Why Least Action?

This Quora Answer by John Fernee gives some useful intuition,

saying:

The principle of least action can be derived simply by asking that each point in a path has enough information to derive the next infinitesimal step.

That is essentially determinism: If you know the complete state of a system at any point in time, you can derive the state of the system at any other point.

The principle of determinism is all that you need to define a unique path. The mathematics of path integrals can then be used to derive the principle of least action using the calculus of variations.

The principle of determinism is actually modified with quantum mechanics. The deterministic aspect of quantum mechanics relates to the evolution of the wavefunction. Whereas measurement outcomes are probabilistic. This lead to Feynman generalising the principle of least action, which resulted in the path integral formalism of quantum mechanics.

As to why does the universe satisfy the principle of least action, you can only state that if it didn’t, we’d be unable to predict anything with certainty. There would be multiple paths to some observed state and we’d be at a complete loss to discover which path was taken. That might sound like the Feynman path integral approach, but it isn’t. The fact that wavefunction evolution is deterministic means that there must be interference between the different quantum paths that result in the deterministic evolution.

A subsequent comment clarifies the meaning of “determinism” as basically referring to being governed by a second order DE in terms of the position coordinates and adds motivation for the fact that L=T-V:

The concept of energy is closely related to the principle of least action, but does not form part of the definition. The definition is far simpler: That a particle will follow a unique path and that path can be determined from position and velocity coordinates.

This is a mathematical process for determining a unique path and it just happens to match with the physical requirements for particles to have deterministic trajectories.

As for energy, the Lagrangian that corresponds to Newtonian physics is simply given by the difference in the kinetic and potential energies; L=T-V. The minimisation is with respect to this Lagrangian and is given by the condition that dL/ds=0, where ds represents the change in path. Given the form of the Lagrangian, we can write, dT/ds=dV/ds. These quantities can be identified as the action and reaction forces of Newton’s third law. So what we’ve really got is a statement that the path is defined as the trajectory where the action and reaction forces are balanced. The simplest example is in circular motion where the centripetal force is equal and opposite to the centrifugal force. Even though the centrifugal force is a fictitious force, it must be equal and opposite to the centripetal force for there to be circular motion.

But I have some quibbles with the references to quantum mechanics which I think are better covered (though still not perfectly) in Andrew Winkler’s answer.

Source: (1001) Why does the principle of least action hold in our universe? – Quora

Source of Uncertainty Principle – Observation or Theory?

A Quoran asks: “Is the Uncertainty Principle in QM primarily a consequence of the mathematics involved (ie. derived from it) or of empirical evidence (which the mathematics has then been constructed to describe)?”

I answered “Both”.

The original idea of the uncertainty principle was motivated by the apparent empirical facts that in order to determine the position of an object very precisely it seems necessary to illuminate it with light of a very short wavelength, and that the empirically observed spectrum of black body radiation suggests that short wavelengths interact with matter not continuously but in discrete ‘quanta’ which transfer more and more momentum as the wavelengths get shorter. In the mathematical theories that were developed to describe these phenomena (but not so much motivated by the uncertainty principle itself) the uncertainty principle (and in fact a more general form of it) can be derived from the fact that measurements of position and momentum correspond to non-commuting operators on the Hilbert space that is used to represent states of the system.

Source: (1001) Alan Cooper’s answer to Is the Uncertainty Principle in QM primarily a consequence of the mathematics involved (ie. derived from it) or of empirical evidence (which the mathematics has then been constructed to describe)? – Quora

Schrödinger’s Cat

Why has Schrödinger’s Cat, the experiment, not actually been performed?
Basically because we have no way of putting a system as complex as a cat into a pure quantum state.
Schrödinger’s cat thought experiment was intended to challenge the idea that a system can have undetermined values of some observables in a way that does not just correspond to a lack of knowledge on our part. If the killing of the cat were the result of a purely classical source of uncertainty such as a coin toss we could just say that the cat is either truly alive or truly dead and we just don’t know which until we look. But quantum mechanics includes uncertainties of a kind that cannot be interpreted as just the result of incomplete information.
An example on a scale much smaller than that of a cat is if a single electron has a known value of its vertical spin component (up or down), and if we subsequently divert it by a device that sends it in different directions depending on its horizontal spin component, then there is a 50% chance of seeing it go in either of the two directions and it appears (from an analysis of the observed probabilities in various other directions) that there was no way of predicting the horizontal component (say from some other initial observations) before we actually measured it.
Schrödinger’s point was that there’s something weird about this business of not having a property until we measure it, and he used the cat as an extreme example. But in practice, if we are good enough detectives, there will always be evidence in the box telling us exactly when the cat died; and so, although there may have been some time before we knew the outcome, we never see anything that looks significantly different from what might have happened if the killing of the cat had been triggered by some classical probabilistic event (such as say the first double six in a series of dice throws).
In order to experience the weirdness of having a cat that is neither alive nor dead (sometimes referred to as being both at once) we would need to keep the system of cat and triggering nucleus (and everything else that they can interact with) in what is called a ‘pure’ quantum state, which requires having complete information about the quantum states of all its constituent elementary particles.
This is obviously impossible for a cat, but experiments have been done in which larger systems such as complex molecules are put into superposition states where something like the shape of the molecule (which perhaps seems more substantial to us than the spin of a single electron) does not have a value until we observe it. These will undoubtedly get larger and more impressively weird seeming as technology improves, but I am pretty sure that they will never reach the scale of an actual cat.

