What interference pattern of double-slit experiment does mean

[vijayantv]

I have a doubt about the interference wave pattern of double slit experiment with light beam.
does it mean the interference of electromagnetic wave (which tells about the frequency causing color) or Spatial probability wave (which was explained by Max Born).

I am not good at math. can someone please explain with math-free manner.

 

[Simon Bridge]

Welcome to PF;
There is no totally math-free manner - the short, non-math, answer is "both" - sort-of.

The distribution of photons is described using the spatial wavefunction.
The spatial probability amplitudes and EM waves are closely related to each other - so either approach is valid.

 

[vijayantv]

Welcome to PF;
The spatial probability amplitudes and EM waves are closely related to each other - so either approach is valid.

Thanks for the information.

double slit with single photon takes long time to get accumulated and shows interference pattern. but stream of continues photons shows interference immediately. Are both different phenomenon? or how both can be related if both are same phenomenon?.

 

[Simon Bridge]

Both are the same phenomenon.
The fact that it works for the single photon indicates that the result is not caused by EM field interference.
Rather the EM field is an emergent phenomenon of the statistics.

On the scale of individual particles, Nature is statistical.
Over many particles, individual effects get averaged out to give you the classical behavior.

 

[vijayantv]

ok. so the interference is caused by two waves regardless of intensity and time, fine.

light has photons of so many different polarizations. but we are seeing the interference with only two wave. can we consider that polarization doesn't matter in this interference pattern?

 

[Simon Bridge]

ok. so the interference is caused by two waves regardless of intensity and time, fine.

Nope - not what I said.
There is only one wavefunction - this is why the better descriptions have math.
But the wavefunction does not depend on intensity.

light has photons of so many different polarizations. but we are seeing the interference with only two wave. can we consider that polarization doesn't matter in this interference pattern?

The effect of different polarizations on the interference pattern is well documented - it's something you can look up.

But the statistics works at an individual-particle level.
You should probably see these lectures: http://vega.org.uk/video/subseries/8

 

[vijayantv]

On the scale of individual particles, Nature is statistical.
Over many particles, individual effects get averaged out to give you the classical behavior.

What is classical behaviour how it differs from statistical?

 

[Simon Bridge]

What is classical behavior how it differs from statistical?

Classical behavior is basically anything that follows Newton's Laws and the rules in his Principia.
Statistical behavior is anything that needs probability math to describe.

For examples - watch the lecture series I gave you in post #6.

But: if you really don't know what "classical physics" or "statistical" means, then I cannot help you because the answers require a much higher education level than you have in order to understand them. It would take years.

 

[vijayantv]

I know something about Classical and quantum physics

So, if the light has very low intensity (1 photon or 2 photons or 3 photons...) then the double-slit will show probability wave interference. on the other hand the high-intensity light is not probability wave under double slit and it is the EM wave (of classical) interference.

Double slit exhibits different wave behaviour based on the light intensity.

is that correct?

 

[DrChinese]

So, if the light has very low intensity (1 photon or 2 photons or 3 photons...) then the double-slit will show probability wave interference. on the other hand the high-intensity light is not probability wave under double slit and it is the EM wave (of classical).

Double slit exhibits different wave behaviour based on the light intensity.

is that correct?

No. There is no appreciable difference between the two.

Higher intensity simply means that there are more photons present at once. The pattern is always a build-up (sum) of the individual photons over time. With higher intensity, the number of photons needed to see the pattern (perhaps around 200* on the low side) will arrive in a much shorter time span. The pattern becomes more clear as the number of photons detected goes up.


*There is no exact minimum. Some people see the pattern earlier, some later.

 

[vijayantv]

No. There is no appreciable difference between the two.
...
*There is no exact minimum. Some people see the pattern earlier, some later.

so did you mean that both the single-photon and stream-of-photons (Light beam) shows only the interference of spatial-probability-wave(of Max Born) and Not EM wave (of Maxwell, which tells wave frequency causing color)?

