@PM2Ring What if the source of the gravity is not the mass, but some inherent, preserving topological defect, causing non-vanishing curvature even without mass?

@PM2Ring Another interesting thing, how can the Universe just end on the Z axis. I think, QFT should work on a way, that every particle path has automatically zero probability if it would have an "out-of-universe" point. I am not sure but I believe, that would result that the "ceiling" and the "floor" of the Universe would behave as mirror.

Beside these, the "borders" of the Universe would behave as infinitely hard walls.

Also such a Univese would not be isotrope: on the axes, we would need to move 20000km to get into our original point. But diagonally, we should move 20000*sqrt(2). That would also eliminate impulse momentum conservation, but only in geological sizes.

On the Z xis, there would be no impulse conservation, if there is an interaction with a floor or ceiling of the Universe.

@PM2Ring In this construction, I can not see anything what would say, where is "down". Some symmetry breaking mechanism should exist, causing that gravity points to a direction.

"In the beginning God created the heaven and the earth..."

@RyderRude There are other fields in cond mat that aren't Schrodinger fields.

man i don't understand photons

no string theory questions for a month then three show up on the same day?

hmmmm

poisson distribution do be like that

Hi

@LeakyNun photons r the energy eigenstates of the free quantum maxwell field theory. Energy eigenstates r the states of definite energy

yeah but more like, how they actually work and stuff

like why does it only interact with one photodetector

U mean position measurements?

yeah

Every particle is detected at a single position.. This is not specific to photons

Upto a position measurement

Quantum mechanics doesnt answer y is it so. This is an axiom called the measurement postulate

It applies to all measurements, not just position

also like when exactly does it hit the photodetector?

like the wave can span more than 1 meter right

when the front half of the wave hits the photodetector does the back half just magically vanish?

Yeah. The probability wave is delocalised always

but like it's basically a changing EM field so you could theoretically "measure" the back half?

by having a test charge or something

I dont think u can predict when it will collapse becuz the collapse is probabilistic. For e.g. let's say u just hav a single detector at a space point and a delocalised wavefunction. At any time, the photon has a probability of being detected there

U cant predict when it will b detected and the wavefunction will get localised at that point aftr the detection

Not exactly localised, it will still have finite spread

But since it's probabilistic, it cud happen at any time

am i having the wrong picture in my mind by imagining it as an oscillating EM field?

Yes, the photon states dont have a definite value of electric and magnetic fields

It's a probbaility wave. The electric and magnetic fields r a much more complicated object in quantum theory. They r no longer "fields" in the usual sense

U can think of them as either operators or a very pathological object like a wavefunctional

The probbaility values of the wave do not mean electric and magneitc fields, i mean

The electric and magnetic fields r themselves probbailistic. U can only speak of their expectation values

so you have a distribution at each point?

Yes, u hav an expected value of electric and magnetic field at each point

So they r probabilistic at each point

and they're highly correlated?

But no one tests these predictions becuz they sound crazy. We only measure energy of the field in scattering experiments

so what exactly travels at the speed of light?

I think some experiments may measure electric field due to a photon but im not surr

@LeakyNun in the quantum theory, the energy states r infinitely delocalised. So those states def dont travel anywhere. These states r what we mean by photons usually

But the same applies to electron field energy eigenstates too

Now u may ask, how do we derive a particle traveling at a definite speed from this

also i feel like there are two kinds of photons that get mixed up as one (maybe they're really the same)

one from attuning a laser beam and one from spontaneous emission

Idk the specifics of lasers

To get a particle with position, u need the position space wavefunctions. U derive these by taking superpositions of energy states

This is only approximate stuff

It's when these position space wavefunctions r relatively localised, they travel at a constant speed given by special relativity for free particles

hi just got in here - the laser's photons and the one emitted by an atom are the same thing in two very different situations. in both cases we start with the vacuum. For the atom spontaneous emission, we have an excited atom, and that time evolves into a groundtate atom plus a state consisting of a single photon in a kind of wavepacket state moving away fromt he atom - like water waves move away from a point where you dropped osmething into the water

in the case of the laser, you have A LOT of photons in a small range of wavevectors. and incidentally once you have a lot of photons you recover classical electromagnetism

so for the emission you would have a spherically symmetrical wavefunction?

