Quora Answers – Physics Notes

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

Is Heat Radiation Just Infrared?

It’s really no different than with any other kind of energy. Heat is basically just energy that is disorganized ie distributed in a way that is too complicated for us to keep track of – like the random motion of molecules in a gas as opposed to bulk motion or pressure waves in that same gas.

With regard to electromagnetic radiation, every hot body emits a broad spectrum of disorganized electromagnetic radiation at all frequencies – not just infrared (though that is the dominant range at the temperatures we normally experience); but it is also possible to produce organized electromagnetic radiation (eg from electronic circuits and lasers) and this is not “heat” even when it is infrared (though it can be used to create heat when absorbed by a body – just as organized mechanical energy can be converted to heat by friction).

Source: (1002) Alan Cooper’s answer to Is all electromagnetic radiation considered heat? Or just infrared radiation? – Quora

Statistical vs Mathematical Physics

Mathematical Physics is the study of what can be really proved about our theoretical models of physical systems. This differs from other kinds of theoretical physics because physicists often take the lack of experimental refutation of a mathematically invalid calculation as “proof” that the result is correct.

Statistical physics (aka Statistical Mechanics) is the study of physical systems having so many degrees of freedom that it is not feasible to measure a complete set of the individual observables (such as the positions and momenta of all the molecules in a volume of gas or the angular momenta of all the electrons in a crystal), but for which some observables (such as temperatures and pressures), defined as averages of those most naturally considered as forming a complete set, are expected to evolve in a way that does not depend on the specific values of all the variables needed for a complete description.

Many of the expected behaviours of these averages are assumed by physicists without any complete proof; and one important area of mathematical physics is the filling in of these missing proofs. A classic text of this sort is the book ‘Statistical Mechanics: Rigorous Results’ by David Ruelle.

Source: (1001) Alan Cooper’s answer to What are the differences between statistical physics and mathematical physics? – Quora

Purely Probabilistic Positions?

What we interpret as the locations of elementary particles can certainly be compared with the predictions of regular mechanics. And they will often be quite close, so the classical predictions are actually useful. But the pattern of (usually small) variations from those predictions, while not “purely” probabilistic, does seem to have a component which cannot be explained in terms of some more precise classical properties that we just have not been able to properly measure. So our idea of a regular particle may just be something that does not really exist and what we interpret as its position may indeed by something that has an essentially probabilistic component.

Source: (1000) Alan Cooper’s answer to Is electron’s location purely probabilistic so its mechanism can’t be compared with regular mechanics, or is it just too small and too fast that the only way we can detect its location and interaction is through probability? – Quora

Maxwell’s Equations for Photons

In the quantum theory of electromagnetic fields Maxwell’s equations play two roles.

One is to describe the behaviour of the actual field observables which measure the combined effects of all possible numbers of photons, and the other is that they are satisfied by something that is as close as possible to being the “wave function” for a single photon.

I say “as close as possible” and put “wave function” in scare quotes because it does not satisfy all the properties of a non-relativistic wave function. In relativistic theories, the concept of strict localization does not exist. It can only be approximated for massive particles in frames where they have low momentum, and cannot be done at all for photons. But nonetheless, (as discussed in this survey article (.pdf) by Iwo Bialynicki-Birula) with appropriate normalization, a function satisfying the complex form of Maxwell’s equations can be used to generate probabilities for detection of a single photon in various experimental contexts.

See also the answer by ‘Chiral Anomaly’ to this question at physics stack-exchange.

Source: (1000) Alan Cooper’s answer to If particle nature of light is involved then what are Maxwell equations? – Quora

Photons in a Refractive Medium

A Quora question asks:Given that light is massless, and that all massless particles travel at the speed of light, it should follow that in a medium with a refractive index >1 (where light slows down), it acquires mass and experiences time. Why is this not the case?

It is not always true that “light is massless”. For example light trapped in a reflective container contributes to the rest mass of the system consisting of the container and its contents.

It is not obvious that massless particles always travel at the speed of light (but unless they are doing so they have zero momentum and so don’t change the momentum of things they collide with).

The speed of a photon is always equal to the vacuum speed of light in between its interactions with matter, but the probability of detecting a photon travelling through a medium is calculated from a sum of probability amplitudes associated with all possible paths including those which involve interacting with atoms in the medium. Since many of these paths are indirect, their lengths are greater than the straight line distance and so the average time taken corresponds to a speed less than that of light in a vacuum.

