i can only see that sakurai's conclusion is true if $V^\dagger$ is injective (which in this case is also equivalent to $V$ being bijective)

also i don't understand...doesn't sakurai's definition of the $S$-matrix implicitly force conservation of kinetic energy and so only describes elastic scattering? from what assumption did this come from?

it seems to come from the defining equation of the $T$-matrix (since this is where the energy conservation enforcing dirac delta comes from), so is this equation not always valid?

oh no i kee p asking questions in here and then realizing they are probably just better to make into a genuine question on the main sitet :P soryr

but in general i see your claims. that is interesting.

@SillyGoose FSDLFJDSKLFJ THIS IS THE BIGGEST LIE EVER lol

u r the largest prescriptivist i have ever seen

i just dont think language is like gymnastics XD i guess im prescriptivist then XD

You're right. Just as muscle cells are not the same as brain cells.

realistically im a mixture of prescriptivist and pragmatist

@Relativisticcucumber i think i am especially presciptivist for mathematics related contexts hehe since i prescriptively think that that is the one place where one must be prescriptivist to make any sense...

@SillyGoose nah

its every context

u pretend like u dont know what is being said if someone uses words "loosely"

"can you bring some soup"

hands a can of soup no can opener no bowl

u DiDnT sAy yOu wAnTeD tO eAt tHe SoUp

XD

LOL

smth only a prescriptivist would ask tsk tsk

ACMist

a follower of ACM

i was gonna say acm enters chat exclaims "well these are not probabilities"

:D

ok time for work bye all

cya pal

Always good to see this cat.

Never seen into Part II, but does this book cover QFT?

@DannyuNDos No, part II contains:

Nevermind; guess QFT is for grad students.

The department of applied physics in my univ is mostly for semiconductors and display devices, so

How "right" is this answer

I really like it

> The fact is, magnetism is nothing more than electrostatics combined with special relativity.

BTW, I don't have a good handle on the history of physics but I was under the impression maxwell's equations as we know them today aren't what they were back in the day?

So this kind of aptness is what physicists seek for beauty.

right, when I read that and saw the explanation I was baffled

if only all of physics was so elegant and beautiful

That said, a question about physics education: Classical mechanics is for 2nd graders. Electromagnetism is for 2nd-and-a-½ graders. Quantum mechanics is for 3rd graders. Why isn't relativity theory a separate class?

@Obliv The current form of Maxwell's equations were written down by Heavyside not Maxwell.

I guess it depends on the curriculum but yeah here it's pretty much sandwiched between QM and EM

Maxwell used a rather clumsy notation. The equations were the same but written in a less clear way.

In my EM class we didn't credit $\nabla \cdot \mathbf{B} = 0$ so were they not a part of the original maxwell-heaviside equations?

I guess no credit is necessary since it's not technically necessary to have $\nabla \cdot \mathbf{B} = 0$ anyway

Well it's just the statement that magnetic monopoles do not exist. Maybe that assumption was not made at the time.

I'm reading on Heaviside's wiki that he was mostly self taught... so we're using his formulation even though he didn't follow a standard education :P

that's kinda ironic

you know when you think about it, lightyears and years are basically treated as the same unit in relativity

so the misconception that lightyears are a unit of time is kind of true in that sense

alright i should probably be sleeping

Regarding that... do physicists ever consider hypervolumes in the timespace? The SI unit would be tesseractic meters.

m⁴, I mean.

well then it should be called a year-year instead :P

The misconception (I think) is more that a lightyear is somehow longer than a normal year or something else entirely I'm not sure

Can someone help me with polarization understanding? If we have linear polarization, then this means that the phase difference between the x and y components for a electromagnetic wave propagating in the z axis is the same?

Linear polarisation means that the x and y components are in phase (but have different amplitudes) so the phase difference is zero.

Circular polarisation is when the phase difference is 𝜋/2.

