Uncertainty principle and double-slit interference?

[xylai]

Both uncertainty principle and double-slit interference are the most important contents in Quantum mechanics. Those days I am thinking about this question: Is there a hidden relation between them? Unfortunately, I couldn't give me a satisfied answer.
How do you think about it?
Thank you!

 

[peter0302]

What do you mean by a hidden relation? Double slit interference, like most quantum phenomina, is a consequence of the uncertainty principle.

 

[peter0302]

xylai, I'm replying to your private message here in the hopes that someone with more experience might chime in with a better answer, but to answer your request for more explanation:

The uncertainty principle tells us that you cannot know the value of two non-commuting observables (such as position and momentum) with perfect accuracy; the more you know one, the less certain the other becomes. In the double slit experiment, when a photon passes through the double slit we know two possible, very precise, positions for that photon: the left slit or the right slit. Those two very precise position possibilities lead to great uncertainty in momentum.

In a single-slit case, this results in a wide gaussian (bell-curve-like) distribution for the photons. In the double-slit case, these two uncertainties interfere with one another to create the beloved interference pattern. But in both cases, the pattern results from the uncertainty in momentum resulting from the certainty (albiet, narrowed down to two possibilities in the double slit case) in position. If not for the uncertainty principle, the double slit experiment would just result in two, well defined bright spots behind the slits.

Why do the two uncertainties in a double slit not simply add up to one, wide blob of uncertainty? That has to do with the Schrodinger Equation and the fact that the wave function represents probability amplutides, not probabilities themselves. The probabilities are derived by taking the complex conjugate of the wave function, thereby eliminating the imaginary component. But in the case of a double slit, the phase difference in the photons going through the two different slits is reflected in the wave function before the complex conjugate is taken (and thereby the actual probability is derived). So that when you do find the probability density function for the two-slit setup, the phase difference between the photons from each slit is reflected in the probability density function as an interference pattern and, sure enough, that's what we see when we do the experiment.

If someone else wants to correct anything I said or explain it more clearly be my guest. :)

 

[Ken G]

That all sounds good to me, though it might not be necessary to think so mathematically if you are not used to complex numbers. I might add that in a sense the "deeper source" to both the uncertainty principle and the interference pattern is just wave mechanics. The only time you need quantum mechanics is if you want to ask what a single quantum will do, but if you just want to understand the overall pattern and think of any spreading as due to "uncertainty", then the whole thing also works with water waves going between two breaks in a jetty. In that case the uncertainty principle comes from the way every excitation of the water can serve as a source of new waves, and those new waves are seen to propagate out unless they destructively interfere (which they do quite a lot if the wavelength is short). That same destructive interference causes the pattern you see.

This connects to quantum mechanics when we connect, as peter0302 said, the wave amplitudes you see on the water with probabilities of the quantum showing up in those places.

 

[peter0302]

Can someone confirm something that's been bothering me?

For double-slit interference to be visible to the naked eye, the light source needs to be a laser, or something collimated, correct? If you try it with a flashlight, you'll just get a blob on the other end, right?

Can we equate the collimated nature of laser light with a well-defined momentum which becomes ill-defined as soon as we simultaneously try to isolate it to a well-defined position (aka the slits)?

 

[DrChinese]

For double-slit interference to be visible to the naked eye, the light source needs to be a laser, or something collimated, correct? If you try it with a flashlight, you'll just get a blob on the other end, right?

There is a different reason that non-coherent light does not generate a double slit pattern when it is something like a flashlight. The beam is not in phase and so there is a tremendous amount of destructive interference. The fringe pattern is canceled out. If you could send those same photons through the double slit apparatus a photon at a time, the pattern would emerge.

 

[Ken G]

Not to be combative, but it's not a question of phase, I would say, it's bandwidth. Lasers have coherent phase, but that's not necessary for a two-slit pattern, because the location of the nulls doesn't depend on the initial phase anyway. In other words, you could superimpose a thousand different laser beams and you'd still get the same pattern-- when they all produce nulls individually, they will also do that in tandem. But if the bandwidth is not tight, you are superimposing different wavelengths that do have nulls at different places, so you won't get the pattern. In short, you need a specific color to the beam, even sharper than the eye can discern. That holds even if you send the photons through one at a time, so the flashlight wouldn't work even at such a low intensity.

