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02:56
I don't understand the last line. They told they would take $g^{-1}$, and instead of that they took $g$ and then suddenly it became $g^T$ . Why would that be true for a GL(2,$\mathbb{C}$) element?
03:21
@imbAF This is correct. In practice we just try to find as many linearly independent quantum states of the same energy (up to some tiny tolerance). It should agree with the statistical thermodynamics measurement of it (This is pretty stringent; very obviously correct or wrong because the thermal behaviour will be very different)
 
7 hours later…
10:23
In a two-level unperturbed system with unperturbed Hamiltonian $H_0$ and corresponding eigenstates $|\phi_1\rangle$, $|\phi_2\rangle$ and eigenvalues $E_1$, $E_2$ of it, any state $|\psi(t)\rangle=a_1(t)|\phi_1\rangle+a_2(t)|\phi_1\rangle$, a stationary state.

Now if a perturbation W is included, such that $W_{11}=W_{22}=0$ (for simplicity) and $W_{12}=W^{*}_{21}\neq 0$, then $H=H_0+W$ and the corresponding eigenvectors and values of this hamiltonian would be $|\psi_{+}\rangle$ with $E_+$ and $|\psi_{-}\rangle$ with $E_-$. An arbitrary state $|\psi(t)\rangle=a_3(t)|\psi_{+}\rangle + a_4|\
11:15
@Obliv there is a consultant company somewhere i live called M-theory who take their inspiration somehow from the string theory M-theory
11:36
@Sanjana This picture makes no sense as written, there are at least 2 typos/errors. Probably a case of someone not catching an error since the result remains correct, since it actually doesn't matter whether you take the $g^{-1}$ or $g^T$ action here - the induced representations are isomorphic
@imbAF 1. Your $\psi(t)$ is not a stationary state if both $a_i$ are non-zero. 2. It's just linear algebra. You have two different bases for your space and you're asking if you can express a vector in both of them. You should know the answer to this - what difficulty do you have in figuring out the answer here?
Perhaps, not being able to see the forest for the trees :P
especially when one finds themselves in unfamiliar territory
11:55
For a consultant company the M probably stands for money 💰 @SillyGoose
::back to lurking mode::
😎
12:11
hi
12:48
do u find the universe to be absurd? why or why not?
 
2 hours later…
15:16
@naturallyInconsistent hello, what if we are talking about crashing into an airbag and without an airbag? The impulse is the same but without an airbag the force will be higher, and without an airbag is more dangerous than with an airbag
@SnoopyKid correct, that is the time that is the issue; you can easily argue in terms of the time
Someone's displaying the 1st starred comment yet again?
@naturallyInconsistent ok so when we are talking in the context of striking with punches and kicks, the logic that "the higher impulse punch or kick despite having lower force will be more damaging" still applies then?
@SnoopyKid in a punch the time is always very short. There would not be a possible situation whereby you give a larger momentum with a much smaller force.
@naturallyInconsistent same goes to kicks, right
15:25
@naturallyInconsistent thank you for clearing the confusion
you're welcome
@RyderRude I find everything as absurd
@naturallyInconsistent Hi,
@LuckyChouhan look at the 1st starred comment
Haha @RyderRude seems a cool guy :)
@naturallyInconsistent from when you're doing a research? I mean how many years
@LuckyChouhan It's been 2 years with this company. But miao miao did other research earlier
@LuckyChouhan look up the chat logs if you want to actually know what happens.
15:33
@naturallyInconsistent Nice, do you have any research advice for me or for one who wants to pursue research? PS: do you call yourself miao miao ? I have noticed this several times
@naturallyInconsistent sure, lemme see
@LuckyChouhan yes. at work, to profs, to friends
What are you currently studying at uni? Which year are you? Where roughly are you?
What research area are you into?
@naturallyInconsistent I am currently doing my bachelor's in mathematics. I'm in 3rd year. I'm from India.
My friends in India are all struggling. I don't know how to help. Do you mind, and do you have the opportunity, to leave?
@naturallyInconsistent I like number theory, real and complex analysis and problem solving. I really want to improve my problem solving skills but you know there are always many problems which I can't solve so I get frustrated so should I stick to that problem or should daily study something new or a book?
