# Two-slit Experiment

1. Feb 20, 2005

### RandallB

Does anyone know the limits on a true two slit experiment?
I see where the use of “Grates” with larger particles (buckyballs c60) is often used.
But when only two real slits are used; what are the largest practical dimensions?
Longest Wavelength?
Widest Slits?
Largest separation of the slits?
Widest readable interference pattern measured from pattern center vs. distance from slits?

RB

2. Feb 22, 2005

### Edgardo

I think that's an interesting question, but nobody answered.
Why can't I send a football through a double-slit?
What's the greatest object we can use for two-slit experiment?

3. Feb 22, 2005

### ZapperZ

Staff Emeritus
It isn't the SIZE of the object, it's the size of it's deBroglie wavelength!

You need to keep in mind that whatever it is that you are passing through must have a significant "spatial spread" so that you cannot tell which slit that damn thing is going through. That is how you get a superposition of all possible path. A football moving at terrestrial speeds will not exhibit enough of a spread of its deBroglie wavelenth so much so that you start seeing it as a fuzzy object with an uncertain position. There is also another issue in which the whole consitituent of the object must be coherent with each other. That's why experiments involving large number of particles are done only when you can get this situation (i.e. such as the superconducting fluid having 10^10 electrons in the Stony Brook SQUID experiment).

Zz.

4. Feb 22, 2005

### RandallB

Most of the time when you construct a viable "football" by creating big Carbon molecule out of 60 or 70 carbon atoms (buckyballs). Experimenters need to use a grating with holes set to a proper size. I guess tuned to the “deBroglie wavelength” (ZapperZ) whatever that is.
The problem with “footballs” going though a matrix of holes is people start to argue that the pattern displayed is just a shadow of the grating.
So to avoid that issue I was only interested in tests that successfully use just two slits.

You got to love the two slit as it indicates a FTL (faster than light) action of some kind taking place.

The question is of course is just how large can we display that affect?
How RED can we get the light to run the test with?
The FM radio band is very “RED” but can they, have they been, used in a two slit?
Is 317.2407 cm wavelength of my favorite FM station the same as it’s “deBroglie wavelength”?
Can we make a Two Slit Test with slits almost a third of a meter wide?
How far apart can we get away with separating the slits?
How wide would the pattern produced be?
If the experiment cannot be done with a wavelength that long, how far red can we push it and still make a detectable pattern?
Can we go further in the red if we drop the desire to do the test one “wave” particle at a time?
Just hoping that someone somewhere has tried to push the limits of the test.

5. Feb 22, 2005

### ZapperZ

Staff Emeritus
SAY WHAT???!!!

Zz.

6. Feb 22, 2005

### Davorak

The two slit and in general interference patterns for some reason are often misinterpreted this way. I do not know where this misinterpretation started catching on either. Take this article posted by a grad student not that long ago sci.physics.research

This person has the same misinterpretation of the two slit experiment and interference patterns.

7. Feb 22, 2005

### Edgardo

I got a question here:

a) What's the DeBroglie wavelength of a football?
b) Whats the largest molecule that I can use to see interference?
That buckyball molecule is relative big. Where does this end?

8. Feb 22, 2005

### ZapperZ

Staff Emeritus
a) http://scienceworld.wolfram.com/physics/deBroglieWavelength.html
You'll notice from the definition that your question cannot be answered since it is velocity/momentum dependent. If you give the football a typical velocity that it can have, then you can find the wavelength yourself. From there, you can figure out why we do not see any "fuzzyness" associated with the motion of a football.

b) Where it ends, no one knows. That's a continuing research front area to see if there is a boundary between quantum effects and classical effects. Refer to Tony Leggett's paper in last week's Science.

Zz.

9. Feb 22, 2005

### Edgardo

Ok, the formula is $\lambda = \frac{h}{p} = \frac{h}{mv}$

Say m is the mass of the football. Then v is the center of mass velocity, right?
BUT then I can choose v close to zero, such that I can make the deBroglie wavelength long enough.
And then I should be able to send the football through the slit.

10. Feb 22, 2005

### Davorak

The center of mass velocity may be small enough but that does not mean all the individual molecules would have a slow enough velocity.

11. Feb 22, 2005

### ZapperZ

Staff Emeritus
This would be fine and dandy if there are no other "competing" effects. First, if v is very close to zero (why not make it zero?), then your object is not moving. Just by saying that, or even when it is moving VERY slow, we have already make the explicit definition that this object's position is well-defined and observable. This isn't the property of a quantum particle.

Secondly, with the wavelength being exceedingly large, it no longer behave as a "wave".

Thirdly, I have already mentioned the need for each part of the object to be coherent with each other, or else even if you can make a 2-slit experiment out of this, you'd get something similar with a non-coherent, non-monochromatic light source.

Zz.

12. Feb 23, 2005

### Edgardo

So can I or can I not compute the deBroglie wavelength of a football?
And what's the meaning of it then, if there are still other competing effects?

Could you explain that a little bit more detailed?
Because I don't understand why a very slow object doesn't have quantum properties. Isn't it in Bose-Einstein-condensates where the particles are made relatively slow, and therefore show quantum properties?
For slow particles (like in BEC) the wavelength is made longer by cooling and they behave like waves.

So size DOES matter, because for interference we have to make
all particles coherent to each other.
The greater the object, the more particles we have, that have to be coherent to each other, and for greater objects it becomes more difficult (due to decoherence effects)

Say, we could cool down the individual molecules of the football to very low temperatures (like in Bose-Einstein-condensation), would it then be possible to observe interference effects?
(I know it's still not possible to cool down a football to such temperatures,
it is even difficult to cool down molecules. But say it's just a gedankenexperiment)

13. Feb 23, 2005

### ZapperZ

Staff Emeritus
Sure you can. But it doesn't mean you can get interference pattern just because you can because of what I said about the consituents of the object. Look here if you want to see the computation for the de Broglie wavelength of a baseball.

http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html#c4

Where did you get the impression that BE condensates are "slow" objects? BE condensates have to be cooled because that's when each of the particles can get close enough to each other so that their individual wavefunction can OVERLAP and thus, causes quantum statistics (indistinguishibility) to kick in. The "wavelength" is CERTAINLY not made longer by this, but rather a "condensation" simply means the whole condensate are coherent.

But that's why larger objects tend to behave classically. That is why I said this is still a continuing area of research. Is it really the size of the object, or the fact that it is more difficult to make a larger object having parts that are coherent with each other? So far, the indication points to the latter (refer to, for example, the Stony Brook SQUID experiment), that it has more to do with making the various stuff that makes up the object coherent.

Zz.

14. Feb 23, 2005

### Edgardo

But what's the sense of computing the deBroglie wavelength of such macroscopic objects then?
Does the deBroglie wavelength give me information about wether it is possbile
to see interference?

The particles in a BEC are slowed down for example by Doppler-cooling, by
Sisyphus-cooling, Raman-cooling, VSCPT etc. They are relatively slow compared to particles at room temperatures.

I saw this picture with the increasing wavelengths at:
http://www.physik.uni-stuttgart.de/institute/pi/5/forschung/lattice/

15. Feb 23, 2005

### ZapperZ

Staff Emeritus
In the case of the baseball, it tells you the SCALE of the "slit size" that one needs to start seeing the interference pattern. But as I've said, you just can't make the velocity arbitrary small just so you can increase the wavelength to comparable, human scale, because other competing factors take over.

Nope. I will challenge you on that. The whole idea of quantum statistics is the INDISTINGUISHIBILITY. You can't pick out an BE particle, and trace its trajectory to be able to say "Oh, that particle is moving with such-and-such a speed". In fact, if you look at the band structure of a superconductor going from a normal state into the superconducting state (where the electrons undergo a similar condensation), you see no dramatic change in the effective group velocity. Again, you are confusing the need to cool the particles so that they make a more efficient overlap of their wavefunction with "slowing down".

I would hate to think you put emphasis on cartoon pictures like this as a basis for physics, and insult me by using it as a counter argument to what I have explained. The picture is trying to convey the meaning of COHERENCE.. that ALL the damn particles that were originally described by their own individual wavefunctions, are NOW, after condensation, can be described by ONE, SINGLE, COHERENT, wavefunction! That is why they showed a picture of just ONE wave!!

I believe I have mentioned this repeatedly. I am SO done with this explanation!

Zz.

16. Feb 23, 2005

### Edgardo

Then, in your opinion, what happens if I apply Doppler-cooling?
I claimed that the atoms are SLOWED DOWN by techniques like Doppler-cooling, Sisyphus-cooling, Raman-cooling, VSCPT etc. What's wrong about this statement.

What's then the meaning of the velocity/momentum distribution of a BEC?
For example here:
http://www.physics.ubc.ca/~birger/lect22/node3.html
http://www.iop.org/EJ/article/1355-5111/8/5/004/qs08005l2.html

17. Feb 23, 2005

### ZapperZ

Staff Emeritus
This is getting sillier by the minute.

First of all, "cooling" means the lowering of thermal vibrations. The reason why this is necessary for BE condensation is because thermal vibrations ADDS to the inability of particles to stay in contact long enough to form a coheresive overlap of their wavefunction. So OF COURSE you have to cool it down! However, by doing that, it doesn't mean the requirement for BE condensation is that they have to be almost "not moving" (as if the uncertainty principle no longer applies). If this is true, then high-Tc superconductors would not have been possible. You have normal superconductors condensing at 9K, while Hg2212 condensing at 150K! The formation of the condensate has less to do with it "cooling" or "not moving", but rather enough to the point that the coherent overlap will occur! If I can find a "glue" strong enough to coherently attach a bunch of particles together, I can make then go a BE condensation at room temperature if I wish!

Maybe you should read up a bit more on QM and figure out what they mean by momentum eigenstates.Or for that matter, read up on quantum statistics of indistinguishable particles. Those are part of the STANDARD college text books, and not from some papers.

Zz.

18. Feb 23, 2005

### Edgardo

There's no need to become unkind. Only because you know
more than me and I ask some questions, you don't need
to call them silly. All I am trying here is to discuss as I want to
learn.

What do you mean by that exactly, could you explain, please?
For me, cooling down means slowing down atoms. Please don't call this silly again, I'd really like to know how you imagine the cooling process (what picture you've got in your head - maybe I got something wrong).

Do I understand something wrong here? In the picture here
http://www.iop.org/EJ/article/1355-5111/8/5/004/qs08005l2.html
where the y-axis shows the population, isn't that the external momentum distribution of the particles in the BEC?
For example in VSCPT, most of the particles have the momentum
$p=\pm \hbar k$.
And please don't call this question silly again :tongue:

19. Feb 23, 2005

### ZapperZ

Staff Emeritus
I call it silly because, considering that we are dealing with BE condensation, I find it ridiculous that I have to go back to freshman thermodynamics to explain these things. I find that an unreasonable, huge backward step to explain something that should already be understood.

If you wish to adopt the principle that the object must not move (inspite of the violation to the HUP), then go right ahead. I appear to have not made any progress in trying to explain why this isn't so.

I can do ONE better than that. My avatar shows the energy versus MOMENTUM of electrons in the superconducting state, meaning they have already undergone a condensation. You can do a horizontal slice and obtain a well-defined momentum peak. But is this really the momentum of an "individual electron"? Or maybe it is the momentum of the EIGENSTATE? But more importantly, I can show you almost exactly the SAME looking curve just slightly above Tc, where there are NO condensation. This means they have roughly the SAME momentum before and after condensation. SHOCK! They didn't need to "slow" down to form the condensate!

Please look at the wavefunction in the form of plane waves, and then figure out what is meant by "momentum" or "k" eigenstate. I can forgive and understand anyone not understanding the language of many-body physics in order to understand exotic phenomena such as BE condensation, but I don't think I have the patience to go back to QM 101 to dig out an explanation, when they are clearly available in standard texts.

Zz.

20. Feb 23, 2005

### RandallB

I SAID it indicates a "FTL -- of some kind"
So what - you can create a "FTL -- of some kind" any time you want.
The popular one in these pages is shinning a laser beam on the Moon and moving it back and forth with a flick of your hand - creating a FTL spot moving back and forth on the surface of the Moon!
SO WHY IS IT THAT: "The two slit and in general interference patterns for some reason are often misinterpreted this way." Because they both do seem to indicate FTL. But seeming that way doesn't proof FTL, just that what's seen needs to be understood (no need to explain in this thread).

The silly thing here is this thread was NOT intended for "general interference patterns" or for Buckyballs & larger "particles". And I have no idea how getting a NFL regulation size football involved could be useful with such a simple question as:

WHAT are the largest limits of a practical two slit interference pattern experiment??
If you don't know or cannot figure it out - that's OK just say so - that puts you and me in the same boat - I ask because I don't know. We can both hope someone does:

So once again - for those that may have missed the point:
What are the max. limits of running a successful TWO SLIT demonstration.

1)Longest Wavelength?
2)Widest Slits?
3)Largest separation of the slits?