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2:05 AM
Aha I’ve found the useful results i was looking for: the adjoint action commutes in a sense with the exponential mapping
 
5 hours later…
7:05 AM
Generators
Oops
Wrong window
Was that meant for a discord server :^)
No I was looking to type in the search bar :p
I see.
🙈
 
1 hour later…
8:44 AM
@Slereah your book is the best GR book that i have seen :D
High praise for a book that's barely written
9:15 AM
@naturallyInconsistent @naturallyInconsistent how does fine structure constant being tiny explains it?
@ShikharChamoli A photon, on average, interacts with only 1 in 10000 electrons because the square of the fine structure constant is about that small. It is thus very unlikely that it would interact twice, because it happen once in 100000000 electrons.
 
1 hour later…
10:24 AM
@ShikharChamoli we do get second and third order diffraction but we don't call it second and third order.
e.g. when we get second order diffraction from 100 we call it first order diffraction from 200.
When we have double slits we write nλ = d sinθ where we take d to be the (constant) spacing between the slits.
But in XRD we write λ = d sinθ where we redefine the 𝑑 to be 𝑑/𝑛
Hello. I think rel. QM is wrongly infamous for allowing negative energy. In the $\psi (x) = \langle 0|\phi(x) |\psi\rangle$ formalism of rel. QM, there is no negative energy becuz $\phi (x) |0\rangle$ annihilates the negative frequency exponential of the quantum field. So there is no negative energy in $\psi (x)$'s evolution
Negative energy only shows up in rel. QM when we interpret Dirac eqn as the eqn of $\psi (x)$ of evolution. But this is a mis-interpretation of Dirac eqn
In the QFT version of rel. QM, there is no negative becuz it gets annihilated when operating on the vacuum using $\phi(x) $
10:45 AM
Now, the exponential term of the creation operator does survive in $\phi (x) |0\rangle$, but creation operators have a positive frequency exponential, for both particles and anti-particles. So when we do $\langle x |\psi\rangle$, this positive frequency gets conjugtaed to negative frequency
Which means we get negative frequency=positive energy for both particles and antiparticles
 
4 hours later…
2:19 PM
@JohnRennie But Sir, for example in a simple cubic lattice, if there is 2nd order bragg diffraction from (001) plane, by this logic we can consider it 1st order from (002) plane; But there's no (002) in a simple cubic.
It's a long time since I did a crystallography course (about 40 years!) so I may have got my terminology mixed up but I remember that in the Bragg equation we basically redefine d to mean d/n so we don't get an 𝑛λ term as we do for a dounble slits or diffraction grating.
@JohnRennie Ok sir, I'll read more about it.
Mad
Mad
2:34 PM
Good day, i am wondering how to reach the defintion of hermetian adjoint in dirac notation starting from the mathematical canon:
$ <Ax,y> = <x,Ay> $ we denote with the braket then
$ <v \vert v' > = \phi_v (v') $ with
$\phi_v (v') = <v,v'> $

How to reach $(A \vert x > )^* = (\vert x > A^*)$ (source sakurai)
I apologize, obviously you need to have the hermetian conjugatre on the right side of the equation
That's a symmetric operator not Hermitian
Mad
Mad
the first equation is only true iff a is hermetian.
* denotes the hermetian adjunct in this notation
@Slereah can you elaborate your answer? what is exactly being refered to with "that is"
i meant to write $ A \vert x > \rightarrow < x \vert A^* $
This is what sakurai states, i would like to translate this into mathematical canon terms.
the arrow indicates taking the adjunct
if we consider the linear mapping i adresse. we can write
$ <x\vert A^* $ as $ \phi_x \circ A^*$
then it is by applying a vector that we have $ (\phi_x \circ A^* )(y)= \phi_x(A^*(y))= <x,A^* y>$
not sure how to procceed
Mad
Mad
3:03 PM
this is what i am refering apprently, hes defining it by using the dual, can someone here explain why is the dual of that vector X(alpha) equal to that?
3:20 PM
@Mad u can also gain some intuition using matrices, instead of thinking in the abstract. For matrices A and B, $(AB) ^{\dagger}= B^{\dagger}A^{\dagger}$. This is becuz dagger means transpose conjugate for matrices
Mad
Mad
I dont think thats the statement being made
the statement is about the dual, not the hermetian conjugate
U can think of kets as column matrices and bra as row matrices. They r Hermitian conjugates of each other @Mad
Mad
Mad
Is dual = hermetian conjguate?
This intuition works for finite dimensional linear algebra, at least
@Mad a dual is defined using inner product stuff, but it's pretty much the same thing as taking transpose conjugate
U can think of bras as row matrices and kets as column matrices that r also transpose conjugates of each other. Then the inner product is just the matrix multiplication of these two @Mad
Mad
Mad
i understand what you mean, in finite dimensions, the hermetian conjugate and the transpose conjugate is the same
I wasnt aware that in finit edimensions, the dual is equal to the transpose conjugate
as far as my memory goes, there was something about transposing the basis vectors to get the dual basis vectors
in finite dimensions
3:26 PM
Yes, it is in Shankar's book. Chapter 1
Bras r identified with row matrices and kets with column matrices
But in the abstract, the definition wud get more nerdy becuz we wud define them in a basis-independent way @Mad
Mad
Mad
is there any proof you can provide for that statement, in sakurai
Or do you mean it is too complicated to show?
Aren't there like very specific and not widely used QM books that actually develop all the math? :) I know to do it properly you need to first introduce Banach spaces, a bit of measure theory, and all kind of wonderful stuff which is often skipped in standard QM texts
I havent read Sakurai
Mad
Mad
@Amit please refrence me to one.
@Mad I think I saw an SE question on that, just a sec
Mad
Mad
3:32 PM
actually
i just proven it
but still refrence me
@Mad - I think that was the one
Mad
Mad
thank you
np, I guess it may be best to use such resources as reference, imho it's probably better to go through QM for the first time without the "full" rigor...
Mad
Mad
Yes good advice
However, i have mahtematical background, i try to connect to whats familiar
thanks anyways for the refrence
3:50 PM
👍🏻
The most abstract definition of dual is the space linear functions on V. But becuz of the inner product in QM, we can map vectors to dual vectors. And aftr choosing a basis, we can think of the dual in QM as a transpose conjugate of a row matrix
But i just directly think using the row matrices and column matrices becuz it's very intuitive
But it doesn't work for GR becuz u have more than two indices there. So there's no point in thinking pictorially about columns and rows. Instead, we just classify the indices into two types
I think Dirac's bra ket notation is only convenient becuz we have only two indices in QM matrices
 
1 hour later…
5:04 PM
@Amit That just happened to me. Quite confusing when you get a red box with an unexpected error message.
lol, i know right...
 