Ok one question

you can have linear polarization in x in y direction or anywhere else in the xy plane

what determines the direction in the xy plane?

the amplitude values of each component i assume, right?

hey

There's something about the fundamental nature of light that I'm trying to understand better.

Hi :-)

I can imagine how my question might sound confused -- because I am confused

Do you know how QFT describes particles and how their creation and destruction is described?

no

is particle creation related to light creation in QFT?

OK. I'm going to use a metaphor that gives an intuitive feel for what is going on, but bear in mind it is a metaphor.

(my question originally said "QFT", but someone changed it to "QED" -- I'm not sure why)

yes, ok

Now imagine we have a wave rippling through the sheet like a water wave moving across a water surface. This wave has some energy associated with it because we had to put energy into stretching the sheet to create the wave. And it also has a velocity beause the wave is moving in some direction at some speed.

Does this make sense so far?

ok

Well this wave describes a particle in QFT.

for velocities not close to c it makes sense, anyway

That is, in QFT we have a field that fills all of space, and particles are waves propagating through this field.

Specifically a particle is an infinite plane wave propagating through the field.

OK so far?

so this is the "wave function"?

yes, ok

In QFT the particle wave function is not a terribly useful concept since individual particles aren't really distinct things. They are just waves in the field and they can be created and destroyed so the number of particles is not constant.

In non-relativistic QM the wave function is useful because there particles cannot be created or destroyed so an individual particle can be described by an individual wave function.

I can't say I have an intuitive sense of particles being infinite waves. The words make sense, but the concept doesn't quite click.

@RickNZ Everyone starts QM with the idea that a particle is like a little ball, and particles colliding is like balls bouncing off each other. Sadly this intuitive notion is highly misleading.

In terms of probabilities, it makes sense, I think. I guess the waves are just the manifestation of those probabilities?

In chemistry, for example, we talk about the probability of an electron's location

Quantum objects are always delocalised so they are more like fuzzy cloud than a ball. The cloud can be small and dense for a highly localised particle or large and diffuse for a highly delocalised particle.

also see this post physics.stackexchange.com/questions/710826/…. particle in accelerators behave classically except for a microscopically small interaction region :)

And this "sheet" is the EM Field?

@RickNZ This is a common misconception. A particle does not have a single location any more than a cloud does. The wavefunction tells us the probability that the particle will interact at that location, not the the partcle's position.

@RickNZ For photons the EM field is indeed the "sheet".

Each particle has its own field so electrons are excitations in an electron field that is different from the EM field.

ok

Suppose you start with an EM field that is zero everywhere. This is the ground state i.e. the state in which no photons are present.

only very high speed quantum particles can sometimes be thought of as rays

ok

If we add energy to the field we create an EM wave propagating through space. And this wave is a photon.

Add more energy and we can create a second photon, and so on.

Likewise if the field loses energy that means a photon disappears.

how does the electron field interact with the EM field?

does anyone have a favorite resource for Chern-Weil theory or maybe a text on characteristic classes that includes discussion of Chern-Weil theory?

@RickNZ When we write down the equations describing fields we can also include a coupling constant that describes how fields interact with each other.

@RickNZ it is described in feynman diagrams and interction terms in equations

The electron and photon fields have a non-zero coupling constant, and that means they can interact and exchange energy.

what are the physical interactions behind the coupling constant, models and diagrams?

That's a meaningless question.

The physical interaction is the coupling constant.

The coupling constant describes how two infinite plane waves interact with each other.

And that leads to the electromagnetic forces we observe at everyday scales.

I don't understand. Doesn't the coupling constant describe how much they interact? You're saying there's no physical process behind that interaction?

yes we dont have a more fundamental theory than qft, and proposals like string theory and lqg themselves rely on their own equivalents of feynman diagrams and stuff. but one thing is that Witten says string theory gives u the form of the interaction using pure mathematics

What I'm saying is the coupling constant is the fundamental origin of the interaction.

