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- Thread starter jerromyjon
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Vanadium 50

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Yes. B0 and anti-B0 mesons produced at CESR, PEP-II or KEK-B (only the last is still running) are entangled.

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So can/do they test for correlation and violation of Bell's Inequality? Or is it just assumed they are entangled...Yes. B0 and anti-Bo mesons produced at CESR, PEP-II or KEK-B (only the last is still running) are entangled.

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Violations of Bell's inequality don't show the presence or absence of entanglement; finding such violations is a sufficient but not a necessary condition for entanglement.So can/do they test for correlation and violation of Bell's Inequality? Or is it just assumed they are entangled...

What Bell's theorem does show is that no local hidden variable theory can violate the inequality. Because some entangled systems do violate the inequality, we conclude that entanglement cannot be explained by such a theory.

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That is exactly what I was trying to get at. If matter/matter entanglement were shown to violate Bell's inequality but matter/antimatter were shown never to violate it could it be concluded that entanglement is a chiral property?Because some entangled systems do violate the inequality, we conclude that entanglement cannot be explained by such a theory.

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Does this mean "static hidden variables"? As in pilot wave theory (which is non-local) it works as a static wavefunction encompassing entangled pairs? Sorry if what I'm asking about is obscure or unclear.What Bell's theorem does show is that no local hidden variable theory can violate the inequality. Because some entangled systems do violate the inequality, we conclude that entanglement cannot be explained by such a theory.

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I have no idea what you mean by "static hidden variables", but you are correct that pilot wave theory is non-local and therefore is not precluded by Bell's theorem.Does this mean "static hidden variables"? As in pilot wave theory (which is non-local) it works as a static wavefunction encompassing entangled pairs? Sorry if what I'm asking about is obscure or unclear.

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The wavefunction calculated by the Schrödinger equation is static is what I mean, it does not change over time, correct? But are there any papers on dynamic local variables which I assume would be deterministic but I can't find anything by searching...pilot wave theory is non-local and therefore is not precluded by Bell's theorem.

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Ah, yes, I overlooked that point. Trying to simplify concepts in my head I tend to miss things now and then. Thanks everyone for your help.they have a changing phase for states with nonzero total energies.

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it could it be concluded that entanglement is a chiral property?

What do you mean by "chiral property" here? I don't think "chiral" means what you think it means.

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I meant (for example) that the spin of atoms might have an asymmetrical component as in clockwise or anticlockwise rotation but I see the flaw in that line of thought.What do you mean by "chiral property" here? I don't think "chiral" means what you think it means.

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I did not think a bound state with an attractive potential and non-identical particles could be entangled.

Why not?

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A bound state is one in which the potential can be chosen to go to zero at infinity and the total energy will be negative.Does this forum have consensus definitions of bound and entangled states?

An entangled state is a state that can be written as a superposition of more than one term of the form ##|\alpha_i\rangle|\beta_j\rangle##, where the ##\alpha_i##s and ##\beta_j##s are eigenvalues of two commuting observables ##A## and ##B##.

These aren't formal definitions intended to survive assault by an army of quibblers, but they're good enough to get you through most discussions of the subject.

Consider that an attractive potential goes to zero at infinity, and that we can take ##A## to be any property of one particle and ##B## to be any property of the other. We can have non-identical particles in an entangled bound state.

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What is Bothering me is that at infinity I end up with two separate wave functions that don't overlap

The wave function doesn't change as the position changes. The wave function is a single function of position--actually, for positronium it's a function of two positions, the position of the electron and the position of the positron. Or you can reformulate it as a function of the center of mass position of the system and the distance between the electron and the positron; the latter is nicer because all of the interesting stuff, like the potential energy between the electron and the positron, depends only on the distance between them, not on the center of mass position.

It is true that the potential energy goes to zero as the distance between the electron and positron goes to infinity, but that doesn't mean the state is somehow separable at infinity but entangled everywhere else. It means you have a single entangled state that is an eigenstate of a Hamiltonian whose potential energy term vanishes as the distance goes to infinity.

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Positronium in the singlet state decays into 2 gammas (each of spin=1, polarized so that spin is conserved) and in the triplet state into 3 gammas. I believe it is the gammas that are entangled NOT the electron and positron in the bound state of positronium. Of course angular momentum is conserved in these decays.

