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One piece closer to a working electronic quantum computer puzzle kit.

Many believe the next generation of supercomputers will be powered by quantum mechanics. Harnessing the strange properties of photons and electrons in special states is often the backbone for quantum computer research. Some of these seemingly exotic properties have already been demonstrated using photons, but until very recently, were not replicated in solid-state systems by electrons.

A group of European researchers, consisting of institutions from France, Spain and Germany, has published their work with quantum entanglement using electron (Cooper) pairs, quantum dots and carbon nanotubes. Quantum entanglement is a quantum state of matter where two particles, typically photons or electrons, form a matched pair based on their physical qualities such as up or down spin for electrons and polarization for photons. When a pair of these particles becomes entangled, quantum mechanics states that measuring one of the pair will instantly force the unmeasured into a corresponding state, regardless of the distance they have been separated by.

In photonics work, researchers used wave guides and polarization filters to form entangled photons, which can then be separated by a beam splitter and measured individually. But for electrons, the work is far more taxing. Measurements are more easily skewed by background noise and leakage from the components of the test device.

The solid-state device used to confirm electron quantum entanglement is fairly simple in design. A superconducting element is used to form Cooper pairs. The pairs then move down the element towards a carbon nanotube. Occasionally the pair is split by the nanotube and each electron moves towards a separate quantum dot. In this time, one electron’s spin can be measured, which infers the spin of its mate instantaneously. These pairs can either be spin-correlated or anti-spin-correlated (spinning in the same direction or opposite directions), but the measurement of one always reveals the properties of the other.

Quantum entanglement could be very useful in theory, especially for quantum computing in the areas of security and data transmission. Theoretically, data can be transferred over any distance instantly and without any risk of security breech, however, the entangled pair still has to be transferred through physical media at this time.

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By GourdFreeMan on 1/13/2010 1:47:50 PM , Rating: 2
Even if you were to construct a chamber that mirrored the effect on the other particle, the communication would still take place at speeds no faster than the speed of light, however. Suppose the photons you use to influence the particles are emitted on your end of the chamber, it will take at least d/c for the second particle in the pair to be influenced (where d is the distance between particles, and c is the speed of light). Suppose the photons are released from the middle of the chamber. Both particles will be influenced at some time at or later than d/2c (relative to a stationary observer at the center of the chamber), however it will also take at least d/2c for you to communicate to the center of the chamber to release the photons. If the target measures his particle before d/c, there will not have been adequate time for your forceful change to have propagated to his particle. The particles are only entangled before either has been acted on, and have a relationship (which can't technically be called entanglement if their state if known) after both have been acted upon (presuming neither has been disturbed while you have been waiting all this time for the communication to take place).

If your chamber only influences the entangled particles as they travel without breaking their entanglement (e.g. without forcing them into a known state), then you aren't communicating anything, as you don't know what state the particles will be in when you measure them until you do. At best you are both getting the results the chamber is designed to force upon the probabilities of the particles being in a given state.

Entanglement is primarily a statement about the observer's knowledge (which influences probabilities), not about the action of forces.

By Spuke on 1/13/2010 2:10:41 PM , Rating: 2
If we knew the entangled states, could we use those states as a "data"? For example, if you have a stream of entangled particles whose states were known, could you parse out the desired particles to form a known string of particles to be used as data?

By Torment on 1/13/2010 4:32:29 PM , Rating: 3

They are entangled *because* they the quantum wave function has not collapsed (which corresponds to collapsing to a state). Say, for example, that an atom emits two electrons simultaneously such that the net spin must be zero. Both electrons are in a superposition of quantum wave states of up and down. When you measure electron1's spin state, you force a collapse of the quantum wave state, thus forcing electron1 to be either up or down. Since the net spin must be zero, this forces electron2 to be the opposite spin--instantaneously, no matter the distance. However, since the spin measurement on electron1 is entirely probabilistic, so is the resulting spin state on electron2. In fact, anyone observing electron2 would not be able to determine if its wave state had been collapsed by its counterpart having already been observed (otherwise, you could still transmit information).

IOW, everything looks completely random at both ends.

By Torment on 1/13/2010 4:38:38 PM , Rating: 2
Beware the errant 'they'

By Spuke on 1/13/2010 4:41:32 PM , Rating: 2
IOW, everything looks completely random at both ends.
Are they in completely random states only or can they be stateless also before observation? Or is that the same thing? Thanks for the explanations.

By Fritzr on 1/13/2010 8:59:34 PM , Rating: 2
In the case of unknown until observation, stateless and random but identical to partner give identical results when observed. If stateless the wave form collapses to one value or the other and that value is observed. Random, the particles have an unknown value until observed and that value is observed. It will require some test of the pre-observation state to distinguish the difference.

By Torment on 1/14/2010 6:38:08 PM , Rating: 2
Your entire post is rather incoherent and nonsensical, but I just want you to think about your last sentence. Do you see the logical fallacy?

By Fritzr on 1/15/2010 5:52:43 PM , Rating: 2
Yep. I did not add a big arrow saying this was a fallacy due to my belief that it is obvious to a reader paying attention.

To clear up the other.
1) If the particles are stateless until caused to collapse by observation, then a discrete value is observed.
2) if the values have a random, but discrete value prior to observation, then a discrete value is observed.

The observed value in case 1 is indistinguishable from the observed value in case 2.

The logical fallacy of the final statement implies that a different test is required to determine if the particles are stateless or possessing a definite state that has a discrete value.

Observation of the discrete value says nothing about the state of the particle before observing it's value.

By Torment on 1/13/2010 8:57:16 PM , Rating: 2
An important distinction is that the electron is not both both up and down at the same. Rather, their wave function is in some admixture of states corresponding to up or down. The electron isn't spin up or spin down until you observe it, the act of which collapses the wave function. This is a probabilistic event, meaning that if the wave functions were in an equal admixture, it's a coin toss whether the electron ends up as spin up or spin down.

So, if I understand your question, the electrons, prior to observation, have no (spin) state. It only has wave functions corresponding to physical states. In the example given, those wave functions have equal probabilities, so that when you observe a stream of electron1's, it will have a random distribution of spin-up and spin-down. Because they are entangled, observing electron1 will force electron2 into opposing state, but to the observer of electron2, it is still purely random, even though it is deterministic at this point.

If, however, you could measure whether a wave-state had been collapsed *prior* to your observing it(and you were careful about timing, as doing so is surely an act of observation and you want observer1 to observe first), observer1 could then send signals in the form of stream segments that were either observed or unobserved on his end. Unfortunately, that is not knowable, so still no action at a distance.

By GourdFreeMan on 1/13/2010 4:36:06 PM , Rating: 2
For Quantum Computing I think they are more interested in using the probabilities to do useful computation, rather than knowing the states themselves.

I'm not sure if that answers your question...

By Spuke on 1/13/2010 4:47:59 PM , Rating: 2
I'm not sure if that answers your question...
It does in a way. It makes my question irrelevant when it comes to Quantum Computing. :) I don't know much (pretty obvious) about this subject but it's really fascinating.

By Torment on 1/13/2010 4:35:26 PM , Rating: 2
This is just gibberish.

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