Shape of the wave of a single photon

[Strange]

 

I mean the intensity of the detector against the position on the detector.

That is a little more complex.

http://en.wikipedia.org/wiki/Double-slit_experiment#Classical_wave-optics_formulation

Edited October 16, 2013 by Strange

[Enthalpy]

 

The decay time will limit where you might expect the photon to be, but that's not the same as any "size" of a photon. I could have the hydrogen atom between two mirrors in a cavity, and the photon would quite happily be emitted toward a mirror, even if they were a few mm apart. The mirrors would not have to be .48m apart for this to happen. If the mirrors were closer than half a wavelength (plus whatever correction is necessary because the atom is there), though, then the photon could not appear.

If the mirrors are closer to an other than the size of the photon, then you have made a laser. This is exactly why laser designers want a long de-excitation time for the lasing transition. A too quick natural transition means spontaneous emission too often, no lasing.

 

A laser is also known (including experimentally) to work with a single photon. This means that the usual explanation of the laser, where one photon stimulates the emission of a sibling, is a bit short. From my electronics background, I prefer to imagine that electrons have a too stiff position in atoms, giving them a high impedance that maches vacuum badly, so that coupling is inefficient and the transitions takes long. In a cavity, resonance increases the impedance seen by the electron (at some places - I've suggested to put the lasing atoms only there for efficiency) and improves the coupling, so that emission is faster. In a Yag, this time can drop from ns fo fs.

 

More photons add their field in the cavity, which also increases the coupling with the atoms. This explains why lasing centers synchronize with an other.

 

By "size" of the photon, I understand the length (and width if useful) of the wave packet. This has very practical implications: these "coherence time" and "coherence length" limit the path difference over which one can get interferences.

Edited October 17, 2013 by Enthalpy

[swansont]

My post had nothing to do with lasers. Without a gain medium, you do not have a laser. My point is a photon half a meter long will not fit into a small cavity. Coherence length also is not related to the "size" of a photon.

[Mike Smith Cosmos]

Did we ever get to the bottom of the rough three x,y,z, sizes of a photon of light was thought to be.

Namely How long, how wide, how high and roughly what shape and how many complete cycles of oscillation does it include?

 

Mike

 

 

 

 

 

I suppose what I have drawn is a copy of the wave packet in swansonts' post ,(hash 49) your inset dvebt123

 

What we need is a few rough values for x, y, z. I guess ! And how many cycles in a typical light wave packet!

 

 

Mike

Edited October 17, 2013 by Mike Smith Cosmos

[DParlevliet]

That is a little more complex.

 

I found the answer on the German wiki: http://de.wikipedia.org/wiki/Interferenz_(Physik)#Interferenz_zweier_Wellen_gleicher_Frequenz_und_Amplitude.2C_aber_unterschiedlicher_Phase. The formula shows the cosine of the interference patteren. Conclusion: the wave is a sine. It is was no sine, the pattern would not be a cosine. Of course we knew already, but it prooves the detector signal shows the wave.

 

If the wave is very long we will have much periods of interference on the detector

But suppose we have a wavelength of only 5 periods:

 

When both waves shift against each other (because of phase shift) lesser parts will interfere with each other, until after 5 periods there is no interference at all. This shows a different pattern. So this shows that the shape of the wave is visible in the interference pattern and when you are smart in math (or write programs) you can more or less calculate from the detector data the original wave of the photon.

 

My guess is that this wave is very long with equal amplitude.

[Strange]

 

My guess is that this wave is very long with equal amplitude.

Not that long. Visible light has a wavelength between about 380 and 740 nanometres.

 

(That Wikipedia page is the German equivalent of the one I linked to above)

[DParlevliet]

Not that long. Visible light has a wavelength between about 380 and 740 nanometres.

 

(That Wikipedia page is the German equivalent of the one I linked to above)

 

I don't mean the wavelength, but the length of the wave(packet) as shown in the graphic.

The German wiki showed the formula I was refering to. That is not on others.

