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How To Build a Quantum Telescope

samzenpus posted about 4 months ago | from the all-the-better-to-see-you-with dept.

Science 60

KentuckyFC (1144503) writes "The resolving power of telescopes is limited by the diffraction limit, a natural bound on resolution caused by the way light diffracts as it passes through a lens. But in recent years, physicists have worked out how to use quantum techniques to beat the diffraction limit. The trick is to create a pair of entangled photons, use one to illuminate the target and the other to increase the information you have about the first. All this is possible in the lab because physicists can use their own sources of light. Indeed, last month, physicists unveiled the first entanglement-enhanced microscope that beats the diffraction limit. But what about astronomy where the light comes from distant astrophysical sources? Now one physicist has worked out how to use quantum techniques to beat the diffraction limit in telescopes too. Her idea is to insert a crystalline sheet of excited atoms into the aperture of the telescope. When astrophysical photons hit this sheet, they generate an entangled pair of photons. One of these photons then passes through the telescope to create an image while the other is used to improve the information known about the first and so beat the diffraction limit. Of course, all this depends on improved techniques for increasing the efficiency of the process and removing noise that might otherwise swamp the astrophysical signal. But it's still the early days in the world of quantum imaging, and at least astronomers now know they're not going to be excluded from the fun."

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TURDS FOR NERDS! (-1)

Anonymous Coward | about 4 months ago | (#46682693)

So a little while ago I felt like I had a great big healthy shit coming on. I mean it makes sense. I have been adding more fiber to my diet lately. So I sat on my white porcelain throne and then ... *unghhnhhghhghh* *PLOP* Ah. Wow that was a big one. Damn it this is a big, wide, long log. It'll probably fold in half and get stuck and clog the fucking toilet. Again. Ah well. I will deal with that if and when it happens.

It was such a wide turd-log with such girth though. Amazing that my otherwise virgin asshole can expand that wide. Hmm that could be a problem. So I start to wipe. Damn it this could take a while. I got feces smeared on the sides of my ass cheeks from the girth of that turd. I wipe some more and ... oh no. I feel something knotty. Yup, what I feared from this turd has come to pass. I have lots of hairs around my asshole. I suppose most men do. Nature's way of saying "I love you! But not that much."

There it is. I can feel it through the single ply paper. A great big DINGLEBERRY. Naturally I try to awkwardly grab it with a folded sheet of toilet paper and dislodge it. I pull. Ouch. I pull harder. OUCH. Wow during its brief passage past the asshair, this turd-let really securely managed to get caught on some ass hairs. I can't just pull the dingleberry out without ripping the hairs right out. I have no idea if that caries the possibility of breaking the fecal encrusted skin and leading to an infection or what, but I know it would hurt so I want other options. I try gathering lots of sheets of toilet paper. I repeatedly wipe the same area over and over, hoping to wear this fucker down. I manage to whittle it down a little but this is taking far too much time. This turd is really determined to stay in its new home!

I can't very well pull my underpants back up now, that would get them all shitty and smelly. If I wanted to smell bad all the time I would let myself get fat. Maybe some water will help. I awkwardly reach for the faucet, turning it to a slow stream, and wet some toilet paper, making a mental note to disinfect the faucet handle later. I can't see the damned thing but judging from the brown stains appearing on the sheets of toilet paper, I am at least making progress. Now my hands are wet and shitty smelling and I am thinking this better be worth it. I use a dry sheet to feel for the dingleberry again. It did shrink but it's still there, dangling from my ass hairs, mocking me. WTF have I been eating lately to produce such a persistent turd? Nature does abhor a vacuum, which is why lots of gas has entered my bowels where the big turd-log recently was. I enjoy a nice after-defecating loud fart while I wonder what to do next. I chuckle because when the fart is your own, you don't think it stinks but you know somebody else would evacuate the area. Ok time to stop laughing, this is a serious predicament.