 

[Simon Bridge]

so did you mean that both the single-photon and stream-of-photons (Light beam) shows only the interference of spatial-probability-wave(of Max Born) and Not EM wave (of Maxwell, which tells wave frequency causing color)?

No. He is saying that there is no difference between the two.
The interference pattern you calculate using probability-amplitude waves is exactly the same as the pattern you calculate using EM waves, and both agree staggeringly well with the results of experiments. When it comes to the interference pattern, there is no difference between the theories.

Have you watched the lecture series in post #6 yet?
It is essential viewing for anyone contemplating this stuff.

I know something about Classical and quantum physics

... what does that mean?
What education level are you doing this at?

If we don't have a good idea of your current state of knowledge, we cannot help you very well.
If you won't follow suggestions, we cannot help you at all.

 

[vijayantv]

I just saw some videos about Quantum Mechanics and I am not Physics student.

I saw that video, and hard to understand.

No. He is saying that there is no difference between the two.

This is the place I get confused.

if both are same then why are you saying that low intensity light(1, photon , 2 photos 3 photons) is statistical and high intensity light (multiple-particles stream) are classical.

I meant statistical that the interference pattern is caused by two Spatial-probability wave. and I meant classical that the interference patterns is caused by two electromagnetic wave.

does intensity determine whether it is statistical or classical?.

 

[bhobba]

does intensity determine whether it is statistical or classical?.

Its just that by the law of large numbers probabilistic things can behave in a deterministic way if the number of observations is so large you don't even notice that are separate observations. In this case it means the photons strike the screen so frequently it looks predictable and continuous.

Thanks
Bill

 

[Simon Bridge]

I just saw some videos about Quantum Mechanics and I am not Physics student.

Ahah ... I had a feeling this would be the case.
It would still help to know where your understanding is at - you realize that the questions you are asking involve some quite deep concepts?

The trouble is, there is a limit to what we can get you to appreciate if you don't have the background.
It's like trying to show the beauty of fibonnachi numbers to someone who has yet to learn to add or subtract.

So where is your science and math at generally - besides what you see in videos (this was online right?).
Don't be imbarrassed if it's quite basic - that's just normal. It's us uber-educated folks who are the weird ones.

This is the place I get confused.

if both are same then why are you saying that low intensity light(1, photon , 2 photos 3 photons) is statistical and high intensity light (multiple-particles stream) are classical.

I did not intend to give that impression.
I was making a generalization between quantum mechanics and classical physics ... classical physics is what happens on average.

In the interference example - while there is no difference in the pattern, there is a difference in the details of how we do the predictions at different scales.

i.e. for many particles, we can work out the intensity of different parts of the pattern by multiplying the probability by the total number of particles. This does not make sense for just one particle - which either gets detected in it's entirety or not at all.

The first approach is really just an average - but it works out so well because there are a great many particles so the variation from the average is usually too small to measure - like you cannot hear someone shouting backstage during a rock concert. It is this first approach that is indistinguishable from the EM approach.

So you can think of it like this: it's all statistics and probability.

 

[vijayantv]

large numbers probabilistic things can behave in a deterministic way

as we are seeing wave in both single-photon and photon-stream scenarios how light can be a deterministic?

can you agree that both are probability waves.

Sorry I am a non physics student, I am a software engineer. somehow get interested in quantum mechanics. it is very hard to me. please don't mind.

 

[vijayantv]

So you can think of it like this: it's all statistics and probability.

Ok. thanks. then my assumed logic will be matched. clear now.

 

[vijayantv]

Simon Bridge.
I need to clarify some doubts in Quantum mechanics. can I use this amed thread?

 

[bhobba]

can you agree that both are probability waves.

Yes indeed they are probability waves - but a funny sort of probability:
http://www.scottaaronson.com/democritus/lec9.html

It will take a while to get the gist - but persevere.

Thanks
Bill

 

[vijayantv]

Yes indeed they are probability waves - but a funny sort of probability:

clear now. thanks

 

[vijayantv]

I am extending My queries.

Assume a photon is passing double-slit. Here the probability wave is splitted into two and interfere with each other. So we are seeing interference pattern over the time period. It is fine and clear.

