Well besides applying for internships this summer I guess I can get ahead on physics/math stuff

besides reviewing old stuff I wanna start a QM book and maybe topology/diff geo

I think I'll finally pick up zee lol

@Obliv think I did decent but it still really sucked

i have an arthritis-like condition, so midway through the latter test my hand was in pretty bad pain from all the writing

just had to power through it for several hours since there were too many questions to rest

@Obliv which book?

Is position an observable in classical mechanics?

@Sanjana i think people associate russian authors with terseness in the context of mathematics/physics. although, in say a literature context it might refer to books like dostoevsky's or tolstoy's in which there are 100+ characters and a rich world of social dynamics and etc.

@SillyGoose yes, for some vaguely defined meaning of what it means for something to be observabke

@SillyGoose Oh, if we are trading in the juicy insults that the Europeans throw at each other, then miao miao can supply that the Russians are stereotyped to be very capable of feats in maths, but are not at all rigorous. Won't be proving everything, and thus can easily accept falsehoods if presented as intuitive.

Pulls up for a drink at the h bar. I think I need to stop asking questions for a minute and read up. Thanks for all the questions answers comments and references on my Torsion tour.

@ACuriousMind ah thats interesting i didnt know that. okay then these are developing cats XD

@Mr.Feynman when i was younger and learned that the human body has a cavity i was extremely concerned. its just very disturbing to think that the inside of the body is actually the outside

wait im not sure if cavity is the right word

what is something you guys have learned recently that shook up the way you look at or think about physics

what im trying to convey is like how the throat is actually the outside of the body not the inside XD

@SillyGoose RR alert instructs me to disengage immediately

@Relativisticcucumber what is RR alert :0

oh xD hey i think it's an interesting question

it is a little dull, structurally, to think of every physical theory as just studying a mathematical structure representing state space, a differential equation over it representing a dynamics relation, and some other miscellaneous features...

lol RR alert is very important

@naturallyInconsistent not sure yet, maybe townsend or something for beginners

@SirCumference dang, sounds difficult. What was the exam on?

@Obliv Have you considered Griffiths? Very beginner friendly

@SillyGoose Recently, a lot. I've known about these topics for quite some time but actually studying it made it surface. The consequences of maxwell-heaviside equations vacuum solution, (I didn't study this, but GR concepts), stat mech being very useful despite fundamentally relying on a priori probability distributions. trying to understand entropy

also magnetism can be thought of as coulomb's law with SR

@naturallyInconsistent I will look into that

thank you

the degeneracy condition for fermion gases is pretty interesting too, like the fermi energy and temperature conditions to basically make pauli exclusion principle change how the system behaves so you can't use boltzmann distribution anymore

thermal physics in general is super interesting. even though i'm done with the class I'm going to keep learning

but i am curious why they say that the metal would liquefy and evaporate

I thought the fermi temperature would be extremely low? Or is it because the density decreases so the boiling point decreases?

The Fermi temperature is the temperature at which kT is about the same as the Fermi energy i.e. it's the temperature at which all electrons in the band are excited out of the ground state and the distribution switches to more like a Boltzmann distribution.

This is a huge temperature. Room temp is about ¹⁄₄₀ of an eV, so for a Fermi energy of 2-3 eV (which is typical) the Fermi temperature would be a hundred times larger than toom temp or about 30000K.

@Obliv just calculate that temperature and compare with the boiling point of the metals that you can simply measure in experiments. It is difficult to boil metals, but it can be done.

@Obliv i would not recommend either of townsends two books for a first look at quantum mechanics. it lacks any ounce of motivation for anything introduced. moreover, townsend essentially is a copy of sakurai's first several chapters, so if you work through townsend you might as well just read sakurai

@Obliv i also hope to learn some stat mech more deeply :D my graduate work i think will be in quantum stat mech...so we shall see hehe

on QM textbooks: I agree with NI that Griffiths is okay as a first look. Mainly, the first two chapters which introduce (1) quantum mechanics in the language of wave functions in position/momentum space (2) solutions to Schrödinger equation with canonical Hamiltonians (Harmonic oscillator, particle in a box). But beyond this content, Griffiths seems to do a poor job. In particular, his discussion of angular momentum is not the best.

One other textbook people have recommended is Ballentine's text on quantum mechanics. It seems pretty nice.

hold up...ballentine's text begins by defining states as density operators :0. now it has even more bonus points in my opinion...

okay Ballentine's text has won my heart.

@SillyGoose yay! I also recommend Ballentine over Sakurai

it really seems to be a wonderful presentation of QM...I will try to read through it this summer :P.

