Understanding the Wave Nature of Photons in Double Slit Experiments

[DParlevliet]

With a double slit measurement a single photon is a wave which goes throught both slits. To extinguish each other at certain places the wave must have the same amplitude at both slits, also at large distance between the slit. The positions also extinguish with 0.5, 1,5, ... periode difference, so also in time the amplitude is the same. If taken to the extreme it would have the same amplitude everywhere in the universe.
What is this wave? Outside the double explanations I don't see it elsewhere in publications. Is this the wave of a photon, freely traveling in space?

 

[mfb]

Is this the wave of a photon, freely traveling in space?

"Yes", but don't extend that answer too far.

If taken to the extreme it would have the same amplitude everywhere in the universe.

It will not. The amplitude is (nearly) the same in the region where you get the interference pattern if the slits have the same width and get the same intensity from the light source, but that won't happen everywhere.

 

[DParlevliet]

How does the amplitude degrade in space-time?

Or is there is formulea for this wave?

 

[jtbell]

The amplitude of the wave decreases as 1/r where r is the distance from the source, if r is large enough that the source "looks" like a point.

 

[Hyrum]

The amplitude of the wave decreases as 1/r where r is the distance from the source, if r is large enough that the source "looks" like a point.

Don't you mean $$\frac{1}{r^{d-1}}$$ where d is the dimension of space (3 in our case)?

 

[jtbell]

I assumed we were discussing the propagation of light/photons in 3-dimensional space.

 

[Hyrum]

I assumed we were discussing the propagation of light/photons in 3-dimensional space.

Right. So it should 1/r² so it expands outward as a sphere, shouldn't it?

 

[DParlevliet]

But: suppose the source is not far above the slits. For detector positions right or left from the slits the waves through both slits always have a different r. If that would mean a different amplitude they can never extinguish each other.

 

[Bill_K]

Don't you mean $$\frac{1}{r^{d-1}}$$ where d is the dimension of space (3 in our case)?

The wave amplitude goes as 1/r. The energy ∝ |amplitude|² ∝ 1/r².

 

[craigi]

How does the amplitude degrade in space-time?

Or is there is formulea for this wave?

There is a forumla for the wave. You should look up the Schrodinger equation:

http://en.wikipedia.org/wiki/Schrödinger_equation

I think you should probably read a book such as this, first:

https://www.amazon.com/dp/1841882380/?tag=pfamazon01-20

Then a more rigorous introductory text. Perhaps some one can suggest a contempory teaching book...?

 

[mfb]

But: suppose the source is not far above the slits. For detector positions right or left from the slits the waves through both slits always have a different r. If that would mean a different amplitude they can never extinguish each other.

That is true.
Single-slit effects (you always have them) are another complication that we did not consider yet here.

 

[craigi]

But: suppose the source is not far above the slits. For detector positions right or left from the slits the waves through both slits always have a different r. If that would mean a different amplitude they can never extinguish each other.

You don't need complete destructive interference to observe an interference pattern, so it is quite forgiving of your experimental setup, in that respect. The human eye is pretty good at picking out contrast. Even if the amplitude through one slit is significantly lower than through the other, the interference pattern is still visible. The peaks just aren't as high and the troughs aren't as deep, on the interference pattern.

 

[DParlevliet]

There is a forumla for the wave. You should look up the Schrodinger equation

I did, and if I had found (or understood) the solution for a photon I would not be on this forum. I am here because I hope someone can explain without all formulae of wikipedia (which therefore is only readable for you guys). If the forum is not intended for that, I wil leave.
Although I have the expression there are more here who don't know yet the answer

 

[DParlevliet]

You don't need complete destructive interference to observe an interference pattern

According Feynman (QED) there is a complete destructive interference. I don't see 1/r2 in his arrows.

 

[Cthugha]

According Feynman (QED) there is a complete destructive interference. I don't see 1/r2 in his arrows.

Please reread the response you have been given earlier, especially the "if"-part:

The amplitude of the wave decreases as 1/r where r is the distance from the source, if r is large enough that the source "looks" like a point.