Assume two photons are passing double slit, simultaniously.
Photn1 and photn2.

so 4 probability waves will be generated after the double-slit.

Wave1-Photon1 , Wave2-Photo1 , wave1-Photon2 and Wave2-Photon2.

Wave1-Photon1 and Wave2-photon1 are colliding each other and show interference pattern this is fine.

Definitely Wave1-Photon2 and Wave2-photon2 are also colliding each other and show interference pattern this is fine.

are Wave1-Photon1 and Wave1-Photon-2 colliding here?
by the same way, are Wave1-Photon1 and Wave2-Photon2 colliding here?

Does probability wave of one photon interfere with the probability wave of other photon?

 

[DrChinese]

Does probability wave of one photon interfere with the probability wave of other photon?

This is a complicated question. Early QM held that photons do not interfere with each other. Over time, it seems that position has changed with a host of opinions one way or the other. The answer also somewhat depends on interpretational issues.

I have not seen a lot of literature on this. If I remember correctly, I believe Glauber comes down on the side that 2 photons can interfere. I believe part of the problem is that to test, you need photons of the same phase. Here is an article, sorry it is abstract only:

http://proceedings.spiedigitallibrary.org/proceeding.aspx?articleid=785438

"We have performed several original experiments, in order to investigate the nature of optical interference. In some contexts, the assumption that light beams suffer perturbations during their interaction is the most plausible. In others, the assumption that they do not is more appealing. Yet, the observable outcomes of both models are compatible with each other, in theory as well as in experiment. We conclude that they work equally well for the purpose of making physical predictions, and that each of them is logically valid. However, their interpretive value is not equal. If we assume that light beams cross each other unperturbed (even at the microscopic level), then we run into theoretical complications and even paradoxes that are not otherwise present."

To the others out there, I welcome your clarification and comments on this.

 

[vijayantv]

In some contexts, the assumption that light beams suffer perturbations during their interaction is the most plausible.

I think it tells about two separate light beam. I am asking about the probability wave of photons in same light beam in double-slit experiment.

if we assume the probability waves of two different photon can collide with each other then why we can not see the interference without double-slit?

so can we assume that the probability wave of one photon can not collide with other photon's probability wave?

 

[DrChinese]

if we assume the probability waves of two different photon can collide with each other then why we can not see the interference without double-slit?

Assuming I understand your question correctly: with a single slit, everything simply reinforces (adds). There is no pattern being formed.

 

[vijayantv]

with a single slit, everything simply reinforces (adds). There is no pattern being formed.

ok let me ask the question in different way.

individual photons are having their own spatial-probability wave. So anyone of these (bellow) two different phenomenons must be true.

1). One photon's probability wave can collide with another photon's probability wave. the two-slit interference pattern is generated by all photons probability waves are colliding with each others.

2). One photon's probability wave can not collide with another photon's probability wave. the two-slit interference pattern is generated by the sum/average of all individual photons' s wave.

let us assume 1st one is true. that means that "Can collide". then the question is that "why they can collide with others and showing interference only after two-slit".

if they (probability waves) are able to collide with each others, the they also could collide and generate interference pattern, without two-slit. right?

so can we consider 2nd is true?

 

[Cthugha]

individual photons are having their own spatial-probability wave. Any one of these (bellow) two different phenomenons must be true.

1). One photon's probability wave can collide with another photon's probability wave. the two-slit interference pattern is generated by all photons probability waves are colliding with each others.

2). One photon's probability wave can not collide with another photon's probability wave. the two-slit interference pattern is generated by the sum/average of all individual photons' s wave.

Putting the question of two photon interference (which is most important if the relative phase of two photons/beams is better defined than the individual phase of the single beams like for entangled photons), this depends on the seemingly trivial question what really qualifies as "individual" photons.

