{H}\cdot\vec{B}) and Poynting's vector S⃗=E⃗×H⃗\vec{S}=\vec{E}\times\vec{H}, respectively.

In homogeneous dielectric media, a monochromatic plane wave of light can be described by its angular frequency ω\omega, wave vector k⃗\vec{k}, D⃗\vec{D}, and B⃗\vec{B}. According to Maxwell’s equations, we have k⃗×E⃗=ωB⃗\vec{k}\times\vec{E}=\omega\vec{B} and k⃗×H⃗=−ωD⃗\vec{k}\times\vec{H}=-\omega\vec{D}. For such a wave, variations of D⃗\vec{D} and B⃗\vec{B} with position r⃗\vec{r} and time tt are given by sinusoidal functions of the phase (k⃗⋅r⃗−ωt)(\vec{k}\cdot\vec{r}-\omega t).

Part A. Light propagation in isotropic dielectric media (1.0 points)

If the medium is isotropic, we have P⃗=χϵ0E⃗\vec{P}=\chi\epsilon_0\vec{E} and D⃗=ϵE⃗\vec{D}=\epsilon\vec{E}, with χ\chi and ϵ=ϵ0(1+χ)\epsilon=\epsilon_0(1+\chi) being the electric susceptibility and permittivity, respectively, of the medium. For a light wave of angular frequency ω\omega in such a medium, a given phase will propagate in the direction k⃗\vec{k} with a velocity (called phase velocity) vp=c/nv_p=c/n. Here cc is the speed of light in vacuum and nn is the refractive index of the medium. One can also use rays to represent a train of light waves. The propagation of a light ray is characterized by the direction and speed vrv_r of the electromagnetic energy flow.

Consider a plane wave of light with angular frequency ω\omega and wave vector k⃗\vec{k} in a homogeneous isotropic dielectric medium.

A.1 Express its phase velocity vpv_p in terms of ϵ\epsilon and μ0\mu_0. [0,4 т.]

A.2 What is the refractive index nn of the dielectric medium for the wave? [0,2 т.]

A.3 What are the direction S^≡S⃗/S\hat{S}\equiv\vec{S}/S and speed vrv_r of its electromagnetic energy flow? [0,4 т.]

Part B. Light propagation in uniaxial dielectric media (4.8 points)

We now assume the dielectric medium to be uniaxial, i.e, it is electrically anisotropic along a special direction fixed in the medium, called the optic axis, which we presently call it the zz direction. In such a case, the displacement D⃗\vec{D} and the electric field E⃗\vec{E} are related by Dx=ϵExD_x=\epsilon E_x, Dy=ϵEyD_y=\epsilon E_y, and Dz=ϵ′EzD_z=\epsilon' E_z, where xx, yy, and zz axes are mutually orthogonal. Consequently, the phase velocity of a light wave is anisotropic and depends additionally on the directions of k⃗\vec{k} and D⃗\vec{D}. Let no=cμ0ϵn_o=c\sqrt{\mu_0\epsilon} and ne=cμ0ϵ′n_e=c\sqrt{\mu_0\epsilon'}, answer the followings questions: B.1, B.2, and B.3.

B.1 Suppose the wave vector k⃗\vec{k} of a monochromatic plane light wave is in the xzxz plane so that k⃗=k(sin⁡θ,0,cos⁡θ)\vec{k}=k(\sin\theta,0,\cos\theta). At each angle θ\theta, what directions of D⃗\vec{D} and B⃗\vec{B} are permissible for the light wave? Find all possible refractive indices and express the refractive indices in terms of θ\theta, non_o, and nen_e. Find the angle θ\theta for which only one value is permitted for the refractive index. [1,5 т.]

B.2 The polarization of a light wave, i.e., the direction of its electric field E⃗\vec{E}, can be either perpendicular (called an ordinary wave or ray ) or parallel (called an extraordinary wave or ray) to the xzxz plane. For each of the light waves you found in B.1, specify its polarization as a unit vector and indicate whether it is an ordinary or extraordinary wave. Also compute tan⁡α\tan\alpha, where α\alpha is the angle between E⃗\vec{E} and D⃗\vec{D} (α\alpha is positive when going from E⃗\vec{E} to D⃗\vec{D} is clockwise). [0,8 т.]

B.3 Extend the results in B.1 and B.2 to the general case when the angle between k⃗\vec{k} and the positive zz direction is still θ\theta, but k⃗\vec{k} is not in the xzxz plane. Find all possible values of the refractive indices and the corresponding polarizations. [0,6 т.]

In a uniaxial medium, the direction of k⃗\vec{k} of a light wave may differ from the direction of the light ray. The phase velocity of the wave is still given by c/nc/n with nn being the refractive index along k⃗\vec{k}, while the ray velocity is defined jointly by the direction and the rate of energy flow.

B.4 Following problems B.1-3, consider a light wave with k⃗=k(sin⁡θ,0,cos⁡θ)\vec{k}=k(\sin\theta,0,\cos\theta). Let the angle between k^≡k⃗/k\hat{k}\equiv\vec{k}/k and the direction of the ray, S^\hat{S}, be αr\alpha_r (αr\alpha_r is positive when going from S^\hat{S} to k^\hat{k} is clockwise). Find all possible values of tan⁡αr\tan\alpha_r, speed vrv_r of the ray and S^\hat{S}. Using these results, express the ray index ns=c/vrn_s=c/v_r in terms of S^\hat{S}, x^\hat{x}, z^\hat{z}, non_o, and nen_e. [0,8 т.]

Consider the propagation of a light ray from A to B through an interface between an isotropic medium, labelled 1, and an anisotropic medium, labelled 2, as shown in Fig. 1. The interface coincides with the yzyz plane, while the plane of incidence is the xzxz plane. Let the angle of incidence be θ1\theta_1. The refractive index of medium 1 is nn, while the refractive indices of medium 2 for axes z2z_2, y2y_2, x2x_2 are nen_e, non_o, and non_o, respectively. Here y2y_2 axis coincides with yy axis. Fermat’s principle states that the propagation time for the path that the light ray goes from A to B is a minimum. For light with polarization parallel to xzxz plane and incident at the angle θ1\theta_1, Fermat’s principle leads to the following equation:

Aˉ(tan⁡θ2)2+Bˉtan⁡θ2+Cˉ=0(1)\bar{A}(\tan\theta_2)^2+\bar{B}\tan\theta_2+\bar{C}=0\qquad(1)

B.5 Find Aˉ\bar{A}, Bˉ\bar{B}, and Cˉ\bar{C} in terms of P1P_1, P2P_2, P3P_3, and nsin⁡θ1n\sin\theta_1, where P1=no2cos⁡2ϕ+ne2sin⁡2ϕP_1=n_o^2\cos^2\phi+n_e^2\sin^2\phi, P2=no2sin⁡2ϕ+ne2cos⁡2ϕP_2=n_o^2\sin^2\phi+n_e^2\cos^2\phi, and P3=(no2−ne2)sin⁡ϕcos⁡ϕP_3=(n_o^2-n_e^2)\sin\phi\cos\phi. From Eq. (1), find corresponding tan⁡θ2\tan\theta_2 to two special orientations: ϕ=0\phi=0 and ϕ=π/2\phi=\pi/2. [1,1 т.]

Interface diagram: a horizontal z axis and a vertical x axis pointing down, with the y axis (circled cross) into the page. Medium 1 is above the interface, medium 2 below. A ray comes from A at angle θ1 to the vertical and continues to B at angle θ2 to the x axis. Dashed axes z2 and x2 of medium 2 are rotated by the angle ϕ from z.
Fig. 1: Propagation of light from A to B through an interface between an isotropic medium 1 and an anisotropic medium 2.

Part C. Entanglement of light (4.2 points)

In a nonlinear medium, the electric field E⃗\vec{E} is related to the polarization P⃗\vec{P} by Pi=(ϵ−ϵ0)Ei+∑j∑kχijk(2)EjEkP_i=(\epsilon-\epsilon_0)E_i+\sum_j\sum_k\chi_{ijk}^{(2)}E_jE_k. Here ii, jj, kk each can be any one of the three components xx, yy, zz, and χijk(2)\chi_{ijk}^{(2)} are constants representing the second-order nonlinear susceptibility of the medium. Non-vanishing of χijk(2)\chi_{ijk}^{(2)} implies that as a light wave travels through a nonlinear medium, it can split into two light waves.

Suppose that because χijk(2)\chi_{ijk}^{(2)} are not all zero, the electric field in the medium is made up of a superposition of three plane waves of angular frequencies ω\omega, ω1\omega_1, and ω2\omega_2, propagating with wave vectors k⃗\vec{k}, k⃗1\vec{k}_1, and k⃗2\vec{k}_2, respectively. Assume ω≥ω2\omega\geq\omega_2 and ω1≥ω2\omega_1\geq\omega_2.

C.1 Find all possible relations (known as the phase matching conditions) between these angular frequencies and wave vectors. Viewing light as composed of photons, what kinds of conservation laws do these conditions imply for the three photons involved? Write down equations expressing these conservation laws for the case that a photon with angular frequency ω\omega and wave vector k⃗\vec{k} being split into two photons of angular frequencies ω1\omega_1and ω2\omega_2, propagating with wave vectors k⃗1\vec{k}_1and k⃗2\vec{k}_2, respectively. [0,8 т.]

C.2 Consider a light wave in a uniaxial medium. Denote an ordinary ray as o and an extraordinary ray as e. There are 8 possible ways of splitting for the light wave: o → o + o, o → e + o, o → o + e, o → e + e, e → o + o, e → e + o, e → o + e, and e → e + e. Assume that the refractive indices non_o and nen_e are both increasing functions of ω\omega. Using the same notations for wave vectors as in problem C.1 and considering the case that k⃗\vec{k}, k⃗1\vec{k}_1, and k⃗2\vec{k}_2 are collinear, indicate which of the 8 ways of splitting are not possible. [0,8 т.]

Consider an incoming e ray traveling along z′z' direction with wave vector k⃗\vec{k} and ω=Ωp\omega=\Omega_p in an uniaxial medium with refractive index ne<non_e<n_o. Suppose that, in a collinear splitting e → e + o, the phase-matching conditions are realized with k1=Kek_1=K_e, ω1=Ωe\omega_1=\Omega_e, k2=Kok_2=K_o, and ω2=Ωo\omega_2=\Omega_o. Here subscripts 1 and 2 refer to e ray and o ray. k⃗1\vec{k}_1, k⃗2\vec{k}_2 and k⃗\vec{k} all point in the z′z'direction. As shown in Fig. 2(a), the optic axis (OA) of the medium lies in the x′z′x'z' plane and makes an angle θ<π/2\theta<\pi/2 with the z′z' axis. Therefore, nen_e is a function of ω\omega and θ\theta, i.e., ne=ne(ω,θ)n_e=n_e(\omega,\theta). For the same incoming e ray with wave vector k⃗\vec{k} and ω=Ωp\omega=\Omega_p, suppose its non-collinear splitting into e + o rays causes the latter two rays to separate but remain on two cones with ω1=ω2=Ω\omega_1=\omega_2=\Omega, k1=k2k_1=k_2, as shown in Fig. 2(b). Note that in the collinear splitting, Ωe\Omega_e is already close to Ωo\Omega_o, and here Ω\Omega is only slightly less than Ωe\Omega_e. In a plane perpendicular to k⃗\vec{k}, two circles on the cones for k⃗1\vec{k}_1 and k⃗2\vec{k}_2 intersect at points aa and bb with the line ab‾\overline{ab} parallel to y′y'axis. As shown in Fig. 2(a), k⃗α(α=1,2 )\vec{k}_\alpha(\alpha=1,2\ ) makes an angle θα\theta_\alpha with the optic axis and has angular coordinates (ψα,ϕα)(\psi_\alpha,\phi_\alpha) with k⃗α⊥\vec{k}_{\alpha\perp} being its projection in the x′y′x'y' plane. Each vector k⃗α\vec{k}_\alpha deviates from z′z' axis only slightly so that ∣(Ω−Ωe)/Ωe∣≪1|(\Omega-\Omega_e)/\Omega_e|\ll1, ∣k⃗α⊥∣/kα≪1|\vec{k}_{\alpha\perp}|/k_\alpha\ll1 and ∣θα−θ∣≪1|\theta_\alpha-\theta|\ll1. Using approximations which agree with the z′z' component of k⃗α\vec{k}_\alpha to terms of the order kα⊥2k_{\alpha\perp}^2 and the angle θα\theta_\alpha to (θα−θ)2(\theta_\alpha-\theta)^2, one finds that k⃗2⊥=(qx′,qy′)\vec{k}_{2\perp}=(q_{x'},q_{y'}) must satisfy M(qx′+N)2+Mqy′2=LM(q_{x'}+N)^2+Mq_{y'}^2=L.

C.3 Let M>0M>0. Evaluate MM, NN, and LL in terms of Ω\Omega, Ωe\Omega_e, Ωo\Omega_o, KeK_e, KoK_o and Ne(ω,θ)=1ne(ω,θ)dne(ω,θ)dθN_e(\omega,\theta)=\frac{1}{n_e(\omega,\theta)}\frac{dn_e(\omega,\theta)}{d\theta} and the group velocities uo=dw2dk2u_o=\frac{dw_2}{dk_2} and ue=dω1dk1u_e=\frac{d\omega_1}{dk_1} for the o and e rays. Estimate the angle between the axis of the cone and z′z', and also the angle of the cone in terms of LL, MM, NN and KoK_o. [1,3 т.]

Two panels. (a) Axes x', y', z' with the optic axis OA at angle θ from z'; vector k_α at angle ψ_α from z' and θ_α from OA; its projection k_α⊥ in the x'y' plane at angle ϕ_α. (b) Axes x', y', z' with OA at angle θ; the incoming k along z' splits into k1 and k2 lying on two cones (dashed lines), whose circles intersect at points a and b.
Fig. 2: (a) Vector k⃗α\vec{k}_\alpha has angular coordinates (ψα,ϕα)(\psi_\alpha,\phi_\alpha) in the x′y′z′x'y'z' coordinate system with k⃗α⊥\vec{k}_{\alpha\perp} being its projection in the x′y′x'y' plane. Note that k⃗α\vec{k}_\alpha makes an angle θα\theta_\alpha with OA. (b) Non-collinear splitting of an e ray into e + o rays that form two cones. Line ab‾\overline{ab} is parallel to the y′y' axis.

Problem C.3 shows that a photon may split into two photons which when passing through points a and b are polarized in perpendicular directions. These two photons are called entangled photon pair because if one photon that passes aa (called aa-photon) is polarized in a direction x^′\hat{x}', the other that passes bb (called bb-photon) will be polarized in the direction y^′⊥x^′\hat{y}'\perp\hat{x}', and if the aa-photon is polarized in y^′\hat{y}', then the bb-photon will be polarized in x^′\hat{x}'. The entangled photon-pair state can be prepared experimentally. It is a superposition of the above two alternative states and can be expressed as 12(∣x^a′⟩∣y^b′⟩+∣y^a′⟩∣x^b′⟩)\frac{1}{\sqrt{2}}(|\hat{x}'_a\rangle|\hat{y}'_b\rangle+|\hat{y}'_a\rangle|\hat{x}'_b\rangle). Here ∣x^a′⟩∣y^b′⟩|\hat{x}'_a\rangle|\hat{y}'_b\ranglerepresents the state when aa-photon is polarized in x^′\hat{x}' direction and bb-photon is polarized in y^′\hat{y}' direction; similar meaning applies to ∣y^a′⟩∣x^b′⟩|\hat{y}'_a\rangle|\hat{x}'_b\rangle. The coefficient 1/21/\sqrt{2} can be viewed as the product of electric field amplitudes (expressed in suitable units) of aa- and bb-photons. As illustrated in Fig. 3, two linear polarizers 1 and 2 have transmission axes at angles α\alpha and β\beta respectively with respect to x^′\hat{x}'. We may use them to perform coincidence measurement on the two photons that pass aa and bb. Let the probability of simultaneously finding two photons passing through polarizers 1 and 2 be P(α,β)P(\alpha,\beta). Alternatively, P(α,β)P(\alpha,\beta) can also be regarded as being proportional to the product of intensities (after appropriate superpositions) of light passing through the two polarizers. Denote α+π/2\alpha+\pi/2 and β+π/2\beta+\pi/2 by α⊥\alpha_\perp and β⊥\beta_\perp respectively.

Two sets of axes x' (up) and y' (right). Left: Polarizer 1 at point a, transmission axis at angle α from x'. Right: Polarizer 2 at point b, transmission axis at angle β from x'.
Fig. 3: Two linear polarizers 1 and 2 for coincidence measurement of photons that pass aa and bb.

C.4 Consider the total electric field projected by linear polarizers. Find the probabilities P(α,β)P(\alpha,\beta), P(α,β⊥)P(\alpha,\beta_\perp), P(α⊥,β)P(\alpha_\perp,\beta), and P(α⊥,β⊥)P(\alpha_\perp,\beta_\perp). [0,8 т.]

C.5 Assign σa=1\sigma_a=1 when polarizer 1 with angle α\alpha finds an aa-photon and σa=−1\sigma_a=-1 when polarizer 1 with angle α⊥\alpha_\perp finds an aa-photon. Similarly, σβ=1\sigma_\beta=1 or −1-1 is assigned when polarizer 2 with angle β\beta or β⊥\beta_\perp finds a bb-photon. If E(α,β)E(\alpha,\beta) denotes the average of σaσb\sigma_a\sigma_b, the quantity S=∣E(α,β)−E(α,β′)∣+∣E(α′,β)+E(α′,β′)∣S=|E(\alpha,\beta)-E(\alpha,\beta')|+|E(\alpha',\beta)+E(\alpha',\beta')| has important meaning. For classical theories of light, S≤2S\leq2. This is a variant form of Bell’s inequality (the Clauser-Horne-Shimony-Holt inequality). Find the expression of SS and evaluate SS for the case α=π4\alpha=\frac{\pi}{4},α′=0\alpha'=0 , β=−π8\beta=-\frac{\pi}{8}, β′=π8\beta'=\frac{\pi}{8}. Indicate if SS is consistent with the classical theories. [0,5 т.]

Отговори

Решение


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