I guess you're asking why is the coupling constant non-zero i.e. what causes the coupling constant?

yes

But there is no answer to that. The coupling constant is a fundamental property of the fields.

That is, it is an experimentally observed property.

seems surprising, but ok. so the electron field interacts with the EM field through the coupling constant, causing it to oscillate, creating light

What happens is that energy can be exchanged between the fields.

and the energy that goes between the two fields is quantized

And it you take energy from the electron field and add it to the photon field that creates a photon i.e. a particle is created.

@RickNZ In general the energy is only quantised in special cases like a hydrogen atom.

and carries a direction/momentum vector and energy amplitude and frequency

If we have two free electrons interacting to create photons the energy exchange is not quantised.

oh? interesting

The name "quantum mechanics" is actually a bit misleading.

It's a historical artefact.

The energy of free particles is not quantised.

ok

What is quantised is the number of particles so we can create one photon, or two photons, but not a fractional number of photons.

I see, sure, makes sense.

so now we have a wave in the EM field, propagating outward at c.

Yes

But it looks different in the inertial frame than the stationary one, right?

can't you have light (excitation in photon field) without interactions with a massive particle (e.g. electron)? or are you specifically asking about this interacting case

@RickNZ I'm not sure what you mean by inertial frame and stationary frame.

I was asking about this specific case, though for me, the source of the light doesn't matter (at least, I don't think it does)

If we are calculating the interaction of two particles we would generally use the centre of momentum frame i.e. the frame in which the total momentum is zero.

In your original question this would be the rest frame of the hydrogen atom.

I'm trying to compare the frame of the electron and the frame of the photon, although I suspect my language is imprecise

There is no frame of the photon

Anything moving at the speed of light, whether it's a photon or some other massless particle has no rest frame.

yes, I've heard that

maybe I don't completely understand the ramifications

So it is meaningless to ask what the interaction looks like to the photon. All we can do is describe the interaction as observed from some frame travelling at less than the speed of light.

Isn't it true that time slows down as an object approaches the speed of light, and that for light itself, time effectively comes to a stop?

> for light itself, time effectively comes to a stop?

yes

This is a meaningless statement because it assumes that light has a rest frame and it does not.

ok, what about an electron moving arbitrarily close to c?

Suppose you are moving arbitrarily close to 𝑐. You look at your writswatch and you see that time is still ticking along happily at one second per second.

from their own rest frame, time passes normally. from the frame in which theyre moving, electron's frame's time can be arbitrarily slow

And this is always true for any speed you can attain relative to me.

yes, that's what I've been trying to say

You will observe my wristwatch to be running slow, just as I observe your wristwatch to be running slow.

But neither of us can ever observe the other's watch to stop completely.

ok so from any other frame, the proper time of light is indeed zero

but we cant talk about the time of light from its own frame becuz that frame doesnt exist

Yes

ah, my imprecise language ... when I say stopped clock, I'm trying to describe how an object moving close to c can travel an arbitrarily long distance without themselves perceiving (almost) any change in time

from their own perspective, the length is contracted so theyre traveling an arbitrarly small distance

yes. I get that part

Careful: we observe the object to have travelled a long way. As far as the object is concerned it's just sitting there motionless.

so theyre able to do it in a short time, doesnt matter what frame u look it from

I understand the spacetime warping aspect of things

it's the behavior of the EM field oscillations moving at c that's bothering me

Are you thinking that it's impossible for the field to oscillate since when moving at 𝑐 time must stop?

since as you said, the distance is compressed -- to arbitrary thinness -- how/why are the EM field oscillations happening with the frequency we observe?

Let me suggest a thought experiment.

Suppose we have a photon coming towards us, and that photon has some frequency 𝑓.

Something like that, yeah. Not really impossible, but it feels like there's a contradiction in here somewhere that's breaking my mental model

ok

Now suppose we start moving at some velocity 𝑣 in the same direction as the photon.

ok

yes

If it's red shifted that means the frequency has decreased i.e. it is oscillating more slowly.

yes

And as we keep increasing our speed the frequency keeps decreasing.