Source: (1001) Alan Cooper’s answer to Why has Schrödinger’s Cat, the experiment, not actually been performed? – Quora

Is Decoherence Reversible?

It depends on what you mean by the word “decoherence”.

The conventional use of “decoherence” to describe part of what happens in a measurement process refers to the interaction of a pure state of an experimental system (which may be a superposition of eigenstates of some observable) with a statistical mixed state of a complex environment (which includes some kind of measurement apparatus for that observable) such that, after the interaction, the relative state of the system is a statistical mixture of eigenstates of the observable, each of which is linked to some indicator state of the environmental apparatus. This is typically NOT reversible for thermodynamic reasons (basically due to the fact that the final state is not known in sufficient detail needed to determine the actions needed in order to reverse the process).

BUT, as another answer notes, if the word “decoherence” is being used to describe interaction of the system with an ancilliary system that is in a pure state, then after the interaction the combined system is still in a pure state and the unitary evolution of pure states is reversible.

Source: (1001) Alan Cooper’s answer to Is it possible to reverse quantum entanglement decoherence? – Quora

Watching Fall Into Black Hole

If we on Earth are now observing an object that we see as near a black hole and in free fall on a path that intersects the event horizon of that black hole, then what we are seeing will be extremely red shifted version of the object that is therefore both very dim and ageing very slowly. So the apparent (to us) progress of everything in the object’s frame (including its rate of fall) is very slow. As time (for us) progresses, we will see the object’s clocks and apparent rate of fall to get progressively slower as it also dims towards invisibility.

Source: (1001) Alan Cooper’s answer to An event horizon is a boundary beyond which events cannot affect an observer. What would we be witnessing if a spacecraft was to cross the event horizon of a black hole? The spacecraft disappearing, or eternally approaching the black hole? – Quora

Definite values and Eigenstates

A Quoran asks: Once we decide that physical states are represented as vectors in a Hilbert Space and observables as operators, can we deduce that states of definite values of an observable should be eigenstates of the corresponding operator?
 
Not in any conventional sense of the word “deduce”. (It is possible to “represent” physical states as vectors and observables as operators in all kinds of ways that have nothing to do with the representation that is useful in quantum theories.)
 
But if you are talking about the usual connection in quantum mechanics that we make between states and vectors and between observables and (self-adjoint) operators, then that includes also the condition that for the observable represented by operator [math]A[/math], the distribution of observed values, when it is measured in the state represented by vector [math]\psi[/math], is such that the probability of a measurement of [math]A[/math] being in any specified interval [math]L[/math] is given in terms of the spectral projector [math]E_{L}(A)[/math] by the inner product [math]\frac{\langle \psi |E_{L}(A) \psi\rangle}{\langle \psi |\psi\rangle}[/math] (or just [math]\langle \psi |E_{L}(A) \psi\rangle[/math] if [math]\psi[/math] is normalized with [math]\langle \psi |\psi\rangle=1[/math]).
 
And if you are requiring that we decide to adopt that representation, then it does follow (essentially immediately from that extra assumption) that a state in which the observable has a definite value is represented by an eigenvector of the corresponding operator.
 
PS The spectral projectors [math]E_{L}(A)[/math] of any observable [math]A[/math] also correspond to question-type observables giving the value 1 when [math]A[/math] is observed to give a value in [math]L[/math] and 0 otherwise, and the expectation value of any question is just the probability of its value being 1 (ie of the answer being “true”). So the assumption made above actually follows from the simpler version which just says that the expectation values of observables are always given by inner products of the form [math]\langle \psi |A \psi\rangle[/math] where [math]\psi[/math] is a corresponding normalized vector and [math]A[/math] is the corresponding observable (which includes the case of spectral questions about other observables).

Source: (1003) Alan Cooper’s answer to Once we decide that physical states are represented as vectors in a Hilbert Space and observables as operators, can we deduce that states of definite values of an observable should be eigenstates of the corresponding operator? – Quora

A Question About Geometrical Optics

How do you draw the image of the light source through the converging lens if the light sources are located higher than the top of the lens (it is not allowed to extend the lens)?

In a conventional diagram of the geometrical optics of any lens or mirror, the line through the source parallel to the axis of the optics does not necessarily represent an actual ray, but rather is included as a way of easily identifying where the image is located. It is used because the optics is assumed to have the property that all rays from the source to any part of the lens (or mirror) are bent in such a way as to converge on some other point (the image) and the mathematical form of the rule describing how this bending occurs makes sense for all points in the plane of the lens (or mirror) regardless of whether or not they are actually on the lens or mirror (and so regardless of whether or not the deflection actually occurs).

