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3ntangled

Can three or more particles be quantum entangled? And if they can, what will the two particles opposite state, like opposite spin, translate to for three (or more) entangled particles?

Certainly! Any n-particle state that cannot be written as a pure tensor product of n one-particle states is “entangled” to some extent, though in many cases the entanglement may be “partial” in the sense of only involving some of the particles, and the question probably wants a state which is “fully” entangled in the sense that there is no way of writing the state as a pure tensor product of states for any partition of the n particles into subsets. [Note: I am using “partial” and “full” here to refer to the breadth or scope of the entanglement rather than to some measure of its pairwise strength or depth.]

A system of three spin 1/2 particles is in a completely unentangled state if and only if the state vector can be written as a pure tensor of the form [math]|\Psi\rangle =|\Psi_1\rangle \otimes |\Psi_2\rangle \otimes|\Psi_3\rangle[/math] where each [math]|\Psi_i\rangle [/math]is a pure state of the one particle system (which may or may not be an eigenstate of some spin direction). The reason for this definition is because if a state can be written this way, then the effective state of the remaining particles after measurement of any one of them can be shown to always be independent of the measurement value obtained on the observed particle.

Note: A state that is given as a sum of several pure tensors may nonetheless be rewritable as a pure tensor and so be unentangled. For example, if [math]|\uparrow\rangle[/math] and [math]|\downarrow\rangle[/math] are the eigenstates for spin up and down in, say, the z-direction, then [math]\begin{align}|\Psi\rangle &=\frac{1}{\sqrt{2}}(|\uparrow, \downarrow ,\uparrow \rangle +|\uparrow ,\downarrow ,\downarrow \rangle)\\ &=\frac{1}{\sqrt{2}}(|\uparrow\rangle \otimes |\downarrow\rangle \otimes |\uparrow \rangle +|\uparrow \rangle \otimes |\downarrow \rangle \otimes |\downarrow \rangle) \\ &=|\uparrow\rangle \otimes |\downarrow\rangle \otimes(\frac{|\uparrow\rangle +|\downarrow\rangle}{\sqrt{2}})\end{align}[/math]

And this is unentangled because it can be written as a pure tensor and we can see that measurement of spin 3 can “collapse” the [math]|\Psi_3\rangle[/math] factor to either [math]|\uparrow\rangle[/math] or [math]|\downarrow\rangle[/math] without affecting either of the others.

But [math]|\Psi\rangle =\frac{1}{\sqrt{2}}(|\uparrow\rangle \otimes |\uparrow\rangle +|\downarrow\rangle \otimes |\downarrow\rangle)\otimes(\frac{|\uparrow\rangle +|\downarrow\rangle}{\sqrt{2}})[/math] is partially entangled because of the connection between the first and second spins. It is not fully entangled though, since we can measure the third spin without affecting the other two.

The state [math]|\Psi\rangle=\frac{1}{\sqrt{2}}(|\uparrow,\uparrow,\uparrow\rangle +|\downarrow,\downarrow,\downarrow\rangle )[/math] is fully entangled because finding any of its spins forces both of the others to be the same. This isn’t the only way to achieve full entanglement though, and it is not necessary to force the other spins to all be the same as the one measured (though of course, as mentioned in the question, more than two can’t all be “opposite” to one another).

For example [math]|\Psi\rangle=1/2(|\uparrow,\uparrow,\uparrow\rangle +|\downarrow,\downarrow,\uparrow\rangle +|\uparrow,\downarrow,\downarrow\rangle +|\downarrow,\uparrow,\downarrow\rangle)[/math] is a fully entangled state because “collapsing” onto the part with any one of the [math]|\Psi_i\rangle=|\uparrow\rangle[/math] gives only cases with [the other two parallel, whereas [math]|\Psi_3\rangle=|\downarrow\rangle[/math] corresponds to them being anti-parallel.

P.S. There are various ways of defining the “amount” or “strength” of entanglement between two particles; and it turns out for some of these measures that in a large class of systems of many particles, the strength of entanglement between any two particles is never more than it could be if they were only entangled with one another. One thing that may have given rise to this question is the unfortunate use of “entanglement monogamy” to refer to this phenomenon. (But as Percy Bridgeman points out in answer to another question, the fact that being fully polygamous reduces the amount of time available for marital relations with any one partner does not actually in any way enforce monogamy.)

fromQuora: Isn’t a ‘probability wave’ simply a statistical function and not a real wave? Does it no more ‘collapse’ than me turning over a card and saying that the probability ‘wave’ of a particular deal has collapsed?

