Category: Quora Answers

More on acceleration in the Twin “Paradox”

There is a possibly interesting comment thread following my answer to How exactly do Minkowski diagrams prove that acceleration is not needed in resolving the Twin Paradox in Special relativity? – Quora

In it Peter Webb passionately (and sadly sometimes rudely) defends the position that “acceleration has nothing directly to do with the Twin paradox”. This is a position shared by a not insignificant minority of apparently competent physicists (on Quora, Brent Meeker comes to mind as a prominent example), but although I continue to feel that the issue is vastly overblown, I find that the “it’s not acceleration” view is misguided and at least some of its proponents are sufficiently intransigent and aggressive that they demand a further rebuttal.

So, with reference to Peter’s final summary of his position, here goes:

“The “paradox”/interest in the TP is that it demonstrates time dilation. The different ages merely demonstrates this.”

To be frank, I have no idea what he is talking about here. The usual time dilation of special relativity, which applies in the case of two inertial observers having a constant relative velocity with respect to one another, only exists in the understanding of each observer regarding the clock of the other. There are many ways of demonstrating this effect (most famously with muons from cosmic ray interactions with the upper atmosphere, but also in many other ways), but it is not something that actually happens to either observer in any objective sense – at least not while they both continue in an inertial state of motion.(We may think its obvious that the muon is the one with a dilated lifetime but from the muon’s frame of reference the Earth looks like a very flat pancake and the distance it has to travel can be covered well within its lifetime.)

The Twin “Paradox”, however, is something different in that it does lead to an objective fact of the matter regarding which twin or clock aged more between two specific events in space-time.

“Maybe 80% of people with some interest in science think that time dilation is a consequence of acceleration. It isn’t, directly. It is a result of changing reference frames and in the Twin’s paradox that involves acceleration. But time dilation still occurs in the absence of acceleration (eg 3 brothers), and it is easy to show that identical acceleration profiles produce very different amounts of time dilation.”

I am not aware of the statistics regarding what “people with some interest in science” think, but I doubt that any significant percentage of them think that the symmetric relative time dilation of unaccelerated motion is a “consequence of acceleration”.

But on the other hand, if they think that the objective observable time difference in the Twins “Paradox” is a consequence of acceleration, then I’m right there with them! (I have no position on how “directly” consequential the acceleration is, but I am happy to see an acknowledgement that indeed having one twin change frames does involve acceleration.)

But now we come to the 3 brothers.

It is true that they provide yet another way of confirming the relative time dilation effect. Indeed, the incoming “brother” brings back to Earth a record of what the outbound one’s clock said when they met  (though of course this is not necessary, and the information could just as easily have been passed by a radio signal). But he (or the radio message) could also inform Earth of the time that the outgoing traveller inferred was on the Earth clock at his idea of when the crossing of paths happened. This would be consistent with the usual unaccelerated time dilation, which is perfectly symmetrical in the sense that while the Earth observer thinks that at the event on the traveller’s path which is concurrent with any time [math]t_{E}[/math] on his Earth clock (measured from the outbound traveler’s departure event), the outbound clock reads [math]t_{O}=\frac{t_{E}}{\gamma}[/math], and the outbound observer thinks that at the event on Earth which is concurrent with any time [math]t_{O}[/math] on his travelling clock (measured again from the shared departure event), the Earth clock reads [math]t_{E}=\frac{t_{O}}{\gamma}[/math].

So, at the particular event where the travelers meet, if the Earth brother thinks this occurs at time [math]t_{M,E}[/math] then the outbound traveler’s clock reads [math]t_{M,O}=\frac{t_{M,E}}{\gamma}[/math] and he thinks that the concurrent time on the Earth clock is just [math]\frac{t_{M,O}}{\gamma}=\frac{t_{M,E}}{\gamma^{2}}[/math].

The inference of the inbound traveler on the other hand, after synchronizing his clock with the outbound, is that during the rest of his trip to Earth, the Earth clock only advanced by the same [math]\frac{t_{M,E}}{\gamma^{2}}[/math].

But when they synchronize clocks the two travellers can also compare notes on what they think is showing on the Earth clock at that time. And their disagreement on that will show them immediately that everything works out and there is no paradox.

So in the 3 brothers case there is clearly no paradox, as everyone is aware that each traveler considers only a small part of the Earth’s experience as concurrent with their own travel time and there is no reason to expect that those two intervals should add up to the whole time on Earth. So no two observers are ever forced to agree that only one of them is right about the time dilation of the other.

Of course there is not a paradox in the single traveler case either, but it is a more powerful (wrong) intuition that the two intervals judged by the traveller to be concurrent with the legs of his trip should cover the entire Earth time interval (even though we now know from the 3 brothers analysis that they do not).

Note: I am not disputing the validity of the 3brothers scenario as the most appropriate explanation of why there is no paradox. But I will continue to insist that it is not significantly counterintuitive in its own right and that the solution it provides to the single traveler version identifies the “missing” Earth time with the traveller’s “frame jump” and that his “frame jump” (no, not just in his head but between two different frames that he is actually at rest with respect to) is nothing more than an integrated acceleration.

This leads to the question of what we can learn from the sudden jump in the traveler’s inferred Earth time at the turnaround. We learn that when a traveller changes frames  then, as his simultaneity space changes, the times he considers concurrent on clocks that are remote from him in his direction of velocity change are advanced (and those behind retarded).

Of course the sudden turnaround is physically impossible and in any real situation there would be a period of finite acceleration. But it is a simple exercise in calculus to approximate a period of finite acceleration with a number of discrete jumps and conclude that while the traveler is turning around his inferred current time on Earth advances more rapidly than when he is not accelerating (at a rate proportional to both his acceleration and his distance from Earth).