I toy with the idea of getting some scissors or something to try and cut the dingleberry out. Then I consider this is a sensitive area, I cannot see what I am doing, and it's too close for comfort to my cock and balls to be wielding a bladed item. I am starting to get angry. I am starting to not care anymore about the consequences of just yanking the damned thing out. I tried the easier ways and they failed. Fuck it, I have places to be and things to do. I can't very well spend all day in the bathroom playing a not-so-fun game with a turd. I double up on toilet paper and get a good secure grip on the dingleberry. Okay fucker, you're going DOWN. *YANK* Yeouch, fuck that hurt as much as I thought it would. And there it is, in my hand, nestled in the folds of toilet paper: my dingleberry! Ha ha ha, you won the battle, dingleberry, but I just won the war! I rub my sore ass cheek. Then I ceremoniously plop that fucker in the toilet bowl, to briefly swim with his big brother log. Oh man, you never heard such a satisfying flushing sound in your life. I rub some aftershave around my ass hairs just to make sure, better to smell funny than to get some kind of infected pimples or something.

I was careful. But not careful enough. Never careful enough. It was no trivial task to scrub the fecal matter completely off my hands and especialy, from under my fingernails. I find it useful to use a bar of soap and dig my nails in it. Repeatedly. I think all the turd particles are gone now. If I eat with my hands and get sick later I will at least know why, but it hopefully will be okay. Man. What an ordeal. Anyone who ever fought against a dingleberry knows how persistently they can cling to life next to your asshole.

So I watch this mixture of feces and paper swirl around the flushing toilet bowl. Sure enough, the very middle of the log gets sucked toward the drain. Not good, not good. It folds in half and the two halves at once is too wide to fit through the toilet. Great. Now my toilet is clogged. But the water continues to rise. Oh man, this doesn't look good. Quick, where the fuck did I put my plunger?!

... the saga continues ...

Re:TURDS FOR NERDS! (-1, Troll)

Anonymous Coward | about 4 months ago | (#46682739)

"... the saga continues ..."

Fuck beta?

Please proofread your post (1)

azav (469988) | about 4 months ago | (#46682789)

> excluded form the fun

from* the fun.

Re:Please proofread your post (1)

Ken_g6 (775014) | about 4 months ago | (#46682995)

Also, it's "early days in the world quantum imaging". So either this only works for seeing exoplanets, or the word "of" is missing.

Re:Please proofread your post (0)

Theovon (109752) | about 4 months ago | (#46683279)

And of course, everyone else on slashdot waits with baited breath to see you and your ilk post grammar complaints. Surely, nobody just gets over the minor typos and actually concentrates on the article, which (unlike so many other articles) is actually really interesting news for nerds.

What astounds me is the arrogance of some people who seem to imply by their behavior that they believe that they themselves never make mistakes. I would assert that losing sight of the forest for the trees (what’s more interesting, quantum telescopes or complaints about grammar?) IS a mistake, which means that you ARE imperfect and therefore might want to stop making yourself look like an ass by unnecessarily complaining about grammar and spelling.

Re:Please proofread your post (0)

Anonymous Coward | about 4 months ago | (#46683715)

Hey hey don't stop there.

Why don't we just overlook rounding in long division... I mean... you get the idea that we are diving 15 by 3 so what the hell do we need all these decimal points for. Let's extend it to variable naming... my_Variable is the same as MyVariable right... I mean why should we concentrate on the minor typos, we should just be concerned with the overall program.

I would assert that you ARE imperfect and therefore might want to stop making yourself look like an ass by unnecessarily complaining about people who complain about grammar and spelling.

Disclaimer: I probably mis-spelt something or used grammar improperly while writing this comment. Please feel free to edit it.

Ich bin ein andere Grammarnazi (1)

Applehu Akbar (2968043) | about 4 months ago | (#46684071)

You mean 'bated breath'. Baited breath is what you get from eating fish.

Re:Please proofread your post (1)

cusco (717999) | about 4 months ago | (#46684813)

It's actually 'bated breath', not 'baited breath', as in holding one's breath in anticipation, not having breath that smells of rotting minnows.

Sorry, just couldn't resist. Actually I agree completely with your point.

Re:Please proofread your post (1)

Soulskill (1459) | about 4 months ago | (#46683783)

I've updated to fix. Thanks.

Re:Please proofread your post (0)

Anonymous Coward | about 4 months ago | (#46690183)

I've updated to fix. Thanks.