[Strange]

I don't mean the wavelength, but the length of the wave(packet) as shown in the graphic.

 

Are you trying to relate this to the size of the photon? I'm not sure that works, unless you are assuming something abut the timing/distance between photons. Remember the experiment works even if you send individual photons through the system one at a time. As far as I know, there is no limit on how separated the photons are; you will still get the interference pattern.

Edited October 18, 2013 by Strange

[DParlevliet]

I show that from the detector data you can determine (roughly) the size/shape/envelop of a wave(packet) which interferes in the double slit when one photon is detected. I don't say anything (yet) about what this wave is.

[Strange]

I show that from the detector data you can determine (roughly) the size/shape/envelop of a wave(packet) which interferes in the double slit when one photon is detected. I don't say anything (yet) about what this wave is.

I'm not sure how you can do that when there is only one photon (and, being a quantum, it is indivisible).

 

Also, how does this apply to things like electrons and molecules (which show the same behaviour)?

Edited October 18, 2013 by Strange

[Enthalpy]

My post had nothing to do with lasers. Without a gain medium, you do not have a laser. My point is a photon half a meter long will not fit into a small cavity. Coherence length also is not related to the "size" of a photon.

An emitter in a cavity is a laser. A gain medium is nothing more than an emitter, even a single excited atom. The very limit to make a laser from any emitter is that the de-excitation time is longer than the cavity travel time.

 

The cavity folds the wave. This is aboslutely necessary to make a laser, and even to make a void resonator.

 

One may define the size of a photon in many ways, as usual with a particle. Once you detect the photon with an atom, you may tell that the photon was the size of this atom, but prior to this, the reasonable definition to me is the extension of the wave packet, which relates directly with the linewidth and the lifetime of the emitting transition.

 

I feel we're back to a previous interrogation: to me, a long-lasting atomic transition does mean that the emitted photon is that long - and not that the short photon has a probability to be emitted over a long time. Only when the photon is detected, or the atom observed, can we tell the position of the photon.

 

I understand as an argument in favour, that emission linewidths are routinely computed by opticians from transition lifetimes. A photon shorter than the transition lifetime would have a broader spectrum.

[DParlevliet]

I'm not sure how you can do that when there is only one photon (and, being a quantum, it is indivisible).

 

Also, how does this apply to things like electrons and molecules (which show the same behaviour)?

 

All photons interfere in the same way, because their waves look the same. So when you add up all, you can see how one looks like. Interference of light is the sum of the behaviour of many photons one by one.

 

For the moment I limit myslef to photons.

[swansont]

An emitter in a cavity is a laser. A gain medium is nothing more than an emitter, even a single excited atom. The very limit to make a laser from any emitter is that the de-excitation time is longer than the cavity travel time.

 

The cavity folds the wave. This is aboslutely necessary to make a laser, and even to make a void resonator.

Amplification is also absolutely necessary, which you don't have with a single atom sitting in a cavity.

All photons interfere in the same way, because their waves look the same.

No, this is not true. Photons of orthogonal polarization do not interfere, for example.

[Strange]

 

All photons interfere in the same way, because their waves look the same. So when you add up all, you can see how one looks like. Interference of light is the sum of the behaviour of many photons one by one.

 

In the classical view, the waves are split and go through both slits, enabling inteference. Is this possible with a single (indivisible) photon? Note that each photon hits a single point on the screen (it isn't spread out across the interference pattern). The inteference pattern only becomes apparent in the distribution of the positions of large numbers of individual photons. I don't think you can explain this in terms of the waveform "in" each photon.

[DParlevliet]

Amplification is also absolutely necessary, which you don't have with a single atom sitting in a cavity.

 

No, this is not true. Photons of orthogonal polarization do not interfere, for example.

 

Perhaps it is better not to mix two topics?

 

I mean the photons which are used for the normal double slit measurements.

 

 

In the classical view, the waves are split and go through both slits, enabling inteference. Is this possible with a single (indivisible) photon? Note that each photon hits a single point on the screen (it isn't spread out across the interference pattern). The inteference pattern only becomes apparent in the distribution of the positions of large numbers of individual photons. I don't think you can explain this in terms of the waveform "in" each photon.