Well, as other answers have noted, the wave function of quantum mechanics is not a probability wave as its values are complex and it is only the squared amplitude that gives a probability density. But the process of “collapse” involved in a quantum measurement does involve something like your playing card analogy.

There are actually two stages in the measurement and observation process. One is the interaction with an incompletely known measurement apparatus which reduces or eliminates the prospects for future interference and basically turns the previously pure state of the isolated system (considered as a subsystem of the larger world) into a statistical mixture. And the second is the collapse of that statistical mixture by observation – with the result that, from the observer’s point of view, of the many possible alternatives only one is actually true.

And if this makes it seem to you that the “state” of the system actually depends on the observer then you are on the right track. (But it is nothing special about the “consciousness” of the observer that is relevant here. Almost any localized system could play the same role relative to the rest of the universe.)

Any configuration history of any physical system can be considered as “seeing” the rest of the universe in a “relative state” which “collapses” when the configuration history in question passes a point beyond which the configuration includes information about that particular measurement value.

Source: (255) Alan Cooper’s answer to Isn’t a ‘probability wave’ simply a statistical function and not a real wave? Does it no more ‘collapse’ than me turning over a card and saying that the probability ‘wave’ of a particular deal has collapsed? – Quora

(354) Is a system of two entangled particles a pure state, and are its subsystems in a mixed state? – Quora

Not necessarily (but the pure state case is often what we study), and not exactly (but kind of).

WITH REGARD TO THE FIRST QUESTION:

I don’t think it is necessary to restrict the concept of entanglement to pure states. One could do so, but it would be odd to describe a state with a 50% chance of being in an entangled pure state as “not entangled”.

I would therefore say that a composition of two systems can be said to be in an entangled state whenever that state is represented by a density operator that includes at least one term which is not a pure tensor product of state vectors for the corresponding component systems.

WITH REGARD TO THE SECOND QUESTION:

In order to make sense of this question we need so decide what is meant by the state of a subsystem of a composite system. The standard approach is to use the “relative state” which is implemented mathematically by a “partial trace” which gives “marginal probability” distributions of observed quantities. And if one does that, then yes, the relative state of a subsystem may indeed be mixed state even for a pure state of the combined system.

Before going on I should point out that it is important to understand both the distinction between a statistical mixture and a linear superposition, and the fact that for a composite system the states that are “pure tensors” are just a subset of all the pure states.

First, with regard to mixtures vs superpositions:

For any system, mixed states (which are classical statistical mixtures of pure states) can be represented by so-called density matrices which are operators of the form  [math]\rho=\Sigma p_{i}|\psi_{i}\rangle\langle \psi_{i}|[/math] where [math]p_{i}[/math] is the probability of being in pure state  [math]\psi_{i}[/math] and the operator [math]|\psi_{i}\rangle\langle \psi_{i}|[/math] is just the projector onto the subspace spanned by the state vector [math]|\psi_{i}\rangle[/math]

The expectation of observable [math]O[/math] in state [math]\rho[/math] is then given by the trace [math]\langle O\rangle=Tr(\rho O)=\Sigma p_{i}\langle\psi_{i}|O \psi_{i}\rangle[/math], which is just the overall expectation from a process which gives the expected value for state [math]|\psi_{i}\rangle[/math] with probability [math]p_{i}[/math] . The case of a pure state [math]\psi[/math] corresponding to a single vector [math]|\psi\rangle[/math] can also be represented by a density matrix in which the sum has only one term and so the density matrix is a one dimensional projector (ie of rank one), and the trace formula gives [math]\langle O\rangle=Tr(|\psi\rangle\langle \psi| O)=\langle \psi| O\psi\rangle[/math] which is just the usual form for the expectation value.