For your next trick, maybe you can fucking spellcheck and proofread BEFORE releasing something for a very large audience. That would be slightly professional.

Try it on for size, you may even like it!

Pixel limit of satellite images? (0)

Anonymous Coward | about 4 months ago | (#46682825)

Would this technology also enable us to get higher definition satellite images of earth? Or is that a different refraction which makes max resolution about half a meter/pixel ?

Satellite imagery on wikipedia [wikipedia.org]

Re:Pixel limit of satellite images? (1)

Joce640k (829181) | about 4 months ago | (#46682955)

I suspect the problem there is lack of demand for higher resolutions.

Selective aerial photography is much cheaper.

Satellite Images are better than that (0)

Anonymous Coward | about 4 months ago | (#46684729)

There are government limits on the quality of satellite images that can be made available.

http://en.wikipedia.org/wiki/GeoEye-2

This commercial bird would of had a resolution for .34 meters per pixel.

Mind blown (3, Interesting)

dargaud (518470) | about 4 months ago | (#46682847)

I already had my mind blown when active mirrors managed to get rid of turbulence. But this is another thing entirely, getting rid of diffraction !?!

Re:Mind blown (0)

Anonymous Coward | about 4 months ago | (#46683159)

Can't we already partially do that by deconvoluting an image?

Re:Mind blown (0)

Anonymous Coward | about 4 months ago | (#46685551)

It works up to a point. The deconvoluting matrix has large entries (positive and negative) which amplify noise. The more details you want to resolve, the less noise you can tolerate in your input image.

That's a bit of a Leap.... (4, Funny)

Lumpy (12016) | about 4 months ago | (#46682857)

Please Please, call the control system "ziggy".

Excited atoms (0)

Anonymous Coward | about 4 months ago | (#46683339)

"Her idea is to insert a crystalline sheet of excited atoms into the aperture of the telescope."

I wonder who is going to excite the atoms and up to what point they should be excited. We know there is a point when they don't take any more excitement and go to sleep.

Can they excite each other?

So many details to figure out...

My theory about photons by Aglae Kellerer (0)

Anonymous Coward | about 4 months ago | (#46682861)

All photons are thin at one end, much, much thicker in the middle, and then thin again at the far end.

The image formation process is still the same (4, Informative)

Anonymous Coward | about 4 months ago | (#46683009)

The field being produced by the telescope optics is still the same, as the same primary mirror, secondary, etc. is being used to form the image. Yes, you can use multiphoton processes, even ones that are promoted by entangled photons, to produce apparently narrowed two-photon wavefunctions. However, this two-photon wavefunction is still derived from the ordinary resolution field created by the telescope optics. Therefore it seems to me that little is to be gained by using photon entangled detection to augment a process of image formation that is still fundamentally limited by the telescope aperture size.

Similar arguments have been used for other forms of imaging (e.g. microscopy and optical coherence tomography) and they all have this issue as the image formation process is still essentially a linear scattering process. There was some excitement around quantum lithography, however, even that has the problem that the probability of two-photon processes can be quite small even with entangled photons. For inherently multiphoton processes, such as two-photon absorption, stimulated Raman scattering, etc. there may be an advantage of increasing resolution and lowering dose, but I don't see much of a benefit to improving an instrument where the image formation process is a linear imaging process.

Re:The image formation process is still the same (3, Insightful)

dak664 (1992350) | about 4 months ago | (#46683631)

Yes, and what's more diffraction causes no fundamental limit to resolution, it just happens to be the distance between the first zeroes of an interference function. For two point sources of equal intensity that leads to an easily seen contrast difference of around 25% but trained observers can detect 5%. On electronic displays the contrast can be cranked up arbitrarily.

The fundamental limit to resolution is signal-to-nose.

Re:The image formation process is still the same (1)

Arkh89 (2870391) | about 4 months ago | (#46685023)

No, there is a limit to resolution as your PSF (Point Spread Function, the image of a quasi-perfect point), is not a point but a pattern with finite size. Which means that the OTF/MTF couple has a limited support and your system CANNOT deliver any spatial frequencies beyond certain point (1.0/(Lambda F), where F is the F-Number of your system).