Coherence is intimately related to indistinguishability. Whether you treat the problem classically or in a quantum way, you always add up all the fields/probability amplitudes leading to detection events at some position and get the modulus squared of them. By definition, interference means that some non-trivial pattern survives after averaging over many repetitions. This corresponds to some cross-product terms between fields surviving. So only fields with well defined relative phase in the average can give some contribution here. However, that also means that it is meaningless to attribute the corresponding detection event to just one of the fields. Both fields contributed to the detection event, so this situation automatically also results in indistinguishability. You cannot in principle trace the detected photon back to one of the initial fields, so talking about individual photons and their individual wave functions is somewhat dangerous. As a rule of thumb, all photons contained within a coherence volume (the volume inside which the phase of fields is so well defined that interference takes place) are indistinguishable.

It may be beneficial to visualize this as a single state of the field occupied by more than just one particle, rather than many individual photons. In the limit of many indistinguishable photons, that is usually well described by a coherent state with large photon number, which is the closest thing to a classical light beam that qm has to offer.

 

[vijayantv]

Putting the question of two photon interference (which is most important if the relative phase of two photons/beams is better defined than the individual phase of the single beams like for entangled photons), this depends on the seemingly trivial question what really qualifies as "individual" photons.

I am lost. Actually I asked about two photons from same light source and not entangled. let me make in more detail.

1). Photon1 and photon2 are from same light source, not entangled and travelling. And have Wave1 and Wave2 respectively..
2). These are passing two-slit simultaneously.
3). now 4 waves are generated from the 2 waves.
4). Wave1 has splitted into WaveP1-1 and WaveP1-2.
5). Wave2 has splitted into WaveP2-1 and Wavep2-2.
6). we are seeing interference pattern in the screen.

which one of the bellow is true?

I).
All 4 waves are interfered with each others (6 combinations). multiple probability waves are generated by Probability waves collision. So these multiple combinations are resulting this interference pattern.

II)
WaveP1-1 is summed with WaveP2-1 resulting waveS1.
WaveP1-2 is summed with WaveP2-2 resulting waveS2.
only two probability wave can cause this interference pattern. So waveS1 and waveS2 are interfered and showing interference pattern.

 

[Simon Bridge]

I am extending My queries.

Assume a photon is passing double-slit. Here the probability wave is splitted into two and interfere with each other.

No. That is not what happens.
The probability wave does not split in two nor does it interfere with itself in the manner you seem to be thinking.

Note: the statistical theory does not say anything about what physically causes the interference pattern. It just predicts one if you make a certain kind of measurement.
If you turn the same theory onto what happens at the slits - it predicts a different pattern depending on how measurements are performed.

This is where you really need the math - this sort of qualitative approach will not work in detail.

Assume two photons are passing double slit, simultaniously.
Photn1 and photn2.

so 4 probability waves will be generated after the double-slit.

No - just one probability wave before the slit and one probability wave after the slit. One photon arrives on one part of the screen and the other ends up somewhere else.

The system is properly described by a 2-particle wavefunction, which can be decomposed into two single-particle wavefunctions and an interaction term. Photons are normally very weakly interacting with each other so the two single-photon terms dominate ...

The bits about individual wavefunctions are just ways to do the math and should not be confused with a physical process.

Just like if you wanted to work out how big 1/8 th of a cake was, you may find it easier to do the math by halving the size of the whole cake, then again, then again ... but that does not mean you are going to cut the cake in half three times (you could, but you don't have to), you may just use the answer you got in math to measure the cake and cut once, or maybe the cake is already cut into 16ths and you just pick out two of them?

Just because we talk about two wavefunctions, one for each photon, does not mean that there are physically two wavefunctions... there might be, but how would we tell the difference?

The best qualitative description of the process I've seen comes from Richard Feynman.
http://vega.org.uk/video/subseries/8
... unfortunately it is rather long because you need the basic concepts first.
It is possible you need more background than that - i.e. probability theory - but it does cover all your questions - like why we need the slits to be able to see an interference pattern.

Feynamns description uses an infinity of terms for each particle. It's just maths.

tldr: the reason we only see a two-slit interference pattern in the presence of two slits is because it is a two slit interference pattern. Any other situation and we see the pattern appropriate to that situation according to how the wavefunction changes.

Do give it a go though.

 

[vijayantv]

Somehow it is making sense to me. Thanks.

but one doubt in your answer.