 

[lightarrow]

You will have an interference pattern (but much more complicated) even with many colours, provided there is a phase relation between all the various colours.

 

[Ken G]

Why would there be a phase relation between the different colors? Perhaps a multicolored hologram? The OPer might think you are saying a flashlight would have such a relation.

 

[peter0302]

Actually what I was getting at was simply whether the light needed to be parellel, and if that had anything to do with a "certain" momentum translating into an "uncertain" position when detected.

 

[Ken G]

Actually what I was getting at was simply whether the light needed to be parellel, and if that had anything to do with a "certain" momentum translating into an "uncertain" position when detected.

Yes, I was referring only to what DrChinese said. You are correct that the light being parallel connects to the pattern we see (and here lightarrow's comments are again relevant-- nonparallel yet patterned sources could yield their own different and possibly more complicated patterns). Also, the overall spreading of the diffraction pattern, regardless of whether one gets fringes or not, is indeed an example of the uncertainty principle. But I would have said it the other way-- knowledge about the lateral positioning as the light exits the slits translates into uncertainty in the lateral momentum, which in turn leads to spreading of the final position.

 

[RandallB]

Actually I think you can generate interference patters with white light. From red to blue is a narrow enough bandwidth provided your source is small enough and far enough away to get out of the near field range and into a far field arrangement. One example; sunlight from the sun is far enough away but the sun is too wide as a source making it a far field source. But in an eclipse during the phase when there is just thin crescent of light, I’ve seen walking interference patterns vibrating within the sunlight on the ground during a eclipse at the ABC islands a several years ago. That’s why they recommend laying out a white sheet so you can see the effect if the sky is clear enough.

I think there are other more down to Earth examples of white light interference patterns as well.

 

[Ken G]

Actually I think you can generate interference patters with white light. From red to blue is a narrow enough bandwidth provided your source is small enough and far enough away to get out of the near field range and into a far field arrangement. One example; sunlight from the sun is far enough away but the sun is too wide as a source making it a far field source. But in an eclipse during the phase when there is just thin crescent of light, I’ve seen walking interference patterns vibrating within the sunlight on the ground during a eclipse at the ABC islands a several years ago. That’s why they recommend laying out a white sheet so you can see the effect if the sky is clear enough.

I think there are other more down to Earth examples of white light interference patterns as well.

That's interesting, the bandwidth is about the same as the frequency, for sunlight, so the variance in the k vector is as large as the k vector itself. To get significant diffraction, one needs a large dot product between the k vector and the slit width, but to get fringes, you also need a small dot product between the variance in k and the slit width. So you can probably use sunlight if you are willing to compromise the spreading of the pattern, but the traditional "two-slit pattern" is probably not possible, I would expect.

 

[Schrodinger's Dog]

Actually I think you can generate interference patters with white light. From red to blue is a narrow enough bandwidth provided your source is small enough and far enough away to get out of the near field range and into a far field arrangement. One example; sunlight from the sun is far enough away but the sun is too wide as a source making it a far field source. But in an eclipse during the phase when there is just thin crescent of light, I’ve seen walking interference patterns vibrating within the sunlight on the ground during a eclipse at the ABC islands a several years ago. That’s why they recommend laying out a white sheet so you can see the effect if the sky is clear enough.

I think there are other more down to Earth examples of white light interference patterns as well.

What you mean like Young's slit experiment, I'm pretty sure he wasn't using lasers.

 

[peter0302]

I was thinking the same thing. The electrons in Young's experiment were neither collimated nor in coherent phase. They were pretty random as I recall. So I think Dr.Chinese is right that if a flashlight emitted one photon at a time, (which, by definition, would be one precise color at a time) you'd get the pattern.

Which is why I continue to be astounded by the fact that you don't get the pattern from entangled photons unless you take affirmative steps to destroy the position certainty of the twin.

 

[Schrodinger's Dog]

I was thinking the same thing. The electrons in Young's experiment were neither collimated nor in coherent phase. They were pretty random as I recall. So I think Dr.Chinese is right that if a flashlight emitted one photon at a time, (which, by definition, would be one precise color at a time) you'd get the pattern.