Another quite different way of combining states is by way of linear combination of state vectors to create what is called a superposition. This is different from the classical mixture because the expected average value of an observable [math]O[/math] in the superposition state [math]|\psi\rangle=c_{1}|\psi_{1}\rangle+c_{2}|\psi_{2}\rangle[/math] is given by [math]|c_{1}|^2|\langle\psi_{1}|O\psi_{1}\rangle+c_{1}^*c_{2}\langle\psi_{1}|O\psi_{2}\rangle+c_{2}^*c_{1}\langle\psi_{2}|O\psi_{1}\rangle+|c_{2}|^2|\langle\psi_{2}|O\psi_{2}\rangle[/math] and the cross terms represent the fact that, if [math]|\psi_{i}\rangle[/math] are not both eigenstates of [math]O[/math], then observation of [math]O[/math] has a mixing effect which produces interference between them.

Second, with regard to composite systems:

If the systems A and B have pure state vectors of the form [math]|a_{\alpha}\rangle[/math] and [math]|b_{\beta}\rangle[/math] in Hilbert spaces [math]\mathcal{H}_{A}[/math] and [math]\mathcal{H}_{B}[/math], then any (normalized) linear combination of pure tensors corresponds to a pure state of the combined system.

The special feature of pure tensors is that they represent states in which the properties of the two subsystems are statistically independent, sometimes called separable states. But in most states (which are formed by taking linear combinations of pure tensors) the properties are correlated (and we say that in such other states, which are not represented by pure tensors, the two systems are “entangled”).

In this setting any pure tensor corresponding to a state vector of the form [math]|a_{\alpha}\rangle\otimes|b_{\beta}\rangle[/math] has a rank one density matrix [math]\rho=(|a_{\alpha}\rangle\otimes|b_{\beta}\rangle)(\langle a_{\alpha}|\otimes\langle b_{\beta}|)=|a_{\alpha}b_{\beta}\rangle\langle a_{\alpha}b_{\beta}|[/math] and is not entangled.

But neither is any classical statistical mixture of such states with density matrix [math]\rho=\Sigma p_{i}|\psi_{i}\rangle\langle \psi_{i}|[/math] with each contributing [math]\psi_{i}[/math] being a pure tensor of the form [math]\psi_{i}=|a_{i}\rangle\otimes|b_{i}\rangle[/math].

On the other hand an entangled state might be pure state represented by a vector of the form [math]c_{1}|a_{1}\rangle\otimes|b_{1}\rangle+c_{2}|a_{2}\rangle\otimes|b_{2}\rangle[/math] (which is a linear superposition of pure tensors ), but it might also be a classical mixture in which one or more of the possible [math]\psi_{i}[/math] is of that form.

In order to make sense of the second question we need so decide what is meant by the state of a subsystem of a composite system.

For an unentangled (pure tensor product) pure state it is natural to take the corresponding factor in the tensor product. For an entangled state it is less obvious what to do, but it makes sense to think of the relative state of system A as giving any observable [math]O_{A}[/math] the expectation value that results from observing the combined state but ignoring the state of system B. This amounts to observing the identity operator in system B so the corresponding observable on the combined system would correspond to the operator [math]O_{A}\otimes I_{B}[/math]

For the pure entangled state  [math]\psi =  c_{1}|a_{1}\rangle\otimes|b_{1}\rangle+c_{2}|a_{2}\rangle\otimes|b_{2}\rangle[/math], if [math]\langle b_{1}|b_{2}\rangle = 0[/math], this gives the expectation [math]\begin{align}&\langle\psi|(O_{A}\otimes I_{B})\psi\rangle  \\&= |c_{1}|^2\langle a_{1}|O_{A}a_{1}\rangle\langle b_{1}|I_{B}b_{1}\rangle + c_{1}^*c_{2}\langle a_{1}|O_{A}a_{2}\rangle\langle b_{1}|I_{B}b_{2}\rangle \\&+ c_{2}^*c_{1}\langle a_{2}|O_{A}a_{1}\rangle\langle b_{2}|I_{B}b_{1}\rangle + |c_{2}|^2\langle a_{2}|O_{A}a_{2}\rangle\langle b_{2}|I_{B}b_{2}\rangle \\& =  |c_{1}|^2\langle a_{1}|O_{A}a_{1}\rangle + |c_{2}|^2\langle a_{2}|O_{A}a_{2}\rangle\end{align}[/math]

which corresponds to the mixed state for system A with probability [math]|c_{i}|^2[/math] of being in state [math]a_{i}[/math] (and density matrix [math]\rho=|c_{1}|^2|a_{1}\rangle \langle a_{1}| + |c_{2}|^2|a_{2}\rangle \langle a_{2}|[/math]).