Going from the standard electrical field picture at high intensities to the single photon level does not change much. You just move on from discussing fields interpreted as real entities to discussing probability amplitudes for the detection of a single photon. When averaging over many of those events, the interference patterns seen will be the same for single photons and bright light fields in the same geometry. The addition you get when discussing probability amplitudes is simply that you cannot detect a single photon twice.

Having said that, the shape of the "wave" part about the photon (the probability amplitude) just depends on geometry. The realistic thing you can get in the lab is a point like source having a single atom or a single quantum dot as the emitter which will give you the 1/r decay. In principle and especially in theory any geometry is possible. You can have a plane wave, something looking like a standard beam or shaped beams. There is no intrinsic shape of the "wave of a photon". So if Feynman does not have any 1/r terms, it is very likely that he considered a plane wave geometry for single photons. To allow us to give a more well defined answer, you might need to quote the exact text in Feynman's book.

 

[craigi]

If the forum is not intended for that, I wil leave.
Although I have the expression there are more here who don't know yet the answer

Don't think for a second that your questions aren't welcome. If they weren't, we wouldn't respond.

You should also expect that many people reading them are familiar with what it's like to try to get to grips with these things. Asking questions on here will give you some good answers, but it's not necessairily the fastest route to understanding the problems. I think many of us know what it's like to be given unsatisfying answers, even if they are well thought out answers given by people who are experts in their field.

Well written books can take you through things in a matter of hours that took others many decades of questioning, because they are written by people who know the right questions to ask to arrive at the level of understanding that you want to get to. That said, I wholeheartedly understand anyone who doesn't find book learning particularly appealing and never let anything discourage your inquisition.

 

[DParlevliet]

Feynman is talking about amplitudes of single photons to follow a certain path (in his book QED). In fig 20 (a mirror) he say (translated back from Dutch):"According quantum theory light has an amplitude for reflection which is equal for every position on the mirror". Different paths give the same amplitude but different phase (direction of the arrow). Fig 5 shows reflection in glas depending on thicknes, which is a cosine between 0-16%, even with glass of more then 50 meters thick.
Then in fig 49 he shows the double slit and mentions (agains translated) "sometimes we get for a certain distance between the holes more ticks then expected, with a somewhat different distance the dector does not tick at all" (the detector is a photon counter).
According Wikipedia: "and when they are in anti-phase, i.e. the path difference is equal to half a wavelength, one and a half wavelengths, etc., then the two waves cancel and the summed intensity is zero"

 

[vanhees71]

I don't know what reference of Feynman's you are referring to. His book on QED is pretty conventional using the old Fermi formulation in the operator formalism. This surprised me a bit, because the path-integral quantization method is most convenient particularly for gauge theories as compared to the operator formalism.

Anyway, one cannot stress often enough that a naive particle picture for massless quanta is almost always wrong. The first thing is that for massless quanta with spin \geq 1 there is not even a position operator in the strict sense.

By definition a single-photon state is an asymptotically free one-particle photon state with definite norm.

The double-slit experiment with single photons is described as any double-slit experiment with single quanta as a scattering process with asymptotically free single-particle states coming in and asymptotically free single-particle states coming out. This leads to the probabilities to register an asymptotically free particle beyond the double slit (e.g., using a photo plate). A single quantum never makes an intereference pattern but only a single dot on the screen. The interference pattern, reflecting the detection probabilities as a function of position, can be found by repeating this scattering experiment several times.

