@Sanjana I assume that RA-declination plot uses a simple equirectangular projection. Sure, a Cartesian straight line will generally not be straight on such a plot (unless the line is on the celestial equator or has constant RA). However, if the line segment is short & not too close to a pole it will look pretty straight on the plot.

Hello Everyone...

@Slereah I think the whole terminology of paradigm shift came from Kuhn, who was after Popper.

@Slereah Feynman in Character of Physical Law makes the argument that scientists would already have consulted all the successful attempts to solve whatever is the new crisis in physics would be, and it must be because the set of all successful new theory generators have all failed, that we would have a new crisis in the first place. This means that the solution to new crises will likely always involve new methods.

@Amit In this statement lies a tremendous overgeneralisation.

@Amit No, I think it is mostly bad presentations that leave students still confused about it. SR is not that much difficult to understand than Galilean relativity. Granted, quite a lot of people are very very very confused by the latter, but it is possible for most people to grasp the central ideas of SR if a tremendously good presentation is being properly followed.

@Amit I was taught it at age 15, and not by my own choice either. I am always appreciative of the award winning teacher who mandated that for our school. I've always been trying to push that age downwards, but it is rare to find students who would not resist the instruction, though.

@Relativisticcucumber no no no, it is not the most shit book ever; they are basically equally bad. Halliday wrote the first such book, it has been downhill ever since.

@Relativisticcucumber After many years getting stuck trying to apply to HEP, it turns out that it is not just a lack of publication that they would reject even considering an applicant, but that if you are not already a student from one of the only two lineages, they would not even consider you. But I have evaded the drumpf years precisely because of this and am not that resentful any more.

@ACuriousMind one has to wonder why a person of your calibre hasnt yet learnt to stay out in the first place and not feed the ...

Hello @JohnRennie sir

what is drumpf

that moment when

im a tiny Slereah

@naturallyInconsistent don't diss haliday Resnick Krane, I had a lot to learn from it in my high school

wow schwartz is big

But my most fruitful endeavours in my high school was studying the red Irodov (fundamental laws of mechanics) the yellow Irodov(basic laws of enm) combined with problems from the green Irodov(problems in general physics)

@Relativisticcucumber Did Schwartz get fatter since the version I got

@SillyGoose The orange menace that was the previous supreme leader of the free world.

@nickbros123 Halliday, again, was the best of the lot. But sadly it is not a very high bar. It is very very very good for high school, though.

@nickbros123 The huge number of Indian students is precisely why Ive been avoiding Irodov. One does not wish to be swarmed

shwank

H O N K

HONK

the maximal torus of $SU(N)$ is the intersection of the maximal torus of $U(N)$ with $SU(N)$.

So, the maximal torus of $SU(N)$ is all diagonal special unitary matrices, right?

(thinking of $SU(N)$ as N x N special unitary matrices)

is the relativistic limit really about speed? in beginning of uni/high school, i was taught that when $v$ is not negligible when compared to $c$, we are in relativistic limit. however, now its seeming like we need relativity when we want to account for effects due to curvature?

@SillyGoose ☀️

:D

@Relativisticcucumber When we say relativistic, we really usually mean SR. Curvature is GR.

does the tensor product of more than two lie algebras generalize in the natural way? $\pi_1 \otimes \pi_2 \otimes \pi_3 (X, Y, Z) = X \otimes \mathbb{I} \otimes \mathbb{I} + \mathbb{I} \otimes Y \otimes \mathbb{I} + \mathbb{I} \otimes \mathbb{I} \otimes Z$?

where $X, Y, Z$ belong to (to be concrete) $\mathfrak{su}(2)$

@SillyGoose Does it work like that for 2 Lie algebras?

yes

@SillyGoose Then it MUST generalise that way.

hm okay i shall work through it tomorrow then

When I cover the generalisation of Pythagorean theorem to 3D for students, this is one thing that I would have to particularly point out. It will never be $\sqrt[3]{x^3+y^3+z^3},$ and the reason is symmetry.

for the harmonic osscilator, it is asked (as a homework) to create a state of $\vert 0\rangle , \vert 1 \rangle$, in which the expectation value of $\langle \hat{x} \rangle $ is maximum We write a vector $\alpha= c_o \vert 0\rangle + c_1 \vert 1 \rangle$ to which we want to calculate $\langle \alpha \vert \hat{x}\vert \alpha \rangle$

as next, they choose $c_o = \vert c_o \vert \exp{(i s_0)}$ and similiar to $c_1$ , i dont understand why would one make such Ansatz?