But the frequency can never become zero. It can become arbitrarily small but not zero.

and at relativistic speeds?

This is true for any speed relativistic or otherwise.

ok

The point is that the EM wave is always oscillating. It never stops oscillating in any frame.

ok

I think I get it now. Thanks for your help.

the rel doppler effect does not reduce to the non rel doppler effect in the non rel limit, right?

@RickNZ what subjects r u interested in

it's changed over the years. space science (sent software to Mars), math, physics, optoelectronics, biochemistry, recombinant proteins, biology, etc. You?

@RickNZ u were involved in mars missions? woww

yes x3

@RickNZ physics, math and some philosophy and history

software of mine has been in orbit there for more than 20 yrs

awesome

ah, philosophy is another interest of mine, too

i like to think about consciousness

OrchOR?

i had never heard of this

oh it's the microtubules thing

it's a quantum physics-based theory of consciousness

yes, microtubules

i dont know the specifics of microtubules. hope to learn it someday :)

there is also integrated information theory

the challenge for me in many theories of consciousness is Qualia.

I prefer consciousness being axiomatic. Like existence.

@RickNZ yeah... i think we have to take those as primitives or derived from some primitives

integrated information theory takes them as primitives

Penrose has a pretty strong argument that consciousness isn't computable -- important these days as it relates to the prospect of true AGI

yes. Penrose uses the Godel theorem to make his argument. i never understood the argument..

yes

and then he relates it to quantum measurements which r also non computable becuz theyre inherently random

i think there is a 60% chance there is some direct relationship between quantum physics and consciousness

The Godel approach is brilliant, IMO. I also have my own version, based on Searle's Chinese Box.

what is it

im familiar with Chinese box

or leave it if it's too long for here ;)

The basic idea is that it's impossible to use the inputs and outputs to the box alone to determine if the box contains a conscious being.

oh

... which in turn is something most can easily understand, but it's not provable mathematically

yeah...i think Penrose turns the intuitive idea into mathematics

yes, exactly

there is also the problem with the girl in the black and white room who can never learn about what red looks like by reading about red for her whole life

yes, related to qualia, I think. there's no way to know if what you see as red is the same as what I see as red.

the other 40% chance i attribute mostly to the Copenhagen interpretation, then some 5% chance to many worlds and others

ah

Witten says this wil never be solved ;)

Ah, Witten.

he also brings quantum mechanics into this discussion, which means he also anticipates some relationship

in the non rel doppler effect, we hav three speeds, $c$, $v_o$ and $v_s$. in the rel doppler effect, the speeds relative to the medium r undefined, so there is only the relative speed of the two observers

so i think these r not related to each other by the non relativistic limit

@Obliv There are some general ideas correct, but it is wrong in a few ways. First of all, it is not a hidden thing; many sources write about this. Even Feynman propagated it. However, it is more pressing that it is technically impossible to convert all magnetism into SR: SR only allows a pure electric field to mix into electric-dominated fields. In particular, $E^2-B^2$ and $\vec E\cdot\vec B$ are conserved quantities of the Lorentz transformations. But we can have purely magnetic fields.

This implies that those fields cannot be coming from just SR. Feynman knows this and stated it in a later part of his books.

The fact of the matter is that, once you finish building up the Maxwell's equations, the correct next step is to take Maxwell's equations as the postulates and derive everything from there. So the rant in the post is just wrong.

@naturallyInconsistent I didn't quite understand your comment about how $k'$ is not just a rotated $k$ in the $S$-matrix described in my question.

