@Relativisticcucumber quite sure ACM is not intending to leave SE, but rather questioning the continued existence of SE in the first place. You've all seen the fires around the site. Soon the chatrooms will also have AI.
some govt organisation should aim to preserve this website at all costs
@naturallyInconsistent that would be saddening, the site's compendium of knowledge has now, in my view, surpassed the size of any organisation that owns it. Its now bigger than them, would be an absolute tragedy if they screw this over
Even the great internet archive succumbs to bit rot. I have lost count of how many things I saved and referenced to the internet archive, and now they are permanently gone
@Sanjana 40 years is an eternity - that same time ago, nothing of the internet as it exists today was there, and many sites came and went. Usenet is functionally dead, the age of decentralised fora and blogs gave way to centralised social media, and SE is already some kind of weird relic, its corporate owners clearly struggling to find some way to extract the kinds of profit from it they desire
@Relativisticcucumber not really, but making such specific plans for how the world will look when I'm 70 seems...optimistic at best
@nickbros123 There are data dumps of the whole site being saved to the archive, and I'm sure plenty of other people mirror them - a legacy of the site being licensed under creative commons. But while you can archive and mirror content, you can't mirror the communities that create it
"the environment is modeled as a set of infinite bosonic modes" -> how should i understand this sentence? i am thinking that we have some system and some environment and the environment is a lot of bosons, but im a bit confused on this "bosonic modes" part
@Relativisticcucumber This is close to how the pioneers thought of the EM wave, as if it were an environment bathing your system. You express the classical EM wave without interaction as normal modes, and then you take each normal mode and convert them into a QHO. Each of them are bosonic modes. Hence infinite set of bosonic modes.
@naturallyInconsistent ok wait backing up a bit -- how should i think of photons as bosons? photons have spin as i understand, but spin seems to be related to rotations via angular momentum, right? so how can i understand a photon rotating?
i guess thats slightly unrelated, but the comment made me think of that q ^
Easy. First of all, photons have spin 1. This is integer spin, and so by spin-statistics theorem, they must be bosons. Secondly, the best way to look at photon spins, is to use their "spin-eigenstates". Those are the circular polarisation states. Technically, those are helicity / chirality eigenstates, not spin; you can learn about the difference if you look up Wigner or Weyl treatments of spin and helicity / chirality.
or like a starting point -- because im interested in experimental physics but feel i have no understanding of relevant techniques
@naturallyInconsistent im kind of struggling to see this. so for the circular polarization states, ive seen with the em field it can be viewed this way, this is how we get the propagating wave picture, right? so you mean the spin states take on this same form/behavior?
so does a photon have zero orbital angular momentum?
@Relativisticcucumber I mean the trivial thing: You have electrons orbiting nuclei and so orbital angular momentum makes sense. Photons don't orbit nuclei and so you dont so often use that concept for photons. Of course, if you choose to use scattering states with definite orbital angular momenta, you can describe photons that way too, but it is very very rare for people to do that.
Whereas it makes intuitive sense that if you have an electron initially in spin up state, bound to some nuclei, and it radiates a photon and jumps to the spin down state, orbital state being kept the same except energy state is reduced, then the emitted photon has to bring away the spin angular momentum of the electron. That would be a circularly polarised light having $+\hslash$ angular momentum.
@Relativisticcucumber Plane waves plus the circular polarisation, yes. If you want to work with radially incoming and outgoing light waves of fixed orbital angular momentum, you could do that to even higher precision, of course, but again, rarely does anybody do that.
Not least because in Schrödinger electron orbitals we use spherical harmonics; the photon equivalent is vector spherical harmonics, which is ugly. Of course, Dirac version, the spinor spherical harmonics, is also a little bit more ugly than the simplest, scalar, spherical harmonics.
Oh. I thought in the sense of "modern physics" as used in Arthur Beiser's book on modern physics courses at univerisities, although the stuff they teach in that course is literally 100 years or so old...
@Relativisticcucumber Not exactly what you want but I was wondering how do people actually compare theory with experiments in HEP. You throw in some particles, but how do you actually measure energy-momentum, mass/spin other q. no.s, what particles are being produced, get cross section data and what not... I wanted to get an overview of the whole process...
I found this book "Deepak Kar - Experimental Particle Physics Understanding the Measurements and searches at the Large Hadron Collider (2019)" which is great for these. I also found some other books focussing on one particular topic: either the experimental design or statistical analysis or something else. But this is the only book which gives a "modern" overview: not just speaking of age old calorimeters and GM counters, but actually how different parts of colliders are built...
@Relativisticcucumber I think you need to be a bit more specific about which field you're talking about. Quantum optics will have wildly different techniques from climate physics or hydrodynamics...
it doesn't need to have a purpose to you, it might have some purpose in the mind of the one writing it, and there is some reason for it to be written because there is a stimulus for it
@Sanjana It comes from expanding the tensor products stepwise - the $l_2\times l_3$ decomposes into $\lvert l_2 - l_3\rvert \oplus \dots \oplus (l_2 + \l_3)$, and then you have to tensor each summand with the $l_1$ and decompose again - that's the outer sum over the $l$, and then each of these decompositions is $\lvert l - l_1\rvert \oplus \dots \oplus (l+l_1)$.
however, the r.h.s. of your formula is only correct for integer spins, for half-integer spins there is something missing as it results in non-integer for $l_1 = l_2 = l_3 = 1/2$
@Sanjana probably easiest to see by just distinguishing the two cases: Note that if $p=0$, then $\lvert l - l_1\rvert = l - l_1$, while otherwise $\lvert l - l_1\rvert = l_1 - l$
that gets rid of the absolute value in each case, and you can just carry out the sums
@ACuriousMind Hmm, Thanks: it works... I did the $p=0$ case, hopefully the other one works out too. But I am still wondering how someone would guess the answer can be written in that form. It looks so nice...
you do the two cases, and then you look at the two results and notice that one of them has no second summand, while the other has it - using the max to express that is just a neat trick
If one could approximate the behavior of a quantum system with that of a QHO, what would be a visible/observable difference in the case when in one scenario the system's state is the eigenstate $|n\rangle$ of its Hamiltonian, and in another instance it's the $|n+1\rangle$ ?