Note that nowhere I have used the idea of position of a single photon but only the possition of a registration of a photon at the screen. This position of a registration of a photon is a well defined physical quantity that can be measured, the position of a photon in the strict sense of an observable cannot even be defined in principle! For details, see

http://arnold-neumaier.at/physfaq/topics/position.html

 

[craigi]

Feynman is talking about amplitudes of single photons to follow a certain path (in his book QED). In fig 20 (a mirror) he say (translated back from Dutch):"According quantum theory light has an amplitude for reflection which is equal for every position on the mirror". Different paths give the same amplitude but different phase (direction of the arrow). Fig 5 shows reflection in glas depending on thicknes, which is a cosine between 0-16%, even with glass of more then 50 meters thick.
Then in fig 49 he shows the double slit and mentions (agains translated) "sometimes we get for a certain distance between the holes more ticks then expected, with a somewhat different distance the dector does not tick at all" (the detector is a photon counter).
According Wikipedia: "and when they are in anti-phase, i.e. the path difference is equal to half a wavelength, one and a half wavelengths, etc., then the two waves cancel and the summed intensity is zero"

Do you remember in the other thread we talked about needing a surface for specular reflection?

Feynman is describing, using path integrals, how the different paths across this surface, superpose to cancel each other out, with the exception of the apparent reflection path from Fermat's law of reflection. He does mention that he is making an approximation when he says that the the amplitude is the same for all points on the section of surface. The approximation holds when the section of mirror that he is treating is very small, hence the distance that the light travels to each point is approximately the same. This is very similar to the approximation used in calculus, where you take 2 points on a curve which are close together to calculate its gradient.

The cool thing about this is that he's demonstrating that the reflection from a mirror can be considered an interference patten itself.

 

[Naty1]

Hi DParlevliet...getting a 'picture' of this stuff takes some time, so be patient...Expect your head to spin a few more times!
I'll see if I can put some pieces together to clarify and summarize some subtle points...no math because I remember relatively little of it!

From the Wikipedia link already provided:

Interpretation of the wave function

...The Schrödinger equation provides a way to calculate the possible wave functions of a system and how they dynamically change in time. However, the Schrödinger equation does not directly say what, exactly, the wave function is.

Here is another very insightful tidbit from another Wikipedia article:

In quantum mechanics, the Schrödinger equation, which describes the continuous time evolution of a system's wave function, is deterministic. However, the relationship between a system's wave function and the observable properties of the system appears to be non-deterministic… A deterministic model will thus always produce the same output from a given starting condition or initial state.

So this IS crazy: a deterministic expression for quantum behavior! Further, after almost 100 years, arguments still ensue about what the wave function 'really' represents.

Post #2:

Is this the wave of a photon, freely traveling in space?
"Yes", but don't extend that answer too far.

[This 'wave' can be thought of as representing some distributed photon behaviors, a probability distribution relating to likely detection location, but says nothing about observables of this mode. When detected, photons are always pointlike as are all particles in the Standard Model of particle physics. As wiki says, oddly enough, it is a deterministic expression, yet measurements/observables based on it are NOT deterministic.]
edit: Particle 'wave' characteristics are always detected as pointlike objects.

post #9:

The wave amplitude goes as 1/r. The energy ∝ |amplitude|2 ∝ 1/r2.

Turns out the probability of locating a particle is also proportional the the amplitude squared.

post #15:

Going from the standard electrical field picture at high intensities to the single photon level does not change much. You just move on from discussing fields interpreted as real entities to discussing probability amplitudes for the detection of a single photon. When averaging over many of those events...

Haven't seen that before...I like it...Bravo!

In Wikipedia terminology, the 'standard electrical field picture' is deterministic, the single photon is a quantum particle, and measurements [averaging over many events ] turns out to be NON deterministic. Nobody knows why.

Analogously, here is what Roger Penrose says:

...The way we do quantum mechanics is to adopt a strange procedure which always seems to work...the superposition of alternative probabilities involving w, z, complex numbers...are an essential ingredient of the Schrodinger equation. When you magnify to the classical level you take the squared modulii (of w, z) and these do give you the alternative probabilities of the two alternatives to happen...

[and THAT changes things from deterministic to probabilistic!]

There is no intrinsic shape of the "wave of a photon"

[It depends on geometry.]

