2:20 AM
honk
2:56 AM
$\{123\}$

4 hours later…
6:30 AM
I am confused...the answers for this question seem to conflate hermitian operators with observables. or perhaps i am misunderstanding something. i don't think it makes sense to say that $|\langle \phi \lvert \rangle|^2$ for arbitrary $\lvert \phi \rangle, \lvert \psi \rangle \in \mathcal{H}$ corresponds to the probability that $\lvert \psi \rangle$ will be found in state $\lvert \phi \rangle$ upon "measuring".
measuring what? it is not necessarily true that $\lvert \phi \rangle$ is an eigenstate of an observable. here is the post: physics.stackexchange.com/questions/765738/…
Silly Goose, you linked to yourself...
oh
i didnt mean to do that
how do i link the post normally :P
1

In the proof of Wigner's theorem, the crucial role is played by the quantity $|\langle\phi|\psi\rangle|^2$ where $|\psi\rangle$ and $|\phi\rangle$ represent two distinct physical states (or more precisely, two distinct rays) of a system. Expanding the states in the orthonormal eigenbasis $\{|a_n\... @SillyGoose No, such overlap integrals do mean the probability that$\left|\psi\right>$will be found in state$\left|\phi\right>$, if you can find a measurement operator that can measure either state directly. but does one exist in general? I think so, because it would be difficult to prepare those states if not 6:47 AM hm i see so e.g. for spin-1/2$1/\sqrt{2}(\lvert 0 \rangle + \lvert 1 \rangle)$i suppose in the context of measuring$S_z$you could compute the probability of getting either spin value by projecting as in other peoples's answers I know you linked to ACM's post that there are superselection rules. But those superselection rules can be taken backwards to mean that some certain states in the Hilbert space cannot be realised. Then there is no point talking about them i have taken back my comments :P @SillyGoose It is easier to think of photon polarisations. You can get any angle's linear polarisation, and you can get circular polarisation. That should give you access to measure basically any state you want and it is nice because you can not just measure any probability you like, but that you can also use the same setup to generate any state you want of course, you will never get fermionic photons this way, but that is a situation that we would not have intended to scrutinise anyway If I have some more time, I would directly answer the OP's question by saying the above. The rest of the answers are mired in the minutiae and are not actually answering the crux of his question, as you are concerned about hm well im still confused because under the assumptions of textbook quantum mechanics any state post-measurement will be in an eigenstate of the observable you measured. so the observable to be measured in this case would have to have eigenstate$\lvert \phi \rangle$, which is what i am caught up on i am not familiar with the photon polarization example because what i am thinking is: is an arbitrary superposition of say energy eigenstates$\Sigma_n c_n \lvert E_n \rangle$really an eigenstate of some observable? but idk .-. @SillyGoose Ok, if that is the case, then you should consider measuring the spin of an electron in an arbitrary direction, and realise that, any$\left|\phi\right>$that you write down, corresponds to spin up in some certain direction 6:55 AM oh i see okay Hm but something bugs me. The spin-1/2 system is almost trivial because it is a dimension two system so for any arbitrary state it is immediate that a post-measurement state will be one of the two eigenstates. And so the understanding of the amplitude squared as given in the current answers follows But youre saying this is generally true for other systems Hm well i think its true for orbital ang because i do recall linear combos of spherical harmonics are eigenstates of a rotated orbital ang operator I guess i have to convince myself for non rotation related things :P 7:15 AM I would just argue that the formalism of QM is internally consistent. Even though it is only trivial to do all these measurements on angular momentum stuff, the formalism is the same, which means that it is in principle possible to do the same kind of stuff with any measurement operator you want. sorry, any state you want as long as it is not forbidden by some superselection rule Hm well i learned something new :D 1 hour later… 8:52 AM Current outline of my book has 110 chapters it is a tad ambitious 2 hours later… 10:46 AM One of them is on Hoyle's theory of gravitation because we need to never forget For an electron in a magnetic field how should the spin orientation be so that the electron has the lowest energy value, hence a more stable state? Aligned with the magnetic field IIRC Landau talks about it lemme check The Hamiltonian for an electron in a magnetic field is $$- \frac{\mu}{s} \hat{\vec{s}} \cdot \hat{\vec{H}}$$ Then, does it mean that the corresponding magnetic moment is anti-parallel with the magnetic field direction? Since the spin and the corresponding magnetic moment are anti parallel ?