My idea was, to write a general vector as i wrote in the first message, than derive $d/d \alpha \langle x \rangle$ furthermore, i can write $x = \gamma (a + a^\dagger)$

are the states orthogonal?

if so i get for $\langle x \rangle = \gamma(c_0^* c_1 + c_0 c_1^*)$

then i need to derive this to find a maximum, but this is my method.

this is a function of c_o an c_1 to C

so the derivative is a gradient right?

so i get $D <x> = <c_1 + c_1^*,c_0 + c_0^*>$

There are several limits you can take

you can change the value of c, G and h

That will give you all the various limits of theories you want

(are you writing to me ? @Slereah)

No.

It is also called the Bronstein cube

Continuing my argument, we must furthermore have $\vert c_o \vert^2 +\vert c_1 \vert^2 =1$ due to normaliziation, however, if we plug in my result, i have $c_o^2+c_1^2 = -1$ which is nonesense

where have i gone wrong

I would appreciate your help guys, i have the exam tomorrow, kind of stuck here

oh i think i forgot $\gamma$

but it doesnt matter, since one gets the same result

@naturallyInconsistent It is a purposeful generalization because I think that the damage done to most kid's natural curiosity and interest by being forced to study stuff they don't want to, is greater than the benefit given fo the few (like you with SR) that bump into a subject they turn out to like in this manner

@Amit I am very certain that you will not be able substantiate this generalisation. There simply never happens in any teacher's teaching career a child, even Wednesday Addams or Sheldon Cooper, coming along claiming that they want to learn about limits today, but as a precursor to calculus it is quite important to talk about the topic. Similarly, students would not even know that they need to learn about Newtonian mechanics, but its teaching is of immense utility for society.

looking at my calcaulation again, i reliaze that the first equation imply the c_0 c_1 must be purely imagenry of the form i b

then i have $b_0^2 + b_1^2 = 1 $ how can i now give an explicit form for c_0 c_1?

it has many solutions

@Mad Use $\hat x=\sqrt{\frac\hslash{2m\omega}}(\hat a^\dagger+\hat a)$ so that it is a constant multiple away from $\begin{pmatrix}&\sqrt1&\\\sqrt1&&\sqrt2\\&\sqrt2&\ddots\end {pmatrix}$ and then you can specialise to the top left 2x2 part.

aaaannnnddd he's gone

@Slereah i see

@Relativisticcucumber When the curvature is small, the Newtonian model is adequate, unless you want to calculate something that simply doesn't exist in the Newtonian model, like the gravitational time dilation. Roughly, the gravitational relativistic effects become significant when the escape velocity is relativistic. Alternatively, when the radius of the gravitating body is only a small multiple of $r_s$ its Schwarzschild radius. Note that for Earth, $r_s$ is around 8.4 mm.

In SR, we can write the equation for KE as $\frac1q mv^2$ where $q=\frac{\beta^2}{\gamma-1}$. For small $\beta=v/c$ we have $q\approx2$.

For more details, see physics.stackexchange.com/a/595175/123208

@naturallyInconsistent hey - you know you can edit your posts for a short time ?

If you catch it quick enough, simply edit the previous statement

I think its 5 minutes of window, and that is checked when you save the change not when you start making the change.

@Criggie Hi :-) Are you responding to a flag or did you decide to look into the physics of cycling?

heh nah it was the chat flag

I did 7th form physics about a thousand years ago (with rounding)

Someone just asked about the benefits of disk vs rim brakes ...

oh god not that one again

Indeed :-)

Pet peeve: chat flags don't get shown in the mobile view.

To be fair they were asking about analysing the forces involved so it is kind of physics.