Post#16:

Well written books can take you through things in a matter of hours that took others many decades of questioning,

a great point...so when you mumble to yourself [as many of us have at times] "That seems crazy." it probably is. It was unlikely not scientists first choice of interpretation...Feynman says something like this about that:

A good physicist is one who has the stubbornness to make all the mistakes possible before finally arriving at the correct conclusion.

 

[Bill_K]

A good physicist is one who has the stubbornness to make all the mistakes possible before finally arriving at the correct conclusion.

"Learning is the process of making progressively subtler mistakes." -- Eleanor Duckworth

 

[DParlevliet]

I refer to "QED, The Strange Theory of Light and Matter", his New Sealand lectures which are often referred to on Youtube.

... He does mention that he is making an approximation when he says that the amplitude is the same for all points on the section of surface. The approximation holds when the section of mirror that he is treating is very small, hence the distance that the light travels to each point is approximately the same. This is very similar to the approximation used in calculus, where you take 2 points on a curve which are close together to calculate its gradient.

He is saying equal, not approximate equal. Using small dx does not matter, because how small it is, it is the same over the whole mirror surface, so 1/r2 (it is about area) still applies. In his figure 24 the paths differ 1.4, which with 1/r2 is about 50% difference in arrow. That is not approximate. All arrows in his drawing are the same.

But has someone references that with double slits the two waves does not cancel out completely because of 1/r2?

 

[craigi]

I refer to "QED, The Strange Theory of Light and Matter"

Yup. I read chapter 2 from it earlier.

Have another read of it and if you still don't find it, I'll get a quote from it for you about his approximation.

 

[birulami]

With a double slit measurement a single photon is a wave which goes throught both slits. To extinguish each other at certain places the wave must have the same amplitude at both slits, also at large distance between the slit. The positions also extinguish with 0.5, 1,5, ... periode difference, so also in time the amplitude is the same. If taken to the extreme it would have the same amplitude everywhere in the universe.
What is this wave? Outside the double explanations I don't see it elsewhere in publications. Is this the wave of a photon, freely traveling in space?

My hunch is you want to understand a wave in general first, whether it is electromagnetic or probability amplitude. For me it was quite helpful to read about the wave equation (not wave function!) as well as the Airy disk. To me the latter was an eye opener, because it seems a single hole already shows the wave properties of photons nicely, why always double slits? Combine what you read there with the Huygens principle and you start getting a good idea what peculiar objects 3D waves are.

Further search for solutions to the wave function on the net to find that indeed a plane wave would have the same amplitude on a plane spanning the whole universe while, interestingly, a spherical wave must have decreasing amplitude as it spreads out.

If a single hole is the wave source, like with the Airy disk, the wave is nearly spherical.

The really tough part for me is to consider the wave spreading from the hole not just to a screen in the lab, but over several light years: the amplitude must decrease to nearly nothing, yet when a part of the wave front finally hits an object, the (probability) amplitude, integrated over thee area hit, tells us how probable it is that the wave leaves its energy as a blip on this object.

(P.S. read wikipedia always in all the languages you understand. The articles have differing content.)

 

[DParlevliet]

Yup. I read chapter 2 from it earlier. Have another read of it and if you still don't find it, I'll get a quote from it for you about his approximation.

I found it, somewhat hidden. That is the risk of using educational text. So: 1/r²

But now suppose case 2: the light source has a curved mirror which gives a parrallel beam. The wave fronts are now flat, no 1/r². The interference will be complete, giving positions on the detector where the waves cancel fully, 0% light. That is conformal with measurements, which try to use a parrallel beam.
But now with one photon. In case 2 you don't know where the photon will be absorbed by the detector, but you do know positions wher it certainly never will be detected. But in case 1 (with a single photon) at these positions the waves does not cancel each other fully, so there is a small change the photon will be absorbed. It seems that the photons in case 1 and case 2 differ: they have a different wave shape. That troubles me.

1/r² is right as propability wave if you only know that a photon is generated but not in what direction. But in reality it went a certain direction, which you know afterwards when the photon was detected. Then there are two possible paths (through slit 1 or 2), so if you reconstruct the wave shape as it was before the slits, what was it then?