Re:The image formation process is still the same (1)

dak664 (1992350) | about 4 months ago | (#46685383)

That's the infinite plane wave approximation for lattices of infinite extent. Scattered spherical waves from finite objects will result in some energy passing through the aperture for every spatial frequency. Although it could be difficult to sort out which frequencies are contributing (aliasing). Analysis of the through focal series can do that, also changing the convergence of incident illumination.

But if the source is known to be two points, accurate measurement of the spacing between the resulting PSFs is limited only by signal to noise.

Re:The image formation process is still the same (0)

Anonymous Coward | about 4 months ago | (#46685611)

Incorrect. Image deconvolution/convolution techniques allow observers to tradeoff between SNR/resolution.

Re:The image formation process is still the same (1)

Arkh89 (2870391) | about 4 months ago | (#46685849)

True... up to the optical cut-off... After that NO optical system will transmit spatial frequencies...
I was talking of this optical cut-off, you are talking of how to improve the system MTF before this cut-off...

Re:The image formation process is still the same (1)

LordVader717 (888547) | about 4 months ago | (#46685481)

You're confusing two different problems, namely locating a point source on a dark background, and wide-field imaging. The former can be improved beyond the diffraction limit, while the latter can't.

Re:The image formation process is still the same (4, Informative)

amaurea (2900163) | about 4 months ago | (#46688205)

The effect of a telescope's point spread function [wikipedia.org] is to convolve [wikipedia.org] the image. A raw image f(x) is turned into f'(x) = int dy f(x-dy) g(y), where g(dx) is the point spread function. By the convolution theorem, a convolution is simply a product in fourier space [wikipedia.org] , so F'(k) = F(k)*G(k), where uppercase functions are fourier-transpformed versions of lowercase ones, and k is the wavenumber [wikipedia.org] . From this you can see that recovering the raw, unblurred image (i.e. overcoming the diffraction limit), is just a question of computing F(k) = F'(k)/G(k), i.e. dividing by by the point spread function in fourier space, or deconvoluting it in normal coordinates.

What limits our ability to do so is the presence of noise in the final image. So a more realistic data model is F'(k) = F(k)*G(k) + N(k), where N is the noise, and when we now try to remove the point spread function, we get F_estimate(k) = F'(k)/G(k) + N(k)/G(k) = F(k) + N(k)/G(k). So we get back our unblurred original image, but with an extra noise term. And since G(k) usually falls exponentially with k [wikipedia.org] (meaning that high wavenumbers = small scales in the image are progressively blurred), N(k)/G(k) will grow exponentially with k, making the noise ever greater as one tries to recover smaller and smaller scales. This puts a limit on how small scales one can recover. But as you can see, it is not given just by G(k). I.e. not just by the diffraction limit. It is given by a combination of the telescope's noise level and the point spread function. In the absence of noise we can fully recover all the small scales. And even with noise one can push down a bit below the diffraction limit with enough data. But due to that exponential rise of the noise with wavenumber, it's an extreme case of diminishing returns. It is much cheaper to make the telescope bigger.

Re:The image formation process is still the same (1)

amaurea (2900163) | about 4 months ago | (#46688225)

I didn't make this explicit, but nowhere does one need to assume that one is looking at a point source (though that helps). I use this for looking at the cosmic microwave background, which is as wide-field as it gets.

Re:The image formation process is still the same (1)

Arkh89 (2870391) | about 4 months ago | (#46691343)

Re:The image formation process is still the same (1)

amaurea (2900163) | about 4 months ago | (#46691883)

Then that particular frequency cannot be recovered. But this usually only happens at zero crossings, which make up a vanishingly small fraction of the frequencies involved. Of course when noise is also present, then it's enough for G to be very small rather than exactly zero, and that would kill all the higher frequencies.

Re:The image formation process is still the same (1)

Arkh89 (2870391) | about 4 months ago | (#46694991)

Look closely at the first graph of the MTF in my previous link (in the EN page, top of the page). It represents your G function for a perfect system (no aberration, only the diffraction introduced by the finite size of the aperture). What you can see is that after the 1.0/(Lambda F) cut-off (500mm^-1 in that particular graph), everything will be equal to 0, thus the optical system is not transmitting any spatial frequencies larger than 500mm^-1.