[Some answers have suggested also delays due to absorption and re-emission but if these really happened with random delays they would destroy the coherence and so in a perfectly clear medium the interactions are all effectively just instantaneous reflections off bound electrons (with minimal energy transfer due to the masses of the nuclei).]

One might be tempted to look for a way of describing the result in terms of effective photons with mass; but we can’t expect any proper Lorentz covariant theory of such particles since the medium is only stationary in a particular inertial frame, and in relatively moving frames it appears contracted which changes the density and so the index of refraction (in a direction dependent way).

Source: (1000) Alan Cooper’s answer to Given that light is massless, and that all massless particles travel at the speed of light, it should follow that in a medium with a refractive index >1 (where light slows down), it acquires mass and experiences time. Why is this not the case? – Quora

Wave Momentum

“How do waves have momentum?” is a very good question, but like many good questions it seems to attract a lot of over-confident incomplete answers.

It is in fact true that many kinds of travelling waves do transfer momentum to anything that actually absorbs or reflects them, and the momentum transfer is often proportional to the energy density and speed of the wave; but just stating that something is true is not an explanation of why it is true, and if the mere fact of carrying energy explained why waves have momentum then a moving charged battery would have more momentum than an uncharged one.

Indeed, it is perfectly reasonable to not be immediately convinced that waves have any momentum at all in the direction of propagation. For transverse waves the primary motions are perpendicular to the direction of motion and for compression waves the motions forwards and backwards mostly cancel out. And the fact that we can get pushed inwards by a water wave doesn’t tell us anything about net momentum transfer, since anyone who has experienced that inward push has probably also experienced the outward suction of the receding wave; and although waves seem to bring flotsam in to the shore it is not obvious that this is due to the waves themselves rather than the wind that gives rise to them.

When it comes to the often mentioned pressure and momentum of electromagnetic radiation, while we can see the effect of light pressure on the tails of comets, the derivation from Maxwell’s equations is rarely given completely. Many sources (such as this one) explain how the perpendicular electric and magnetic fields lead to a force on any charged particle that is perpendicular to both of them, but don’t give any proof that this is in the forward direction of wave propagation rather than backwards; and even when such a proof is given it is usually shown just as a formal calculation without any physical motivation as to why it is working.

A google search for “wave momentum” is unfortunately overwhelmed by ads and reviews for a popular brand of volleyball shoes, but if we change the order and/or add words like “electromagnetic” or “water” we do get a lot of useful hits. The best I’ve seen so far is

https://as.nyu.edu/content/dam/nyu-as/as/documents/silverdialogues/SilverDialogues_Peskin.pdf

This gives pretty complete arguments for the momentum content of various kinds of waves, (and also includes examples of waves that carry energy but do not have momentum – which shows that your skepticism is not at all unreasonable). But it is at a fairly high mathematical level and so takes a pretty advanced reader to see the physical motivation for why its results are true.

So what I want to do in the rest of this answer is provide a bit of a handwavy argument to give some physical motivation for the momentum content of one particular kind of wave. It is not to be taken too seriously, but just as a hint of what might actually be shown by a proper detailed analysis.

Consider a rope tied to a wall or post at one end, with you holding the other end and moving it up and down to project a wave along the string. If the wave carries momentum then during at least part of the cycle your hand must be applying a forwards push (or at least a reduced amount of tension compared to the starting situation) – and I suspect that, even before thinking about this, it has indeed felt that way when you tried it. That may of course just be a psychological effect rather than anything real, but perhaps we can think of an actual physical reason for it. When your hand is at the extreme top of its motion the rope near the end you are holding is bent up a bit, and as you move it back down the tension in the rope tends to straighten it (even if you just let it go free rather than pulling it down). This pulls up the lower part of the bend, and to counter that pulls down the part nearest your hand. But this swinging down of the end would, if tension were maintained, cause it to project outwards a bit – and so maintaining the original distance from the far end would require a bit less tension (or equivalently a slight push forward relative to the starting level of tension). As I said, this is not a real argument, but it’s the best I can do short of a proper mathematical proof as given in the paper linked to above.

Source: (1000) Alan Cooper’s answer to How do waves have momentum? – Quora