Concluding from the measurement result I suppose what I mentioned before: a wave with flat wave fronts, everywhere in the universe (if without matter), with equal maximum amplitude. It is of course not a propabality wave, but is does determine propability at the moment the photon absorbed (or perhaps direct the photon).

 

[mfb]

I found it, somewhat hidden. That is the risk of using educational text. So: 1/r²

But now suppose case 2: the light source has a curved mirror which gives a parrallel beam. The wave fronts are now flat, no 1/r². The interference will be complete, giving positions on the detector where the waves cancel fully, 0% light. That is conformal with measurements, which try to use a parrallel beam.
But now with one photon. In case 2 you don't know where the photon will be absorbed by the detector, but you do know positions wher it certainly never will be detected. But in case 1 (with a single photon) at these positions the waves does not cancel each other fully, so there is a small change the photon will be absorbed. It seems that the photons in case 1 and case 2 differ: they have a different wave shape. That troubles me.

The shape of the wave depends on the light source. What is surprising here?

1/r² is right as propability wave if you only know that a photon is generated but not in what direction. But in reality it went a certain direction, which you know afterwards when the photon was detected.

Before the detection, it is not reasonable to say "the photon went in a specific direction". And after the detection, there are no probabilities to consider, as you have certainty then.

Then there are two possible paths (through slit 1 or 2), so if you reconstruct the wave shape as it was before the slits, what was it then?

Depends on your light source.

 

[craigi]

I found it, somewhat hidden. That is the risk of using educational text. So: 1/r²

But now suppose case 2: the light source has a curved mirror which gives a parrallel beam. The wave fronts are now flat, no 1/r². The interference will be complete, giving positions on the detector where the waves cancel fully, 0% light. That is conformal with measurements, which try to use a parrallel beam.
But now with one photon. In case 2 you don't know where the photon will be absorbed by the detector, but you do know positions wher it certainly never will be detected. But in case 1 (with a single photon) at these positions the waves does not cancel each other fully, so there is a small change the photon will be absorbed. It seems that the photons in case 1 and case 2 differ: they have a different wave shape. That troubles me.

1/r² is right as propability wave if you only know that a photon is generated but not in what direction. But in reality it went a certain direction, which you know afterwards when the photon was detected. Then there are two possible paths (through slit 1 or 2), so if you reconstruct the wave shape as it was before the slits, what was it then?

Concluding from the measurement result I suppose what I mentioned before: a wave with flat wave fronts, everywhere in the universe (if without matter), with equal maximum amplitude. It is of course not a propabality wave, but is does determine propability at the moment the photon absorbed (or perhaps direct the photon).

The 1/r² is what you'd expect from an omni-directional radiation source, in classical physics. You don't need quantum physics to explain that. The surface area of a sphere increases as r², so the intensity at any location on that sphere, from a radiation source at its centre, is proportional 1/r².

I think you're still misunderstanding that complete destructive interference still occurs with an omni-directional light source in the double slit experiment, providing the source is equidistant from the 2 slits.

You're right that the photon wave functions are different with different geometrical set ups. In your second case where you place a mirror around your omni-directional light source to collimate your light, that changes the wave function. The section of Feyman's book that you were reading demonstrates that. The other notable example in what we've been discussing is that placing an opaque barrier with 2 slits in it changes the wave function of the photon. It is exactly this change in the wave function that gives rise to the interference pattern.

Sure, once a photon is detected, its full intensity is found at that location. This is what is often referred to in as wave-function collapse. I should re-iterate that the wave determines the probabilty of finding a photon at a particular location. It is that probability that is proportional to 1/r² for an omni-directional light source.

You say "in reality it went a certain direction". Nope it really didn't. There is no such reality. The double slit experiment, amongst others, demonstrates that. "It" takes all possible paths. These paths interfere with each other to determine the probability of where it will be found in the future. This is what we call wave-particle duality. It's not a wave and it's not a particle at a single location. It has properties of both.

