User:DParlevliet/sandbox

Polarization in the quarter waveplates

In the classical wave description Q1 and Q2 are quarter-wave plates which have perpendicular a "fast" and "slow" axis. The slow axis has a π/2 phase delay compared to the fast axis. Q1 and Q2 are mounted mutually perpendicular (see figure). In setup 2 the incoming photon has a wave sinωt with a random polarization. This polarization (shown as red) can be resolved into two waves, parallel (green) and perpendicular (blue) to the optical axis of the waveplate (see figure): y = sinαsinωt and x = cosαsinωt if α is the angle between incoming polarization and the waveplate axis. Because the phase of the slow axis is delayed π/2 against the fast axis the output of the wave plates will be y₂ = sinαsinωt, x₂ = cosαsin(ωt-π/2) and y₁ = sinαsin(ωt-π/2), x₁ = cosαsinωt. At the detector both waves superimpose by interference. Outside the center of the detector there will be also a phase difference φ because of path length difference between slits and place of detection, so:

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

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

Composing again to one wave gives: I = square root (y² + x²).

The probability of absorption of a photon is in quantum mechanics: P = ʃ(I²)dωt = ʃ(y² + x²)dωt integrated from 0 to 2π = ... = 0.5(1-sinφcos2α) (normalised)

• A: If α = 0 (incoming polarisation parallel to the fast axis of Q1) then P = 0.5(1-sinφ)
• B: If α = π/2 (incoming polarisation parallel to the fast axis of Q2) then P = 0.5(1+sinφ)
• C: If α = random then P = ʃ(0.5(1-sinφcos2α)dα integrating from 0 to 2π = 1 (normalised)

So if the incoming polarization is parallel to a fast axis of a wave plate a sinφ pattern is visible on the detector but with a phase difference of π. If α is random there is no sinφ dependence.

In the Walborn e.a. experiment the incoming photon is the signal photon of a BBO. In setup 2 the polarization is random, so the measured result is as C above. The BBO results in a quantum entanglement, causing the photon pair to have mutually perpendicular polarization. So if the idler photon is forced by POL in a certain direction, the signal photon polarization wil be perpendicular. According the article in setup 3 POL was set to the angle of the fast axis of Q1, so the signal polarization will be perpendicular, parallel to the fast axis of Q2. Indeed is the measured result the same as above B. In setup 3 POL was rotated π/2, wich gives the same result as A.

Classical waves explanation

The Hong-Ou-Mandel interferometer is an experiment in which two entangled photons interfere. A 351 nm laser radiates a KDP crystal, causing spontaneous parametric down conversion, which generates now and then two entangled ~702 nm photons. Those are mixed in a 50% reflecting mirror, filtered by a bandpass interference filter IF and detected with a coincidence of 7.2 s ns.<ref>. After the 50% mirror there is 50% blue wave, which is not reflected, minus 50% red wave, which is reflected. The minus is caused by the 180º phase shift during reflection. If both waves are the same, they will extinguish each other and there will be no absorption of the photon. Any phase shift, caused by delays or different frequency, will increase the absorption possibility. In the experiment the 50% mirror is shifted, causing a time shift between both photons, so a phase shift, so increased absorption. The result is a FWHM of 19 μm.

With classical wave formula after the 50% mirror $I=-0.5sin(\omega t)+0.5sin(\omega t+\varphi )$. According quantum mechanics the probability of absorption $P=\int _{0}^{2\pi }\!I^{2}\,d\omega t=0,5(1-cos\varphi )$ (after normalising). If the phase shift is caused by a different frequency between both entangled photons, then $(\omega t+\varphi )=(1+p)\omega t$, so $\varphi =2\pi cpt/\lambda$ and coherence length $FWHM=1.7.10^{-9}\lambda /p$. For this experiment with p = 0.01 and 702 nm, FWHM = 120 fs. This is caused by a 50% mirror shift of 120 fs * c / 2 = 18 μm (factor 2 because displacement of the 50% mirror changes both wave paths). This agrees with the result (but only as approximation, because in reality there is a mix of 0 ≥ p ≥ 0.01).

So according classical waves the HOM dip is the result of the coherence length of each photon, determined by the band pass filter.

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I have added a chapter which solely shows the results and bare conclusions from the experiment. Both real mesurement and thopught measurements. It is intended to show in a simple and graphical way the remarkable proterties of photon-wave, so it also clear for those who are no familiar with the experiment and quantum mechanics. So please don't mix it with other parts or add quantum theory and interpretation. That is a anopther subject. DParlevliet (talk) 19:55, 17 September 2013 (UTC)

Results of the experiment

The double-slit experiment is exceptional in the way it shows the particle-wave properties, before using quantum theory and interpretation. Below shows the results of real experiments and thought experiments with one photon.