@SillyGoose There was no restriction on $k^\prime$. That is, there was no restriction that $|k^\prime|=|k|$

Hm then how am I misunderstand the presence of that Dirac delta

because I am reading it as enforcing that $E_k' = E_k$, which implies $\lvert k' \lvert = \lvert k \lvert$ to my knowledge

since either $k = k'$ or $k \neq k'$ and we get $0$ probability amplitude

there is a wonderful quote by Witten, which is something to the effect of "Spin(3) $\times$ Spin(1) is a subgroup of Spin(5) because 3 + 1 is less than 5"

@Relativisticcucumber yes - because my parents were below a certain income threshold, I got enough money to pay for housing and food from the government. Usually this is an interest-free loan you have to pay back once you earn enough money, but I had a scholarship that meant I don't have to pay back anything.

honk

There are some wonderful quotes in here also

ACM: have you found characteristic classes to be at all illuminating to any area of physics :P

it seems like a potentially fun subject

@SillyGoose they're what many of the "topological" terms in physics are - many of them are Chern or Pontryagin classes

@ACuriousMind ooh nice okay i see. thanks for the info

is there an operator that does not conserve particle number?

@Relativisticcucumber sure - all the Hamiltonian operators that allow for particle creation/annihilation for one

@ACuriousMind hm so is it incorrect to associate operators with observables?

and if i take the expectation value of an operator similar to, say the displacement operator, what does this actually even mean?

@Relativisticcucumber I'm not sure what you mean - what did I say that gave you the impression it's incorrect?

oh you did not

i asked the OG question partially because of a confusion i have encountered about operators

initially i thought operators in QM are associated w observables (namely their eigenvalues are the observables) and that expectation values should show how that observable looks as the system evolves in time

but then i became confused because there are operators such as the displacement operator that i am not really sure how to interpret the expectation of

@Relativisticcucumber That is correct (if the operators are self-adjoint)

so is it that we only look at the expectation of hermitian operators generally?

Yes - non-self-adjoint operators are not guaranteed to have real eigenvalues and cannot correspond to observables

ok makes sense thanks

That doesn't mean you can give any such operator an actual physical meaning - e. g. the sum of two self-adjoint operators is self-adjoint, but there's no generic meaning to the sum of position and momentum

i think i was getting tripped up because i am confused about how to physically "implement" a nonhermitian operator

but i think theres no reason in principle this cannot be done by some experimental methods?

in particular this operator is what i was trying to understand en.wikipedia.org/wiki/Displacement_operator

im not sure if this is just mathematical formalism or an actual implementable thing since operators are usual physical

Well, first, that's a unitary operator - so it can in principle occur as the time evolution of some system

But you don't need to "implement" operators for them to be useful: that's just like in classical mechanics where we can talk about (symmetry) transformations but it is irrelevant if you can "implement" e. g. an actual rotation of the system, the abstract transformation exists regardless

@ACuriousMind do you mean that, since unitary operators do not change probability amplitudes, a state can experience action of unitary operators as it evolves ? like a state sporadically can undergo a unitary operation?

because in the case of hermitian operators the "point" is clear to me -- we establish quantities of interest in a system and we look at them as time evolves but i guess im confused about the "point" of other types of operators in qm. i have used them, but its more like in abstract problems

@ACuriousMind oh no symmetries are a long standing confusion of mine

@Relativisticcucumber I just mean that time evolution is unitary, and you can just choose the self-adjoint operator in the exponent of any unitary operator as Hamiltonian to get a (theoretical) system whose time evolution is that unitary operator

@Relativisticcucumber they're just maps (like operators are linear maps) and they don't need any physical "manifestation" to be valid mathematical objects

is it correct to say in qm we only care ab unitary or hermitian operators?

To a good approximation, yes

okay i see. i was also wondering -- for HO, we have that the position operator and momentum operator can be written in terms of the CA operators. i know this can be shown mathematically, but conceptually this doesnt make any sense to me. why should the CA operators give you the position of a state?

or the momentum

@SillyGoose you are not misunderstanding. There is an overall Dirac delta that conserves energy, and it is this that sets $|k^\prime|=|k|$; if you have more than one particle, then they can exchange energy-momentum. The enforcement of this conservation is not at the matrix elements level. It is at the overall Dirac delta level.

