# Shape of the wave of a single photon

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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.

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I don't think there's a definite answer to this. Any wave that satisfies Maxwell's equations is a valid "shape" for a photon.

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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.

WHAT YOU ASK IS A VERY GOOD QUESTION!

We speak about photons as waves but just what quite do they look like ?

I have pondered the following. illustration.

Also on top of this I believe I saw an example of quantum wave production.

First though the possible illustrations

Edited by Mike Smith Cosmos
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What are the shape and the dimensions of a typical sunlight photon wave?

Couple of centimeters long. Amplitude is largest at the middle, amplitude decreases towards the front and the rear. Wider than the earth, after travelling from sun to earth.

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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)?

Edited by DParlevliet
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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 tried to produce an artists illustration of what I thought was going on accoding to the various quantum explanations .

attached

The photon is the bit shooting out at the speed of light after the electron has jumped from higher to lower orbit .

Hope of some help. just a visualisation ( maths boffs dont like these, but I need them to remain sane ! )

Mike

Edited by Mike Smith Cosmos
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ya its right according to quantum physics that electron emits energy as photon when jumps from higher to lower orbit but can you suggest that how much energy or how much photon it can release at the time of orbit change??? and how much time it continue the jumping process???

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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

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Can you elucidate further what you are expecting?

I mean the 'shape' of what exactly?

The photon is better modelled as a wave packet. It is not a wave in the sense of a sine curve, though it is often drawn as one.

In fact a sine curve is one it can never be, simply because the sine curve periodically equals zero and the so called probability 'wave' is never zero anywhere in the universe.

Edited by studiot
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My understanding is that photons (and similar) are not *actual* waves but sometimes it's useful to think of them as *like* waves. In the same way they are not *actual* particles, but sometimes it'se useful to think of them as being *like* particles.

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Finally someone who think its important

Saying that there is no definitive answer is not the same thing as declaring it as unimportant.

I tried to produce an artists illustration of what I thought was going on accoding to the various quantum explanations .

Make no mistake — this is a guess, nothing more.

I reiterate: the physics that the photon must follow is the wave equation as described by Maxwell's equations. That's what you get, even in QM, when you try and describe light's behavior when it's not interacting with anything. Without boundary conditions to apply, you can't narrow this down to a specific wave shape. Any wave that satisfies the equation is a possibility.

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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)?

The length is just a some fact I happen to remember very faintly. Maybe the length was actually couple of meters.

Why that length? The length depends on how long time the emission process takes.

An excited hydrogen atom, decaying spontaneously to the ground state from the 2p state decays in 1.6 x 10^-9 seconds, or 1.6 nanoseconds.
1.6 x 10^-9 seconds * speed of light = 0.48 meters

0.48 m is some kind of average length of these photon waves.

Edited by Toffo
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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.

Why that length? The length depends on how long time the emission process takes.

The question was about one photon.

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The question was about one photon.

Edited by Toffo
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The length is just a some fact I happen to remember very faintly. Maybe the length was actually couple of meters.

Why that length? The length depends on how long time the emission process takes.

An excited hydrogen atom, decaying spontaneously

to the ground state from the 2p state decays in 1.6 x 10^-9 seconds, or 1.6 nanoseconds.

1.6 x 10^-9 seconds * speed of light = 0.48 meters

0.48 m is some kind of average length of these photon waves.

Um, no. The hydrogen 2p-1s transition has an energy of 10.2 eV, which gives it a wavelength of 121.5 nm, just as your link says. There's nothing there that implies it has a length of half a meter.

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.

And that's correct — the Schrödinger equation only works for massive particles and is non-relativistic. If you want to talk about photons, the relativistic wave equation is the one from Maxwell's equations. As I have already stated.

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I seem to recall that the speed of light, say a photon of light, is roughly a

Foot in a nanosecond.

A nanosecond is a billionth of a second (1/1,000,000,000 second). A foot is roughly the length of a big foot (approx 1/3 meter)

So this photon (whatever it looks like) will slip by in front of you " through 1 foot in billionth of a second.

By the time you think about it it will be two thirds of the way to the moon. (1 second).

The frequency of light must be up in the Tera hertz (1,000,000,000,000 cycles or single waves per second))

So swansont is saying hydrogen orbit change is 1.6x 1/1.000,000,000 seconds

So 1.6 x 1/1,000,000,000 X. 1,000.000,000,000 Is 16,000 cycles waves oscillations in that hydrogen photon

Or is my reasoning all to pot !

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And that's correct — the Schrödinger equation only works for massive particles and is non-relativistic. If you want to talk about photons, the relativistic wave equation is the one from Maxwell's equations. As I have already stated.

- 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?

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- 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?

That's a pretty wavelet picture.

But you need to label your axes.

What are you plotting against what, which was in effect my orignal request about your question and the proper basis (I think) for your answer.

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

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- And that wave equation results in a sine I suppose?

Sine is one possible solution. But so is a superposition of sines, and you can construct any arbitrary waveform from that.

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StudioT: But you need to label your axes.

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

I'm really sorry to rain on your parade, but how many axes did your picture have and how many did you label?

Edited by studiot
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Sine is one possible solution. But so is a superposition of sines, and you can construct any arbitrary waveform from that.

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.

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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.

The interference pattern depends on the wavelength and the separation of the slits. But the wavelength is not the same as what the "shape of the wave" is.

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To figure out the typical envelopes of photon waves in a light beam:

1: Set up double slit experiment were the path length difference is adjustable.

2: Start adjusting the path length difference.

2: Record the percentages of light that interfered and light that did not interfere at different path lengths.

3: The path length difference were half of photons did not interfere with itself is such a path length difference that makes half of the photon waves non-overlapping photon waves. Half of the photon waves are shorter than this length.

Edited by Toffo
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Am I wasting my time or do you want an answer to your original question?

Swansont et al have told you the truth about waves and slits but their discussion is not about shape.

The vibrating string analogy is useful in some respects but fallacious in others. And 'shape' is one of these others.

The wave variable in a vibrating string is displacement. Displacement in say the y direction if propagation is in the x direction. So both x and y are distances.

This allows the equation to directly display the two dimensional shape (in the x-y plane) taken up by a vibrating string.

In sound waves, however, the wave varaible, is pressure at a point. This does not take on a shape in space in the same way.

Can you imagine the 'shape' of a sound wave? You can only draw pressure contours of the sound field.

The Schrodinger equation is similar in this respect to the sound field equation. The wave variable is a derived quantity, not even as recognisable as pressure. The axis variable is certainly not a distance, so again the solution does not take on a 'shape' in space.

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