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Perhaps it is better not to mix two topics? I mean the photons which are used for the normal double slit measurements. 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.

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.

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 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.

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.

In the double slit the propability is measured of detecting a photon on a certain position on the detector. Because for each photon this is the same propability, you can use multiple photons to get a full curve. That is what I think. The question is now: what is the formula of this not-constant amplitude. I would already be happy with a approximation of (for instance) its FWHM.

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

Do you know the interference equation? Please read above, perhaps my use of shape is wrong and confusing. Better is perhaps to call it just the wave itself. Or how does the wave look like when you plot E (in an EM wave) against time.

A bit. Perhaps the word shape is chosen wrong. Englisch is not my native language and certainly not the scientific use. Therefore later on I use "looks like", which is of course not scientific either. For instance if my previous used is an EM wave with x = time and y = E, then I would have called that the shape of a sine with amplitudes as a gausian curve. So perhaps must I call it a curve, or the envelope of the amplitudes. Do you mean with the bubbles the particles in their indeterminable position?

But I also explained how the double-slit measurement shows how the waves which interfere look like. Then it is just simply calculating back with interference rules. So if anyone knows the formula of the interference result on the detector, then we know the shape of the wave which did cause the interference.

The graph image has: x-axis is time, y-axis is "the real part of the wave function" according Wiki. I don't know what that is. In my question X-axis is time, y-axis is what interferes in the double slit experiment, whatever that is.

I am sorry but I am not always availble to answer (my company would not allow that) so I cannot answer quickly. Also it seems that I don't understand you questions, because I really try to answer them. So let me start again, first keeping it simple. If you search on the web for double slit images and animations you will always see flat (sinus) wavefronts moving to the slits. My question is: how does that look with only one photon? Then seen from the double slit: this experiment is based on adding two parts of the same wave, which are separated in space (by the distance between the slit) with increasing phase difference. See the image before and picture in two of them with increasing phase. In the middle the phase difference is zero, so both add to double value. Going to the right the phase difference increases until the waves cancel each other. But when going much further to the right the shift between both waves becomes so large that only the tails interfere. Those are lower, so will cause smaller interference. Even futher to the right both waves misses each other completely and no interference pattern will be visible. So my argument is, that the interference pattern on the detector is a kind of image of the original wave(packet).and can be calcuated back to the incoming wave. The fact that this pattern is clearly visible means that most photons which are detected has the same wave/packet/shape

According Wiki "A wave function that satisfies the non-relativistic Schrödinger equation with V=0. In other words, this corresponds to a particle traveling freely through empty space. The real part of the wave function is plotted here." But I used it just to show the shape. Of course my next question is about the other space axes, but I can easier proove that the time sinus exist (by the interference pattern)

But the interference pattern of the double slit does not show an arbitrary waveform. With precise date from the detector one could calculate back how wide the sine is. I don't have that data. But I supposed there would be an (appoximation) formula for the envelop of such a basic wave.

I mean in time, so along the axis in which the wave moves

- And that wave equation results in a sine I suppose? - For one photon, does the wave have a certain length (for instance FWHM) in time? - Is the quadrature here also the propability?

Look to the double-slit interference: there are two parts added of the same wave (of a photon) which are sliding against each other because of the varying phase shift given by the instrument. If it is a sine then the original wave must me a sine. The same with the envelope of the amplitudes. That is the wave I mean. In QM a photon is ór a wave (packet) ór a particle (not both), depending on what you measure. That is the wave I mean In the Copenhagen interpretation "A system is completely described by a wave function, representing the state of the system, which evolves smoothly in time" (Wiki). That is the wave. Normally the wave is a solution of Schrodinger equation, but in another forum it was told that Schrodinger has no solution for photons. The quadrature of the modus of the wave is the propability (as it is for solutions of the Schrodinger equation). Wiki gives for Schrodinger solution for a particle with mass: I suppose a photon will look a bit like this. The question was about one photon.

The question is not about the production of a photon, but about its shape when free running in space, or better when it arrives at the double slits experiment

Finally someone who think its important Why a few centimeters long? Where is that based on? After travelling from the sun to my garden, then it is wider then my garden (and smaller then my town)?

I have a question about the shape of the wave of a single photon entering the double-slit experiment, in time and space. Although I thought it was basic, it seems to be difficult because I did try other fora and Wiki, but nobody could give an answer. This is my last shot. Of course the interference pattern shows the shape, the envelop, in time. But because I don't have accurate measurement data (most pixtures of the pattern are simple and idealised) I cannot construct it from there. So I wonder if there is the (rough) formula, only as an approximation. Not exactly, because that will probably be of a high mathematical level.