This procedure of mapping the density matrix [math]\rho=|\psi\rangle \langle \psi|[/math] which is an operator in [math]\mathcal{H}_{A}\bigotimes \mathcal{H}_{B}[/math] to an operator on just [math]\mathcal{H}_{A}[/math] is often referred to as taking the partial trace.

 

Source: (354) Is a system of two entangled particles a pure state, and are its subsystems in a mixed state? – Quora

When you solve the Schrodinger equation for an electron, does it tell you the state of the electron in terms of quantum numbers?  – Quora

tl;dr Yes – but read on for more details.

The Schrodinger equation for any system has many solutions, of which only those which are square integrable (aka “normalizable”) correspond to possible states of the system. And to determine a particular solution for any differential equation requires the imposition of “boundary conditions”, which for a Schrodinger equation might take the form of a specification of the function at some particular time (say t=0)

Quantum numbers are just possible values for observables which may be discrete (such as the energy levels for a confined system like an electron in the potential well of a nucleus), or continuous (such as the position and momentum coordinates). And once a particular solution has been identified it does tell us all we can know about what those values will be.

For example, the time-dependent Schrodinger equation for a free particle has “plane wave” solutions, which do not actually correspond to any physical state because they have infinite norm and so to normalize them would reduce them to zero – which means that the particle has zero chance of actually being found anywhere in any finite region. [This is a special case of the uncertainty principle which says that having a precisely defined value of the momentum quantum number implies a completely unknowable position and the other side of this coin would be the “delta function solution” which (is not really a function but) somehow represents the concept of having a precisely defined value of the position and has a momentum-space representation that is completely uncertain with zero probability of being in any finite range of values.]

The evolution of a normalizable “wave packet” solution then does tell how the probability distribution of position quantum number varies over time (and its fourier transform gives the corresponding information about momenta).

For the case of a particle confined by a box or potential well (such as that of an electron in an atom) there may be observables such as the energy (at least when it is low enough) and angular momentum that can only have discrete values. And finding the general solution of the Schrodinger equation will identify these.

So when you solve for the general solution it tells you all the possible values of the quantum numbers but doesn’t say anything about any particular electron. And when you solve for a particular solution (say with some particular initial conditions) then its time evolution will allow you to calculate how the probabilities of having different quantum numbers varies over time.

Furthermore, since energy is conserved, the energy quantum numbers don’t change over time; but once we have calculated all the possible fixed energy solutions, [math]\psi_{n}(x,t)[/math] with energy [math]E_{n}[/math], we can use them to shortcut the solution for a general initial condition by expressing it as a linear combination [math]\Psi(x,0) = \Sigma c_{n}\psi_{n}(x,0)[/math] of the eigenfunctions and multiplying each by the corresponding complex exponential phase factor to get [math]\Psi(x,t) = \Sigma c_{n}\psi_{n}(x,t)=\Sigma c_{n}e^{iE_{n}t}\psi_{n}(x,0)[/math] which allows us to see how the probability distribution for the position quantum numbers evolves over time.

Source: (252) Alan Cooper’s answer to When you solve the Schrodinger equation for an electron, does it tell you the state of the electron in terms of quantum numbers? If this is wrong, what does the Schrodinger equation give you when you solve it? – Quora

The theory of relativity states that as you approach the speed of light, time slows down. At what speed would we be able to notice this? – Quora

Whose time? As I approach the speed of light (relative to you), even though what you see tells you that my clock is running slow, it’s actually your time that appears (to me) to be slowed down (by a factor of [math](1-\frac{v^2}{c^2})^{-1/2}\approx 1+\frac{1}{2}(\frac{v}{c})^2 )[/math].

The speed needed for either of us to notice the effect depends on how accurately we can measure time intervals. But if we can do it to one part in a million (ie to 6dp accuracy) then for us to notice the difference would require [math]\frac{v}{c}\approx 10^{-3}[/math], so roughly a few hundred km/sec. But if you wanted to notice it on anything not moving fast enough to escape the solar system then you’d need a couple more sig figs of accuracy (ie about 0.1 sec in a year). And if you want to see it on a low orbiting satellite you’d need another couple of places (ie to notice about one millisecond delay over a year).

Source: (248) Alan Cooper’s answer to The theory of relativity states that as you approach the speed of light, time slows down. At what speed would we be able to notice this? – Quora

Einstein’s train embankment thought experiment (showing relativity of simultaneity)

The fundamental observed fact that underlies this is that all unaccelerated observers get the same result when measuring the vacuum speed of light in any direction.

