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The wave of a photon

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According the Copenhagen interpretation matter exhibits a wave particle duality which states that a quantum exhibits the properties of both particles and waves. Because the nature of a quantum is unknown, it does not mean that the particle or wave are real physical entities. But until a certain limit one can calculated as if it were classical particles or waves, with classical formula.

For quanta with mass, like the electron, the shape of this wave is a solution of the Schrödinger_equation. The result is a small wave packet around the position of the electron, which in distance degreases fast, but never becomes zero. So in principle the wave is present everywhere in the universe, but very small. According quantum mechanics the wave determines the probability of a particle to be observed at that location, by squaring the wave:


For the massless quanta the Schrödinger equation has no solution, so there is no formula for the shape (envelope) of the wave of a photon. Therefore there are no clear descriptions and a lot of opinions about this shape. The general view is that there still is a wave"packet", which length is determined by the source. In time this would be the coherence time, for which typical time scales are mentioned of nanoseconds for atoms, picoseconds for semiconductors or femtoseconds for light from the sun. Also for photons the wave is supposed to determine the probability, calculated with the same formula.

Although there is no formula, the wave of photon shows often with various experiments, especially with the double slit. By interpreting the results it is possible to draw conclusion about its shape. For the convenience of explanation in the following it will look a real physical wave is mentioned, but that is never the case. In all examples the wave is used as a tool to calculate what the experiments show, so will sometimes also show different behaviour then a normal wave.

The double slit experiment

The double slit experiment is especially suitable for showing the wave properties of photons because the wave is split in two slits in two waves, introduce a time (so phase) difference while compared both by interference.


If light passes a single slit (1) it will diffract, resulting in a wide "wave" on the detector. With a second slit (2) there will be two separate waves, which add up at the detector. At places outside the centre of the detector there will different path lengths of both waves causing a phase difference. The sum of both waves will show an interference pattern. With a single photon the effect will be the same: in time the same wave will be build up with single photons. So what the wave shows, is the probability of absorption of a photon at the that position of the detector, according the quantum mechanical formula above.

Formulation: at the detector the two waves add up as: I = sin(ωt-φ/2) + sin(ωt+φ/2), if φ is the phase shift between both waves. Then the probability P = 0.5(1+cosφ)

Another classical formulation in wikipedia.


If with a single photon is measured, the detector it will show a certain energy. If one slits are narrowed, the interference pattern will change, but not the measured energy. Conclusion:

  1. The amount of the wave does influence the absorption probability pattern
  2. The amount of the wave does not influence the energy. So the wave does not contain energy in its volume (although its frequency is a measure of the energy)

Spatial coherence

In (2) the waves through both slits cancel each other completely in B, so are equal in size in both slits. Still the particle must have gone though one of the slit. That is by quantum definition (indivisible particle) and it has never been measured to split up in two particles. If detectors are placed just after the slits, they always measure one particle in on or the other slit, while the waves are the same at avery slity. Conclusion:

  1. the wave has the same amplitude at a spatial distance of the photon.

Coherence length

In (2) the interference can extend to the left and right over many periods. Suppose in C the phase difference is 10 periods. If the particle went through slit 1, its wave will interfere with the wave which comes 10 periods later out of slit 1. But if the particle went through slit 2, its wave will interfere with the wave which came 10 periods earlier out of slit 1.

  1. In direction of propagation (and time) the amplitude is about the same, as much wavelengths before and after the particle as the interference pattern shows. This is the coherence length of the wave.
  1. According Feynman with a point source the wave amplitude degrades with 1/r2, with r is the distance to the source.

Wave carrier

With a single slit (1) the wave is differacted, according the Huygens-Fresnel principle. However the principle is calculated for a wave with carrier. It is the interference of every point of the carrier radiating spherical which result in the diffraction pattern. This not caused by the particle because the diffraction behaviour depends on the wavelength and in the double slit the slit which passes only the wave gives the same diffraction as the slit which passes particle + wave. Conclusion:

  1. The wave behaves like a wave with carrier (this certainly not a classical carrier, like the ether in the past)

Point source

Measuring the particle

In principle in quantum mechanics is that when the path of the particle is known, there will no interference. This could be done by measuring the particles with detectors in the two paths. With photons that is quite difficult, because detecting always destroys the photon.

With particle detectors at the slits


The most simple and most often mentioned measuring is placing a detector before each slit.[1] However the experiment is probably not possible in this way, because photons cannot be detected without absorbing. There is also no publication where this experiment has been actually done.


  1. Feynman, Richard P. (1965). The Feynman Lectures on Physics, Vol. 3. US: Addison-Wesley. pp. 1.1–1.8. ISBN 0201021188. {{cite book}}: Unknown parameter |coauthors= ignored (|author= suggested) (help)

Marking the wave, Quantum erasers

Marking the wave, Quantum erasers


Mach-Zehnder interferometer

Diagram of mirrors and detectors

In a Mach-Zehnder interferometer a light beam is split by a 50% mirror in two beams (red and blue). Both beams are reflected by a mirror and cross each other. Finally each beam ends in a separate detector. In (1) each detector can detect photon from the red or blue path and no interference is measured. In (2) a second beam splitter mixes the red and blue beam. Now both detectors show interference, but cannot detect if the photon followed the red or blue path.