(1) shows the result if parallel light uniformly illuminates a detector. When a screen with a slit is added (2), diffraction will cause scattering of the light resulting in a wide single wave on the detector. When a second slit is opened (3), the different lengths the waves travel through both slits causes interference. At A both lengths are the same, the waves run in phase and add up to a double height peak. At B the light from slit 1 travels longer and becomes opposite to the wave from slit 2, so extinguish each other. With single photons the effect will be the same: in time the wave in (3a) will be build up with single photons.

Suppose the experiment is repeated with one photon with a certain energy, which is the same in every measurement (although in theory impossible). Also suppose that without a screen (1a) and with one slit (2a) the photon will go to B.

• If in (2a) the slit is narrowed the result will not change: the energy is the same. Conclusion: the energy of a photon is independent of the area of the slit.
• Open slit 1 (3a). Comparison with (3) shows that no photon will arrive B. So the photon turns off and arrives somewhere else. Conclusion: it appears that something goes through slit 1 which changes the track of the photon.

What does the experiment show about the properties of this "something"?

• The interference pattern in (3) show that this "something" has the shape of a sine wave.
• If both slits are narrowed, the result will not change: the energy and detection position of the photon are the same. Conclusion: the energy is independent of the amount of wave through the slits. So the wave does not contain energy in its volume.
• If only one slit is narrowed, the result will be a mix between (2) and (3). The photon still has the same energy. Conclusion: the amount of wave through the slit effects the position where the photon is detected.
• Back to (3). The waves through both slits cancel each other completely in B, so are equal in size in both slits. Conclusion: the waves have the same amplitude at a distance of the photon.

• If in (3b) the wave from slit 2 arrives at B and cancels fully a part of the wave which went earlier through slit 1. Also the wave from slit 1 cancels a part of the wave which went through slit 2 at a later time. Conclusion: thre is a wave before and after the photon, withn the same amplitude. So the wave has above shape: wave fronts, equal in amplitude, travelling in the same direction and with the same (light) speed as the photon.

• Move the detector to the slits and note the places B where no photon arrives. These are the red lines in (3c). But how can a photon travel to A through the "forbidden" zone, that is the zone without photons when the detector would be placed there? Suppose the photon travels as in (3a) and will cross the red line at B'. Place the detector at B'. Now the photon will not arrive at B', so will follow a different track. Conclusion: also the distance between slits and detector, so the place where the photon is detected later on, does influence the track of the photon.
• A photon travels between a sun and a planet. That is also a kind of slit, so has effect on the track of the photon, also after it travelled light years. The effect will be extremely small, but present
• Place a mirror above the slits (3d). The effect will be same as (3), with a single photon equal to (3a). In a mirror a photon is not reflected, but absorbed by an atom, which transmits a new photon. The atom does not know that it is part of a mirror on a certain angle, and transmit the new photon in an arbitrary direction. How does the photon know that it has to go downwards? (3d) implies that not only the photon, but also the wave is reflected. The wave is wide and hits all atoms in the mirror. Conclusion 1: the atoms reflect also the wave in an arbitrary direction. According Huygens principle these combine again to parallel wave fronts, but now mirrored. Conclusion 2: the new wave directs the new photon, not the other way around.

• Place detectors in slit 1 and 2, to measure through which slit the photon travels (3e). The photon will be detected in slit 1 or 2. But the interference as in (3) disappears. Conclusion: if a situation is created in which a photon is forced to become visible (detected), the wave cannot pass that.
• To be able to reach B the wave must be diffracted in slit 1. To get the result of (3), the wave must diffract all around, as in (3f). This effect is known for waves with a carrier, like a sound wave. Then every air molecule in slit 1 is struck by the incoming sound wave and radiates it all around. For this reason in the past the ether was proposed, as carrier of the light wave. However it was proven that this physical ether does not exist. But then diffraction should not be possible and one would expect the wave travels as (3g). Conclusion: there still must be "something" in split 1 which diffracts the wave. If this is true, it its likely that also outside the slit this must be the carrier of the wave, with unknown properties.
• Above explains also why in (2) the photon diffracts. Also through slit 2 comes a wave which is diffracted all around. Because the mirror shows that the photon follows the wave, the photons will also diffract in all directions.