I'm sure it can be answered from a purely theoretical point of view, and while interesting, it may not reflect real-world usage.

We start by assuming the bicycle is a sphere ...

I have rim brakes on my road bike, and its the middle of winter here. So they pick up a heap of sand and debris from the road, and make horrid grinding noises. So disk brakes have the braking surface another ~250mm away from the road and are therefore much cleaner.

That can't really be modelled by theory

@JohnRennie if you say "frictionless" Imma leaving :)

:-)

The comedy physics stack is called Worldbuilding... :D

crap this is turning into economics.

sorry

@Criggie Yes, and I was just seconds away from it. It was definitely a lot shorter than 5 min though

All good :)

Thanks

@naturallyInconsistent i did just that

@Mad and the rest?

well, i dont understand what you wrote

i have this equation which must be true $c_0 = c_0^*$ and $c_1=c_1^*$

Setting $c_0= \vert \sqrt{1-\vert c_1\vert^2} \vert \exp (i\phi_0)$ and $c_1= \vert c_1 \vert \exp (i\phi_1)$

not sure how to procceed

Sorry the first equation has a - sign on the c_i side

if i put the second condition in the first one

i have then soimething in the lines of $ exp (2 i \phi_0 ) = -1$

not possiber

actually possiber me dumb

then i have smth like $\phi_0 = ((2n+1)/2 )\pi$?

Is the averaging smooth test function used in Russokof averaging have it's range (the radius after which the function tapers off like a bumb function) to be the distance of many thousands molecules?

@Mad What part of taking $\left|\alpha\right>=e^{-i\varphi}\cos\theta\left|0\right>+e^{+i\varphi}\sin\theta\left|1\right>$ do you not understand?

the part where you dont say why you do that

@Mad It is the most general representation of a normalised state between two basic states

you are giving an explicit form for the coefficients c_0 c_1

either you have tijme travel and know how they look like

or you are not explaining this correctly to me

You already know that $|c_0|^2+|c_1|^2=1$, which is obviously to be parametrised by the cosine and sine part. The phase part is just the leftover possibility, and since a global phase is undetectable, it is thus possible to always pick for their phases to be opposite each other.

@nickbros123 The thousands of molecules part is due to the need for unit cells to represent something almost spherical without too much error, so yes.

@naturallyInconsistent I am all for finding ways to familiarize kids with a wide range of subjects, but not stick to any rigid curriculum but make class time more like a conversation. For example I admire teachers that start class with the words "Are there questions?"..

My only point is that the education system is traditionalist and lacks imagination / courage

Well the entire point of education in math n phys is to solve cool (sometimes not so cool) problems. So the best way to educate in such cases is to solve cool problems!

@Amit that is a statement that is coming from totally not understand how education is supposed to work. What you see as a very organic and flexible teacher, is the product of a tonne of practice and simulation and preparation. The result of making things up on the fly, i.e. not following a rigid curriculum, is to keep forgetting details and thus very patchy coverage

@nickbros123 Are limit questions cool? Are solving the 10000th principle of moments question cool?

I mean, these issues that you are coming around to, are the most basic of basic issues that a teacher has to deal with. As long as the teacher is even interested in reliable transmission of knowledge, they would have thought about such things.

@naturallyInconsistent The words "supposed to work" imply you accept very specific goals to education, which may be different from what I consider to be important ;-) For example, I want to help kids become happy individuals more than I want to help them be useful to society

There are, obviously, many countries where the education system, the curriculum or syllabuses, are mired in horrendous bureaucratic nonsense, and thereby produce garbage. That is happening too often to be dismissing. However, when you actually get to sit in, or even just consider what would go into a curriculum planning panel, the results will be a totally different way to look at what should be taught.

@Amit Happy, like Bhutan? i.e. irrelevant to reality? Look, when one is responsible for the future of a society, the pressure is necessary

@naturallyInconsistent no obviously. But if that's what you think books like Irodov or, krotov, say SL looney coordinate geometry, or spival calculus, Hall and knight algebra or David Burton number theory do, you'd be dead wrong. This is why JEE students and Olympiad students generally are much better equipped to face harder / challenging problems, for they basically learn from them

You dont get to "oh, the kids will be happier to know this or that"

I didn't say that

@nickbros123 You made that leap of logic yourself. I am very clear what such books do, and they serve a good purpose.