It takes a while to get your head around, but that's the way the world works. Amazing isn't it?

Let me give you an analogy that might help. If I were to show you an animal that you'd never seen before, a bat, you might at first think it's a bird. If you looked a bit more closely you might think that it's a mouse, but the reality is, that it's not a bird or bat, or a mouse stuck to a bird. It's an entirely different thing that you'd never seen before, that happens to be a bit like a bird and a bit like a mouse. Particles are the same, they're a bit like a wave and a bit like a point-particle, but they're neither. In many ways, it's shame that we kept the word particle for them, because I expect that it leads to a lot of confusion when learning about them. Perhaps a duck-billed platypus would be a better analogy.

 

[DParlevliet]

No, I described that there is no complete destructive interference with an omni-directional light source in the double slit experiment. I agree with that.

But in case 2, the omni-directional light source is replaced by a source with a parrallel beam. All photons goes one direction. What is the shape of the wave of one photon when it approaches the slits? I suppose a flat wave because there is no 1/r2 in parrallel light (the surface area does not increase).

 

[mfb]

No, I described that there is no complete destructive interference with an omni-directional light source in the double slit experiment. I agree with that.

There can be, it depends on the setup.

But in case 2, the omni-directional light source is replaced by a source with a parrallel beam. All photons goes one direction. What is the shape of the wave of one photon when it approaches the slits? I suppose a flat wave because there is no 1/r2 in parrallel light (the surface area does not increase).

Right.

 

[craigi]

No, I described that there is no complete destructive interference with an omni-directional light source in the double slit experiment. I agree with that.

Then you still misunderstand the experiment. If the intensity at each slit is equal, from a single light source, then complete destructive interference occurs at all points where the difference in distance from each slit is one half wavelength.

Again, this isn't specific to quantum mechanics. You'll get the same result with classical water waves and a double slit.

 

[DParlevliet]

Then you still misunderstand the experiment. If the intensity at each slit is equal, from a single light source, then complete destructive interference occurs at all points where the difference in distance from each slit is one half wavelength.

I thought you told before that there is no complete destructive interference because there must be difference in r: one path l, second path l + half wavelength. And difference in wavelength is difference in 1/r2

 

[DParlevliet]

But in case 2, the omni-directional light source is replaced by a source with a parrallel beam. What is the shape of the wave of one photon when it approaches the slits? I suppose a flat wave.

Right.

- How far to the left and right does this flat wave extend (compared to the position of the slits)?
- The photon was generated upwards in the parrallel beam, did it directly after emission also have this flat wave?

 

[jtbell]

- How far to the left and right does this flat wave extend (compared to the position of the slits)?

It extends across the width of the beam, which in turn depends on the size and construction of the apparatus that produces it.

 

[craigi]

I thought you told before that there is no complete destructive interference because there must be difference in r: one path l, second path l + half wavelength. And difference in wavelength is difference in 1/r2

Apologies. Now I see what you mean.

The visible spectrum has wavelengths of 380-750 nm. Half a wavelength is say, 300 nm, this does introduce a very small difference in amplitude from the components of the wave function from each slit, as we move from the central point of the interference pattern.

If the distance between the slits and the screen is about 1m, the intensities from each component at the first minimum on the interference pattern, have a ratio of about 1+6x10^-7. This difference is less than 1 part in 1 million and is negligble, but strictly, you are correct this isn't complete destructive interference. In reality, you would never be able to get the light source equidistant from the 2 slits to this level of accuracy.

You should be aware that this is also true of a parallel light beam, as in your other example, since the slits diffract the light and you need to measure the distance from the slit to the point on the interference pattern.

 

[DParlevliet]

It extends across the width of the beam, which in turn depends on the size and construction of the apparatus that produces it.

So if (as thought experiment) the beam is a kilometer wide (still parrallel), the wave would extend for a kilometer with the same amplitude?

 

[craigi]

So if (as thought experiment) the beam is a kilometer wide (still parrallel), the wave would extend for a kilometer with the same amplitude?