Which is why I continue to be astounded by the fact that you don't get the pattern from entangled photons unless you take affirmative steps to destroy the position certainty of the twin.

Just to be sure are you talking about the double slit quantum eraser experiment?

You would it is called the http://www.sciam.com/article.cfm?id=how-can-a-single-photon-p" which produces interference patterns when you fire one photon at a time.

 

[peter0302]

Just to be sure are you talking about the double slit quantum eraser experiment?

I am indeed. That and the Dopfer thesis as well.

 

[Schrodinger's Dog]

I am indeed. That and the Dopfer thesis as well.

https://www.physicsforums.com/showthread.php?t=158413&highlight=double+slit+eraser

Yeah its been done here and its explainable in quantum theory.

http://arxiv.org/PS_cache/quant-ph/pdf/9903/9903047v1.pdf

Still very interesting though.

https://www.physicsforums.com/showthread.php?t=106070

Mathematically, it turns out that the integral of the wave is well explained by theory. So the behaviour of the entangled photons is exactly what we would expect, given the conditions in the experiment.

It's much more adequately explained on those links and in much better detail.

Even in the case of the normal delayed choice quantum eraser setup where the which-path information is erased, the total pattern of photons on the screen does not show any interference, it's only when you look at the subset of signal photons matched with idler photons that ended up in a particular detector that you see an interference pattern. For reference, look at the diagram of the setup in fig. 1 of this paper:

http://xxx.lanl.gov/PS_cache/quant-ph/pdf/9903/9903047.pdf

In this figure, pairs of entangled photons are emitted by one of two atoms at different positions, A and B. The signal photons move to the right on the diagram, and are detected at D0--you can think of the two atoms as corresponding to the two slits in the double-slit experiment, while D0 corresponds to the screen. Meanwhile, the idler photons move to the left on the diagram. If the idler is detected at D3, then you know that it came from atom A, and thus that the signal photon came from there also; so when you look at the subset of trials where the idler was detected at D3, you will not see any interference in the distribution of positions where the signal photon was detected at D0, just as you see no interference on the screen in the double-slit experiment when you measure which slit the particle went through. Likewise, if the idler is detected at D4, then you know both it and the signal photon came from atom B, and you won't see any interference in the signal photon's distribution. But if the idler is detected at either D1 or D2, then this is equally consistent with a path where it came from atom A and was reflected by the beam-splitter BSA or a path where it came from atom B and was reflected from beam-splitter BSB, thus you have no information about which atom the signal photon came from and will get interference in the signal photon's distribution, just like in the double-slit experiment when you don't measure which slit the particle came through. Note that if you removed the beam-splitters BSA and BSB you could guarantee that the idler would be detected at D3 or D4 and thus that the path of the signal photon would be known; likewise, if you replaced the beam-splitters BSA and BSB with mirrors, then you could guarantee that the idler would be detected at D1 or D2 and thus that the path of the signal photon would be unknown. By making the distances large enough you could even choose whether to make sure the idlers go to D3&D4 or to go to D1&D2 after you have already observed the position that the signal photon was detected, so in this sense you have the choice whether or not to retroactively "erase" your opportunity to know which atom the signal photon came from, after the signal photon's position has already been detected.

This confused me for a while since it seemed like this would imply your later choice determines whether or not you observe interference in the signal photons earlier, until I got into a discussion about it online and someone showed me the "trick". In the same paper, look at the graphs in Fig. 3 and Fig. 4, Fig. 3 showing the interference pattern in the signal photons in the subset of cases where the idler was detected at D1, and Fig. 4 showing the interference pattern in the signal photons in the subset of cases where the idler was detected at D2 (the two cases where the idler's 'which-path' information is lost). They do both show interference, but if you line the graphs up you see that the peaks of one interference pattern line up with the troughs of the other--so the "trick" here is that if you add the two patterns together, you get a non-interference pattern just like if the idlers had ended up at D3 or D4. This means that even if you did replace the beam-splitters BSA and BSB with mirrors, guaranteeing that the idlers would always be detected at D1 or D2 and that their which-path information would always be erased, you still wouldn't see any interference in the total pattern of the signal photons; only after the idlers have been detected at D1 or D2, and you look at the subset of signal photons whose corresponding idlers were detected at one or the other, do you see any kind of interference.