You should limit the time spent daily on any specific problem. If in 2 hours you don't get progress, you should start on another topic entirely. Come back another day.
15:41
@naturallyInconsistent Oh no miao miao, I can't leave India for obvious reasons. I think you're from USA or Europe.
@naturallyInconsistent I see,
no, im much closer than you think
@naturallyInconsistent Are you also an Indian? I don't know but it may be the case that you're from China.
Middle-East??
pineapple island
@naturallyInconsistent Shri Lanka
but then again, miao miao is not from pineapple island
15:47
Haha,
@LuckyChouhan no...
@naturallyInconsistent Then Maldives?
You have a big chicken land neighbour to the North. They always want the pineapple island
@naturallyInconsistent So where do you live now?
@LuckyChouhan pineapple island
15:48
@naturallyInconsistent Taiwan?
Gaemi just passed by
ding ding ding
I don't know
you got it
like, just
Now I see, because I know about the tensions between China and Taiwan. Few weeks ago china did extensive military exercise nearby Taiwan
@naturallyInconsistent That was the hint. So you live in Taiwan?
15:52
@naturallyInconsistent okay The h bar acknowledges the fact that you like *cats.
16:19
do you have a different problem to work on today?
16:38
@ACuriousMind I am aware of the process of switching between different basis. The difficulty here was that the basis considered, belonged, to what it looked to me, two different systems. The system unperturbed and the system that is perturbed. And whenever we did basis switch, linear algebra, we would of course consider a single system. Unless considering the unperturbed and perturbed system as two different ones is wrong. In that case eveyrthing falls into place
@imbAF they are different systems because they have different Hamiltonians, but that doesn't change that their state space is the same - the whole idea of perturbation theory is that you only change the Hamiltonian, not the state space
But we did change the state space, didn't we? From $\phi_1$ to $\psi_+$
Isn't that considered a state space change?
$ \phi_1$ and $\phi_+$ are just state vectors, not spaces
no, still the same complex 2D space
you're literally just changing your preferred basis, the vector space itself remains $\mathbb{C}^2$
16:43
But they span the space?
Spans the same space
the 2D space?
i.e. $\left|\psi_\pm\right>=\left|\phi_1\right>\!\left<\phi_1|\psi_\pm\right>+\left|\phi_2\right>\!\left<\phi_2|\psi_\pm\right>$
But if that's the case, then I can consider two random systems, with a 2D state space, which have nothing to do with each other, are at different points in space and time, and use their basis kets for a state?
well, not exactly, but you're not considering two random systems here - one is a perturbed version of the other
16:46
So, what systems we are considering does play a role, not their state space dimensionality. Is that what you are saying ?
you would not be able to write $H = H_0 + H_I$ for the perturbed Hamiltonian if all these $H$ operators were not operators on exactly the same Hilbert space
Of course not, I am not opposing this
If I may
When we were introduced to basis switch, it was only explained from a mathematical pov, how to perform it. And I thought, that this occurs (change of basis to express some random state), when at a system, we consider two different observables, each with its own eigenstates, that span a basis. So one could choose either basis to express a state. I never though that you could pull the same when considering two basis kets in two different systems
@ACuriousMind And from what you said there are some limitations or conditions such as
you're probably imagining too much under "different system" here. One of the simplest examples of "perturbing" a system is turning on an external field. So the unperturbed system would be a particle in a trap, and the perturbed system the same particle in a trap but with a background electric or magnetic field
why would the kinds of states the system can occupy change just because you turned on a background field?
Or take a particle freely moving through space, the space of states is just the space of square-integrable wavefunctions; you can turn on various fields here or put some potential barriers or whatever, but in each of these "different systems" you will still describe the state of the particle by a wavefunction from $L^2(\mathbb{R}^3)$, the state space doesn't change just because there's different stuff acting on the particle
Because the systems energy would change, and as such the corresponding eigenstates
what do the eigenstates of the Hamiltonian have to do with the space?