Education in India, atleast the entrance exam prep sect do one thing right, subscribe to these books and base their teachings in them. creative problems that apply concepts from many chapters. If you go to the website of IITJEE questions, you'll see a bunch of questions which are conceptually elementary yet nonetheless makes one think

@Amit Not to mention that you imply with that statement that my students are not "happy individuals"; on the contrary, my lessons are filled with laughter, and from both sides.

@nickbros123 I am not denying that.

@naturallyInconsistent That's saying we have no trust that people can find their own way..

@naturallyInconsistent No, the students in such higher education already passed through a lot of filters

@Amit You mean you believe that the random Tom, Dick or Harry can reproduce the last 2000 years of progress just if they try things out themselves?

@naturallyInconsistent No, I didn't say rediscover stuff yourself. I say become sensitive to the interest of a kid and teach what he wants to study

@Amit I'm teaching age 10 upwards, every step of the way. I just got off a call with my age 15 kid that was taught since age 10, and he just learnt some basic vectors.

@naturallyInconsistent how many students do you teach at once, sir?

@Amit And how is that inimical to a standardised education system that allows kids flexibility? I come from a privileged place that has such a thing, I know, but I think it is a lack of imagination of what can happen more than anything else.

@user858770 It can be one-to-one, it can be a small or big class, and at uni level it can go into hundreds and even exceed thousand.

A thousand? :O

@naturallyInconsistent BUT I have to add, for all the good things that the entrance exam prep sect does to the ones truly interested, the country screws everything up when it comes to higher education. These people cannot curate a good course if their life defended on it

MOOCs go into the millions.

@naturallyInconsistent If it allows flexibility it's the right direction, I was saying usually it isn't so. That's just one element though, the other is good teaching (Not letting the kid feel he disappoints by understanding slowly, etc.)

@nickbros123 That is part of why rote learning is incredibly detrimental to a society, and academia actually knows that quite well.

@user858770 Yes, but rarely do we get to 1) be so fortunate as to make them, 2) get that kind of outreach, 3) check that they are actually good.

@Amit If you are involved in adult education, especially part-time studies people who are coming in after full-time work, you will know that the successful educators have a tonne of patience.

But then again, I see you have quite a lot of patience with students yourself

@naturallyInconsistent rote learning, as much as the country's finest exams try to mitigate it (by creating novel questions, generally pushing ones knowledge level etc), still allows some students to go through these exams by "memorising" instead of breaking their heads over like a million questions, and however creative one can get, I'm sure most of the questions would come into some bracket of this million. It's a massive loop hole that cannot be fixed

Most good students know better than to rote learn, some like me are gifted with the amazing gift of poor memory, but most students are hardworking but decide to apply it in a wrong manner, like this. The system cannot mitigate this

@nickbros123 I'm particularly ok with rote learners passing on along. I do not see why we have to plug that loophole. I have better things to do, like helping other students.

@naturallyInconsistent yes I agree, and also, I don't think one can actually plug it by any means. There are only a finite combination of concepts that can be used to create a good problem, say from mechanics; if students are ready to work freakishly hard (and there are many of them), they can absolutely memorise these large but finite possible problems and that's what happens in most cases

What I don't get is, it's much easier to just do it the normal way

@nickbros123 If you look at the actual exam questions in national exams here, you can actually see a pattern. There is a smart thing that is being done to discourage people from doing stupid things, but it has to be kept secret.

what is the relationship that allows us to use the metric to go between vectors and dual vectors? what makes a tensor the proper tensor to undergo that mapping?

There's nothing special about the metric tensor

Any symmetric tensor of rank (0,2) allows you to do that

@Relativisticcucumber The metric is the tensor that enacts the scalar product. The dotting. To do that, it takes two vectors and finds the scalar product. Anything that reacts with a vector to form a scalar product is a dual vector. Hence, the metric acting with just one vector alone, forms a dual vector.

