# The wave of a photon/Calculations
### General

Propability P = (ʃ(I$^2$)dωt)/2π, from 0-2π
ʃsin$^2$(nωt+φ)dωt, from 0-2π = ʃcos$^2$(nωt+φ)dωt = π (if n = integer)

### Double slit

With interference of two waves sin(ωt+φ$_1$) and sin(ωt+φ$_2$)
The result is I = √(a$_1$$^2$ + a$_2$$^2$ + 2a$_1$a$_2$cos(φ$_1$-φ$_2$)) sin(ωt+φ)
Propability P = ʃ(I$^2$)dωt, from 0-2π = (a$_1$$^2$ + a$_2$$^2$ + 2a$_1$a$_2$cos(φ$_1$-φ$_2$) ʃsin$^2$(ωt+φ)dωt = (a$_1$$^2$ + a$_2$$^2$ + 2a$_1$a$_2$cos(φ$_1$-φ$_2$)π
If a$_1$ = a$_2$ = 1 and φ$_1$-φ$_2$ = φ: P = 2π(1+cosφ). Normalised to 1: P = 0.5+0.5cosφ
If a$_1$ = 0.5, a$_2$ = 1 and φ$_1$-φ$_2$ = 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 = √(x$^2$ + y$^2$)
Propability P = ʃ(I$^2$)dωt, from 0-2π = ʃ(x$^2$ + y$^2$)dωt = cos$^2$αʃsin$^2$(ωt+φ/2)dωt + sin$^2$αʃsin$^2$(ωt-φ/2)dωt = (cos$^2$α + sin$^2$α)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 = x$_2$ + x$_1$ = cosαsin(ωt-π/2-φ/2) + cosαsin(ωt+φ/2) = 2cosαcos(π/4+φ/2)sin(ωt-π/4)
y = y$_2$ + y$_1$ = sinαsin(ωt-φ/2) + sinαsin(ωt-π/2+φ/2) = 2sinαcos(π/4-φ/2)sin(ωt-π/4)
Combined: I = √(x$^2$ + y$^2$)
Propability P = ʃ(I$^2$)dωt, from 0-2π = ʃ(x$^2$ + y$^2$)dωt = (4cos$^2$αcos$^2$(π/4+φ/2) + 4sin$^2$αcos$^2$(π/4-φ/2))ʃsin$^2$(ωt-π/4))dωt = 4(cos$^2$α (0.5+0.5cos(π/2+φ)) + sin$^2$α (0.5+0.5cos(π/2-φ))π = 2π(cos$^2$α + sin$^2$α + cos$^2$αcos(π/2+φ) + sin$^2$αcos(π/2-φ)) = 2π(1 - cos$^2$αsin(φ) + sin$^2$αsin(φ)) = 2π(1 + (-cos$^2$α + 1 - cos$^2$α)sin(φ)) = 2π(1 + (1-2cos$^2$α)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:
I$_u$ = -0.5sin(ωt+φ/2) + 0.5sin(ωt-φ/2) = -cosωtsin(φ/2) and I$_l$ = 0.5sin(ωt+φ/2) + 0.5sin(ωt-φ/2) = sinωtcos(φ/2)
Probability P$_u$ = ʃ(I$^2$)dωt), from 0-2π = sin$^2$(φ/2)ʃcos$^2$ωtdωt = (0.5-0.5cosφ)π and P$_l$ = cos$^2$(φ/2)ʃsin$^2$ωtdωt = (0.5+0.5cosφ)π.
Normalised: P$_u$ = (0.5 - 0.5cosφ) and P$_l$ = (0.5 + 0.5cosφ)
and coincident: P = P$_u$ * P$_l$ = (0.5 - 0.5cosφ) (0.5 + 0.5cosφ) = 0.25 - 0.25cos$^2$φ. Normalised: P = 0.5 - 0.5cos$^2$φ
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 D$_u$, then the phase difference can be written as:
φ = φ$_2$ - φ$_1$ = 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 P$_u$ is equal within p = -d/2 - +d/2 (and zero outside). When used in the formula of P$_u$ above:
P$_u$ = 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)
P$_u$ = 0.5 - 0.5cos(4πΔx/λ) sin(πd(2k+2Δx/λ)) / πd(2k+2Δx/λ)
and P$_l$ = 0.5 - 0.5cos(4πΔx/λ) sin(πd(2k+2Δx/λ)) / πd(2k+2Δx/λ)
and coincident: P = P$_u$ * P$_l$
The general form of the second sinus part is:
Δx$_{FWHM}$ = cΔt/2 = 0,00148 λ/d = 0,000443 λ/d μm (λ in nm)