# The wave of a photon/Calculations

### General

[edit | edit source]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

[edit | edit source]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) [1]

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

[edit | edit source]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

[edit | edit source]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

[edit | edit source]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**= (0.5 + 0.5cosφ)

_{l}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)