(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

If photons travel at the speed of light in a vacuum, and are said to “slow” when they travel through different media but do they really “slow” or do they just bounce off more atoms and thus travel farther and seem to slow? – Quora

Photons don’t really “travel” at all. They are just units of energy exchange between the electromagnetic field and other fields such as that of the electron (or any other charged “particle”) and only “exist” at the events (positions and times) of energy transfer, by “emission”, into or, by “absorption”, out of the EM field.

It is only after an emission and absorption event have both been observed that we describe the process as corresponding to the motion of a photon, but really all that has happened might just as well be described by saying that a wave of changes in the EM field propagated from the emission event, and that at any point in spacetime (given the presence of a suitably excitable atom), the probability of an absorption event is proportional to the magnitude of the appropriate frequency component in the EM field at that point in space and time.

But the propagation of the EM field obeys Maxwell’s equations in empty space, and through a medium which can be approximated as consisting of classical particles it is governed by the same equations but with reflected waves from all the particles interfering with one another in such a way that the overall effect is like a wave moving more slowly. So there is something like the “bouncing of all the electrons idea” going on even in the classical case.

Also, the propagation of disturbances in the EM field from one point at one time to another point at a later time can be calculated by a process which amounts to considering all possible paths by which an imagined particle could end up at the right place and time (taking into account all of the ways that imagined particle could bounce off other things in its environment), and using something like a sum of contributions from all such paths to give the field values at the later event. So although there isn’t actually any particular path followed by any real photon, the effect we see is kind of like the average over all possible paths, and the presence of all those electrons allows longer paths to contribute in such a way that the overall effect is as if there was a photon travelling more slowly than it would have if it had only been able to go in a straight line.

Or something like that.

Source: (310) Alan Cooper’s answer to If photons travel at the speed of light in a vacuum, and are said to “slow” when they travel through different media but do they really “slow” or do they just bounce off more atoms and thus travel farther and seem to slow? – Quora

Do thought experiments really uncover new scientific truths? | Aeon Essays

Well, I’m late to the party here, as the article linked below is now five years old; but I do have two comments.

Source: Do thought experiments really uncover new scientific truths? | Aeon Essays

One is that Norton is right. ALL valid thought experiments are just logical deductions of consequences of things that we have learned from experiment seem to always be true. Case in point being Einstein’s deduction of Lorentz transformations from the assumption that speed of light is invariant which is not itself deduced but extrapolated from observations (either Michelson-Morley or tests of Maxwell’s equations). In fact anyone who designs a machine is doing the same thing – thinking about logical consequenses of the assumption that previously observed patterns will continue to apply. And if the assumptions have been well tested then we strongly expect the derived laws to apply in just the same way as we expect the machine to work.

My other point is that Galileo’s beautiful argument is NOT valid. It is basically circular in that there is no reason other than prior experience to expect that two cannonballs will not fall faster when close together than when far apart. In fact it is easy to imagine scenarios where combining objects (such as water droplets) does increase their rate of fall and there is no a priori reason why all space might not be filled with a permeable but resistive medium in which the drag effect is reduced by proximity (as for cyclists in a peloton for example).

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

Entanglement

Entanglement is just a word we can use to describe a situation where knowledge of some property of one object gives us information about some (possibly different) property of another.

The term is rarely used in the classical case, because we take it for granted. If we separate a pair of gloves for example and pack them up in identical boxes and then choose by a coin toss to send each in one of two opposite directions, then we are not surprised by the fact that if someone who knows how they started out but does not know the result of the coin toss opens one box and sees a left glove, he or she knows immediately that whoever opens the other box will see a right glove.

There is often similar classical chance-based uncertainty in our knowledge of quantum systems; but for such systems, even in the most precisely prepared “pure” states the knowledge of some properties makes it impossible for us to know others. This residual uncertainty is expressed by representing the state of the system by a “state vector” in a Hilbert space and the “mystery” of quantum entanglement is that the correlations between systems (like the gloves) that were once together but are now far apart can sometimes be greater than would be possible for any way of randomly assigning the properties at the outset.

The reason this extra correlation is sometimes considered “spooky action at a distance” is because the change of state vector (often called “collapse”) that occurs when we open one box seems to trigger a simultaneous collapse at the other box – and in a way that can change what the remote observer will see when looking at different properties from the one that obviously has to be opposite. At first sight it may seem that this effect might be used to send a signal where what the second observer sees might depend on what the first one chose to measure, but that turns out not to be the case.

Whether or not this bothers you may depend on whether you consider the state vector to be a property of the system itself or rather of the way it appears to a particular class of observers.

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.