 

[peter0302]

Oh, I know what the math and quantum theory say. I just still find it remarkable. :)

 

[Ken G]

Young did use a coherent source of light, though not a laser:
http://micro.magnet.fsu.edu/primer/java/interference/doubleslit/ says
"In order to test his hypothesis, Young devised an ingenious experiment. Using sunlight diffracted through a small slit as a source of coherent illumination, he projected the light rays emanating from the slit onto another screen containing two slits placed side by side."
As I said, the phase relation in a laser is irrelevant to the interference pattern, but the sharp color and the collimated direction are very relevant. As for entangled photons, the behavior you cite is exactly what quantum mechanics predicts. It would be more astounding if our macroscopic intuition trumped quantum mechanics in that single application but not all the other surprising things particles do.

 

[Schrodinger's Dog]

Oh, I know what the math and quantum theory say. I just still find it remarkable. :)

Yes it is, but only remarkable in as much as it confirms quantum theory in that no matter how we set up the experiment the results are consistent with quantum theory. Which I admit is pretty cool. But damn it it works as Ken G just said.

 

[lightarrow]

What you mean like Young's slit experiment, I'm pretty sure he wasn't using lasers.

...and not even the interferometer in Michelson-Morley experiment.

 

[peter0302]

I thought Young used electrons.

 

[Ken G]

He used sunlight, but diffracted through a small slit to achieve a result similar to the collimation of a beam. The key requirement is to have the wave function for each photon to give a nearly constant phase over the whole slit, which means that you want parallel rays (such as from a tiny source) and a narrow bandwidth for each photon, and then also a fairly narrow bandwidth for the ensemble, but this last requirement seems the easiest to compromise on (as lightarrow said, it only alters the pattern more so than destroying it). So you can get away with sunlight as long as you pass it through a narrow slit to collimate what impinges on the two slits.

 

[peter0302]

Ah, I was a little off - looks like the double slit experiment wasn't performed with anything other than light until 1961.

http://en.wikipedia.org/wiki/Double-slit_experiment

 

[xylai]

You say when a photon passes through the double slit we know two possible, very precise, positions for that photon: the left slit or the right slit. However, I think the core of the double-slit experiment is that the particle passes through the two slits simultaneously. Or we can't get the pattern.

xylai, I'm replying to your private message here in the hopes that someone with more experience might chime in with a better answer, but to answer your request for more explanation:

The uncertainty principle tells us that you cannot know the value of two non-commuting observables (such as position and momentum) with perfect accuracy; the more you know one, the less certain the other becomes. In the double slit experiment, when a photon passes through the double slit we know two possible, very precise, positions for that photon: the left slit or the right slit. Those two very precise position possibilities lead to great uncertainty in momentum.

In a single-slit case, this results in a wide gaussian (bell-curve-like) distribution for the photons. In the double-slit case, these two uncertainties interfere with one another to create the beloved interference pattern. But in both cases, the pattern results from the uncertainty in momentum resulting from the certainty (albiet, narrowed down to two possibilities in the double slit case) in position. If not for the uncertainty principle, the double slit experiment would just result in two, well defined bright spots behind the slits.

Why do the two uncertainties in a double slit not simply add up to one, wide blob of uncertainty? That has to do with the Schrodinger Equation and the fact that the wave function represents probability amplutides, not probabilities themselves. The probabilities are derived by taking the complex conjugate of the wave function, thereby eliminating the imaginary component. But in the case of a double slit, the phase difference in the photons going through the two different slits is reflected in the wave function before the complex conjugate is taken (and thereby the actual probability is derived). So that when you do find the probability density function for the two-slit setup, the phase difference between the photons from each slit is reflected in the probability density function as an interference pattern and, sure enough, that's what we see when we do the experiment.

If someone else wants to correct anything I said or explain it more clearly be my guest. :)

 

[peter0302]

You say when a photon passes through the double slit we know two possible, very precise, positions for that photon: the left slit or the right slit. However, I think the core of the double-slit experiment is that the particle passes through the two slits simultaneously. Or we can't get the pattern.

