Light as a Probability Wave — Question 10

Section 38-4 · Physics pages 6330–6331

Question 10 Interference phase sensitivity follows from both slope and counting noise A balanced interferometer detects N independent photons with plus-port probability p(ϕ) = (1 + cos ϕ)/2. The estimated fraction has standard deviation σp = p p(1 −p)/N. Use local linear propagation σϕ = σp/|dp/dϕ| where the slope is nonzero. ϕ p π/2 Schematic; labels, not drawing lengths, specify the data. (a) Derive the phase uncertainty away from fringe extrema. (b) Find it at ϕ = π/2 for N = 104, and state why direct linear propagation is invalid exactly at ϕ = 0.
Show solutionHide solution
Question 10 - Solution Interference phase sensitivity follows from both slope and counting noise ϕ p π/2 (a) Method and calculation dp dϕ = | sin ϕ| 2 , p(1 −p) = sin2 ϕ 4 . σϕ = 1/ √ N (sin ϕ̸ = 0). (b) Interpretation and further calculation ϕ = π/2, N = 104 : σϕ = .0100 rad . ϕ ≃0 : p = 1 −ϕ2/4 + O(ϕ4), p′(0) = 0 . Physical check The limiting formula does not license dividing zero by zero at an extremum. A nonlinear likelihood analysis is needed there, and a single fringe also has phase ambiguities.

Question 10

Interference phase sensitivity follows from both slope and counting noise

A balanced interferometer detects N independent photons with plus-port probability p(ϕ)=(1+cos⁡ϕ)/2p(\phi)=(1+\cos\phi)/2. The estimated fraction has standard deviation σp=p(1−p)/N\sigma_p=\sqrt{p(1-p)/N}. Use local linear propagation σϕ=σp/|dp/dϕ|\sigma_\phi=\sigma_p/|dp/d\phi| where the slope is nonzero.

Diagram shown in the original SVG worksheet.

  1. Derive the phase uncertainty away from fringe extrema.

  2. Find it at ϕ=π/2\phi=\pi/2 for N=104N=10^4, and state why direct linear propagation is invalid exactly at ϕ=0\phi=0.

Question 10 - Solution

Interference phase sensitivity follows from both slope and counting noise

Diagram shown in the original SVG worksheet.

(a) Method and calculation

|dpdϕ|=|sin⁡ϕ|2,p(1−p)=sin⁡2ϕ4.\left|\frac{dp}{d\phi}\right|=\frac{|\sin\phi|}{2},\quad p(1-p)=\frac{\sin^2\phi}{4}.σϕ=1/N(sin⁡ϕ≠0).\boxed{\sigma_\phi=1/\sqrt N}\qquad(\sin\phi\ne0).

(b) Interpretation and further calculation

ϕ=π/2,N=104:σϕ=.0100rad.\phi=\pi/2,\ N=10^4:\quad\boxed{\sigma_\phi=.0100\,\mathrm{rad}}.ϕ≃0:p=1−ϕ2/4+O(ϕ4),p′(0)=0.\phi\simeq0:\quad p=1-\phi^2/4+O(\phi^4),\quad\boxed{p\prime(0)=0}.

Physical check

The limiting formula does not license dividing zero by zero at an extremum. A nonlinear likelihood analysis is needed there, and a single fringe also has phase ambiguities.