This is not just "a couple of points", this is a hard physical limit to the resolution of your system.

Re:The image formation process is still the same (1)

amaurea (2900163) | about 4 months ago | (#46698655)

I think I found the figure you you're referring to. It's this one [wikimedia.org] , right? I don't think that figure lets you distinguish small from zero due to its very poor dynamic range. A logarithic second axis would be much more informative.

Here is an example of an MTF [folk.uio.no] from an experiment I've worked on. It looks quite similar to the figure on the Wikipedia page, and by eye one might think that's it's reached zero by 18000 or so. But consider the logarithmic version of the same graph [folk.uio.no] . As you can see, the graph had only fallen by about 20 dB by that point, and even at the end of the figure it's only down by 45 dB or so. So I don't think the Wikipedia figure supports your point.

Re:The image formation process is still the same (1)

Arkh89 (2870391) | about 4 months ago | (#46699655)

Ok, let's go to the Maths then : the OTF gives you how spatial frequencies are transferred through your optical system. You understand that it is equivalent to an auto-correlation of the aperture, right?
Well, if the aperture is circular (a disc function, for a perfect system), the auto-correlation is equal to the area of the intersection between two shifted discs (of equal radius). This shift represents the spatial frequency : at 0 spatial frequency (the DC component), the discs are aligning perfectly and you get the highest transfer (=1); at high frequency, the two circle are shifted a lot and you have only the area corresponding to a very small cat's eye shape; at the cut-off, only a single point is common between the two discs; finally, after the cut-off the two discs are not intersecting anymore and thus the transfer function is EQUAL TO 0.

In the case of your experiment, what was the cut-off frequency? (1.0/(Lambda F) where F is the F-Number of your optical system)
Did you made measurements after that frequency?

You can also read that kind of resource [spie.org] .

Re:The image formation process is still the same (1)

amaurea (2900163) | about 4 months ago | (#46703197)

The thing that's the self-convolution of the pupil function is the point spread function (g(theta) in my example from a few posts back). For the case of an ideal, top-hat shaped pupil function, the point spread function will fall to zero, and stay zero, at theta = 2*theta_pupil. But the optical transfer function (G(k) in my post) is the fourier transform of the point spread function. And the fourier transform of a function with limited support has unlimited support in fourier space [wikipedia.org] . Hence, while the optical transfer function will never fall to zero in this case (or any other case with a sharp edge to the pupil function), except for occasional zero crossings.

Re:The image formation process is still the same (1)

Arkh89 (2870391) | about 4 months ago | (#46705377)

The thing that's the self-convolution of the pupil function is the point spread function (g(theta) in my example from a few posts back).

Wrong. The Point Spread Function (in the case of incoherent imaging) is the magnitude squared of the Fourier Transform of the pupil function.

You can read : Born & Wolf, Principle of Optics, Chapter 8, Section 5 : "Fraunhofer Diffraction at apertures of various forms" (see here, starting p436 [google.com] ).

For the case of an ideal, top-hat shaped pupil function, the point spread function will fall to zero, and stay zero, at theta = 2*theta_pupil.

Wrong. In that case, the PSF will be similar to the Airy pattern [wikipedia.org] (circular aperture) or a Cardinal Sine function (square/rectangular aperture). Both have unlimited support (see previous book reference).

And the Fourier transform of a function with limited support has unlimited support in Fourier space [wikipedia.org] .

Right. And the Fourier transform of a function with an unlimited support in the direct space has a limited support in Fourier space.

Hence, while the optical transfer function will never fall to zero in this case (or any other case with a sharp edge to the pupil function), except for occasional zero crossings.

Wrong. The PSF has unlimited support, thus the OTF (and MTF) has limited support.

Re:The image formation process is still the same (1)

amaurea (2900163) | about 4 months ago | (#46708943)

Thanks for the book reference (though the missing pages were very annoying). You have convinced me. I guess it's good my job isn't designing telescope optics.