We can get the pattern only when we have no idea, even in principle, which slit the photon went through. Mathematically, it's _equivalent_ to saying the photon went through both slits, but you whether it _really_ went through both is a matter of interpretation.

My point is that the two possibilities, which one was right we can never know, need to be very precise position measurements. If it was a "double punch hole" experiment instead of a "double slit" it wouldn't work.

 

[Schrodinger's Dog]

We can get the pattern only when we have no idea, even in principle, which slit the photon went through. Mathematically, it's _equivalent_ to saying the photon went through both slits, but you whether it _really_ went through both is a matter of interpretation.

My point is that the two possibilities, which one was right we can never know, need to be very precise position measurements. If it was a "double punch hole" experiment instead of a "double slit" it wouldn't work.

Well yeah we have to open the box to see if the cat is dead. That's a given in the two slit experiment with single photons. thus we demonstrate the measurement problem and particle/wave duality in one neat experiment.

 

[RandallB]

If it was a "double punch hole" experiment instead of a "double slit" it wouldn't work.

Why not?
As long as the two punch holes were designed with same rules as the two slits (small size, close together). That is what Afsher used, there they are called http://en.wikipedia.org/wiki/Afshar_experiment" in wikipedia.

 

[peter0302]

Why not?
As long as the two punch holes were designed with same rules as the two slits (small size, close together). That is what Afsher used, there they are called http://en.wikipedia.org/wiki/Afshar_experiment" in wikipedia.

I should be more specific. When I say "punch hole" I mean something wider than a couple of millimeters. Like a hole punch in a piece of paper. Afshar used pinholes, much smaller. Point is making the holes/slits small is what makes the position measurement certain, leading to the uncertain momentum measurement.

 

[Ken G]

But photons can defy SR, and can still show finite expressions even though this should not occur at C.

I'm not sure what you are trying to say here. SR was essentially built around photons, as kind of the asymptotic limit of a "relativistic particle". Photons do not "defy" SR, though they are outside of the constraints we put on "observers" in SR (an observer cannot zoom past another observer at a relative speed of c, so a photon is not allowed to "bear witness" to reality. Good thing, they'd have some weird notions-- nevertheless they'd be limiting examples of SR in the way that the concept of "infinity" is a limit of the concept of all the other real numbers).

 

[peter0302]

Agreed. Photons don't defy SR at all. The finite expressions siphon is referring to are special cases of more general equations.

For example, take the general Energy-Momentum relation:

E^2=(pc)^2+(mc^2)^2

For photons, m=0, therefore E=pc.

Another cool derivation from the above:

For a photon:
E = pc
ymc^2 = ymvc
v=c

So you can use the momentum-energy relation to show how V=C for a photon without even having to assume the constancy of the speed of light! So, photons don't defy relativity at all - they're a consequence of it.

 

[RandallB]

I should be more specific. When I say "punch hole" I mean something wider than a couple of millimeters. Like a hole punch in a piece of paper. Afshar used pinholes, much smaller. Point is making the holes/slits small is what makes the position measurement certain, leading to the uncertain momentum measurement.

I see what your saying, but don't forget "punch holes" are small if the source is far enough away to be "Far Field".

 

[xylai]

We can have an imaginary experiment. If a particle passes through one slit, left or right, then we can close the other slit, because it does not influence the partcle. However, after this experiment, we can get the diffractive pattern, but not the interference pattern.

We can get the pattern only when we have no idea, even in principle, which slit the photon went through. Mathematically, it's _equivalent_ to saying the photon went through both slits, but you whether it _really_ went through both is a matter of interpretation.

My point is that the two possibilities, which one was right we can never know, need to be very precise position measurements. If it was a "double punch hole" experiment instead of a "double slit" it wouldn't work.

 

[peter0302]

The _possibility_ of passing through either slit is what creates the interference pattern. When you close one slit, you remove that possibility. When you peek behind the slits, you remove that possibility.

You don't have to believe that it _actually_ passed through both, for one reason because no one can agree on what "actual" or "really" mean. All we can say is that possibilities influence the wave function, and the wave function influences where you'll end up finding the particle. What happened to the particle before it's detected is not something on which there is universal agreement. Quite the contrary.