It is actually possible to calculate the angle of deflection for each ray (either by an algebraic formula or a geometric construction), and so to just use rays which pass through the lens (such as those through T and B, which I have shown as solid lines in the picture) without any reference to the imaginary ray (shown as a dashed line in the picture) which would bend at H if the lens was big enough; but the formula or construction is simplest for that imaginary ray and it works just as well as any other for determining the image point.

Source: (1003) Alan Cooper’s answer to How do you draw the image of the light source through the converging lens if the light sources are located higher than the top of the lens (it is not allowed to extend the lens)? – Quora

Block Universe Nonsense

A Quora question asks: Is it true that the past still exists and the future is predetermined? Is the ‘’block universe’’ correct, since Special relativity is scientifically proven? – Quora

No. It’s not true that in the context of Special Relativity “the past still exists”.

In fact, according to Special Relativity (which does appear to be correct for situations where the effects of gravity are negligible – though nothing outside of mathematics is ever “scientifically proven”), the whole idea of the (unique) past is meaningless since there are events that you consider past that are not in the past for a remote relatively moving observer who is at an event which you consider simultaneous with your present.

Also, again according to Special (and maybe* General) Relativity, the word “still” is also meaningless as its use at any time refers to events that are simultaneous with the statement, but that criterion will correspond to different sets of events for different observers.

*- I have added the word “maybe” since Viktor Toth# has mentioned that in General (but not Special) Relativity it is possible to identify a preferred set of inertial frames (in which the redshift due to expansion is the same in all directions), and these do share a common concept of simultaneity.

But of course, even in a Galilean universe, there is also the question of what you mean by saying that the past “exists”. Yes it exists as the past (and even in SR the past of my worldline exists as my past) but that does not mean it is happening now. And the same applies to the future. Even in a non-deterministic universe the future “exists”, as what will turn out to have happened, regardless of the fact that we have no way of predicting it.

#PS I am not linking to Viktor Toth’s answer because, despite having some good bits, I feel that it is marred by the inclusion of some very misleading (and quite unnecessary) statements about Quantum Mechanics.

Source: (1003) Alan Cooper’s answer to Is it true that the past still exists and the future is predetermined? Is the ‘’block universe’’ correct, since Special relativity is scientifically proven? – Quora

Another Quora Question

A Quora question asks: “How does quantum physics know that if a system were not measured it would be in multiple possible states (without measuring it), and that when measuring it collapses into one definite state?”

Quantum theory doesn’t “know” anything. All it does is describe what we know and makes predictions about what we may find out in future. What we call the “state” of a system is just a summary of what we know about it, and a system on which we can gain no new information without losing some of what we already have is said to be in a “pure” state.

An example is the case of a single electron whose position we are ignoring (so it can be considered fixed) and whose only measurable property is the direction of its spin. If we first measure the spin component in the vertical direction (say that of the z-axis), then we will always find that the spin is pointing straight up or straight down; and if we repeat that measurement we will see that the direction is unchanged. But if we then follow that with a measurement in any perpendicular direction then we have a 50% chance of finding that the spin is now pointing in that new direction and 50% chance of its opposite. And if we now return to the original direction we find equal chances for pointing up or down.

Here, the spin up state is a pure state because we cannot determine the sideways component without losing information about the vertical component, but being in the up state is not the same as being in both left and right states at the same time, and it is also not the same as being in a statistical mixture (ie in one or other of those two states but we just don’t know which).

P.S. In either quantum or classical physics, a system that is not measured or which we have only measured incompletely may be in any one of several pure states with different probabilities, and we call the resulting state a statistical mixture; but that is not the same as being in multiple pure states at once.

Source: (1002) Alan Cooper’s answer to How does quantum physics know that if a system were not measured it would be in multiple possible states (without measuring it), and that when measuring it collapses into one definite state? – Quora

Bergson vs Einstein

After reading this article twice, and yet again the paragraph where the author purports to show that “it’s wrong to think that Bergson’s idea of duration can be assimilated into the idea of psychological time”,

I am still unable to find any explanation of the difference between our internally experienced psychological time (which, by the way can not necessarily always be “aligned with external clock time”) and “the first-person experience of (Bergson’s unmeasurable) duration” (which they appear to identify as the “lived time” in terms of which “An hour in the dentist’s chair is very different from an hour over a glass of wine with friends”).

On the other hand Steven Savitt’s “solution” does not address the subjective nature of duration and appears to just identify it with the non-subjective proper time associated with a possible observer’s world line – which seems to be just giving up on the idea of any special “philosophical” time as this has always been the only kind of time that is ever discussed in relativistic physics.

Source: Who really won when Bergson and Einstein debated time? | Aeon Essays