If beam were perfectly collimated and evenly distributed, then yes, the wave would have exactly the same amplitude everywhere.

 

[DParlevliet]

Apologies. Now I see what you mean.
You should be aware that this is also true of a parallel light beam, as in your other example, since the slits diffract the light and you need to measure the distance from the slit to the point on the interference pattern.

... first I thought that I indeed did overlook that in my earlier post. But what is r with a parrallel beam? With an omni-directional light source this is the distance between source and position of absorption. When you move the light source further away, r becomes longer, r / (r + halve wavelength) closer to 1. When the source is infinite far away, the beam is parrallel... and waves equal

 

[DParlevliet]

If it were perfectly collimated and evenly distributed, then yes, the wave would have exactly the same amplitude everywhere.

And just after emission of this photon (suppose it is electronicly triggered emission),
- are there waves in front of the emittor?
- do they have equal amplitude towards the slits?
- If so, how far (in the direction of the experiment) does the waves extend?

 

[craigi]

... first I thought that I indeed did overlook that in my earlier post. But what is r with a parrallel beam? With an omni-directional light source this is the distance between source and position of absorption. When you move the light source further away, r becomes longer, r / (r + halve wavelength) closer to 1. When the source is infinite far away, the beam is parrallel... and waves equal

This isn't strictly true. You need to follow the path. In the case of a point source. It's the distance from the source to the slit plus the distance from the the slit to the point on the screen.

For a parallel coherent beam hitting the slits, you can just measure r from the slits.

The thing to remember here is that the slit spreads out the light through diffraction. The spread is not uniformly omni-directional, the idea that the intensity is proportional to 1/r² from that point forward, is a crude approximation. When the wave from an omni-directional point source his the slits, the intensity at the slit, is actually proportional to 1/r².

And just after emission of this photon (suppose it is electronicly triggered emission),
- are there waves in front of the emittor?
- do they have equal amplitude towards the slits?
- If so, how far (in the direction of the experiment) does the waves extend?

-yes, but the common terminology would be "a wave". Though waves can be made from a combination of other waves. Remember superposition? We'd tend to start off with the notion of one wave per particle, but can consider that going on to produce 2 new waves when it hits the slits for example, but they can equally, be considered all the same wave.
-yes
-it goes off to infinity until it hits something. It propogates forward at the speed of the particle (the speed of light, in the case of the photon).

 

[DParlevliet]

The thing to remember here is that the slit spreads out the light through diffraction. The spread is not uniformly omni-directional, the idea that the intensity is proportional to 1/r² from that point forward, is a crude approximation. When the wave from an omni-directional point source his the slits, the intensity at the slit, is actually proportional to 1/r².

Alright, now I see. It has two steps. First the r between source and slits, then a new r between slits and absorption. The slits are then new sources (which in fact they are).
So if the source is in the middle the somewhat less destructive interference will be the same for an omnidiractional source and parrallel beam.

 

[DParlevliet]

If the wave has traveled half way (calculated with emission time and light speed
- is there also a wave behind this (half-way) position?
- does it have the same (maximum) amplitude backwards?
- if so, does it extend backwards until the emittor?
- does it proporgate in the same direction as the above discussed "in front" wave?

 

[craigi]

If the wave has traveled half way (calculated with emission time and light speed
- is there also a wave behind this (half-way) position?
- does it have the same (maximum) amplitude backwards?
- if so, does it extend backwards until the emittor?
- does it proporgate in the same direction as the above discussed "in front" wave?

Remember that the wave represents the probability of finding the particle and a photon always travels at the speed of light.

 

[DParlevliet]

Remember that the wave represents the probability of finding the particle and a photon always travels at the speed of light.

Yes, therefore I am interrested in the answers. With the wave in front I tought I got you in my trap, but now you become suspicious :)

Interference takes place between waves which are shifted 0.5, 1.5... in phase, so shifted in time, so there must about equal wavelengths before and after the absorbed photon. On the other hand the propability of the photon being before or after its average (light speed) calculated position becomes smaller. What shape will saftisfy both demands? I hope you will find an answer.