No - just one probability wave before the slit and one probability wave after the slit. One photon arrives on one part of the screen and the other ends up somewhere else.

assume single photon scenario.

one probability wave before slit is fine.
but after the two slit it is splitted into two wave and and then the collision of two is causing interference pattern. is that not correct?

 

[Simon Bridge]

but after the two slit it is splitted into two wave and and then the collision of two is causing interference pattern. is that not correct?

No - that is not a useful way of thinking about it at this stage.
For instance, there is nothing to "collide" - it's just maths, probability amplitudes, no substance there at all.
In the maths, which you want to avoid, the probability amplitude is a complex number - it involves the square-root of minus one.

Perhaps we need to be careful...
The setup is as follows:

We define a x-y coordinate system in a plane.
We put a pair of slits width a at the origin so one slit is centered on y=+b/2 and one slit is centered at y=-b/2. We need b>a to get two distinct slits.
The source is placed on the -x axis at position x=-S
A detector may be placed anywhere - D=(x,y).

The theory tells us the probability that the detector will go off (detecting a photon) given the properties of the source along with the type and (x,y) position of the detector.

We can get cute and work out the probability p(x,y) that the detector will go off, sometime during the experiment, given that a single particle has been sent out from the source. This is worked out using the single particle wavefunction - and some other maths - and depends on the detector position (and the exact setup).

If n non-interacting particles are sent out, then the probability p_n(x,y) that at least one will hit the detector can be worked out from the single-particle case in the usual way for the probability of independent events.

The average number expected at the detector is np(x,y).

The probability function for such a setup does not change with time - it is static.

Now try to ask your questions within that framework.
A lot of them stop making sense. i.e. it makes no sense to talk about a probability wave passing through the slits because the probability function does not move about. You could picture it as standing waves if you like - that's close. We can decompose standing waves into two interfering traveling waves if we like ...

Things can get tricky for other situations - it depends on how much we control the photons that get sent out by the source and exactly how we build the slits. (This statement included because I know there are people here who have had a major cringe-reaction at the simplifications I just made.) Let's not deal with that right now - no entanglement, no funny polarizations, nothing like that. Let's get you familiar with just the basics.

I have left out the details of how we find the probability distribution before and after the slits ... that's maths. Instead I've concentrated on how we shift from the single particle case to the multi-particle case - for the same setup.

 

[vijayantv]

this is the problem to me.

I am lack of this kind of math. I came to know that single photon is showing interference pattern in two slit experiment by traveling into both slit at a time. and it causes two wave from each slit and these two waves are interfering and showing interference pattern.

if that is wrong, then my entire assumption is wrong.

ok thanks for the help.

 

[Simon Bridge]

this is the problem to me.

I am lack of this kind of math. I came to know that single photon is showing interference pattern in two slit experiment by traveling into both slit at a time.

Then you "came to know" something that is not true.
That is a very common misunderstanding though.

 

[Cthugha]

I am lost. Actually I asked about two photons from same light source and not entangled.

My comment was not about entangled light. Is is just important to note that your typical light source does not emit well defined individual, unentangled photons. Only highly specialized light sources in few physics labs around the world can do that.

I am lack of this kind of math. I came to know that single photon is showing interference pattern in two slit experiment by traveling into both slit at a time. and it causes two wave from each slit and these two waves are interfering and showing interference pattern.

Well, if you just have one wave moving outward radially and damp it everywhere except at two points, what is left over is just a small part of the initial wave. Putting a double slit basically does that.

 

[vijayantv]

Then you "came to know" something that is not true.
That is a very common misunderstanding though.

If we let one photon pass through two-slit, photon will be detected some location at screen. if we continue the same for long time and sending one photon again and again, we can see the interference patterns in the detector screen.

how it is happening? how single photon is showing interference pattern over the time period?

 

[Jilang]

Well, if you just have one wave moving outward radially and damp it everywhere except at two points, what is left over is just a small part of the initial wave. Putting a double slit basically does that.

Are we then in some sense preselecting the photons that get through?