In the classical wave description the wave of the photon is split in two by the 50% mirror. In (1) each detector sees only one wave, so there is no interference with the other wave. In (2) both waves are mixed, so each detector sees both waves, causing interfere.

Particle-path detectors and entangled photons (Kim et al)

Double split with prism and lens.

A high intensity laser radiates though the "red" and "blue" part of a wikipedia:beta barium borate crystal (BBO), which generates now and then two entangled photons with half the energy of the laser photons. These photons follow two paths by a wikipedia:Glan-Thompson Prism. One of the photons, the "signal" photon, goes upwards, through a lens, to the target detector D0. The other photon, the "idler" photon, goes downwards and is deflected by a prism that sends it along divergent paths, depending on whether it came from the red or blue BBO area. Beyond each path a 50% mirror acts as a wikipedia:beam splitter (green blocks), resulting in a 50% chance to pass through and a 50% chance to reflect to detectors D3 or D4. The photons which pass through are reflected by 100% mirrors (gray-green blocks) to the detectors D1 or D2. Where both beams cross, a 50% mirror is placed, which mixed both beams.Because of this arrangement, if the photon is recorded at detector D3 (resp. D4), it can only be a blue (resp. red) photon. If the photon is recorded at detector D1 or D2, it can be both a red or blue photon. A coincidence counter selects from D0 only those events which coincide with a chosen other detectors. This includes a delay of 8 ns to compensate for the 2.5 meter longer path to the other detectors. the result was:

  • When events were counted which coincided with D3 or D4, there was no interference.
  • When events were counted which coincided with D1 or D2, there was an interference pattern.

With classical wave physics a set of photons is emitted in the red or blue area of the BBO, no in both at the same time. There with ill be only one wave, to interfere with D3 or D4. But the interference with D1 or D2 cannot be explained classical, because of the absence of two waves.

Linear polarizers[1]


In this experiment both waves are differently (perpendicular) polarized . Then no interference pattern is visible. A polariser in front of the detector, with its axis 45° to the other polarisers, erases this information, causing the interference pattern to reappear.

Classical this effect is know as the Fresnel and Araglaws laws, which state that perpendicular polarization does not interfere. The 45° polarisers force both perpendicular polarizations to become parallel polarizations, which can interfere. Also classical calculation gives the same result.

Circular polarizers and entangled photons (Walborn et al)[1]


A beta barium borate (BBO) crystal, radiated by a strong laser, will generating now and then two entangled photons. One photon (yellow) goes through a double slit to signal detector Ds, with rotating polarizer Q1 or Q2 in each path. Q1 and Q2 axis are pendicular, producing a opposite rotating polarization. The other photon (green) goes to detector Dp, with a linear polarizer cube POL in the path. Only photons in Ds are registered which coincidence with photons in Dp. Dp is situated closer to the BBO then Ds, so photons are first detected by Dp. The results were:

  1. Without Q1/Q2 and POL there was an interference pattern in Ds
  2. With Q1/Q2 there was no interference.
  3. With Q1/Q2 and POL, adjusted on the Q1 fast axis, there was interference.
  4. With Q1/Q2 and POL, adjusted on the Q2 fast axis, there was interference, but 180º shifted with 3.
  5. 1-4 with Dp on a larger distance then Ds gives the same result. This is called a delayed eraser, because the photon in Dp is detected later then in Ds
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. In setup 2 the incoming photon has a random polarization (shown as red) which can be resolved into two waves, parallel (green) and perpendicular (blue) to the optical axis of the waveplate (see figure). Depending on the axis the wave on x or y are delayed π/2. At the detector the x and y vectors interfere (add up) and can be composed to one vector. Calculation shows that the probability is:

P = 0.5 - 0.5cos2αsinφ

  • A: If α = 0 (incoming polarisation parallel to the fast axis of Q1) then P = 0.5 - 0.5sinφ
  • B: If α = π/2 (incoming polarisation parallel to the fast axis of Q2) then P = 0.5 + 0.5sinφ
  • C: If α = random then integrating α from 0 to 2π and normalised gives P = 1. This of course the same as A + B: P = 0.5 - 0.5sinφ + 0.5 + 0.5sinφ = 1

If the incoming polarization is parallel to a fast axis of a wave plate a φ interference pattern is visible on the detector. Between A and B there is a phase difference of π. If α is random then there is no φ pattern. C can be explained as having no interference pattern or being the sum of two interference pattern A and B which have a phase difference of π, which when added does not show an pattern. According classical rules both explanations are right. Also the Walborn article mentions that the result of setup 2 is the sum of setup 3 and 4.