@SillyGoose The part that sold for meow is that he covered the NR symmetries, which would map cleanly to the SR version that we use in QFT. So we get to see a simpler, yet functionally equivalent, introduction

yes! very nice to see Wigner's theorem and then explicit and concrete discussion of galilean relativity

i am reading through ch 9 right now...interested to understand his take on how to interpret a "state"

i can't even tell what interpretation of quantum mechanics I am thinking in 😂

That is ok; the problem is when every tom dick and harry has their own peculiar version of copenhagen, and then they exclaim loudly that everybody else is wrong and that everything is so simple. They cannot even agree with each other what copenhagen should mean, and then we are supposed to stop arguing about interpretations because they have sorted it out

I don't understand. Isn't the text above literally true? As a mathematical object, a state (vector or operator) is a (proto) probability distribution.

I say proto because you need to do another step of maths for a state vector to obtain an object that is like a probability distribution

The proto part isnt the problem. The problem is that, if you change the type of measurement that you are doing, the new detection's probability distribution may include interference terms, and those are underdetermined. Any single measurement, which has just one single probability distribution, necessarily will suffer from such underdeterminism

what do you mean by underdetermined?

You might also have to navigate the different definitions; those who only consider pure states will have a different definition from those that cover mixed states.

@SillyGoose I specifically said interference terms right before that...

im still not sure what underdetermined means in that context

@SillyGoose Consider that you have prepared a Stern-Gerlach system that gives you probability $I_\uparrow$ that goes through spin up, and $I_\downarrow$ spin down, then if you rotate the detector to measure spin +x, say, you can get $I_\uparrow+I_\downarrow+2\sqrt{I_\uparrow I_\downarrow}\cos\vartheta$ for some interference phase $\vartheta$

(you are much more likely to have seen it using $|\alpha|^2$ and $|\beta|^2$ and so forth. The point is that just knowing the $I$ distribution is not sufficient. You might need to augment with $\vartheta$ when you change your detection.)

@SillyGoose Oh...is it?

@Sanjana did ya see meow's addition right after that?

@naturallyInconsistent I did rn...

:P

btw...one of my former high school students was asked a question in a grad school interview where he had to qualitatively plot the ground state wavefunction for a particle in a box where the particle was actually not free in a box but also had a Gaussian potential added to it (positive strength).

He could plot the two peaks towards the end and the point where the wavefunction became zero and also the wiggly region in between.

Apparently the interviewer asked for the number of peaks in the wiggly region, qualitatively!

Is it possible to give the number of peaks in the intermediate region?

Does anybody know where I can find a full fledged perturbation theory description of Raman scattering?

All the resources I am getting are more on the spectroscopy side: not the theory

@NairitSahoo Bransden and Joachain books on atomic and molecular physics has a brief discussion, but yeah it is not "full fledged".

@Sanjana Sorry, what is the shape of the potential again? This looks quite crazy.

It will be extremely difficult to get a "full-fledged" description. Raman is quite difficult to theoretically model!

Any quick way to see that $$A(n+1)=\frac{(n+2)A(n)-(2n+1)A(1)}{n-1}$$ admits a closed form $A(n)=cn^3+dn$ for $c,d$ constants to be determined?

I tried plugging the recurrence relation into itself iteratively but the second term is a bit nasty to deal with

By the way, this is the recurrence relation arising in the quantum anomaly of the Virasoro algebra

@naturallyInconsistent Gaussian potential but the space is limited to $[-1,1]$ say i.e. hard walls at $x=\pm 1$

@Mr.Feynman I don't recall having the $A(1)$ term. Can't it be set to zero somehow?

I read this part from Blumenhagen's CFT book which is pretty detailed in most calculations. Pg. 16

i think i am having a misunderstanding about the pauli matrices. so i have seen that the spin operator is proportional to the pauli matrices for x,y, and z. but what basis are these in (the pauli and thus sigma matrices)? if i write a "spin state" $ \vert + \rangle = \vert 0 \rangle + \vert 1 \rangle$ (ignoring normalization), then this is specifically in a basis, say the $z$ basis, so how can i act on it with $S_{x,y,z}$ / $\sigma_{x,y,z}$?

@Mr.Feynman If you set the 2nd term to zero, you do get the central extension term: $n^3-n$, and the normalization is fixed by requiring $c=1$ for free boson etc

@Relativisticcucumber I think the basis is the one in which $\sigma_z$ becomes diagonal. So if you want to act the spin operators in terms of Pauli matrices on a generic state, you have to express it in terms of eigenvectors of $\sigma_z$

@ACuriousMind Raman please

what?

Raman scattering in QFT language, do you know?

Books

In what world is pinging me with "Raman please" a remotely acceptable way to start a conversation?

