The wave of a photon/Calculations

General
Propability P = (ʃ(I2)dωt)/2π, from 0-2π

ʃsin2(nωt+φ)dωt, from 0-2π = ʃcos2(nωt+φ)dωt = π (if n = integer)

Double slit
With interference of two waves sin(ωt+φ1) and sin(ωt+φ2)

The result is I = √(a12 + a22 + 2a1a2cos(φ1-φ2)) sin(ωt+φ)

Propability P = ʃ(I2)dωt, from 0-2π = (a12 + a22 + 2a1a2cos(φ1-φ2) ʃsin2(ωt+φ)dωt = (a12 + a22 +  2a1a2cos(φ1-φ2)π

If a1 = a2 = 1 and φ1-φ2 = φ: P = 2π(1+cosφ). Normalised to 1: P = 0.5+0.5cosφ

If a1 = 0.5, a2 = 1 and φ1-φ2 = φ: P = π(1.25+0.5cosφ). Normalised to above: P = 0.5+0.2cosφ

Linear polarizer
x and y are parallel to the polarisers axis, both polarisers are mounted perpendicular, α is the angle of incoming polarization, φ is the phase difference caused by path length difference

The two waves are x = cosαsin(ωt+φ/2) and y = sinαsin(ωt-φ/2)

Combined: I = √(x2 + y2)

Propability P = ʃ(I2)dωt, from 0-2π = ʃ(x2 + y2)dωt = cos2αʃsin2(ωt+φ/2)dωt + sin2αʃsin2(ωt-φ/2)dωt = (cos2α + sin2α)π = π. Normalised: P = 1, so no φ interference pattern.

Walborn
x and y are chosen parallel to the axis the quarter-wave plates. Both plates are mounted perpendicular, α is the angle of incoming polarization, φ is the phase difference caused by path length difference

x = x2 + x1 = cosαsin(ωt-π/2-φ/2) + cosαsin(ωt+φ/2) = 2cosαcos(π/4+φ/2)sin(ωt-π/4)

y = y2 + y1 = sinαsin(ωt-φ/2) + sinαsin(ωt-π/2+φ/2) = 2sinαcos(π/4-φ/2)sin(ωt-π/4)

Combined: I = √(x2 + y2)

Propability P = ʃ(I2)dωt, from 0-2π = ʃ(x2 + y2)dωt = (4cos2αcos2(π/4+φ/2) + 4sin2αcos2(π/4-φ/2))ʃsin2(ωt-π/4))dωt = 4(cos2α (0.5+0.5cos(π/2+φ)) + sin2α (0.5+0.5cos(π/2-φ))π = 2π(cos2α + sin2α + cos2αcos(π/2+φ) + sin2αcos(π/2-φ)) = 2π(1 - cos2αsin(φ) + sin2αsin(φ)) = 2π(1 + (-cos2α + 1 - cos2α)sin(φ)) = 2π(1 + (1-2cos2α)sinφ) = 2π(1 + (1 - 2(0.5+0.5cos2α))sinφ) = 2π(1 - cos2αsinφ)

Normalised: P = 0.5 - 0.5cos2αsinφ

If α = random: P = 0.5ʃ(1-cos2αsinφ)dα from 0 to 2π = |α - 0.5sin2αsinφ| = (2π - 0 - 0 + 0) = 2π. Normalised: P = 1

Hong–Ou–Mandel
It is supposed that the beam splitter is a glass plate with reflector on one side, causing (only) the reflection from the normal input having a phase shift of π. Both photons start in the KDP with the same phase. Then at the upper (u) and lower (l) detector the waves are:

Iu = -0.5sin(ωt+φ/2) + 0.5sin(ωt-φ/2) = -cosωtsin(φ/2) and Il = 0.5sin(ωt+φ/2) + 0.5sin(ωt-φ/2) = sinωtcos(φ/2)

Probability Pu = ʃ(I2)dωt), from 0-2π = sin2(φ/2)ʃcos2ωtdωt = (0.5-0.5cosφ)π and Pl = cos2(φ/2)ʃsin2ωtdωt = (0.5+0.5cosφ)π.

Normalised: Pu = (0.5 - 0.5cosφ) and Pl = (0.5 + 0.5cosφ)

and coincident: P = Pu * Pl = (0.5 - 0.5cosφ) (0.5 + 0.5cosφ) = 0.25 - 0.25cos2φ. Normalised: P = 0.5 - 0.5cos2φ

Phase shift The phase shift between both waves can be caused by the different wavelength (1+p) and the displacement of the beam splitter causing a Δt in only one path. If t = time between the KDP and Du, then the phase difference can be written as:

φ = φ2 - φ1 = ω(1+p)(t+Δt) + π - ω(1-p)t = ωt + ωpt + ωΔt + ωpΔt - ωt + ωpt + π = 2ωpt + ωΔt + ωpΔt + π = ωΔt + ωp(2t+Δt) + π

This is the formula for a fixed p, but in practice there is a mix of different wave lengths. With an ideal bandwidth filter Pu is equal within p = -d/2 - +d/2 (and zero outside). When used in the formula of Pu above:

Pu = ʃ (0.5 - 0.5cos(π + ωΔt+ωp(2t+Δt)) dp / d from -d/2 - +d/2  = ʃ (0.5 + 0.5cos(π + ωΔt+ωp(2t+Δt)) dp / d

= 0.5 | p - sin(ωΔt+ωp(2t+Δt)) / ω(2t+Δt) | / d

= 0.5(d/2 - sin(ωΔt+ωd(2t+Δt/2))/ω(2t+Δt) + d/2 - sin(ωΔt-ωd(2t+Δt)/2)/ω(2t+Δt) ) / d

= 0.5(1 - 2cos(ωΔt) sin(ωd(2t+Δt)/2) / ωd(2t+Δt)

= 0.5 - cos(ωΔt) sin(ωd(2t+Δt)/2)) / ωd(2t+Δt)

ω = 2πf = 2π/T = 2πc/λ, so T = λ/c, Δt = 2Δx/c (Δx has double effect), so Δt/T = 2Δx/λ. Suppose t = kλ. Then:

= 0.5 - cos(2πc2Δx/λc) sin(2πd(2kT+Δt)/2T)) / (2πd(2kT+Δt)/T)

Pu = 0.5 - 0.5cos(4πΔx/λ) sin(πd(2k+2Δx/λ)) / πd(2k+2Δx/λ)

and Pl = 0.5 - 0.5cos(4πΔx/λ) sin(πd(2k+2Δx/λ)) / πd(2k+2Δx/λ)

and coincident: P = Pu * Pl

The general form of the second sinus part is:



ΔxFWHM = cΔt/2 = 0,00148 λ/d = 0,000443 λ/d μm (λ in nm)