Now none of this made any use of General Relativity. The acceleration effect we have discovered is purely from Special Relativity and I do agree that it’s a monstrous abuse to address the Twin “Paradox” by invoking acceleration=>GR=>”it’s like being in a gravity well which we all know (from the movies!) causes time dilation”.

But the real beauty of all this is that it DOES go the other way!

In SR, acceleration=>time dilation (relative to clocks that are remote in the direction of acceleration), and so, using only the equivalence principle, we learn (without any of the harder GR analysis) that Matt Damon really does outlive his grandchild!

Given Peter’s earlier answer about the Ehrenfest “Paradox” I am totally surprised and disappointed that he doesn’t seem to appreciate this.

What is Schrodinger’s equation? Is it deterministic or not? If it is, how can we prove that? And what conditions must be satisfied for it to be non-deterministic? 

Schrodinger’s equation was originally just the partial differential equation satisfied by the position-space wave function of a particle (or more general system) in non-relativistic quantum mechanics. The same name is sometimes also used for the equation [math]\frac{d}{dt}\Psi(t)=iH\Psi(t)[/math] satisfied by the state vector in any NRQM system regardless whether or not a position-space representation is being used (or is even available).

It is deterministic (in the sense of determining [math]\Psi(t)[/math] uniquely for all [math]t[/math] if given an initial condition [math]\Psi(0))[/math], so long as the Hamiltonian [math]H[/math] is self-adjoint (symmetry is NOT enough!).

The proof of this involves more analysis than I could fit into a Quora answer, but in the general case it follows from the fact that for any self-adjoint operator [math]H[/math] on a Hilbert Space, the equation [math]\frac{d}{dt}\Psi(t)=iH\Psi(t)[/math] is uniquely satisfied by [math]\Psi(t)=e^{iHt}\Psi(0)[/math] where the complex exponential of [math]H[/math] is defined in terms of its spectral resolution; and for the PDE special cases it might be done by various theorems involving greens functions or Fourier analysis and convergence properties of improper integrals.

It may be non-deterministic if [math]H[/math] has not been specified on a large enough domain to be essentially self-adjoint (as sometimes happens if boundary conditions are omitted from the specification of a problem in which the particle is confined somehow – either by an infinite potential or by living in a single cell of a crystal lattice for example). But such cases are normally just due to inadequate specification of the problem rather than to a real physical indeterminacy.

So I would say that in a properly defined quantum theory model the Schrodinger equation is indeed almost always deterministic.

[N.B. It wasn’t part of the actual question, but I should perhaps add that the reason this does not make quantum mechanics deterministic is because even complete knowledge of the quantum state of a system is not sufficient to predict the outcomes of all possible experimental measurements. For any a state which happens to produce a predictable value for one observable there will be other observables for which the outcome is uncertain.]

Source: (1000) Alan Cooper’s answer to What is Schrodinger’s equation? Is it deterministic or not? If it is, how can we prove that it is deterministic? If it isn’t, what conditions must be satisfied for Schrodinger’s equation to be non-deterministic? – Quora

What is the definition of an eigenstate of a hermitian operator?

An eigenstate of a quantum observable is a state which results from a measurement of that observable which has produced a precise value; and according to quantum theory this means that it is represented by a normalized eigenvector for the corresponding self-adjoint operator (whose eigenvalue is equal to the observed measurement value).

An eigenvector of an operator [math]A[/math] is a vector [math]\Psi[/math] for which [math]A\Psi=\lambda\Psi[/math] for some number [math]\lambda[/math] (which is called the corresponding eigenvalue).

Source: (1000) Alan Cooper’s answer to What is the definition of an eigenstate of a hermitian operator? – Quora

Falling from Space

Source: (1000) Alan Cooper’s answer to If a Geostationary Satellite runs out of fuel, how long will it take to descend to earth in free fall? Galileo calculated that Moon would descend to earth in under 4 hours if it is blocked in orbit and allowed to fall free to the earth. – Quora

If it runs out of fuel then its free fall is just the current orbit in which it will stay essentially forever (since it is well above any atmospheric drag that might slow it down). But in the unlikely event that it has the extremely large amount of fuel needed to kill all of its orbital velocity, (and if it actually uses the fuel for that purpose), then it will fall straight down to the earth in a time that one can calculate – either with a bit of calculus, or by the clever trick of applying Kepler’s law for a very thin ellipse of major axis length equal to the orbital distance.

[Kepler’s third law says “The square of a planet’s orbital period is proportional to the cube of the length of the semi-major axis of its orbit” so the period is proportional to the 3/2 power of the major axis; and the full axis of a narrow ellipse from the satellite down to the earth is just the radius (ie half the major axis) of its previous circular orbit. So the period of that ellipse, if completed, would be the geostationary period of 24 hours divided by a factor of 2^{3/2}. But the satellite only gets (a bit less than) halfway round before crashing, so the actual time for the fall is a bit less than 24/2^{5/2}=3\sqrt{2} or about 4 hours.]

P.S. For the Moon, since the orbital period is 28 days, the time to fall would be more like 5 days, but I think the claim attributed to Galileo may arise from a misreading of his argument that the Earth is spinning on its axis (basically something like: If the Earth was NOT rotating and instead everything on the celestial sphere was going round once every 24 hours, then the Moon, Sun, and stars might all be at the geostationary distance but only those near the equator would actually be in balance and those at the pole would just fall to Earth in 4 hours)

P.P.S. It is not clear that Galileo actually had a good understanding of orbital mechanics and the inverse square law, so if he estimated a falling time of 4 hours that may have been little more than a lucky guess

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