Oh sorry. I thought you read all text when you come

You are the caretaker of this room

I literally got online like a minute ago

You very fast I thinked

@NairitSahoo even if these two statements are true (they are not), you are being extremely rude.

@Sanjana so if i have, say, $\sigma_x \vert + \rangle$ with $\vert + \rangle$ defined as above, then what i get is truly a value that can be interpreted as spin in the $x$ direction? i dont need to change the basis of everything?

@Relativisticcucumber What do you mean by "the spin operator is proportional to the Pauli matrices"? There's no "the spin operator", you have a family of spin operators: the "spin in direction of the unit vector $\hat{n}$" is $\hat{n}\cdot \vec \sigma = \sum_i n_i \sigma_i$, and then e.g. choosing $\hat{n}$ as the z-axis you get back that $\sigma_z$ is the spin in that direction

@Sanjana do we even have a solution to the problem when the potential is Gaußian and not restricted to a box? But this is a red Herring: If we have a suitable ground state, then it will have no nodes. If the ground state energy is smaller than the peak of the Gaußian, then it will be a symmetric state with two peaks, one on each side of the Gaußian. If the Gaußian is a tiny perturbation, then just one peak in the middle.

of course everything here is basis dependent in the sense that if you choose a different z-axis then you need to transform everything that you've expressed in terms of the components $n_i$ or $\sigma_i$, but that's how it is with vectors

@Relativisticcucumber What ACM said + If you study the Weyl and Wigner ways to classify spin states, you will realise that, to even begin, you have to align the x y z axes of every point in spacetime before you start. So then it makes sense that all the Pauli matrices are fixed and pre-determined.

@ACuriousMind oh i had seen that for instance $S_x = \frac{\hbar}{2}\sigma_x$ i think

but hm i am still confused. so this would tell me that if i choose $\hat{n}$ as the $x$ axis, then i get $\sigma_x$ i assume. but then should i rewrite $\vert + \rangle$ in the $x$ basis so that when i do $\sigma_x \vert + \rangle$ the result i get is sensible?

@ACuriousMind you wouldn't say that if you were a barista and someone popped up saying "coffee please"

By the way, my first read was "Ramen please", which made more sense :P

i had in mind someone at a bar ordering a drink called "raman"

@Relativisticcucumber What do you mean by "i do $\sigma_x\lvert +\rangle$"?

measuring an operator is not applying it to the state

@Relativisticcucumber also this equality really only makes sense if you have an independent definition of what $S_x$ is supposed to be, otherwise this is just a definition

@ACuriousMind oh no but that is how i was taught to determine the $S_{x,y,x}$ matrix forms D: to see when i act on the state if i get the eigenvalues i want

unless i am remembering very wrong which is also possible

I think I'm confused about what the setup and the goal here is

what do you have given, and what question do you want to answer?

i am also confused BLEH. so i have the state $\vert + \rangle$ and would like to calculate the expectation values of $\sigma_{x,y,z}$. i did so for $\sigma_z$ but then i was unsure about $\sigma_{x,y}$ since the state is written in the $z$ basis

what's the definition of the state $\lvert +\rangle$?

$\vert 0 \rangle + \vert 1 \rangle$ with these being down and up in the $z$ direction

and normalization of $\frac{1}{\sqrt{2}}$

so what do you know about $\sigma_y$ and $\sigma_z$?

@Relativisticcucumber I just told you that you have to first align the axes everywhere in spacetime. Part of this means that the standard $\sigma_x$ and $\sigma_y$ operators are already being represented in the $z$ basis. You can just use the standard representations and be done with it.

if you knew their matrix forms in the z-spin basis this would be trivial, but I don't quite understand what information you want to deduce the values here from if you don't know them

I concur with ACM: It seems like you are either given a very abstract introduction to QM, or you are yourself imposing upon yourself a treatment of QM that is way too abstract and unhinged from how we would be doing physics, both experimentally and theoretically. If you do not have an understandable introduction to QM, it would be no wonder that you get confused by such things.

@Mr.Feynman When KPop meets metal, you get Ramenstein. :)

lol

@PM2Ring I don't know what that thing is but it's the second time in a week I read that word

I mean, the original word without pun

@ACuriousMind i have their matrix form also but this is where i get lost because i was told they are written in the "standard basis that does not have a name" so i dont really know what this means

well that's just unnecessarily confusing :P

@SillyGoose GR

if your matrices look like the ones on Wiki they're in the $\lvert 0\rangle,\lvert 1\rangle$ basis

also note that the Pauli matrices are a basis for su(2), so one needs to know all three to be able to construct an arbitrary agebra element as a linear combination

@Relativisticcucumber That is why I had to emphasise that your x y z axes are already aligned. Then you can see why it is that the Pauli matrices are the same everywhere, all expressed in the z basis

Why are generalized velocities assumed to be constant during a virtual displacement?(See hand and finch analytical mechanics where he defines virtual displacements) I'm asking this question since Goldstein nowhere uses this assumption in his derivation of Lagrange equations in his 1st chapter

By generalized velocities I mean $\dot{q_j}$

the usual derivation of Euler Lagrange equation is using the principle of least action

in this derivation, we change by a small path $\epsilon (t)$ and then set the functional derivative to zero

@RyderRude The "usual derivation" of the principle of least action is in turn derived from d'Alembert's principle of virtual displacements and Newton's laws (cf. e.g. physics.stackexchange.com/a/131392/50583), and that's clearly what Arjun is asking about. If you don't know what a question is about, you don't need to pretend to.

yes. im aware of this

i thought Goldstein might have directly assumed the principle of least action, like most texts do

Goldstein, chapter 1.4, title: "D'Alembert's Principle and Lagrange's Equations"

oh

this is why i wrote "usual derivation", in case they were using some other derivation

@Arjun The whole derivation in Goldstein makes no sense if the virtual displacements would also change the $\dot{q}_j$, for instance the expansion of $\delta r$ in terms of $\delta q$ and the appearance of $\dot{p}$ in d'Alembert's principle, etc. It's not a single step that doesn't work, but the whole idea falls apart if you try to talk about changing velocities, too (also the name "displacement" wouldn't fit anymore in such a case)

note that while Goldstein doesn't spell out the "velocities are constant" part explicitly, it's there in that he starts with a virtual displacement as being a change of the coordinates $r_i$ - he doesn't talk about changing velocities, so they don't change

@ACuriousMind GAH this makes sense.

@naturallyInconsistent i was confused because i thought i was told the pauli matrices are NOT in the $z$ basis but if they are then i see BAH

@Relativisticcucumber Consider the $\sigma_x$ Pauli matrix. If it were in the $x$ basis, then $\sigma_x=\begin{pmatrix}+1&\\&-1\end {pmatrix}$ which is what you would think $\sigma_z$ would be. Needless to say, it is extremely confusing.

i should have known

i was just confused because when i tried to look this up i couldnt find a solid answer which made me thing there was smth more to this also

that is why I made it absolutely clear that you have to first align the x y z axes. This implies an implicit agreement to measure everything in the z basis

@naturallyInconsistent bah i see now

@Mr.Feynman I can make one up. However, it is quite convoluted. I mean, there is a step in the middle that is like, if you didn't tell meow that it is of this form, miao miao would not have considered trying to search for it. Alas, miao miao is going to rush to gym in a moment and will thus be gone for hours.

@Relativisticcucumber Ah, maybe I should have explained why it has to be of this form. Remember that the spin half operator is proportional to the Pauli matrices. Now, you know that the only result of measuring spin half is that you get $\pm\frac\hslash2$ as the result. You also know that these are two-dimensional, i.e. two non-degenerate states. Thus, in its own basis, it must be a diagonal matrix of that form.

@ACuriousMind If $\dot{q}_j$ changes by an infinitesimal amount as q changes by $\delta q$ i fail to see why $\delta r$ or $\dot\dot{r}$ would be affected since r is only a function of q_j and not of $\dot{q}_j$

blebs consider a two level system, maybe an atom and 2 spin 1/2 items like electrons. i have seen that a triplet state is a state with two unpaired electrons. i have also seen that the three triplet states can be written as $\vert s = 1, s^z = 1 \rangle, \vert s = 1, s^z = 0 \rangle, \vert s = 1, s^z = -1 \rangle$. my interpretation is that, in the first case, we must have both electrons [...]

[...] be spin up, and in the third case, both are spin down. however, to satisfy the middle case of $s^z = 0$, we must have one up and one down. in this case, they can fill the same level, and we do not have a triplet state? so how is the middle case a triplet state?

@Arjun I'm not sure what you mean

@Relativisticcucumber and I'm not sure what you mean either :P

@ACuriousMind well do you agree with the three states i wrote being triplet states? or am i wrong on that

The two electron have a four-dimensional space of states with a basis of state of definite spin for the individual electrons $\lvert \uparrow \uparrow\rangle, \lvert \uparrow\downarrow\rangle, \lvert \downarrow\uparrow\rangle,\lvert \downarrow\downarrow\rangle$. The "middle" state of the triplet is $\lvert 1,0\rangle = \lvert \uparrow\downarrow\rangle + \lvert \downarrow\uparrow\rangle$. What is the question?

@ACuriousMind how is this a triplet? i thought a triplet state is a state with 2 unpaired electrons?

I don't really know what you mean by "unpaired"

@ACuriousMind I'm trying to understand why an infinitesimal change in velocity would affect the $\delta r$ expression in any way..and also it is not clear for me why a change in velocity during the displacement would lead to the crumbling of Goldstein's derivation

the 4d space of states for the two spin-1/2 objects splits into a 3d triplet and a 1d singlet subspace

@ACuriousMind im referring to the intro/image on wiki here en.wikipedia.org/wiki/Triplet_state

I have no idea what that picture is trying to depict :P

so how should i understand the difference between singlet and triplets?

the definition of a triplet state isn't "2 unpaired electrons", it's what the intro sentence of the article says: It's any of the $s=1$ states where you have three possible values for $s_z$, hence "triplet"

compare that to the $s=0$ state where you only have a single choice $s_z=0$, hence "singlet"

@ACuriousMind Should'nt the expression for $\delta r$ remain the same since r is a function of q_j 's and not the velocities..so a change in velocity shouldn't really affect $\delta r$?

OH

so their uniting characteristic is not a physical one

@Arjun but how do you "change velocity" without changing $r$? The velocity is the derivative of $r$! Again, it's not that any single step here wouldn't work, it's that the whole setup doesn't really allow you to talk about independent changes in velocity at this stage

at some point we will be allowed to treat $q$ and $\dot{q}$ as independent in some sense, but this is before that (and also many texts don't explain that part very well)

@ACuriousMind i always wondered ab this

@Relativisticcucumber I don't know what this means, either :P

@ACuriousMind i never got a satisfactory answer :,(

@ACuriousMind its ok i think i get it

triplets having angular momentum 1 and singlets having angular momentum 0 sounds pretty physical to m

@ACuriousMind i was thinking too much about the name triplet :P

but it was right in front of me

LIKE A CHILD

if there are triplets we call each a triplet

wow

damn that word was probably chosen bc it would be obvious

@Relativisticcucumber see the last paragraph of this answer of mine and also physics.stackexchange.com/q/428990/50583

i could be ashamed or amused and i choose amused

@ACuriousMind NO

i clicked on the answer

and i have upvoted it already

what is happening

lol

@ACuriousMind wouldn't the holonomic constraint on the system guarantee that $r_i =r_i (q_1,q_2,...t)$ so any $\delta r$ will only have $\delta q$ dependance and this expression would hold good even if $\dot{q}$ is not zero? Also i fail to see how $\dot{q}$ not being zero would affect any step in his derivation

Who's talking about $\dot{q}$ being zero?

what's zero is any potential independent $\delta \dot{q}$

I'm sorry..I meant its change not being zero

in terms of the principle of least action this is just because we're varying a path and not positions/coordinates independently

I'm not really sure what your problem here is - you do agree that this derivation of the principle of least action works, right?

what's the point in considering some strange more general form of displacement of velocities - we already got the principle we wanted to get

@ACuriousMind my question is if the above derivation of Lagrange equations hold even if $\delta \dot{q}$ is not zero during a virtual displacement

I have not yet seen the derivation of the principle of least action..

but, uh, that's the following chapter in Goldstein as d'Alembert's principle?

did you start asking questions before actually finishing reading the derivation?

Nope in his third edition it's in the next chapter i think

yeah, so the whole point of this derivation here is just that Lagrange's equations will result in the principle of least action/Hamilton's principle etc.

@ACuriousMind No I've seen the derivation he gave for Lagrange equations using the principle of virtual work as given in his third edition..

after that you will never hear from d'Alembert or his displacements again :P

so it's not really worthwhile to obsess over the nature of virtual displacements all that much, this concept literally never appears again after we've gotten to the action principle

yes. action principle is the default starting point

it works for field theories too

no D' Alembert or Newton's laws for field theories

i cant focus on studying anything... it's like my head is jammed

@naturallyInconsistent sometimes formal power series can be used but you get weird stuff here D:

@Relativisticcucumber What ACM is trying to say is that the uniting characteristic of spin states is necessarily physical. Your complaint that it is not "physical" is just misplaced. You certainly can claim that it is un-something, but that something is not "physical".

@Mr.Feynman Give meow more time.

@Mr.Feynman I think the clearest derivation of the form of the Virasoro central charge is chapter 5.3 [pdf link] of Schottenloher

Thanks, Imma read it :)

Looks like they use the condition $A(1)=0$ as @Sanjana was saying

Ok, I think GSW do that too when they write that $$\langle 0;0\lvert [L_1, L_{-1}]\rvert 0;0\rangle=0$$

Just that they write it after closing that recurrence relation but I guess it's just a disordered presentation in the way physicists like :P

Hmm, maybe it is only strongly suggesting that $A_n=-A_{-n}$ rather than directly showing, but I'm too tired and in a rush to disentangle this right now.

After all, this working is definitely already sufficiently motivating that this would be a fruitful method.

do y'all prefer eating more expensive food or it doesnt matter?

Ehem, it is nonsense that $A_n\propto n$ doesnt work. It has to work, since we can pick $c=0$. But you get the idea.

@Obliv first one was on QM, second was on cosmology

@naturallyInconsistent oh, that thing with parity was a good trick

@Mr.Feynman this is the most straightforward derivation of the Virasoro algebra I can find

@Mr.Feynman If you can see that $A(-n) = - A(n)$, then you know it has to be odd, so you are guessing odd polynomial solutions of a recurrence relation, which is a natural possibility. The easiest odd possibility is $A(n) = dn$ is too easy, so you try an odd cubic $A(n) = cn^3 + dn$

@bolbteppa yeah, I didn't think about it before but it's fairly obvious that if it is a polynomial it has to be odd

Why would we exclude other possible behaviours though?

Like, sometimes recurrence relations give rise to exponentials (it can't be the case here due to parity but it's just an example)

The first derivation I linked to just avoids this, all it amounts to is writing out the normal ordering explicitly in one of the $L_n$'s in the commutator, and realizing one of the results when you work it out is not normal ordered, so you just put it back in normal ordered form, which results in the central term

@Relativisticcucumber did you say bundles ;)

@Relativisticcucumber the pauli operators $\sigma_i: \mathcal{H} \to \mathcal{H}$ where $\dim \mathcal{H} = 2$ are basis independent. the pauli matrices $[\sigma_i]$ are indeed written in the so-called $\sigma_z$ basis in which $[\sigma_z] = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, which is written using eigenstates of $\sigma_z$ as basis vectors for Hilbert space $B = \{ \lvert + z \rangle, \lvert -z \rangle \}$.

if you would like to keep things straight in general, an operator $O : \mathcal{H} \to \mathcal{H}$ where $\dim \mathcal{H} = d$ is basis independent. Its matrix representation $[O]: \mathbb{C}^d \to \mathbb{C}^d$ is, well a matrix.

@Sanjana maybe if you knew the green's function for the particle in a box equation :P

is particle physics really stalling in terms of progress?

@Mr.Feynman the eigenfunction of a differential operator is an exponential. The discrete analogue thereof is a power. However, the resulting polynomial tends to grow unwieldy. That is, remember that the RHS is a constant multiple of (2n+1), i.e. it is at most affine, and this is a tremendously serious constraint. It is unlikely for higher order polynomials than quadratics and cubics to be able to zero the unwanted terms; If you try a quartic or quintic, you ought to find that there is no solution.

some of the better teachers in the maths department would cover differential equations, and (discrete) difference equations, with techniques emphasising these relationships, thereby giving a general overview of how people would go about guessing solutions.

@SillyGoose probably depends on how you define progress

in terms of further validating what we already thought to be true, it's been very productive

also, I am quite confused what one does in a course of general relativity if one can hardly ever solve Einstein's field equations

there i couldn't tell you---i never took a GR course myself

i would guess a substantial amount is just learning the math required to -state- the field equations

and maybe some of the solvable cases

@Semiclassical this is correct

but then shouldn't physicists just take differential geometry XD

i wonder what a GR course looks like having assumed the diffe g pre-requisites

at that point you're probably looking at more of a "GR for mathematicians" course

@Semiclassical hm i see

It takes a full lecture timeslot (cannot remember if it is 1h or 2h) to start from EFE and the Schwarzchild ansatz, and derive just the first non-zero term (and thus the full solution) of the Schwarzchild black hole.

There is no way to make it faster.

It's pretty rare in university physics to need differential geometry outside of GR

not such an elegant theory after all :P

So commonly the differential geometry course is the GR course

oh i see

is GR usually a "shorter" course? at my consortium GR is a half-semester course lol

I had to warn my students in the previous lecture that I would have no extra time to entertain questions. That the EFE is where I start, and then it is just straight up deriving the Christoffel symbols, the Riemann curvature tensor, the Ricci tensor, etc.

My GR course was appaling rly

i was like what are you going to do in half a semester XD but maybe diffe G is a pre-requisite

By "GR course" I really mean cosmology course

It is much easier to teach the topic via tensors than via diff geo

Focused on FRW

didn't even do the EFE

what is FRW?

The madman known as Hartle introduced GR by starting from line elements. It is ridiculous, and none of the students understood anything.

