Light as a Probability Wave — Question 5

Section 38-4 · Physics pages 6320–6321

Question 5 A random phase mixture differs from a stable dark-port state A balanced interferometer has P+(ϕ) = (1 + cos ϕ)/2. In ensemble A every photon has ϕ = π. In ensemble B the phase is independently uniform on [0, 2π) for each photon. Both ensembles contain exactly N photons. ϕ P+ 1/2 Schematic; labels, not drawing lengths, specify the data. (a) Find the mean plus-port count in each ensemble. (b) For ensemble B, find the count variance and explain why averaging amplitudes would give the wrong rule.
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Question 5 - Solution A random phase mixture differs from a stable dark-port state ϕ P+ 1/2 (a) Method and calculation ⟨P+⟩A = 0, ⟨P+⟩B = 1 2π ∫2π 0 1 + cos ϕ 2 dϕ = 1 2. ⟨KA⟩= 0, ⟨KB⟩= N/2 . (b) Interpretation and further calculation KB ∼Bin(N, 1/2) ⇒ Var KB = N/4 . ⟨|c+|2⟩= 1/2̸ = |⟨c+⟩|2 = 1/4 , c+ = (1 + eiϕ)/2. Physical check Different members of a statistical ensemble are averaged as probabilities. Coherent alternatives within one member are combined as amplitudes.

Question 5

A random phase mixture differs from a stable dark-port state

A balanced interferometer has P+(ϕ)=(1+cos⁡ϕ)/2P_+(\phi)=(1+\cos\phi)/2. In ensemble A every photon has ϕ=π\phi=\pi. In ensemble B the phase is independently uniform on [0,2π)[0,2\pi) for each photon. Both ensembles contain exactly N photons.

Diagram shown in the original SVG worksheet.

  1. Find the mean plus-port count in each ensemble.

  2. For ensemble B, find the count variance and explain why averaging amplitudes would give the wrong rule.

Question 5 - Solution

A random phase mixture differs from a stable dark-port state

Diagram shown in the original SVG worksheet.

(a) Method and calculation

⟨P+⟩A=0,⟨P+⟩B=12π∫02π1+cos⁡ϕ2dϕ=12.\langle P_+\rangle_A=0,\quad\langle P_+\rangle_B=\frac1{2\pi}\int_0^{2\pi}\frac{1+\cos\phi}{2}\,d\phi=\frac12.⟨KA⟩=0,⟨KB⟩=N/2.\boxed{\langle K_A\rangle=0,\qquad\langle K_B\rangle=N/2}.

(b) Interpretation and further calculation

KB∼Bin⁡(N,1/2)⇒Var⁡KB=N/4.K_B\sim\operatorname{Bin}(N,1/2)\quad\Rightarrow\quad\boxed{\operatorname{Var}K_B=N/4}.⟨|c+|2⟩=1/2≠|⟨c+⟩|2=1/4,c+=(1+eiϕ)/2.\boxed{\langle|c_+|^2\rangle=1/2\ne|\langle c_+\rangle|^2=1/4},\quad c_+=(1+e^{i\phi})/2.

Physical check

Different members of a statistical ensemble are averaged as probabilities. Coherent alternatives within one member are combined as amplitudes.