It's also common to define such mappings with symplectic forms

If you have seen in computer science the functional programming concept of currying, this is currying.

and separately, in wald, he says "given a spinor $\psi^A \in W$, we can construct a vector $k^{AA'} \in V$ by $k^{AA'} = \psi^A \bar{\psi}^{A'}$" but i do not understand how to even interpret this

But of course in the context of GR it usually makes more sense to do it with the metric tensor

@Slereah that is skew-symmetric

@Slereah oh really?

@naturallyInconsistent Same principle, you just have to watch out the order you're using if it's not symmetric

@Relativisticcucumber Sure

Should be nondegenerate too

@Relativisticcucumber Have you seen the Weyl construction of spinors?

@naturallyInconsistent i dont think so, this section of wald is my first formal encounter with spinors

Any such tensor $T$ acting on a vector $v$ will define a one-form $\omega$ such that $$\theta(w) = T(v, w)$$

$\theta$ is usually denoted by $T(v, -)$

@Relativisticcucumber Then you had better ignore this part. To be fair, between Wigner and Weyl, they make the concept of spin (and helicity and chirality) so much more sensible, in the context of QFT. Spin in standard QM is a mess.

@Slereah ok that clears up one thing

@Relativisticcucumber hey, I just said that earlier!

@naturallyInconsistent but i am reading this section to understand spinors D:

@naturallyInconsistent yes i think what you said above also makes sense

you fool, nobody understands spinors

LOL

@Relativisticcucumber You should try to understand spinors in the flat spacetime before attempting the curved!

but i am

this is spinors in minkowski !

i heard this section of wald is the best resource for spinors? maybe im confused

You should even start with spinors in R3

ive been spending quite awhile on it and im v lost

I never once thought Wald taught anything well. Too mathematical to be for beginners.

@Slereah do you have a resource recommendation ?

@naturallyInconsistent i quite liked his curvature chapter

but thats the only one i read in depth

@Relativisticcucumber Jigglypoof

@Relativisticcucumber idk, Landau's book on QM?

every QM book has a section on spinors

the wave functions in QFT are spinors, right? the dirac spinor?

@Slereah hm i have yet to see it in sakurai but ill check again

they transform under the spinor rep anyway, yes

Or do you mean the operators

@Slereah what do you mean

@Relativisticcucumber QFT have wavefunctionals. The operators that look like wavefunctions are not wavefunctions at all.

What he said

@Slereah That is true, but they tend to be very rudimentary and cursory.

oh no

wait the dirac spinor is not a wave function?

of a qft version of a wave function?

What do you mean by the Dirac spinor

@Relativisticcucumber In the horrible nonsense that is RQM, the Dirac spinor is a wavefunction, yes, but RQM should simply not exist.

i think in sakurai he says the solution to the dirac equation is a vector with 4 solutions and i thought this is the dirac spinor but now i am questioning

what is RQM

In full QFT, the operator that is supposed to classically solve the wave equations, be it Maxwell's or Dirac's, is not a wavefunction at all.

relativitic quantum mechanics

The relativistic theory of point particles

@Relativisticcucumber It is a Dirac spinor, but in the SR QM framework, and that is just bad.

In RQM the wavefunction is spinor-valued, yes

$\psi \in L^2(\mathbb{R}^n) \times \mathbb{C}^2$

@naturallyInconsistent hm and what do you mean by "operator that is supposed to classically solve the wave equations"

What Wigner did was to systematically work out how to construct every Poincaré invariant wave equation, arbitrary spin, excite them individually up from scratch, and so forth. Unifies the understanding of spin and (helicity/chirality).

Weyl's method is to deconstruct the Dirac spinor, so that the chiral/helicity parts that are mixed up in Dirac spinor by mass term, are studied separately. Then you work out that they rotate the same way but boost in opposite ways. Then you build them up to get the tower of possibilities that Wigner did.