@SillyGoose Cosmology metric

@SillyGoose Google it. It is the most basic thing in cosmology

oh

how do i get an idea of what people mean by saying that general relativity is supported experimentally

look up the experiments I guess?

pbs.twimg.com/media/GNJXqnIWIAAY13q?format=jpg

True when applied to theoretical physics

(Why doesn't it show the picture :\ )

@naturallyInconsistent mhh I know what you mean but I mean like this

However, I think it might just happen that doing the polynomial method you end up adding terms indefinitely, so it builds up the power series

Does anyone know of applications of algebraic geometry to quantum mechanics? Or QFT? Not string theory/quantum gravity.

@naturallyInconsistent Ok sorry for rudeness. He is not caretaker so

@SillyGoose Algebra and geometry are different. They are both used in quantum mechanics. QFT is different from quantum mechanics. String theory is false

en.wikipedia.org/wiki/…. @NairitSahoo

4 messages deleted

Ah, ACM you baddie. After 50 years of research on the topic we finally have the opportunity to know why ST is false and you dare accuse this user of trolling? :P

they can come back in 24 hours if they are interested in having an actual conversation :P

never would i have thought to see someone to call ACM a baddie XD

Please, tell me there nothing weird with it

I'm not very skilled with double meanings in English

i think in american vernacular a relatively new meaning for baddie has emerged

oh no

"A Baddie is a bad girl who is always on fleek. slaying the game ."

I'm sorry, ACM. I see you as an AI friend

@ACuriousMind lol...What was he even saying!

I have never seen such things happening in here. Is this automatic or the mod has to intervene in such situations?

@Sanjana you may argue that ACM himself is automatic :P

that was all manual; the only automatic moderation mechanism on chat is that if a spam/offensive flag is validated by reviewers, the flagged user gets suspended for 30 minutes

All his messages were weird after some time. I wonder what else he said in those deleted texts :p

what is the distinction between aether in old physical theories and our conception of spacetime now? isn't spacetime just the medium through which things propagate and etc. now?

The aether is more than just space

a place where things can be

It also has mechanical properties

the old aether notions usually define a notion of absolute rest (i.e being at rest w.r.t. the aether)

spacetime alone gives you no such notion

Also see this

Aether is the thing that propagates in aether theories

it's just a magic fluid

@naturallyInconsistent It appears that the ground state energy is smaller than the peak of the Gaussian...but no even that gaussian potential on $\mathbb{R}$ is not given. The student told about the nodes also though

@Sanjana Then there must be two peaks and no more. No nodes.

Oh oh...No I think, I was asking something else. There will be some wiggles in between..I was asking how many "wiggles" we have?

But you cannot have wiggles in the ground state.

Those increase the energy

I mean, won't there be wiggles which don't cut the $x$ axis?

There can only be one, the one that connects from one peak to the other. There cannot be more.

With some nice placement, we can make it appear obvious.

can you see it?

imma head to sneeppuuu. if ya want meow to type up why it has to be so, say so, and miao miao will come back to it tmr

is there a precise definition of a subsystem or even of a system in quantum mechanics?

If your Hilbert space splits into tensor products of Hilbert spaces

assigning each tensor factor to be a subsystem doesn't seem to be a nice definition because a change of basis on the composite system will in general mix up states in one tensor factor with states in the other tensor factors

Those are entangled states

You have an injection from each of the subsystems into the total Hilbert space

are theories of physics self-referential?

it seems that textbook quantum mechanics is. but it does not seem like something like textbook classical mechanics is.

textbook quantum mechanics is in the sense that its formalism seems to require the notion of a measurement by a measurement device. but this measurement device presumably is a quantum mechanical system itself. so quantum mechanics presumes quantum mechanics.

where's the difference to classical mechanics being measured by classical mechanical systems?

i wasn't sure if textbook classical mechanics references measurement devices at all to set up the physical theory's formalism

well, it doesn't need to so explicitly because classical measurements don't change the state

so you can just have your state and compute all the observables on it and you know that's what a perfect measurement device would measure without perturbing the state

what i mean is: it seems textbook classical mechanics is set up without reference to making measurements. in this case then, measurement devices are classical mechanical systems, but the set up of classical mechanics does not depend on the existence of measurement devices. so there is no self-reference

in the end, all physics is about measurement!

it's just that ideal classical measurements are so straightforward we don't really need to talk about them

but if you've ever had to do an actual real measurement you'd know classical mechanics very much needs to talk about measurement devices

the simplest instance is probably how real voltage or current measurements are via resistors that themselves have an effect on the voltage/current

i do not claim that discussions on how to construct a proper measurement device need not be had. or that talk in general of measurement devices need not be had. but more like if i axiomized classical mechanics in the sense that I set up a precise mathematical framework for it then set up a dictionary that isomorphizes the mathematics into the real world, then would anywhere in this set up be a dependency on the notion of a measurement device.