So if an observer, A, riding at the middle of a long train sees signals emitted from the front and back of the train at the same time, since the distances and speeds are equal he or she infers that they were emitted at the same time.

If A raises a flag if and only if he or she receives the signals at the same time, then an observer B on the bank also knows that the signals reached A at the same time. But when raising the flag, A has moved ahead in B’s frame from where he or she was when the signals were sent. So B sees that the signal from the back travelled further than that from the front and so must have taken longer (since B sees exactly the same speed of light in all directions as A). So from B’s point of view, the signals which A saw as being emitted simultaneously were actually not simultaneous (with the one from the back having been sent earlier).

Source: (152) Alan Cooper’s answer to Does the train embankment thought experiment of Einstein really demonstrate relativity of simultaneity? Is there actual experimental proof of RoS? See comment attached to the question for details. – Quora

How can time be dilated for both observers without contradiction? – Quora

Because it’s not the timing of the clock of either observer that is dilated but the other observer’s interpretation of their view of it. It’s really no more of a contradiction than the fact that if we are facing one another then when I ask you to raise your right hand you raise the one on my left. (Note: I didn’t say it’s the same, just “no more of a contradiction”. If it seems mysterious to us that’s just because our experience of comparing notes at low relative velocities leads us to wrongly guess that there is an absolute standard of simultaneity for spatially separated events.)

Note: It’s only when both observers are in fixed inertial frames that they both see the other’s time as dilated. In the travelling twin scenario (appended as a comment to the actual question) one or other of the twins must undergo acceleration (or impulse) in order for them to get back together, and whoever is accelerated (or jumps from one frame to another) sees much of the time of the other as very much compressed (or collapsed to a single point of his or her own time in the physically impossible case of an instantaneous turn-around).

This is a real asymmetry because in the absence of gravity (ie in Special Relativity), the twin who is accelerated actually feels a force which the other does not. So they do know which one was accelerated.

In General Relativity an observer in free fall can accelerate without feeling any forces (and if the traveler is turned around by a gravitational slingshot then the time difference is explained by the time spent at the bottom of a very deep gravitational potential well – or equivalently the proximity of a very large mass – which is again an asymmetry between their experiences); but that’s another matter form the question of whether it’s a “paradox” in SR. In SR one cannot accelerate without being pushed by something like the reaction forces of a rocket and one twin knows he or she is in a rocket ship and feels the force while the other does not.

Source: (129) Alan Cooper’s answer to How can time be dilated for both observers without contradiction? – Quora

What is the problem associated with solving Klein-Gordon partial differential equations? – Quora

There’s no problem with solving the Klein-Gordon Equation. It’s just that some of the solutions have a form for which [math]H\Psi=i\frac{d\Psi}{dt}=E\Psi[/math] with arbitrarily large negative values of [math]E[/math]. If we were to try to interpret the solution as a quantum wavefunction this would correspond to having no lower bound on the energy – which is physically unstable.

Source: (128) Alan Cooper’s answer to What is the problem associated with solving Klein-Gordon partial differential equations? – Quora

Is special relativistic time dilation a real effect or just an illusion? Given two inertial frames each observer finds that the clock of the other runs slower than that observer’s own clock. So who is right? 

This is a pretty good answer except that I wouldn’t say either of them is right if they think that their perception of relative slowness represents something that is objectively true for all observers.

Time dilation is a real effect on the perceptions of observers (with regard to the rates at which one another’s clocks are ticking). Neither of them is “right” if they think there is any real sense in which the other’s clock is objectively slower. But neither of them is wrong about how it appears to them, so it’s not really an illusion any more than the fact that if they are looking at one another then their ideas of the “forward” direction are opposite to one another. What turns out to be more of an illusion is the sense we all have that there is some absolute standard of time which determines which of two spatially separated events occurs before the other.

Are the electric and magnetic fields always perpendicular to each other?

Source: (164) Alan Cooper’s answer to Are the electric and magnetic fields simultaneously perpendicular to each other? – Quora

Not always. The rate of change of either is perpendicular to the other, but there is no such general relation between the actual values. In fact for static fields it is easy to make them parallel – as, in the picture below, at the centre of a current carrying loop with positive and negative charges above and below: