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Quantum random-walk puts single atom into observable superposition of states Answered

http://physicsworld.com/cws/article/news/39802

Single atoms have been spotted doing the quantum version of the random walk by physicists in Germany. This sighting of a "quantum walk" could help in the design of quantum search algorithms, or in the understanding of the transition from the quantum, microscopic world to the classical, macroscopic world.

Update 15 July 2009: The actual paper appears in this week's Science (pay-per-view, unfortunately). Here's a nice diagram from the article, contrasting (left) the quantum superposition with (right) one of the classicial random walks. The histograms below show their actual measurements in the two situations.

Discussions

I don't understand. Can you put this in Lamen terms?

And here's what my lamen looks like; I wear it whenever I'm at SLAC...

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No, that's a sample badge from our ES&H; Manual, as is the dosimeter image.

I've been diagnosed with GERT too, oh wait, that's GERD never mind ;-)

By the way, you look so much younger on your badge ;-)

According to Kelseymh in reply to my comment, Scientist's have observed Atom's being in two places at once.

Ya hoo! Thank you very much. You've got the basic point right there. We've been doing the superposition trick with photons and electrons for years. This experiment (and other recent ones), is extending the wacky features of QM up to whole atoms, and even molecules!

(bump) I grabbed a figure from the published paper, which might (or might not) clarify what was being measured, and why the quantum superposition is so funky compared to the simple classical case.

Some of this sounds familiar.....is it possible that some of it was being discussed on (for instance) the Science Channel (cable) ? I don't remember reading about it, but I think I remember hearing about it.

It probably was. The published article is in this week's Science magazine, and it's one of those cool "macroscopic quantum" things that science journalists love to pick up.

What specifically is the fluorescence that they're observing? L

I found the published article in this week's Science (access to the article is by paid subscription). Here's the paragraph which answers your question:

We realize a quantum walk with single laser-cooled cesium (Cs) atoms, trapped in the potential wells of a 1D optical lattice with site separation of λ/2 = 433 nm (here, λ is the wavelength of the lattice laser light). The atoms are thermal with a mean energy of kB x 10 µK, whereas the optical potential depth is kB x 80 µK (here, kB is the Boltzmann constant). They are distributed among the axial vibrational states with a mean occupation number of Formula. Initially, the atoms are prepared in the |0> = |F = 4, mF = 4> hyperfine state by optical pumping, where F is the total angular momentum, and mF its projection onto the quantization axis along the dipole trap axis. Resonant microwave radiation around 9.2 GHz coherently couples this state to the |1> = |F = 3, mF = 3> state.

Could you give me this in terms of:
energy in = energy out (fluorescence)?
They are pumping a system with optical frequency, and observing the re--emission of this? Didn't do Physics to a very high level.

L

Unfortunately, the authors don't specify the detector details of their fluorescence imaging. It's just being used as a tool to read out which bucket of the optical lattice "contains the atom" after each iteration of the quantum walk process.

I don't know. Widera, the author (post-doc) referenced in the Physics World report, doesn't have anything on his official Web page, so it's not published yet (or he just hasn't updated). There's no preprint in arXiv, either.

"I don't know"?! Any ideas? I'm not that familiar with this area, a bit too extreme for my education... L

Yup, I don't know. The problem is that most atoms fluoresce, if you put them into the appropriate state. And lots of atoms can be held in optical-lattice traps. So there are just too many possibilities for me to be comfortable guessing.

So I'm guessing this is useful for determining the location of particles? If so, cool (no pun intended).

No. The point is that the location is not determined. They put one single atom into their optical trap. An atom (if you think classically) is a little billiard-ball like thing. It's an object which is sitting at some location. You can see where it is by shining a laser on it and taking a picture of the light (fluorescence) it gives off. If you give the atom a kick (say with an electric or magnetic field), it will move. You can watch it move by taking a series of pictures of the fluorescence.

So far so good? Pretty simple, huh?

Now we reveal that atoms are really these weird quantum-mechanical wave functions, and not little billiard balls after all. If you give it a kick such that it could randomly go either left or right, what happens? Well, the wave function takes account of both possibilities, and the atom ends up in a superposition (look on Wikipedia) of states going both left and right. At least, that's the theory and mathematics that we get taught in freshman physics.

What Widera and his group at Bonn have done is to take that math, and turn it into an actual experiment! The put a single atom in a trap, kicked it with these ambiguous (left or right) kicks, and took pictures. Lo and behold, their pictures showed the atom in a superposition of being in two places at once.

Freaky and bizarre? Yeah. But exactly what we expect to see if the last 100 years of quantum mechanics is correct.

Wait...it's literally in both places at once? I don't understand. :-(

Um....yes. You've just triggered a discussion of "The Measurement Problem" and "Collapse of the Wavefunction." I'll let you read the Wiki's on those two topics, then let me know if you'd like a "lecture" on the issues involved.

Alrighty, I'll look at it tomorrow (sorry for the stupid question :-) ).

It's not a stupid question. The issue of how you get from a wavefunction (which is a distribution in space and time of probability amplitudes) to an observation (which is a definite value of something at a specific location in space and time) is really the deep philosphical problem.

So In the very basic concept of vibrating atoms doesn't that complicate matters if atoms do physically exhibit superposition behaviour, or am I simply looking at this with one oversimplified concept and one not so oversimplified concept...

Yeah, I think you're sort of mixing apples and monkeys here. If you try to describe the atom as a classical "vibrating spring," then just do that. You'll have a nice, concrete picture in your mind, but none of the quantum stuff. If you want to deal with superposition, then you need to work with complex-valued wavefunctions, and don't think of things as little billiard balls.

Actually I think of them as tiny objects composed of various types of billiard balls, Hydrogen's like someone's having trouble with black ball... I understand the concept of superposition, but not complex wave functions and I doubt you're here to give a science lesson...

If you understand superposition, you're 90% there. The only thing about the wavefunctions having complex amplitudes is that it allows for destructive interference without introducing negative probabilities. The probability is obtained by taking the modulus-squared of the wavefunction.

Yeah, I got what you were saying, but misinterpreted the sentence when they said that the particle was likely to be farthest away from the original point. I thought that meant that it would help identify where the particle would be according to probability, eg. The particle was at A, so in a minute it is most likely to be as far away as possible from A. Does that make sense? :\

Take a look at the second picture in the article. As the atom random walks, there's a small, but not zero, chance it will end up far from where it began. That could be either to the left or to the right (and that either is the key, here!). Of course, random walking can also take it back into its starting point.

What happens with the superposition of the wavefunctions is that (a) both the left and right excursions become part of the atom's state; and (b) the different possible visits back to the center interfere destructively and cancel out. That means the wavefunction has a "hole" at the atom's starting point, and peaks on both the left and right sides of large excursion.

Thanks for explaining that, I was just confused from the sentence I mentioned.

Here the sentence I was talking about:
"...means that the eventual position of the particle is most likely to be farthest away from the starting point."

Quantum mechanics are cool! I like the superposition stuff.

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