To answer your question about the telescope: It's f-number is 2.5 at Gregorian focus, and we observe at 148 GHz. The optical transfer function plots I showed had multipole number on the horizontal axis.

Re:The image formation process is still the same (1)

Arkh89 (2870391) | about 4 months ago | (#46709405)

No problem. I know how difficult it is to "bridge" the multiple scientific domains required for today's projects.

Re:The image formation process is still the same (1)

Raenex (947668) | about 4 months ago | (#46688781)

The fundamental limit to resolution is signal-to-nose.

Otherwise known as the Pinocchio ratio.

Re:The image formation process is still the same (1)

Immerman (2627577) | about 4 months ago | (#46684565)

I'm not an expert on optics, but from what I understand aperture size determines two things: the number of photons collected, and the angular resolution. And provided you've got plenty of photons to work with it's really only the second that matters. The implication is then that if you want more detail you need a larger aperture. That's fine for handheld telescopes, but once your aperture starts being multiple meters across you start introducing serious practical (including financial) difficulties, both in manufacturing a large and perfect enough lens and, if we're talking space telescopes, the size limits of the lift vehicle and the orbital cross-section (for collision risks with minute orbital debris)

It sounds to me as though this technology has the potential to add a second way to increase the resolution - the microscope example mentions being able to increase the angular resolution by 35% with the same optics, and it sounds like that's only a proof of concept device. Consider the implications if you could triple (or more) the resolution of a telescope without changing the size.

Re:The image formation process is still the same (0)

Anonymous Coward | about 4 months ago | (#46689327)

The field being produced by the telescope optics is still the same, as the same primary mirror, secondary, etc. is being used to form the image. Yes, you can use multiphoton processes, even ones that are promoted by entangled photons, to produce apparently narrowed two-photon wavefunctions. However, this two-photon wavefunction is still derived from the ordinary resolution field created by the telescope optics. Therefore it seems to me that little is to be gained by using photon entangled detection to augment a process of image formation that is still fundamentally limited by the telescope aperture size.

Similar arguments have been used for other forms of imaging (e.g. microscopy and optical coherence tomography) and they all have this issue as the image formation process is still essentially a linear scattering process. There was some excitement around quantum lithography, however, even that has the problem that the probability of two-photon processes can be quite small even with entangled photons. For inherently multiphoton processes, such as two-photon absorption, stimulated Raman scattering, etc. there may be an advantage of increasing resolution and lowering dose, but I don't see much of a benefit to improving an instrument where the image formation process is a linear imaging process.

Ya I agree with you, this paper is nonsense. I guess that's why it's not published in a peer-reviewed journal.

Entanglement aint what people are made to believe (0)

Anonymous Coward | about 4 months ago | (#46683245)

Its simply somehow 2 'quantum' objects are generated the same way (that gfilter turns 1 photon into 2 with the same quantum properties and they can go down different paths and hold that quantum 'value' to be used in parallel fo different things.

There is no OH we change this one over here and THAT one changes too.

Of course a typical problem is what states you can have for a 'photon' (like energy level) - Im not so sure how many others their coul be. Setting it (or a pair of them) to that state (with any accuracy) is often a seperat problem. Co-emitting pairs of photons of the same energy state is what they are doing.

Another issueis that 'Reading' one of these split 'photons' is destructive, so if you can generate that 'copy' you have a spare taht once one is used up you still have the other to do something else with (a 'copy' that you can reference (once))

Redirecting a photon to stear it a different direction --- shouldnt that change the 'state' of one photon versus another ???

"to increase the information you have about the first" weird phrase meaning 'read its value so you know what the other one was BUT NOT what changes the other subsequently goes independantly thru.

Seriously this quantum stuff is becoming the same old HYPED BS like so many things weve had in the past.

Re:Entanglement aint what people are made to belie (0)

Anonymous Coward | about 4 months ago | (#46684439)

You describe the Hidden Variables description of Quantum Mechanics, which was wiped by Bell's Inequality [wikipedia.org] ...
Sorry...

Re:Entanglement aint what people are made to belie (1)

Immerman (2627577) | about 4 months ago | (#46684895)

No.