@naturallyInconsistent hm this seems more satisfying but what do i know

@Relativisticcucumber The $\hat\psi$ that solves $i\gamma^t\partial_t\hat\psi-i\vec\gamma\cdot\vec\nabla\hat\psi=m\hat\psi$ (dont quote meow on details now) in QFT is an operator, not a wavefunction.

okay i see -- yeah i was curious about the term operator there bc i have not heard that yet

but hopefully soon !!! i dream of qft everyday

@Relativisticcucumber Inside that operator is an annihilator of electron and creator of position states. It is so much of a mess.

(And that is in free particle interpretation only. In interacting QFT we don't really even know what it is)

ok i will abandon wald spinors for now and use another reference for them thank you for the help @Slereah @naturallyInconsistent and hope i can investigate the operator soon

@Relativisticcucumber Have you seen how Feynman lectures did spinors? He started with Stern-Gerlach expt like what Sakurai, Ballentine, Townsend, and all, did, but he properly covered that if we wanted it to work sensibly, then we needed to explain how to deal with rotations of the SG apparatus, and how the spinors would transform if so. It is a tedious but good understanding of Pauli spinors (which form the basis for Weyl later)

@naturallyInconsistent interesting. i havent seen that.

ill check it out

@Relativisticcucumber It is too tedious, but at least it is something very doable. Takes pages upon pages, just to reproduce the $\frac12\hat{\vec n}\cdot\vec{\hat\sigma}$ formula and its eigenvectors.

@Relativisticcucumber Spinors are realizations of the fundamental representations of $SL(2,C)$, i.e. vectors on which the (fundamental representation of the) generators of SL(2,C) act. From this fundamental representation we can build higher rank representations, e.g. forming direct products of the vectors etc.

A tensor which transforms like $\psi^{A \dot{A}} = \psi^A \psi^{\dot{A}}$ has $2 \times 2 = 4$ independent components, the same number as a 4-vector so a 2 by 2 matrix transforming under $SL(2,C)$ in this way, i.e. as $P \to M P M^{\dagger}$ should be equivalent to a 4-vector, and it is since $P = p_{\mu} \sigma^{\mu}$

@bolbteppa bah i will keep endeavoring to understand

@Relativisticcucumber Seen from a suitably arranged and organised introduction, it would make a lot more sense.

@Slereah i think this schwartz that i have might be bigger bc i think its a fraud copy. its like a thing in china that the only english textbooks you can get are actually just printed versions by some people/companies, but they are p good copies actually and like 1/4 the price of the US versions so in consider it a win

@naturallyInconsistent i found a youtube series by eigenchris that im watching now !

@Relativisticcucumber oh nice! and now im shocked watching his latest video

@Relativisticcucumber Nevermind special relativity, do you get $SO(3)$ spinors in QM

Might even want to start with SO(1) spinors

any idea how he derives those relationships?

apparently something like writing F and G as potenzseries

and then repeatedly applying commutations

After reading so much philosophy of science I'm a bit sick of reading about Galileo

@Mad see physics.stackexchange.com/a/87039/50583

thanks!

one more question

(pls)

any clue how he did write that for x(t) and p(t)

Sorry if its simple and i am not seeing it, i am burned out very much rn, got exam tomorrw

@Mad It is something stupid. He expanded a(t) in terms of x(t) and p(t), and a(0) in terms of x(0) and p(0), and then multiplied the exp + or - i omega t out.

i dont understand

@Mad $a(t)=a(0)e^{-i\omega t}$, just expanding, $x(t)+\frac{ip(t)}{m\omega}=\left[x(0)+\frac{ip(0)}{m\omega}\right]e^{-i\omega t}$

ohhhhhhhhhhhhhhhh

Thanks man

is it possible to determine the eigenkets of an operator, if we only know its eigenvalues, but not its full representation?

For example, if an operator fullfills A^2 - 3A +2 Id = 0. the eigenvalues are 1 and 2

Can we also construct the eigenvectors without knowing the form of A?

Furthermore, is it possible to show that A is self adjoint?