I don't know what "isomorphizes the mathematics into the real world" means

the relation of physical mathematics to the real world isn't "isomorphisms", it's measurements :P

an assignment to each mathematical object what it is meant to model in the world (if the object is meant to model anything)

I reject the idea that physics works by assigning "mathematical objects" (whatever exactly those are) to real world objects

this reeks too much of platonism for me :P

well i am making no claims of "how the real world really is" just claims about what a physical theory does

in particular, thinking of a physical theory as a mathematical model of the real world.

@SillyGoose no, when you are saying that there should be "maps" from mathematics to the real world you are very much making a claim about "the real world", namely that its nature can be captured by talking about "maps into it"

@SillyGoose the point of contention is precisely what it means for a mathematical model to be "of the real world"

you seem to assume that this means somehow mapping mathematical structure to "real world structure"; I would stress relating the input and output of the model to the results of measurements!

i guess i mean to use "real world objects" somewhat loosely since i am not sure what i would actually define such objects to be

hm but i think i see what you are saying

a prerequisite for talking about the relationship between mathematics and reality is a notion of what reality is ;)

i don't know if i'll ever have one of those :P

i am trying to see a or several fundamental structural differences in classical versus quantum mechanics. not at the detailed level of particular structures, but like one level higher. so for instance, both theories have a state space and a dynamic relation, so at this level of consideration both theories are the same.

I mean then looking at measurements is the correct thing to do - the difference is not that classical mechanics somehow doesn't have measurements, but that in classical mechanics there is a unique assignment of definitive measurement outcomes to each state, while in QM you only get probabilities

the whole reason people get so upset about Bell's theorem is because it shows that this difference is really fundamental :P

i was also curious if there is some paradoxical logic that would arise from either theory being self-referential through using measurements to set up their formalisms

also what exactly was the objection to quantum mechanics being non-local (local in the sense as used in the context of stuff like the EPR paradox)? I was trying to look into the discussion surrounding EPR and entanglement in general recently, but it is sometimes difficult to judge if what someone is saying is nonsense or actually true

I don't know that "self-referential" is the word you're looking for; it sounds to me more like you're fishing for a version of Kuhn's incommensurability argument that the experiments by which we claim to "test" physical theories are already interpreted through the lens of the dominant physical theory (i.e. you can't build a detector at CERN without assuming large parts of particle physics are already correct)

and hence it is difficult to imagine interpreting experiments testing truly radically different theories until those theories have gained social traction

@SillyGoose People don't like non-local theories because it means the "communication" between the non-local parts violates our usual notions of causality, particularly since relativity put a speed limit on how "things can affect each other". If you do the math (with results like no-communication etc.) you find that there's no technical problem with this, but that doesn't make the interpretational discomfort people have with non-locality go away

the catch is that the non-realism you have to alternatively embrace is also uncomfortable for many, and that's why we got to have 100 years of debates about quantum interpretation :P

@ACuriousMind hm i guess i see this as strange. i would have thought that the discomfort is that there could be a technical problem (i.e. faster-than-light communication). but given that there are no technical problems it is interesting to imagine people still feeling uncomfortable with it.

well i suppose that is what physics gets for rewarding intuition :P

@ACuriousMind i was more thinking of why quantum mechanics has all these interpretational problems that classical mechanics does not seem to have

or it would also be interesting if classical mechanics has the same interpretational problems but you can just give reasonable answers to all the problems posed

well yeah about the non-locality stuff. I don't understand why someone would blindly think a physical theory is incorrect solely based on being uncomfortable with a concept in the physical theory which itself does not actually lead to any technical problems. the reasonable course of action would rather seem to be that one is misunderstanding their own definition of "things affecting other things". or using an inappropriate definition of such a phenomena.

but this makes me think that i misunderstand the objection as made by people like Einstein :P

well, I say "uncomfortable" in a bit of a pejorative way; the more neutral way to say this would be that people already have a pre-existing philosophical commitment to things like an idea of causation, and the problem is that they cannot square Bell's theorem with those other beliefs about reality

Einstein famously said "God does not roll dice" - this isn't about a technical problem with the probabilities of quantum mechanics but with those probabilities not being compatible with the rest of his belief system (don't fixate on the word "god" there, Einstein wasn't a theist, belief here has no necessarily religious connotation)

i wonder where how i think about causation came from :P

the problems of language already found in physics multiplies a billion in passing to interpretations of the physics...

unrelated but lol

@nickbros123 have you ever looked at maxwell's original treatise on E&M? perhaps you'd enjoy it