 

A photon can show wave or particle properties, not both at the same time (according Copenhagen). Here we measure the wave, so cannot analyse the particle. The particle is anyway another discussion.

The interference pattern is visible with many photons, but every photon acts according the same pattern. For instance there are black lines with no light. Also a single photon will never hit these black lines.

Edited October 18, 2013 by DParlevliet

[swansont]

I still don't see how you conclude you know the waveform based on detecting interference.

[DParlevliet]

See my earlier image of the wave with 5 periods. Let us discus only on the length of the wave. If the wave had infinite periods then interference will be the dashed sine: full interference on all distances of the middle of the detector. Black lines will occur at 0.5, 1.5, 2.5, ... periods phase shift between the waves through slit 1 and slit 2.

But suppose the wave has only 5 periods, what happens after the phase shift between both is 5 periods and larger? Then both (short) waves don't see each other anymore, so they cannot interfere. At the detector position of 5.5 period phase shift first wave 1 arrives while there is no yet wave 2, and only after wave 1 has passed wave 2 arrives. So there will be light of cause, because there are waves, so propability of absorption, but there will be no interence pattern anymore.

I don't say that the interference pattern looks the same as the wave, but you can calculate it (more of less)

[Mike Smith Cosmos]

I am not convinced either my understanding, or everybody else are talking about the same waves.

as my understaning goes , which i stand to be corrected:- 3 different waves

 

A ) There is a fundamental constant amplitude Sine Wave frequency born out of the jumping of an electron from one energy band to another. This energy is precise and gives rise to the precise colour light or photons of light , coming away from a light /photon source. This on a spectrum would appear as one distinct line. by fourier analysis the fundamental would be the only line all other coefficients would be zero from plus to minus infinity.

All-be-it ,that this sine wave is chopped into bits by the Plank Quantum issue The sine wave is no doubt reassembled ,( without gaps ) in the eye or detector before observation I assume . ( on the grounds of a class C style of amplification, where incomplete parts of the sine wave cycle are reintroduced under a resonance principle)

 

This sine wave is surely not the same as the two other waves.

 

B ) De Broglie wave associated with all mass objects from quarks to worlds,to galaxies. The only significant ones being at Atomic sizes (quantum)

 

C) any Interference pattern (apparent wave shape) caused by Two parts of a continuous constant amplitude sine wave arriving from different routes ( say two slits in an experiment or two different paths in an optical mirror experiment.

 

 

Any lumpy waves (by fourier analysis ) would throw up coeffiients in the fourier series of frequencies giving rise to additional frequencies on the em spectrum from plus to minus infinity. ( which I thought was not the case in Pure Light coming from an Atom)

 

I stand to be corrected.

 

 

mike

 

 

 

Mike

Edited October 19, 2013 by Mike Smith Cosmos

[DParlevliet]

A ) There is a fundamental constant amplitude Sine Wave frequency born out of the jumping of an electron from one energy band to another.

 

That is the question of this topic. Is it really constant amplitude (of one photon), also a kilometre ahead of the photon? For instance the wave of a mass particle I showed before (#13) does not have constant amplitude, but is short.

 

Because the consequence is: the quadruple of the modulus of this wave is the probability. So a constant amplitude would mean that the probability of absorption of the photon right after emitting is these same near the emitting atom is a kilometre ahead.

 

Now I think it: the Born rule (propability is the quadruple of the wave) is for matter particles. Is that also for photons?

 

 

Edited October 19, 2013 by DParlevliet

[swansont]

See my earlier image of the wave with 5 periods. Let us discus only on the length of the wave. If the wave had infinite periods then interference will be the dashed sine: full interference on all distances of the middle of the detector. Black lines will occur at 0.5, 1.5, 2.5, ... periods phase shift between the waves through slit 1 and slit 2.

