the double slit experiment (again)

[michel123456]

A deep question.

The answer is that it behaves as if it does.

You still get a diffraction pattern

You are influenced by the known result of the experiment.

 

The photon path is a straight line.

Does the laser shoot randomly? In different directions? So that the light ray one time goes through the left slit, the other time through the right slit, the next time bounces back because it misses the slits?

Edited February 10, 2014 by michel123456

[John Cuthber]

Thankyou.

Being influenced by known results of experiments is called doing science.

 

As I said, one of the classic ways of illustrating this is with a blackened photographic plate.

The notable thing about about blackened plates is that they are black.

So light doesn't bounce off them if it misses the slits.

[swansont]

Ah. Then i understand that a single photon is considered as a line (a straigth path)

In this case, how does it come out that the scientist does not know where the photon goes (in the left or right slit)

The photon goes where the laser is pointed to. No?

If the laser is pointed to the left slit, there goes the photon. If the laser is pointed to the right slit, the photon goes there.

Or do I miss something?

 

If you look at the laser beam, i.e. with many photons, the illuminated area will be generally wider than the slits. So pointing the laser doesn't tell you which slit it goes through. For any given photon, you don't know exactly where it's going. However, if you can determine it went through one slit, you destroy the interference pattern. Since you get interference for a single photon, the implication is that the photon passes through both slits and interferes with itself.

[Sensei]

 

That is the whole point of the quantum double-slit experiment: you can't know that. If you know which slit it illuminates / goes through then there is no interference pattern. If you don't know, then there is. You cannot know anything about where a photon is until you detect (and thereby destroy) it.

 

Imagine that we have photon emitter, and polarization filter or half reflecting mirror, that's reflecting 50% of photons in one direction, and 50% in other direction (or passing through filter).

We are performing experiment close to black hole, and one direction is pointing directly to it.

Then firing one photon at a time.

If we see photon in our detector, we know that this photon went through mirror/filter, and didn't go to black hole.

If we don't see photon we know it went to black hole.

Just a thought.

Edited February 10, 2014 by Sensei

[michel123456]

Thankyou.

Being influenced by known results of experiments is called doing science.

 

As I said, one of the classic ways of illustrating this is with a blackened photographic plate.

The notable thing about about blackened plates is that they are black.

So light doesn't bounce off them if it misses the slits.

[out of the subject] I don't believe that is correct. When I use a laser beam and light a black surface, I observe a red dot (my laser is red). If I can observe a red dot, that means some photons have bounced back to my retina. And since I can observe the red dot from any point in the rooom, it means photons have bounced in all directions.[out of the subject]

If you look at the laser beam, i.e. with many photons, the illuminated area will be generally wider than the slits. So pointing the laser doesn't tell you which slit it goes through. For any given photon, you don't know exactly where it's going.

here i will use an analogy:

You want to study a drop of water. But for some mysterious reason you are not able to study drops of water, the only thing you have at your disposal is the Amazon river.

You know from your Theory that a drop of water must have a xx charachteristic. So you study the xx charachteristic of the entire flow of the Amazon river and divide it by a very very (very) small interval of time such that the result corresponds at the xx charachteristic of a drop of water. But do you obtain a real drop of water? Or something unphysical which is a tiny flow of water over the entire width of the Amazon?

IOW, are you really sure that you are looking at a single photon?

 

However, if you can determine it went through one slit, you destroy the interference pattern. Since you get interference for a single photon, the implication is that the photon passes through both slits and interferes with itself.

That happens when you try to measure at the slit, IOW when you intercept or disturb the photon at the slit. Does that also happen when you shoot surely at one slit? Then at the other?

[Strange]

 

When I use a laser beam and light a black surface, I observe a red dot (my laser is red).

 

Therefore, the surface is not black, just nearly black.

 

IOW, are you really sure that you are looking at a single photon?

 

Because individual photons are detected, one at a time.