@Relativisticcucumber The 1D QHO case is pretty special. The entire Hamiltonian spectrum is non-degenerate and the CA operators can access them all. This means that all other operators must be expressible in terms of CA operators because, again, they already access the entire spectrum.

@naturallyInconsistent sorry, i am not following the logic of "This means that all other operators must be expressible in terms of CA operators" -- can you elaborate?

@Relativisticcucumber CA operators access ALL the states. Then all other operators must be a combination of them

Think of it like the Casimir operators argument.

If you commute with enough important operators, then at some point, you end up being the identity operator

Can anyone explain to me how The angular spectrum method is different than Fourier optics? angular spectrum method considers waves of a field as a superposition of plane waves of different propagation direction. Fourier optics is the study of classical optics using Fourier transforms (FTs), in which the waveform being considered is regarded as made up of a combination, or superposition, of plane waves.

How are they different exactly ?

If we consider an electromagnetic wave. How can it happen that $k_z=\sqrt{k^2-k_x^2 + k_y^2}$ is imaginary. How can it happen that componets of the wavenumber are larger than the wavenumber itself? How is that physically possible? Because, this is a condition for evanescent waves. But I don't understand how two of the 3 components of the wavenumber can give a value larger than the wavenumber itself

?

@imbAF you should look up what evanscent waves are, in the simplest case. in 1D.

damping waves I guess

But I don't understand how did the professor claims this : $k_z=\sqrt{k^2-k_x^2 + k_y^2}$

If i square both sides then I get k^2= k_x^2 +..

and that is somehow smaller

@user70432 it's really good

@imbAF trivially the case since you already know that imaginary components subtract

I see

But there is one additinal thing

In my lecture it was said that for a wave, which has an imaginary k_z component, which I believe implies, that the z component of the electric field is damped in the z direction, only waves whose spatial frequency is smaller than k=\frac{2\pi}{\lambda}n can propagate

Is it talking about wave for which $k_x^2+k_y^2<k^2$?

There are some conditions. Cannot tell without specifics

Same

but the thing that confuses me the most is that we follow that statement with:

$\Delta k_T\le \frac{2\pi}{\lambda}n$

and $Delta k_T$ is the bandwidth of spatial frequencies

but this makes no sense

If you keep telling us this little, nobody can determine if your question even makes sense or not

This is what I have as information

Basically we have a wave, which has constant z=const. Which is a stupid convoluted way of saying we consider a plane wave with propagation in z

and we were considering the cases when k_x^2+k_y^2<k^2 and k_x^2+k_y^2>k^2

In the first case, we are dealing with a plane wave propagating in the z direction

in the 2nd an evanescent wave

and from this point, what I wrote above, is what was given in the halfassed lecture from the person in charge

@naturallyInconsistent ooh i see what you’re saying

yay

meow/honk

🐈/🦆

M I A O ~

does SR predict coulomb force increasing/decreasing based on relative velocity in same axis of motion

based on what I read from that post before, SR explains $\mathbf{B}$ as the coloumb force for relative motion around the moving charged objects

@Obliv what's the point of this specification? The Maxwell's equations describe the entire behaviour including SR, so it is not particularly meaningful to ask what it is like just within SR alone.

Hm? I'm asking whether the coulomb force increases/decreases due to lorentz contraction of a current

moving in the same axis

see this answer explaining $\mathbf{B}$ in terms of just SR and coloumb force

Oh I see what you mean now

but it seems like SR is more fundamental.. but if they both describe the same thing shrug

I guess it doesn't matter

the first comment to that answer btw sounds incredibly sarcastic to me but i think it was genuine which is very confusing lol

Special relativity is not more fundamental, it is a special case of relativity.

See here

or "relativity principle"

relativitäts prinzip