 

[craigi]

Yes, therefore I am interrested in the answers. With the wave in front I tought I got you in my trap, but now you become suspicious :)

Interference takes place between waves which are shifted 0.5, 1.5... in phase, so shifted in time, so there must about equal wavelengths before and after the absorbed photon. On the other hand the propability of the photon being before or after its average (light speed) calculated position becomes smaller. What shape will saftisfy both demands? I hope you will find an answer.

What was the trap and what should I be suspicious of?

 

[DParlevliet]

And just after emission of this photon (suppose it is electronicly triggered emission),
- are there waves in front of the emittor?
- do they have equal amplitude towards the slits?

-yes
-yes

Then just after emission of this photon the propability that this photon arrives at the slit (is able to be absorbed there) is the same as the photon is still near the emittor. That cannot be true, because as you mentioned later on a photon always travels at the speed of light, so needs some time to arrive at the slit.

But if there is no trap (it is still QM): what are the answers on my last questions of the backward wave part?

 

[craigi]

And just after emission of this photon (suppose it is electronicly triggered emission),
- are there waves in front of the emittor?
- do they have equal amplitude towards the slits?

Then just after emission of this photon the propability that this photon arrives at the slit (is able to be absorbed there) is the same as the photon is still near the emittor. That cannot be true, because as you mentioned later on a photon always travels at the speed of light, so needs some time to arrive at the slit.

But if there is no trap (it is still QM): what are the answers on my last questions of the backward wave part?

I think I see what you mean now.

What I meant was that the wave infront of the emitter got there by propogating at the speed of light from the source and that for an evenly distributed source, the amplitude is equal at each slit when it gets there.

What you're talking about is an equal amplitude across the length of a beam for a single photon. This doesn't happen, in the observer's rest frame. At risk of confusing this thread even further, a photon in a vacuum travels no distance in its own rest frame. The entire universe is contracted along its direction of travel.

Where did you get the idea of the backwards wave from? Why would there be a backwards wave? Did this originate from the idea that the wave for a single photon has the same amplitude across the entire length of the beam?

To be clear, QM waves must overlap in space and time for interference to occur.

 

[DParlevliet]

I think I see what you mean now.
Why would there be a backwards wave? Did this originate from the idea that the wave for a single photon has the same amplitude across the entire length of the beam?
To be clear, QM waves must overlap in space and time for interference to occur.

I did not mean backwards, but at the back. I don't expect the same amplitude. I expect something like this: [PLAIN]http://upload.wikimedia.org/wikipedia/commons/b/b0/Wave_packet_%28dispersion%29.gif. So to come back to the start of this topic: what is the shape.

 

[craigi]

I did not mean backwards, but at the back. I don't expect the same amplitude. I expect something like this: [PLAIN]http://upload.wikimedia.org/wikipedia/commons/b/b0/Wave_packet_%28dispersion%29.gif. So to come back to the start of this topic: what is the shape.

The animation that you have there is commonly used to decipt the shape of a photon.

Is there something about it that you're not happy with?

 

[DParlevliet]

It is the shape of a particle with mass (solution of SV in Wikipedia). Probably a photon will look alike, but different in length. I suppose there is a formulea for the envelop of the peaks?

I started the topic with the assumption that at all "cancel" positions on the detector both waves cancel fully, and then the wave of a photon must be equal in size everywhere. If that is not the case, the wave (tops) decrease in time, and I like to know how fast.

 

[craigi]

It is the shape of a particle with mass (solution of SV in Wikipedia). Probably a photon will look alike, but different in length. I suppose there is a formulea for the envelop of the peaks?

I started the topic with the assumption that at all "cancel" positions on the detector both waves cancel fully, and then the wave of a photon must be equal in size everywhere. If that is not the case, the wave (tops) decrease in time, and I like to know how fast.

The "wave tops" being the amplitude of the wave?