 

 

That happens when you try to measure at the slit, IOW when you intercept or disturb the photon at the slit. Does that also happen when you shoot surely at one slit? Then at the other?

 

Obviously, if you only allow photons to go through one slit, there will be no interference pattern.

Edited February 11, 2014 by Strange

[michel123456]

Therefore, the surface is not black, just nearly black.

So you say that when shooting with a laser a REAL black surface, you will observe no dot?

Because individual photons are detected, one at a time.

The Amazon river has a width of several kilometers. The divided stream in nanoseconds wets several kilometers of width. That is what you observe.

Does that mean that a droplet of water can wet several kilometers?

[Strange]

So you say that when shooting with a laser a REAL black surface, you will observe no dot?

 

Correct. But I doubt any such really black surface exists in reality.

 

Correct. But I doubt any such really black surface exists in reality.

 

 

The Amazon river has a width of several kilometers. The divided stream in nanoseconds wets several kilometers of width. That is what you observe.

Does that mean that a droplet of water can wet several kilometers?

 

I have no idea what you are asking. A single photon is detected at a single location (the atom that it interacts with in the photo-detector). But, until it is detected you don't know where it is or where it has been.

Edited February 11, 2014 by Strange

[swansont]

 

The Amazon river has a width of several kilometers. The divided stream in nanoseconds wets several kilometers of width. That is what you observe.

Does that mean that a droplet of water can wet several kilometers?

 

It means that you don't know where a specific drop of water is in that range. Since we're discussing QM and not some classical system, the analogy has its limits; the drop would be everywhere in that area until you actually measure its location.

That happens when you try to measure at the slit, IOW when you intercept or disturb the photon at the slit. Does that also happen when you shoot surely at one slit? Then at the other?

 

There are other ways of detecting which path it takes. Any of them will destroy the interference pattern.

[StringJunky]

It's probably not a practically realisable question because an atom is so small, but according to theory would a single photon arriving at a photographic plate in the experiment only interact with one electron in an atom on it?

[Strange]

It's probably not a practically realisable question because an atom is so small, but according to theory would a single photon arriving at a photographic plate in the experiment only interact with one electron in an atom on it?

 

Yes. The whole point of the quantisation of light (aka photons) is that each photon is indivisible: it either reacts with one electron in an atom or none. (This is basically what Einstein got his Nobel prize for.)

[StringJunky]

 

Yes. The whole point of the quantisation of light (aka photons) is that each photon is indivisible: it either reacts with one electron in an atom or none. (This is basically what Einstein got his Nobel prize for.)

So it takes lots of them to make the wave pattern just as if the photons were exchanged for electrons?

[swansont]

So it takes lots of them to make the wave pattern just as if the photons were exchanged for electrons?

 

Yes. The single-photon (or electron) interference experiments take many iterations to discern a pattern.

[StringJunky]

 

Yes. The single-photon (or electron) interference experiments take many iterations to discern a pattern.

Thanks. That brings me a little closer to getting to grips with the experiment.

[Sensei]

So it takes lots of them to make the wave pattern just as if the photons were exchanged for electrons?

 

 

If you have laser with 100 mW max output and 532 nm wavelength, and all energy will convert to photons without loses (idealistically), you will have:

 

100 mW = 0.1 W = 0.1 J/s

 

E = h*c/wavelength = 6.62607e-34 * 299,792,458 / 532e-9 = 3.734e-19 J each photon energy

0.1 / 3.734e-19 J = 2.678 * 10^17 photons emitted per second

Edited February 11, 2014 by Sensei

[John Cuthber]

 

[out of the subject] I don't believe that is correct. When I use a laser beam and light a black surface, I observe a red dot (my laser is red). If I can observe a red dot, that means some photons have bounced back to my retina. And since I can observe the red dot from any point in the rooom, it means photons have bounced in all directions.[out of the subject]

 

If the surface was truly black, you wouldn't.

The reason you can see it is that the "black" surface you use reflects a small percentage of the light.

