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The way that I best understand quantum entanglement (and maybe someone like jessriedel can correct me if I'm wrong) is thus: You may be familiar with the Schro
by Xcelerate 14y ago
The way that I best understand quantum entanglement (and maybe someone like jessriedel can correct me if I'm wrong) is thus:
You may be familiar with the Schrodinger equation (HΨ = ih' dΨ/dt) or the more accurate time-dependent Dirac equation. In each of these equations is a function called the wavefunction (denoted with Ψ). This function represents the "quantum state" of your system -- in other words, all the information that exists within a system. Ψ evolves deterministically in time. You can apply an operation to this function, and when you do, you get the original function back multiplied by some value. There are different operators, each corresponding to an "observable" (the thing you measure in the lab). For example, you can measure momentum, position, energy, spin etc... and each of these observables has a different corresponding mathematical operation that you perform on Ψ to get XΨ, where X is the mean value of the observable.
Now in QM textbooks, you'll frequently see Ψ written as Ψ(r, t), where r is a position vector and t is time. r denotes the position of whatever particle constitutes your system. So what if you have a two particle system like hydrogen (proton and electron) or positronium (positron and electron)? Well, a QM textbook will write the state of your quantum system as Ψ1 * Ψ2 and completely gloss over the fact that this approximation does not apply in all situations. It is physically inaccurate to say that Ψ of two particles is two individual Ψ's multiplied together. For some systems, it is a good approximation and makes things easy to calculate, but in reality, there is really only one wavefunction Ψ and it is a function of all the particles in the universe.
So now you can see where entanglement comes in. If Ψ is a wavefunction of all extant particles, then surely there will be correlations between every measurement of Ψ that you take! Now most of the time, you can't find correlations -- too many particles are affecting too many other particles (decoherence). But if you prepare two of them together and keep outside particles from interfering with them, then you can observe the correlations between two particles no matter what the distance!
So now for the whole "information transfer" business. I said earlier that applying an operator to Ψ gives you the value of an observable -- what you measure. The weird thing though is that what you measure isn't always exactly this value. Instead, the mean value of many measurements will be this value. You can also compute the standard deviation of these measurements using Ψ, but that's about it. Nobody knows where the random "noise" in measurements comes from. So far, it seems as though our universe just has some randomness inherent to it (and we're quickly ruling out all remaining superdeterministic theories; Gerard t'Hooft seems to be a hold-out: http://physics.stackexchange.com/questions/34217/why-do-people-categorically-dismiss-some-simple-quantum-models http://physics.stackexchange.com/questions/34217/why-do-peop...)
So you can't control or predict the individual measurements to as much precision as you'd like. Sucks, huh?
Anyway, you can plot and analyze this data, and what you'll notice for two entangled particles separated by thousands of miles or more is that there are statistical correlations between the two sets of data (again, data you can't control -- if you can't control it, you can't send information with it).
Obviously, you need both sets of data to notice that there are correlations.
- jessriedel 14y agoBeing requested by name is too ego-boosting to pass up, so let me take a crack at clarifying at least one apparent confusion. > For example, you can measure momentum, position, energy, spin etc... and each of these observables has a different corresponding mathematical operation that you perform on Ψ to get XΨ, where X is the mean value of the observable... >The weird thing though is that what you measure isn't always exactly this value. Instead, the mean value of many measurements will be this value. You can also compute the standard deviation of these measurements using Ψ, but that's about it. A good QM textbook will actually say something much more precise. It says that (a) the set of possible outcomes of the measurement is equal to the spectrum of the observable being measured and (b) the chance of getting a particular outcome is given by the squared inner product of the wavefunction with the appropriate eigenvalue. (The whole business of calculating means and standard deviations is confusing unless you understand that; unfortunately, this is allowed to happen often in into QM courses.) This means that QM doesn't just predict some statistical properties of the outcome distribution, it completely specifies the distribution. Also, I figure you know this, but I want to mention that when you say > you can plot and analyze this data, and what you'll notice for two entangled particles separated by thousands of miles or more is that there are statistical correlations between the two sets of data it's important to emphasize that these are non-local correlations (in the Bell sense). You can generate mere local correlations using everyday classical systems.
- Xcelerate 14y agoGreat, thanks for the clarifications!