First off entangled particles aren't "copies", they're linked complements. I.e. in an entangled electron pair one will have an up spin, and one a down spin. And there very defintiely *is* an "Oh, we change this one over here and THAT one changes too", or at least that's the conclusion of every experiment thus far constructed to try to prove or disprove your proposition, that there are "hidden variables" set at the moment of entanglement. As best we can tell the entangled particles become a single interlinked wavefunction and doing something to particle 1's quantum state *will* change particle 2's state instantaneously, regardless of the intervening distance. And yes, at first glance this does appear to violate special relativity, but doesn't actually do so for two reasons: 1 - it appears to be impossible to meaningfully transmit information in this way, and 2 - particles don't actually possess locality. The wavefunction of every "particle" in the universe exists simultaneously at all points in the universe, even if it's almost entirely focussed within a single small volume. Yes, it's brain-bending stuff, but all the experiments and math so far point strongly to the fact that it's the truth.

Secondly you only destroy the entanglement if you do something to one "particle" that depends on the quantum state. A lens that treats every photon the same will not disrupt the quantum state, and thus entanglement will be preserved.

And finally there's nothing weird about "to increase the information you have about the first" - so long as entanglement is maintained the particles quantum state *cannot* change independently. You need to make the measurements pretty much simultaneously (i.e. "second" refers to "the particle labeled as 2", not "the particle we measured later"), but when you do so you know that particle 1's detection will have a certain amount of quantum distortion to it, while particle 2 will have a complementary quantum distortion. By comparing the two you can then determine the nature of the distortion (to within certain limits) and computationally remove it from your measurements.

Oh, great idea... (1)

Vrallis (33290) | about 4 months ago | (#46683427)

Just wait until we deal with the giant starfish, sentient lobsters and mirror girls.

Path of Light and Clarity (1)

ZahrGnosis (66741) | about 4 months ago | (#46683851)

I remember reading somewhere (and I've spent nearly two minutes searching on Google, so you know it's somewhat obscure) that some people were concerned that our images of distant galaxies were TOO CLEAR. Their reasoning was that any given photon will take a (likely highly biased) random walk through quantum foam, or that the clarity actually helps disprove quantum foam theories (some information here: http://www.scientificamerican.com/article/is-space-digital/).

I realize I'm light on details, and that's due to my memory and weak goggle-fu (I'm on a lot of [legal] drugs today) so go gently on ripping the science apart, but if anyone had actual references to the real theory, I'd be interested in its current efficacy, and how it relates to the /. article.

A question about the microscope (4, Informative)

interkin3tic (1469267) | about 4 months ago | (#46684125)

I must have been actually working last month because I haven't heard about the entanglement-enhanced microscope. Does it do fluorescence microscopy? The was no mention of fluorescence in the article. It sounds like this is just better DIC imaging, which is of limited use in biology. Electron microscopy has (literally) been around since before the internet and has better resolution than anything you're going to get with light. Light microscopy seems to be primarily important today for basic stuff like whole tissue imaging (generally not requiring the resolution described here) or fluorescence microscopy, which it doesn't sound like this microscope can do. Fluorescence is useful because with most applications, you're trying to visualize a small thing in a much much much bigger volume of stuff. Like you're trying to see a protein within a cell within a tissue. Looking at cells with light for a small thing doesn't tell you much, you just see a blur. When the small thing is basically emitting it's own light, as happens sorta with fluorescence, you can see it.

There are also already fluorescence based microscopy techniques which surpass the diffraction limit. [wikipedia.org]

I'm not going to say it doesn't sound useful, since most of the time, you only realize how useful a thing is once you already have it. I'll just say that if the microscope mentioned here doesn't do fluorescence, I can't think of anything one would use it for that they wouldn't be able to do better with EM or fluorescence microscopy.

Re:A question about the microscope (1)

Charliemopps (1157495) | about 4 months ago | (#46685521)

There is a big push now to create ultra-cheap microscopes for the developing world. This might be related to that. If you could get a powerful microscope that was the size of a postage stamp, that might help a lot of people. With this technique you could not only improve resolution, you could instead keep the same resolution and shrink the microscope by 25%

Re:A question about the microscope (0)

Anonymous Coward | about 4 months ago | (#46687189)

this is just better DIC imaging

Am I the only person who giggled when reading this?