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I believe it is the gammas that are entangled

That's correct, they are, because their total spin is fixed by conservation of angular momentum.

NOT the electron and positron in the bound state of positronium

That's not correct. See my previous post.

Also, even aside from the point I made in my previous post, if angular momentum is conserved, then the total angular momentum of the positronium must be the same as the total angular momentum of the gammas that it decays into. But that means the spins of the electron and the positron can't be independent, which they would have to be if the electron and positron were not entangled.

In other words, the states you are describing as the "singlet" and "triplet" states are different possible entangled states of a bound electron and positron, corresponding to the different possible values of the total angular momentum of the positronium system.

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The good thing with science is that you have to make the definitions really clear. To say "I have an entangled state" is not accurate enough. It doesn't tell you what's entangled. So we have to say what is entangled!

The most simple way to define it is to consider two compatible independent observables ##A## and ##B##, represented by self-adjoint operators ##\hat{A}## and ##\hat{B}##. Then there is a complete orthonormal set of simultaneous eigenvectors ##|a,b \rangle## of ##\hat{A}## and ##\hat{B}##, where for simplicity I assume that both ##\hat{A}## and ##\hat{B}## have only true eigenvectors and eigenstates (i.e., both have a discrete spectrum like, the spin-##z## operators of two particles). The meaning, according to Born's rule is: If the quantum system is prepared in a state represented by the Statistical operator

##\hat{\rho}=|a,b \rangle \langle a,b|rangle,##

then both observables take the determined values ##a## and ##b##, respectively.

Now we define any state that is a proper superposition of these basis states as a state, where the observables ##A## and ##B## are entangled. For simplicity we assume that the common eigenstates of ##\hat{A}## and ##\hat{B}## are only one-dimensional, i.e., the determination of both observables with certain values determine the state completely. Then, if you define

$$|\psi \rangle=\sum_{a,b} c_{ab} |a,b \rangle,$$

where

$$\sum_{a,b} |c_{ab}|^2=1,$$

and at least two of the ##c_{ab}## are not 0, the state represented by

$$\hat{\rho}_{\psi}=|\psi \rangle \langle \psi|$$

is a state for which ##A## and ##B## are entangled.

The fascinating thing is that ##A## and ##B## can be any observables of this kind you can define within QT. It can be two compatible observables of one particle. E.g., you can use position and spin-##z## component of a non-relativistic particle. With a Stern-Gerlach apparatus you can get states, where the position and spin-##z## component are entangled, i.e., if you have a beam of particles entering the inhomogeneous field of the Stern-Gerlach apparatus, after some traveling there the beam will split in two parts (position), and each partial beam consists of particles with a definite spin-##z## component. Take the original setup with spin-1/2 silver atoms. Then the wave function after running through the magnetic field is a superposition of exactly this form

$$\psi(\vec{x})=\psi_1(\vec{x}) |\sigma_z=1/2 \rangle + \psi_2(\vec{x}) |\sigma_z=-1/2 \rangle,$$

where ##\psi_1## and ##\psi_2## are wave packets (e.g., Gaussians) which are peaked narrowly in clearly distinct regions of space. The wave function doesn't describe particles with a well-defined position and spin-##z## component, but you have 100% correlation between position and spin-##z## component, i.e., if you find a particle in region 1 (region 2), then with 100% probality the spin-##z## component will have the value ##+1/2## (##-1/2##).

The two entangled observables can also be related to two different particles. Often you have experiments with two photons, whose polarization states are entangled. The fascinating thing with such a case is that the two photons may be detected on very far distant places but still can be in a polarization-entangled state. As in the case of the Stern-Gerlach example above the single-photon polarization can be completely indetermined (i.e., each of both observers just sees a stream of unpolarized photons when you send them such prepared photon pairs), but on the other hand they are 100% correlated, i.e., if you put the polarization filters for each of the photons in the same direction, always one will go through and then necessarily the other will be blocked and vice versa, when the state is given by the polarization-entangled state

$$|\psi \rangle=\frac{1}{\sqrt{2}} (|HV \rangle-|VH \rangle).$$

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