yeah... I intentionally avoided saying that because (A) I don't know and (B) I think if the original atom was in a definite m_l state it wouldn't be spherically symmetric. I.e. more likely to be up or down but not to the side

but really i'm not sure one way or another

@AXensen i'm referring to the case where you use filters to decrease the power of the laser down to "single photons"

yeah okay. im really not sure how to best describe the EM field's wavefunction in that situation

somehow I know that the end result is your photon counter will give counts that look like shot noise - every once in a while it will report it has recieved a photon. but I'm not sure how best to think of the wave function

good question

so it what sense are all electrons the same? is this different from photons?

uhh its only in the sense that they're fermions. if you exchange the positions in the wavefunction you get a factor of -1 for electrons and a +1 for photons

not sure what level of physics education you're at.

this is a statement from your first or second semester of college quantum mechanics

i'm a math major haha

if you aren't there, then maybe the best answer is... no its not different. photons are all the same the same way electrons are all the same

idk what i'm doing in this room

accepting that physics is more interesting

(thats a joke)

anything that is not your main job is more interesting

@AXensen ok but like from your description it sounds llike those two kinds of photons would have different wavefunctions

for sure they would

also does it make sense to say that the wavefunction spans the whole universe?

i think the dirac equation actually restricts the wavefunction to be zero outside of the causal cone (things cannot move faster than the speed of light)

I'm not sure about that statement

could be a decent physics stack exchange question

because someone set up an experiment to show that after splitting a photon and separating their path length for 1m "they" can still interfere

by "they" i mean the two "copies" (???) of the photon interact with "themselves"

Huygens Optics?

yeah

That kind of suggested the opposite to me, though I don't really understand it at the moment. There's no interference at all.

i g2g good luck with that one

bruh after all these years we still don't understand photons??

U wud need an infinite potential barrier 4 the probability of a particle to be detected outside a boundary to vanish. In practice, probability doesnt vanish, it only falls off exponentially

@WaveInPlace do they have a picture showing the result path?

I haven't gone through it yet. There's a set of paths shown, but I don't know how sensible they are.

Even right after a position measurement, the particles arent localised. Delta functions arent physical

You can look at the path of quantum particles in the limit of infinitely many measurements

also like popsci makes it sound like "photons make up light" the same way that atoms make up stuff

It is a fractal curve

Unrelated, but where should I go to get some math checked over? I think I've derived the magnetic moments for the electron, proton and neutron in the same framework.

I modeled them as physical charge loops. Which...shouldn't have worked.

@LeakyNun its not an accurate statement. Classical waves need not b eigenstates of the quantum energy

We can derive classical light from the free quantum theory by using Ehrenfest's theorem and constructing quantum states with sharp-ish values of Electric and Magnetic fields

This same method holds when deriving Newtonian mechanics using non. Rel. QM. We just assume the uncertainties r low

waveInPlace you've piqued my curiosity

it seems implausible given that the neutron is uncharged, and both the neutron and proton's magnetic moments can only be predicted with a very complicated lattice QCD computation

the electron's can come from perturbative QED, which is described in a lot of detail in some of muon g-2's papers

unless you just mean that with a sufficiently chosen loop radius/current you can achieve the same as the experimental value of the magnetic moments. I wouldn't classify this as "deriving the magnetic moments"

Where's ACuriousMind?

@AXensen I don't 100% know why the math worked. It pretty much has to be an abstraction of something deeper, because there's no wires at the subatomic scale.

The Bohr magneton is simple enough to write here though. μ = ecr = ecλ/4pi.

If you want to see the formal writeup I can upload it.

Do the quantum numbers of momentum etc. stay the same upon dressing the particle?

Lamba = compton wavelength for the electron.