Because a laser is very bright you can still see the small fraction of it that bounces off the "black" surface.

 

However, and this is the important point made earlier.

If you use a blackened photographic plate, very few photons will be reflected.

If you use two razor blades and a piece of copper wire, quite a lot of the photons will be reflected.

But the diffraction pattern is pretty much the same in both cases.

 

So the amount of reflection can not be important.

[michel123456]

Correct. But I doubt any such really black surface exists in reality.

 

 

 

I have no idea what you are asking. A single photon is detected at a single location (the atom that it interacts with in the photo-detector). But, until it is detected you don't know where it is or where it has been.

Ok. But after it has hit the plate, you can make estimation about its path.

 

There are several possible situations:

Schematically

 

 

 

1. the photon passed through the left slit exactly and hit the plate in a straight path

 

 

 

 

2. the photon passed through the right slit exactly and hit the plate in a straight path

 

 

 

 

3. the photon hit the plate outside the direct straight path (it is far left or far right). In this case it is believed that the photon has interfere with itself.

In red all the paths I could figure.

 

 

 

 

4. the photon does not hit the plate because it is believed that it has interfere with itself and acting as a wave it has destroyed its own.

 

 

 

My question is: in situation 3 and 4 above, what is the thing that made the photon changes its direction? Why has the photon changed its path when going out of the slit?

Edited February 12, 2014 by michel123456

[Sensei]

4. the photon does not hit the plate because it is believed that it has interfere with itself and acting as a wave it has destroyed its own.

ScreenShot078.jpg

 

Interference doesn't destroy wave..

 

It looks more like this (it's small version of mine animation of interference effect):

 

 

Edited February 12, 2014 by Sensei

[John Cuthber]

The light isnt destroyed, it just gets moved.

The fringes have light and dark bands but the light bands are brighter than you would expect without the interference effects.

[swansont]

 

My question is: in situation 3 and 4 above, what is the thing that made the photon changes its direction? Why has the photon changed its path when going out of the slit?

 

You're approaching this classically by assuming there's a trajectory, like with a ball. That's not what is happening, which may be why this is confusing.

[Strange]

Ok. But after it has hit the plate, you can make estimation about its path.

 

But, in fact, you can't. You can't imagine photons to be like little billiard balls.

[michel123456]

You're approaching this classically by assuming there's a trajectory, like with a ball. That's not what is happening, which may be why this is confusing.

Before hitting the shield, the sum of all photons are restrained into the stream of the laser, no?

 

I mean if you put a screen anywhere else in the room out of range of the laser, the screen will detect no photon.

[Strange]

Before hitting the shield, the sum of all photons are restrained into the stream of the laser, no?

 

Strictly speaking, no.

 

 

I mean if you put a screen anywhere else in the room out of range of the laser, the screen will detect no photon.

 

It is just very, very, very unlikely:

http://vega.org.uk/video/subseries/8

[swansont]

Before hitting the shield, the sum of all photons are restrained into the stream of the laser, no?

 

I mean if you put a screen anywhere else in the room out of range of the laser, the screen will detect no photon.

 

The stream of the laser has a wave behavior, typically that of a Gaussian beam. They aren't confined, like water in a pipe. A Gaussian beam's intensity drops off exponentially away from the propagation axis, so it never actually goes to zero.

[michel123456]

The stream of the laser has a wave behavior, typically that of a Gaussian beam. They aren't confined, like water in a pipe. A Gaussian beam's intensity drops off exponentially away from the propagation axis, so it never actually goes to zero.

O.K.

If I put a water pipe around the laser stream, does that change the experiment? I don't think so.

 

 

We know that the photon has been emitted from the laser. Point A.

We know that the photon passed through one or the 2 slits. Point B.

We know that the photon hit the screnn somewhere. Point C

 

No matter the real or quantic way, path, wave or how-do-you-call-that, we know that tthe photon went through points A B C.

My question is the following:

When points A B C are not aligned, the photon must have changed direction, no?