@Mad $\begin{pmatrix}1&\\&2\end {pmatrix}\neq\frac12\begin{pmatrix}3&-1\\-1&3\end {pmatrix}$ both satisfies that matrix polynomial and have the same eigenvalues, but disagree in the form of the eigenvectors. Separate but related question: if you just rearrange the eigenvectors in a $A=PDP^{-1}$ decomposition, the ordering of the eigenvalues in $D$ will change; do you consider that as a different matrix with different eigenvectors?

as a homormorphism it works the same way

the operator i wrote should be adjoint

i am trying to write 1 = ... and work with that

no dice

@naturallyInconsistent in my case i doooo

@Mad I am not fluent in the maths terminology (I think you know the chasm between theoretical physicist and mathematical physicist) to be fully certain what you meant there, but the direct counterexample definitely kills it. And both matrices are Hermitian.

@SillyGoose do do dooo do dooo

@naturallyInconsistent is this like four momentum?

@SillyGoose I have no idea how your question could be related to what I was writing there. Could you ask in more detail so I can get some context and give a more useful reply?

well you mention the operators which look like wave functions. which makes me think of vectors of operators, i.e. four momentum in relativistic qm

which is $p^\mu = (H, p_x, p_y, p_z)$

@SillyGoose No, I meant the $\hat\psi$ in QFT, which look like they are $\psi$ of QM, are not the latter at all.

oh

@Mad A = [[1, a], [0, 2]] solves the equation and is a counterexample to all the claims

@fqq grrr I knew one of Jordan normal form shit would be correct here. meowwww

hmm, how fast do yall think P.SE would close my question if I ask why it is so often the case that my washing machine would flip my underwear inside out?

@naturallyInconsistent You really wanna ask this?

More of a question for the engineering stack exchange

so we can not give form of the eigenkets?

it is an old homework question

and it is said to show its adjoint

so i am not sure how

but your counter example kills it

i mean its not even said which dimension the space has

its only written as "let V be a quantum mechanical state space" A: V to V

i mean we can see that the determinant is none zero (its either 1 or 2)

so it has full rank.

So the number of eigenvectors need to be equal to the dimension of the space

right?

Furthermore the given polynomial is according to cayley hamilton its charachterstic polynomial

So we can rewrite it as CP(A)= (x-1)(x-2)

Which means that its diagonalizable

Hello fellow physicist. I have a multidisciplinary question on physics and English language and I would ask you to advise me where to post it.

Because nobody seems to be present at the moment, when you have time, please visit the question I need answer to here: physics.stackexchange.com/questions/771797/…

I mean, anything goes in physics when it comes to use of language, it seems a bit unlikely you're going to be able to ask a non-opinion-based question like that.

As someone who has been writing books on physics for years, I find this question legitimate and interesting, but I do not know who to address it to. Do you have any ideas?

I think it is interesting, physics.se definitely isn't the place as it's opinion based. I'm afraid I can't really say where on the se network would take it though

@Mad look at the counterexample again ;)

the issue is, you do not know it is a 2v2 matrix @fqq

I am aware this is looks to all of you too pedantic, but I really put enormous effort into my books and want to get every detail right. Maybe I just have to accept the defeat in this detail...

@Mad yeah, so?

The matrix you wrote has a 2 v 2 form

@Pygmalion i find the argument, that the mass is not inherit to the object, to be adequate.

Btw you can generalise the counterexample to any dimension easily

@Mad Isn't that the same I wrote?

@Pygmalion yes.

@Mad OK, so why isn't the same argument applied to cross section or ground plan? They are also not inherit to the object?

Because people are not too literal

unless one is a mathematician

In the limit of none relativstic physics, you might argue that the mass becomes inherit to the object

But at this point its philosophical

it's inherent, not "inherit", and I think most people would consider both formulations "correct" or at least not worthy of criticism

Rigor cleans the window through which intuition shines.

Well bye everyone

@Pygmalion thats a very decent statement

i am going to steal that one

small (hypothetical) question

suppose we have a system composed of two energy eigenkets $\vert 1 \rangle, \vert 2 \rangle$ and lets suppose the system posess another observable W and expremintally one finds: $\langle 1 \vert W \vert 1 \rangle = 1/2 , \langle 1 \vert W^2 \vert 1 \rangle = 1/4$ can we deduce anything about the eigenvalues of W

any help ?