3 hours later…
7:41 PM
@Mad I think there is also "Spectral Theory and Quantum Mechanics" by Valter Moretti and "Quantum Mechanics for Mathematicians" by Brian C. Hall to my knowledge.
oh oops it looks like these are included in the second answer to the post amit linked
@Slereah that depends on how many GR books they've seen :P
I'm not sure if the completed book will be a pleasure to read
I feel it may end up with MTW levels of organization
8:00 PM
Hello everyone, Does anybody know what is the role of Helium in an He-Ne laser. Everywhere, I find written that directly transferring energy to Neon atoms by electron collisions is inefficient, so first the electrons excite the Helium atoms and then those He hit neon atoms.
But I don't understand why is it inefficient. Someone kindly guide.
@ShikharChamoli You want to excite the 4s and 5s states of Ne. but there's a bunch of 3s (and other states) in between. So if you try to excite the neon directly, you potentially lose a lot of energy because your collision might get the neon "stuck" in the 3s states instead of getting it all the way to the metastable states you want. But for Helium, the two states you want are the lowest-lying ones - no states in between the atom can get stuck in,
so it's more efficient to excite the He as you cannot get these "half-way" transition to the state you actually want
@Slereah If it gets too out of control, you can try to imagine planning a course based on that book and condensing it into lecture notes :D It's more of a mental trick but maybe it can work
The ideal textbook is always somewhere between "Holy bible/sacred reference" level and "lecture notes" level, lol
Another technique is to decide what to move into an appendix
8:18 PM
Well part of it is just that there isn't any need for more introductory textbooks
I'm not really aiming for that
maybe you develop a new textbook art form for your book :-)
Nah big textbook of everything is far from new
i would like to do a footnote based book/lecture note style
something pedagogically written but with footnotes on any sentence that is actually not really true
@Slereah Ah that's cool then. So if it's aimed at an advanced audience it's really more tricky to decide about the format, 'cause everyone is looking for something different... at least I think so.. that the advanced audience is naturally more "heterogeneous"
But as I mentioned before, if you write for yourself you can't be wrong :D
@SillyGoose at which level of footnotes-on-footnotes-on-footnotes are you gonna start a new book that's just a footnote to the main book :)
8:23 PM
@ACuriousMind Ok sir I got your point. But now I am stuck in thinking that how a helium in n=2 can provide sufficient energy to Neon ( since bohr energy levels are proportional to z^2) to get excited into 5s or 4s. That would require energy very larger than He in 2s.
Well like every physicist, I believe deep in my heart that my exposition of the topic is clear and obvious, even though that is false
lol... "just give me 5 minutes and I'll teach you GR k?"
@ACuriousMind Lol a 10 volume series, 9 of which are filled from front to back with annotations
@Amit sign me up :)
Sure... first lesson... "this is gonna be 5 minutes in some frame of reference but not in this class's frame of reference..." 😂
don't dislike the timelike!
The first lesson is teaching u why spacetime is a set
hopefully for good reasons
8:35 PM
I like it, sets are more friendly than manifolds
although i guess i get why such a footnote styled book does not exist. either you don't care for such (perhaps needlessly) thorough exposition, or I assume by the time you realize you would like such an exposition you can just fill in the holes of your knowledge yourself just by jumping from source to source :P
but it would be nice to have everything in one place
@ShikharChamoli Energy levels don't work like that: The proportionality to $Z^2$ from the Rydberg formula holds only for hydrogen-like atoms, for atoms like Neon the computation is not so simple
@Amit But is space a set
We'll find out
Maybe space turns out to be gunky
@SillyGoose Maybe history teaches us that by the time a physical subject is perfectly codified it is superseded by a more general theory... I am thinking about Lagrangian and Hamiltonian mechanics... so maybe it really is a good idea, we gotta do this because it must happen before GR is superseded :)
@Slereah lol, we need @RyderRude for this line of thinking
No I think I'll go with someone who actually studies philosophy
8:38 PM
@Amit i have been led to think that it is being able to precisely formulate one's questions which is 90% of the difficulty of solving a problem :P which agrees with this
@Slereah what if it's a generalized space where the underlying set of points does not encode all relevant information D:
spacetime as an $\infty$-stack
@ACuriousMind I think the cool modern framework for gunk is locale theory
oh yeah, my locale is very gunky
@SillyGoose Yes, that does make sense. But often the remaining 10% looks insurmountable too :)
8:41 PM
that is true :)
also is there a reason a priori to think that there should be an way of writing something in terms of an integral vs. in terms of usual algebraic operations
@Slereah Nice roast, lol. JK :)
like the bessel functions have an integral representation and more recently i found that the derivative of the exponential mapping has an integral representation
One thing I want to do is some exceedingly complicated framework to work out the classical limit of GR
Because boy that's a lot of different things
There's like 3 or 4 different factors leading to the classical limit
@SillyGoose differentiation is mechanics, integration is art
this is the paper if anyone is interested: pubs.aip.org/aip/jmp/article/26/4/601/227573/…
8:44 PM
@Slereah You mean recovering Newton's gravitational law?
@Amit That is the Ultimate one, yes
But that's kind of a lot of different factors there
I did notice textbooks kind of do it sloppily one way or the other
And introduce unmotivated unclear assumptions to get to the answer they need lol
It's apparently not even guaranteed that the spacetime is causal if you just do $c \to \infty $
and course it's not even true that it reduces to the Newtonian limit
The FRW metric does not reduce to it
Isn't that related to how they used to think the electric field reacts backward in time?
@ACuriousMind perhaps now more than ever is time to try and pick up the art of integration :P i have neglected calculus for a very long time
8:46 PM
I don't think so
nah, integration sucks
just buy a big book of integrals and look stuff up :P
@Slereah I thought it's a feature of all "action at a distance" formulations, like Newton's gravity law
i like me some algebra and algebra only :P
perhaps also topology but they seem kind of similar in how the textbook theory is set up
8:48 PM
@Slereah I guess this is what I was referring to... but I read about this stuff only in a non technical manner, idk if the same thing happens in a gravitational field classically
Wrong type of causality issue :p
ok :)
I think for Newton you basically need like integrable spatial section + linearization + Wigner contraction
or something like that
Besse has the rigorous version of linearized gravity
Gotta use the moduli space
9:18 PM
0
Q: Has activity on Physics Stack Exchanged decreased over the years?

EEHWhen I look through questions on stack exchange, it appears that the community used to be more active. It feels like the most often years I see are 2013-15. The majority of questions I come across appear to be older than say 2016. Is this just a sampling bias or did stack exchange use to be more ...

the reference given is... Riemannsche Flachen and Langenspectrum vom Trigonometrischen Standpunkt aus
Damn Germans
here, you lost these up there: ::
I took them down to use as ellipsis
fortunately another reference is from the best land
Déformations localement triviales des métriques Riemanniennes
Atlantis?
oh, you just mean French :(
Ironically I can find neither paper
Thank u Google Scholar

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