I started from the equation for the magnetic moment of a macroscopic wire, μ = nIA, with the path of the wire two layered on top of each other (or a figure 8 folded over, if you prefer). A = πr^2, I = 2ec/λ.

almost at 15k reputation

I think at that level I am allowed to murder bad posters

I mean, sure? Have I stepped over a line?

Apparently it's actually "mark questions as protected"

@WaveInPlace it seems what you have discovered is dimensional analysis

using the "relevant quantities" of (1) the speed of light (2) the charge of the electron (3) hbar (4) the mass of the electron, there is only one quantity that can be formed that has the units of a magnetic moment

(e*hbar/2mc)

so ANY MODEL, no matter how silly, which predicts the magnetic moment of a particle, whill produce a result that is proportional to that value, and you might be tricked into thinking that model was accurate

True, though I'd argue that cuts both ways. ;)

the real challenge is to get the right value of the constant out front, in other words mu=some sonstant* e hbar/(2mc).

I remember there was some super short derivation of maxwell Boltzmann distribution which basically uses isotropy and something else to claim only a gaussian could be the distribution.

Anyone has any idea?

i suspect your model doesn't get that value right at all for the neutron or the proton

and if it gets it right for the electron (constatnt=2), that's just a coincidence

99.7% experimental for the proton, 99.96% for the neutron.

and I bet it doesnt get the leading order correction from QED, which is 2+alpha/2pi

alright send me your document [email protected]

i g2g ill respond to ur email if you send it

It's sent. Cheers.

I don't remember 3

But is there a nice way to justify it:

?

@imbAF Why would it be wrong? For a charged particle, the probability density is essentially the charge density, so the probability current is the current.

@DIRAC1930 I'm not in here 100% of the time :P What exactly do you mean by "dressing the particle", and what's a "quantum number"? For instance, mass and charge both get renormalized, so if those are "quantum numbers" for you, no, they don't stay "the same", even though we again run into the general issue with renormalization that the bare and dressed quantities don't make sense at the same time

what is this method of solving differential equation called

thes are the steps while solving shrodinger equation for hydrogen atom

Asymptotics?

method involved here?

@PrateekMourya I think it's asymptotics. There's some wonderful lectures of Carl Bender on this subject as well

@DanielUnderwood hey!

@PrateekMourya this is the so called Power series method. First they isolate the asymptotic behaviour though

Hey @Mr.Feynman

Hey

hows things going?

I'm propagating myself home rn

Where I'll read the latest chapter of CSM, have dinner and collapse on the bed

@Mr.Feynman CSM?

Chainsaw Man

ah epic manga

I use that acronym a lot and I tend to assume everyone knows it

I'm a Tog fan

Like when I talk to non-physics people and write "QM", "QFT" and so on

Also does this question (post edit) make sense?

@MoreAnonymous I mean, it's pretty ironic that I'm asking but... What is it?

@Mr.Feynman Tower of God XD

a webtoon

ew, webtoons

Just kidding, I'm not into those

@Mr.Feynman I was gonna rant XD

I miss my university days

alright i gtg

cya!

Can any1 help me with a simple multivar volume problem

@ACuriousMind Does the momentum quantum number (and others) of a free particle stay the same upon switching on the interaction?

That GR answer I did is doing numbers

nice

unfortunately delete votes on answers are at 20k

although you can now protect questions, hurray

So maybe we aren't interested in the asymptotic states being free but the fact that the 'adiabatically switched off interaction' state has the same quantum numbers

@DIRAC1930 I'm not sure what you mean by "switching on the interaction"

which of the constructions here do you think depends on the assumption that the interaction is switchable

The Haag-Ruelle construction of asymptotic states depends on no such assumptions, and Epstein-Glaser uses switching functions in intermediate steps but in the end wants to take the limit $g\to 1$, i.e. turn the interaction always on

thinking about QFT interactions as being "turned on/off" is just a handwave you have to do if you don't want to construct the asymptotic states carefully (and also might mislead you to think the asymptotic states have bare mass)

Given that I would like the ability to one day understand quantum theory in full mathematical rigor, what topics out of Brian C. Hall's Lie Groups, Lie Algebras, and Representations should I prioritize?