But suppose the wave has only 5 periods, what happens after the phase shift between both is 5 periods and larger? Then both (short) waves don't see each other anymore, so they cannot interfere. At the detector position of 5.5 period phase shift first wave 1 arrives while there is no yet wave 2, and only after wave 1 has passed wave 2 arrives. So there will be light of cause, because there are waves, so propability of absorption, but there will be no interence pattern anymore.

I don't say that the interference pattern looks the same as the wave, but you can calculate it (more of less)

 

At the detector you get a particle-like interaction — the photon is localized. You get a flash on your phosphor, or a dot on your CCD (or film). That tells you nothing about the wave.

[Mike Smith Cosmos]

 

That is the question of this topic. Is it really constant amplitude (of one photon), also a kilometre ahead of the photon? For instance the wave of a mass particle I showed before (#13) does not have constant amplitude, but is short.

 

Because the consequence is: the quadruple of the modulus of this wave is the probability. So a constant amplitude would mean that the probability of absorption of the photon right after emitting is these same near the emitting atom is a kilometre ahead.

 

Now I think it: the Born rule (propability is the quadruple of the wave) is for matter particles. Is that also for photons?

 

 

 

I am of the belief that B) is the wave of probability associated with all matter and is in this sort of shape similar to your #13 .

And that the wave coming out of the atom is more like a chopped but nonetheless constant amplitude smooth sine. as in A)

 

However I never seem to get to the bottom of this. Everybody talks about a wave then jumps at the nearest image of a wave. [ either probability wave or light wave sinusoidal ] But people never seem very definite as to which they are talking about.

When I chatted with a professor coming out from CERN to visit schools to give talks on quantum physics. He uhmed and ahhed then said they are sort of part of the same thing, then moved on.

 

I can not see it myself, i thought it was more what you got from your german wickepedia quote [ # 55]. I can see those interfering as they go through the slits. The resultant pattern looking more like a probability wave . as shape of my B )

 

Now where the particle probability wave fits in all this ( in the context of a Photon) I am not sure. but I can NOT see each photon being a probability wave with a frequency of the light [ but maybe it is ?]. It maybe has a probability wave associated with it [ in addition to the straight forward light wave sine (A) ] giving its position a probability as per this probability wave (B), WHILE maintaining its ongoing constant amplitude sine wave , yet plank interrupted nature. [my A) ]

 

If they are the Same thing, Then we would have these Humpy bullets firing along , which with Fourier Analysis would be a very messy spectrum of harmonics and goodness knows what. I thought we have a pure line spectrum.

 

BUT I STAND TO BE CORRECTED Mike

Edited October 19, 2013 by Mike Smith Cosmos

[DParlevliet]

 

At the detector you get a particle-like interaction — the photon is localized. You get a flash on your phosphor, or a dot on your CCD (or film). That tells you nothing about the wave.

 

But there are parts of the detector where the photon is never localized. These are the places where two waves from both slits arrive which cancel each other fully. That is only possible if they have opposite sign (180 degree phase shit) and equal amplitude. If not then the cancelling is not complete.

[Strange]

But there are parts of the detector where the photon is never localized.

I don't see how that is relevant.

 

IF the photon hits the screen then it does it within the areas where you would expect constructive interference. But it does it at just one point in that area. So you cannot detect the interference pattern from one photon. Because there isn't one. You can't even detect the pattern from 2 or 10 photons.

 

 

These are the places where two waves from both slits arrive which cancel each other fully.

But, in the case of a single photon, there aren't waves from both slits.

[DParlevliet]

IF the photon hits the screen then it does it within the areas where you would expect constructive interference.

....

But, in the case of a single photon, there aren't waves from both slits.

 

How do you get constructive interference.when there is only one wave from one slit?

[Mike Smith Cosmos]

 

Of course then there is the matter of axis and dimensions......

 

E -M (electromagnetic ) waves are X Y Z coordinates X say electric field sine wave Z say magnetic field in phase sine wave Y say time.

 

Then probability wave must be some form of distance in two dimensions and time in the third dimension. I think !

Edited October 19, 2013 by Mike Smith Cosmos