$\mu_z=-2m_s\mu_b$and since$E=-\mu_zB_z=2m_sm_b B$. So for$m_s=1/2you'd have a positive energy value for spin aligning with the magnetic field. Whilem_s=-1/2you'd have a negative value for when spin is anti parallel to the magnetic field, so this implies that you get the smallest energy, for when the spin is opposite to the magnetic field, because the corresponding magnetic moment aligns with the B field. But isn't this wrong? 11:26 AM is this a mistake Should it be - - ? Which QM book will have a brief introduction to relativistic QM? Just the Klein-gordon and Dirac eqn. parts.. Armin wachter. ok got it. Thanks 11:38 AM in addition, why he writes that the expectation value of the spin operator = root of 0.5? what is momentum momentumm 2hk here all i know is to solve it like this but how to invoke this extra info of momentum you have an initial starting position You need to apply the fourier transfom to get the function in K space as given in your pciture, equation 2 After doing that, you apply equation 1 in the picture to get the time dependent schrödinger wave okkk yesss but how to use this momentum 2hk the probability density is given then by the wave function * the conjugate 11:42 AM yes but what is the function of 2hk here @PrateekMourya The question is worded very poorly. I think what it wants you to do is to compute the probability to measure the momentum of the particle as2\hbar k$at a time$t$for a particle whose initial wavefunction at$t=0$is the$\psi(x)$from the question. ok...... Hello @ACuriousMind can you check my picture? 11:48 AM @Mad It is not a mistake. The equation says exactly what the text says; I don't know why you'd think it was a mistake. they did this wierdly like this hi -- i have a question. :D somebody asked me recently why$v^ie^i$is not correct notationally in GR. my answer was that vectors need to have contravariant and covariant components such that the overall vector is preserved and this is not how we write such components -- should be$v_ie^i$or$v^ie_i$depending on the basis, but then they asked what$v^ie^i$would actually represent and i was unsure @ACuriousMind he defines the expectation value as follows, he choses a state. Not two states I dont understand what +- should mean. In addition, i am not sure i understand the concept of the expectation value and why he equates it to 1/root 2 @Mad Why is this relevant? @Mad which book is this 11:50 AM Your first quote talks about probabilities, not expectation values. @Relativisticcucumber it represents an error :P if the indices don't match, you've made a mistake @ACuriousMind ok so its just nonsense right? yes but be careful: In non-GR contexts this notation can be correct. @ACuriousMind @ACuriousMind lovely XD okay thanks 4 mins ago, by Prateek Mourya they did this wierdly like this 11:52 AM @ACuriousMind i didnt notice. he did not define how one calculates the probability of an operator. he just used this for the first time. he only defined the probability of a state$\vert < a,a> \vert ^2$To be in state a @PrateekMourya I have no idea what is going on in the solution you posted, I'm afraid Oh i am very very sorry i misread the equations completly @Mad And so what is the problem?$\lvert S_x,+\rangle$is a state.$\lvert +\rangle$and$\lvert -\rangle$are states, too. Yes honestly, i blame the terrible notation and the terrible Bra ket notation. Nevermind, then i understand very well what he means @PrateekMourya what is there to not understand? In the first picture, they calculate the wave function in K space to time zero Then find the time dependent wave function in K space so basically if f(k,0) then they apply time deveolopment to find f(k,t) then the density of momentum is given by f . f* where * is the conjugate 11:58 AM oh, the question actually just wants you to compute$\phi(k,t)$and its square ah.. yes i.e. the wavefunction and probability in momentum space and that's what they're doing yes it is very badly worded by saying momenutm is 2hk "the probability density of the particle with momentum$2\hbar k$" is a strange way to say that 11:59 AM thwy are asking to find probability density in momentum space ok thanks @mad b/w what book was that do you know why they wrote 2 times that Because the particle velocity is twice the group velocity yes in the third part thwye are asking for mean energy it should the expectaiton value of H but they calculated something more wierd H in ort space they are calculating it in K space but psi should be k space htem it is or is it not 12:03 PM no i cant understand this terrible handwriting Okay well then its not problem do you know how to write H they used psi(x,t) how? the hamiltonian is given without the correspondenzprinciple to be: H = p^2/2m + V this is a free particle. V = 0 P is substuted to be hk ok... but to be honest, ia m not sure why they are not using the correspondenceprinciple to write p as an operator it probably has to do with calculating the "mean" 12:07 PM also And not the Expectation value does this question makes any sense? asking about mean energy Whelp, this turned ugly fast. 12:48 PM Wait until the AIs get robot bodies 12:59 PM > Wang (1951, 1952) and Tits (1955) classified two-point homogeneous Riemannian manifolds 1 hour later… 2:06 PM i have another question about the lorentz transformation. it says in carroll that the velocity in reference in this coordinate transformation is$\frac{x}{t}$where$x,t$are the original coordinates. then it says the transformation is$t' = \gamma (t - vx)$and$x' = \gamma (x - vt)$then it says that in this transformation, the light cones stay the same. i think i am misunderstanding smth because the line$x = t$seems to become the origin in the new frame so ? how is that the light cones staying the same @Relativisticcucumber no, the line$x = vt$becomes the origin aw man that message will forever say gamme XD so sad and "the light cones stay the same" means that in every frame, the light cones will be at 45° @Relativisticcucumber not while I can abuse some mod powers :) but$x = vt$doesnt make sense if$v := \frac{x}{t}$? @ACuriousMind omg thanks or, it makes sense but not as a line 2:13 PM @Relativisticcucumber I don't think that's the definition of$v$but in any case even if it was, then setting$x' = 0$(the origin in the primed frame) in your transformation is simply equivalent to$\gamma(x-vt) = 0\implies x = vt$. i think im still not understanding this notion of$v$. on the one hand i was told this is the velocity of the one frame wrt the other frame, but then in this book it says$v = \frac{x}{t}$and i am not sure how these are the same thing. the later one is slightly more sensible to me they're both the same but you need to interpret correctly what$v = x/t$is supposed to mean for instance, by saying that these are the$x$and$t$coordinates of the origin$x' = 0$of the primed frame but doesnt the notion of$\frac{x}{t}$exist solely within the$(x,t)$system whereas the notion of frame velocity relies on there being two frames?$x' = 0$defines a line in spacetime. You can look at the points on that line from any frame and think about how it looks from there when you look at it from the frame with the$x$-$t$coordinates, then the line looks like the line a point moving at$x/t = v$traces out conversely, when you look at the point$x=0$from the primed frame, you'll find it looks like it moves with$-v = x' / t'$! i see! thanks !! 2:51 PM why? And isnt oservables associated with operators? I dont understand this passage @Mad why not? (I don't see anything to explain about that equality) how does the term inside <> become that? @Mad the expectation value is linear,$\langle A + B\rangle = \langle A + B \rangle$for any two operators$A,B$and$\langle cA\rangle = c\langle A\rangle$for any number$c$and operator$A$okay i didnt know that thanks! that erxplains the math, can you explain the physics? @Mad This is a strange definition, most people don't define$\Delta A$as an operator and this is actually called the variance or uncertainty not "dispersion" 2:56 PM this is sakurais book it is a measure of the statistical concept of dispersion, but "dispersion" is not the common term for this I dont understand in any case, this is just the specific quantum version of the general concept of variance is it possible to report a university professor for plagiarism he literally copied the whole lectureout of the book Word for word and he didnt even explain it any better And he did not refrence the book as a source for his lecture Rather as "suggestion" along many others that sounds like a question better suited for Academia 3:05 PM Also that sounds like a pretty typical course My solid state class was based on Kittel Sounds more like lazy it is immoral in my opinion Atleast mention your base your lecture on it Ask him. It is absloutely ridicioulous to the extent, that when the book says "the proof is trivial" or "left as an excercise" The professors literally write the same thing in his lectures @Slereah yeah, but usually the notes then start with something like "This course is mostly based on Kittel", no? 3:11 PM It did not! I say don't snitch its immoral lets see what people think 3:24 PM Does the link in the comment answer your question? Just look at all the "plagiarism" that is on YouTube. 3:41 PM I guess I am missing something very trivial. If for a ferromagnetic material, the magnetisation v/s temperature relation is given by$\frac{M_s(T)}{M_s{0)}=tanh(\frac{T}{T_c}$, then why is this tan hyperbolic curve going down?? (Figure from Kittel) 3:59 PM what does one mean with the commutator but with index n ? [A,B]_(n) in what context quantum mechanics hello -- for the vector transformation rule$V^{\mu^{\prime}} = \frac{\partial x^{\mu^{\prime}}}{\partial x^{\mu}}V^{\mu}$, does this solely refer to the components of the vector? it doesn't make reference to the bases? 4:14 PM The basis has the opposite transformation ok got it and this one is for covariant only? wait no contravariant I have never learned which index position is "co" and which is "contra" and I never will those are silly words bleh i feel i need to better understand whats important xD there should be 4 relevant transformation rules for vectors, right? 