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
@ACuriousMind hm so when you say you didnt pay for it, does this mean you got money to go there? like for food and housing and whatnot? so there was literally no financial barrier to attending/there isnt the issue of some students having to work full time jobs alongside school to afford housing and food?
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
@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...
II Applications 7 Time-Independent Perturbation Theory 7.1 Nondegenerate Perturbation Theory 7.2 Degenerate Perturbation Theory 7.3 The Fine Structure of Hydrogen 7.4 The Zeeman Effect 7.5 Hyperfine Splitting in Hydrogen
8 The Varitional Principle 8.1 Theory 8.2 The Ground State of Helium 8.3 The Hydrogen Molecule Ion 8.4 The Hydrogen Molecule
9 The WKB Approximation 9.1 The “Classical” Region 9.2 Tunneling 9.3 The Connection Formulas
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?
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?
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?
Using QED, what happens, exactly, when light is created?
Let's say an electron loses energy by dropping to a lower orbital, causing quantum of energy to be "emitted." What is that energy quantum? A massless, size-less "particle" called a photon? But what does that actually mean?
I'm looking for a...
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.
Imagine we have an elastic sheet like a drum skin. The ground state, i.e. the lowest energy state, of this sheet is for the sheet to be just sitting motionless. OK so far?
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.
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.
@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.
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.
@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.
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
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
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.
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.
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.
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
since as you said, the distance is compressed -- to arbitrary thinness -- how/why are the EM field oscillations happening with the frequency we observe?
it's changed over the years. space science (sent software to Mars), math, physics, optoelectronics, biochemistry, recombinant proteins, biology, etc. You?
Knowing nothing about consciousness puts me in a perfect position to speculate :-) And I like the idea that consciousness is illusory i.e. we only think we think.
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
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.
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.
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
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
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
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?
@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?
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.
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
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 $k_x^2+k_y^2<k^2$?
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
first of all we are comparing an interval with a single value. 2nd the statement above is implying that the condition for propagation is that the transverse spatial frequencies must be so that satisfy the condition written above
so how do we come up with this: $\Delta k_T\le \frac{2\pi}{\lambda}n$ And how is this bandwidth defined? How does it make sense to compare it with a single value? Doesn't it makes more sense to say that waves, whose transverse spatial frequency belongs to this interval, those waves propagate ?
@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.
> Just a note, Einstein didn't name it "Special Relativity". Originally, he just called it relativitätsprinzip. After GR was published, everyone started referring to his first theory as special relativity – Jim Sep 7, 2014 at 16:30