Re:A question about the microscope (2)

czert (3156611) | about 4 months ago | (#46691151)

Well, for one thing, electron microscopy is a destructive process, so if you actually want your sample back, you will likely use light microscopy. Also, electron microscopy cannot capture a moving thing (a living cell, for example), as it takes a lot of time and goes "pixel by pixel", instead of capturing a whole frame at once. I'm sure there are a lot of other reasons why light microscopy could be far more practical.

Patient of Patience (1)

Tablizer (95088) | about 4 months ago | (#46684419)

...entangled photons, use one to illuminate the target and...

Taken at face value, that would be sending photons to nebulas millions of light-years away, and then waiting another million+ years for them to bounce off the target and arrive back here.

At least the cockroaches will have these great space-themed calenders.

Re:Patient of Patience (2)

commander_gallium (906728) | about 4 months ago | (#46684691)

...entangled photons, use one to illuminate the target and...

Taken at face value, that would be sending photons to nebulas millions of light-years away, and then waiting another million+ years for them to bounce off the target and arrive back here.

At least the cockroaches will have these great space-themed calenders.

Maybe if there was a way to bounce photons off of the second half of the summary you'd have been able to read it.

i really didnt understand that (0)

Anonymous Coward | about 4 months ago | (#46684733)

but it sounds pretty cool! yea quantum physics bitch!

Holy Crap! I remember this ....... (1)

sgt_doom (655561) | about 4 months ago | (#46685631)

..... it was called Tom Swift and His Megascope Space Projector --- never thought it would actually work, though?

Re:Holy Crap! I remember this ....... (0)

Anonymous Coward | about 4 months ago | (#46688393)

Tom Swift and His Megascope Space PROBER. How could you mess that up?

Promises, promises (1)

sgt_doom (655561) | about 4 months ago | (#46685711)

I promise that How to Build a Quantum Telescope will go on my bookshelf right next to How to Stuff a Wild Bikini 'cause I'm soooo serious......

Mirrors (1)

Katatsumuri (1137173) | about 4 months ago | (#46686199)

I thought most modern telescopes used mirrors instead of lenses to avoid diffraction. Well, maybe there are still some lenses left in the system, or maybe we can switch back to lenses if this works. Also, the research is interesting anyway.

Re:Mirrors (0)

Anonymous Coward | about 4 months ago | (#46689313)

I thought most modern telescopes used mirrors instead of lenses to avoid diffraction. Well, maybe there are still some lenses left in the system, or maybe we can switch back to lenses if this works. Also, the research is interesting anyway.

Light travels all paths of an optical surface simultaneously, and the resulting electric field at the detector is the integral over all paths. The intensity (what you measure with a detector) is the absolute value of the electric field squared. Both mirrors and lenses rely on a difference in optical path length between different parts of the optical surface to create an image, and thus neither "avoid diffraction." The advantage of mirrors is that glass for lenses has chromatic abberation (light of different wavelengths experiences a slightly difference refractive index and thus path length). But the "diffraction limit" still holds for both mirrors and lenses. Stay in school kid.

Re:Mirrors (1)

Carnildo (712617) | about 4 months ago | (#46689391)

No, modern telescopes use mirrors instead of lenses for two reasons:

1) Once a lens gets more than about a meter across, it starts deforming measurably under its own weight (and the direction and amount of deformation changes as you shift the telescope). A mirror can be supported across its entire width and does not have this problem.

2) A lens experiences chromatic aberration, causing different frequencies of light to focus at different points. You can reduce (but not eliminate) this by using achromatic doublets or other optical tricks (such as absurdly long telescopes), or you can take the easy way out and just use a mirror.

How do you get... (0)

Anonymous Coward | about 4 months ago | (#46689001)

... a crystalline sheet of excited atoms? Do the electrons dance for them? Dance of the Seven Wave functions perhaps?

Unecessary (0)

Anonymous Coward | about 4 months ago | (#46690391)

Physicists are already working on meta-materials with a negative diffraction index and work without the need for glass at all.

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