In the 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 emits an entanglement photon pair with mutually perpendicular polarization. According the article in set-up 3 POL was set to the angle of the fast axis of Q1. So the idler photons which passed through must have a signal photon with perpendicular polarisation, so parallel to the fast axis of Q2. These will show an interface pattern according B, which confirms with the measured result. In set-up 3 POL was rotated π/2, which gives the result of A. In other words, Dp detects only idler photons with a signal photon which polarization parallel to Q1 or Q2, which show an interference pattern according A or B. Dp selects those photons by coincidence from the output of Ds.

In above formulation the time the idler is detected in Dp is not important (as long as delays are used to compensate for path length difference). The "delayed" effect is in the classical wave explanation the effect that POL determines the polarisation of the signal wave before Q1/Q2, although POL is position on a much larger distance of the BBO then Q1/Q2.

Wheeler's astronomical experiment


Wheeler's delayed choice experiment is a thought experiment in which the type of detector (particle or wave) is changed after a photon passes the double slit. [2] In a second version the scale is magnified to astronomical dimensions: a photon has originated from a star and its path is bent by an intervening galaxy, so that it could arrive at a detector on earth by two different paths images. The detector could be a screen (as wave detector) or two telescopes, each focused to either side of the black hole. In both versions Wheeler expected that the screen will measure an interference pattern, while two telescopes will observe the photon only in one of them, without interference. If the outputs of both telescopes are optically combined, the interference pattern would be back, but losing the information of the path of the photon.

This result follows the classical waves: interference is caused by adding two waves during detection. If a telescope is directed at one wave, so no interference. If both telescopes are optically combined there are two waves again, which will interfere. The experiment shows that a photon always keeps full wave- and particle properties, until they combine during absorption.[3]. The first version has been confirmed in practice by Jacques e.a. with a Mach-Zender interferometer. The astronomical version is not yet done and probably too difficult in practice, because for an interference pattern the photons must be coherent when they arrive on earth, after such a long time.

Other experiments

Hong–Ou–Mandel Interferometer [4]

Hong–Ou–Mandel diagram

The Hong-Ou-Mandel interferometer is a variant of the Mach-Zehnder interferometer in which two simultaneous photons are compared. A 351 nm laser radiates a KDP crystal, which because of SPDC emits now and then two entangled photons of about 702 nm. The KDP is propably a type I, in which both photons have the same polarization and the same phase. Both waves are mixed in a 50% reflecting mirror, filtered by a bandpass interference filter IF and detected with a coincidence of 7.2 ns.

It is supposed that this 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. All other have no phase shift. At the detectors the waves are: Iu = 0.5sin(π+ωt+φ/2) + 0.5sinαsin(ωt-φ/2) and Il = 0.5sin(ωt+φ/2) + 0.5sinαsin(ωt-φ/2). With an ideal bandwidth filter d = Δω/ω, the probability is calculated:

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)

These formula predicts the result of the measurement if the cosine term is ignored and k = 0. Then the formula will be:

Pu = 0.5 - 0.5sin(πd2Δx/λ) / πd2Δx/λ

Pl = 0.5 + 0.5sin(πd2Δx/λ) / πd2Δx/λ

coincident: P = Pu * Pl

Beam spliter displacement compared to norm

In the experiment ω = 2πc/702nm = 2.7 1015. The bandwidth filter is 3 1013, d = 3 1013 / 2.7 1015 = 0.011. Then FWHM = 113 μm.


  1. S.P. Walborn (2002). "Double-Slit Quantum Eraser". Physical Review A. 65 (3): 033818. doi:10.1103/PhysRevA.65.033818. {{cite journal}}: Unknown parameter |coauthors= ignored (|author= suggested) (help)
  2. Mathematical Foundations of Quantum Theory, edited by A.R. Marlow, Academic Press, 1978.
  3. The fabric of the Cosmos, by Brian Greene, 2009, chapter 7
  4. Hong, C. K.; Ou, Z. Y.; Mandel, L. (1987). "Measurement of subpicosecond time intervals between two photons by interference". Phys. Rev. Lett. 59 (18): 2044–2046. Bibcode:1987PhRvL..59.2044H. doi:10.1103/PhysRevLett.59.2044. PMID 10035403. {{cite journal}}: Unknown parameter |lastauthoramp= ignored (|name-list-style= suggested) (help)




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(φ12)) sin(ωt+φ) [2]

Propability P = ʃ(I2)dωt, from 0-2π = (a12 + a22 + 2a1a2cos(φ12) ʃsin2(ωt+φ)dωt = (a12 + a22 + 2a1a2cos(φ12

If a1 = a2 = 1 and φ12 = φ: P = 2π(1+cosφ). Normalised to 1: P = 0.5+0.5cosφ

If a1 = 0.5, a2 = 1 and φ12 = 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.


Polarization in the quarter waveplates

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



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)