2 components and 2 basis? and one co and one contra for each? @ACuriousMind The contra components and indices go up. @ACuriousMind Co-variant "varying with the basis" 4:21 PM I don't think about these transformation rules as something one should learn by heart. Every time I need to know how something transforms, I just re-derive it from the idea that vectors have$\partial_\mu$as their basis and covectors$\mathrm{d}x^\mu$as their basis Contra "opposed to" (physics convention. The maths convention is often the opposite, because they would label the basis) Vector components are contravariant since they have the opposite transformations from the basis @ACuriousMind In which case you have also thus remembered which is contra and which is co. is it the covectors that transform covariantly? 4:22 PM @ACuriousMind oh and we can just use the chain rule for this, right? really, I don't see the need for these words :P @Relativisticcucumber yes oh i see -- ok thats easier :P so a vector is$v = v^\mu \partial_\mu$and I know the full expression doesn't change under coordinate change, so the chain rule gives how$\partial_\mu$transforms and hence the$v^\mutransforms in the opposite way right i see 4:41 PM @ShikharChamoli People are free to help or not help. If you want people to help you, you have to try to find some way so that people are more motivated to help you. Spamming the chat with calls for help will only make things worse. 4:51 PM @naturallyInconsistent In a chat room, what would be some way to motivate people to help other than directly asking for help? It seems that you commonly have questions that are really out of the blue, especially when people are hotly talking about something else, leading to your questions being buried quickly. And quite often the questions you are asking seem to require too much work, and it does not seem like you put in enough effort to answer them yourself. For example, this question from Kittel that you brought in, is very well-treated in Kittel if you would just read it. The graph is obtained from the tanh But it is not a graph of the tanh itself Ok, I'll wait for a better moment from the next time onwards. @naturallyInconsistent This is the relation, given in Wahab. Honestly I was reading from multiple books. This is what's plotted and the relation is of tan hyperbolic, isn't it? I have thought enough but couldn't understand what's happening The graph you are looking for appears in page 19 It is not the tanh graph, which appears in page 18 The argument for these things appear a little earlier, and also in the captions on the graph in page 19 5:13 PM ok got it, got it. thanks. 5:37 PM \begin{align}\frac{\mathrm d\ }{\mathrm d\varphi}\left[\left(\frac{\mathrm dw}{\mathrm d\varphi}\right)^2+w^2\right]&=0\tag{SHO equiv.}\\\therefore\exists\varepsilon\in\mathbb C\ |\qquad\left(\frac{\mathrm dw}{\mathrm d\varphi}\right)^2+w^2&=\left(\varepsilon\frac{GM}{\ell^2}\right)^2\tag{15}\end{align} Oh goody, Chatjax can deal with equation alignment 6:36 PM @ACuriousMind i had a follow up to this. sorry this might be pedantic, but you wrotex \rightarrow f(x)$but in a simple case of polar coordinates, we would do something like$x \rightarrow r\text{cos}\theta$, right? so notationally, this seems to not match up? am i missing smth yesterday, by ACuriousMind @Relativisticcucumber What do you mean by "a means of doing that"? Spherical to Cartesian is just the map$(r,\theta,\phi) \mapsto (r\sin(\theta)\cos(\phi), r\sin(\theta)\sin(\phi), r\cos(\theta))$. ok i see thanks you are right that$x = r\cos(\theta)$, but I am talking about the map$f$that sends a tuple of Cartesian coordinates$(x,y)$to the corresponding tuple of polar coordinates$(r,\theta) = (\sqrt{x^2 + y^2}, \mathrm{atan2}(x/y))\$
yes i see. that makes sense ┏(・o･)┛
always happy to make sense :)
6:59 PM
@Qmechanic They're aware that with the new buttons it can be hard to tell if you've voted, especially on meta sites, with their grey scale themes. Allegedly, they've made some improvements... meta.stackexchange.com/a/389771/334566
7:52 PM
@SillyGoose Don't know why you deleted that but how rep for deleted answers works is described at the end of this very long faq
8:15 PM
oh because i noticed it got removed after i refreshed
@PM2Ring The latest update on the Physics meta addresses the issue quite well, dunno if you've seen it yet
Oh, first time I clicked that link it for some reason sent me to another answer. The one I am seeing now is indeed exactly how it looks
Maybe I just scrolled my mouse unintentionally :)

3 hours later…
10:51 PM
@Amit It's still very hard to see on my phone. But Samsung & its browser do some odd things in Dark mode, and I can't blame SE for that.
Eg, in an SVG with a filled rectangle & a filled polygon using identical colours, in dark mode they become not identical. :( It's rather annoying. But the Dark theme is much less harsh on my old eyes, so I (mostly) tolerate its quirks.
11:14 PM
Yeah, I get that. I used to use dark mode too, but gave up because of issues like that. Instead I set the color balance to be slightly "warmer" so that white looks more yellow'ish
What I am saying is about my PC however, not phone