your Hamiltonian is a different operator, so it has different eigenstates, that doesn't change the vector space
16:55
Also, there is a huge assumption in perturbation theory, which is that your state does not change toooooooo much. That is part of why we spend so much effort on energy eigenstates and their enumeration. Technically, it is possible for a low energy state, upon perturbation, to get a large contribution from a high energy state. We are somewhat assuming that this does not happen, justified by experience, and by perturbation theory showing that the contributions go by $\frac1{E_n-E_i}$
Because every state belong to that space, and can be expressed via the basis kets, which are different. When I mean the state space changes, I don't mean it's dimensionality but the basis kets do
That's what I believe, well, believed, a state space change was
@imbAF When the dimensionality stays the same, then any basis is good to expand any state you want
@imbAF Again, this is just linear algebra: Any given vector space has infinitely many different possible bases, and your choice of which basis to use preferentially does not change the space
you should know that there is only one vector space of dimension 2 up to isomorphism
I know that
there is simply no mathematical possibility for the space to "change" - we have only one
16:58
Ok so, the space stays the same, the difference is that the basis changes, when perturbation is intruduced
so one could switch between the two basis
right?
and again: By assumption the operators $H$ and $H_0$ act on the same space here. You could not equate them as $H = H_0 + H_I$ if they did not.
I feel a bit trolled by you claiming to understand that and then asking follow-up questions showing you didn't
This is very important and very much used: The $\left|\ell,m_\ell,s,m_s\right>$ basis does not commute with the Hamiltonian but is easy to state and enumerate, whereas $\left|\ell,s,j,m_j\right>$ basis commutes with Hamiltonian but is not so easy to find, so we expand it in terms of the stuff we can easily state.
@ACuriousMind How exactly?
@imbAF when I mentioned this argument the first time, you said "Of course not, I am not opposing this" and then everything you've said since then was still you arguing that there should be two different spaces here
The idea that the space does not stay the same for these two system is opposing the claim that $H = H_0 + H_I$
And I do believe, I did clarify, what different spaces meant for me
Prior to your explanation of what it actually means
17:04
technically you cannot say that you understood it when you have "what different spaces meant for me"
Yeah, what they meant for me before, is not what "different spaces" mean for me now.
17:33
@ACuriousMind oh ok. This was one of the books you recommended btw. Woit's QM and rep. theory book. The problem is that on the next page he derives representation of the Lie algebra and the expressions will change a lot...
@Sanjana why would you fix this in the way that changes all subsequent expressions and not just fix the representation to be $g^T$ from the start? :P
The 2nd yellow highlighted equation will be changed because some Pauli matrices are not symmetric
@ACuriousMind So I would define the action by acting with a $g^T$? But that's not what is mentioned in the beginning of the last screenshot I sent you, right?
Oh you said there must be some typo, that's why?
Did I not say that you can choose either $g^{-1}$ and $g^T$ and those both lead to the same representation in the end?
 
3 hours later…
20:48
As I am progressing with my notes, I now am dealing with the angular momentum, its algebra, eigenvectors and values etc. In the notes we generalized by considering an angular momentum $\vec J$, that can represent the orbital angular momentum, the spin or any other type. The following claim is made:
$\vec J^2$ and $J_z$ build no complete set of commuting observables, therefore an additional index is needed to distinguish the different joint eigenvectors belonging to the same pair of eigenvalues $j(j+1)\hbar$ and $m\hbar$. An eigenstate can be written as $|k,j,m'rangle$.
Or said simply: How do I know or how can I prove that $\vec L^2$ and $L_z$ do not form a complete set of commuting observables ?
The claim as stated is incomplete; you need to specify on which space of states it is made
As you already said, it's true if we consider the space $L^2(\mathbb{R}^3)$ for particles moving in 3d like in the hydrogen atom
That was the claim
@ACuriousMind since I reached this point. Could you tell me how to read this:
$L^2(\mathbb{R}^3)$ ?
it's the standard notation for the Hilbert space of square-integrable functions on $\mathbb{R}^3$, i.e. wavefunctions $\psi(x)$ where $x$ is a 3d vector.
The "L" comes from Lebesgue, see $L^p$-spaces
Aha so it's the same thing we write like $\psi(\vec r,t) \in L^2$
But regarding my initial post. You can't say anything right?
Cuz the claim feels like it tries to prepare one for when we study wavefunctions of electrons in H-like atoms
I've already said everything I can say - without a given space on which the operators $J^2$ and $J_z$ act, it's not really possible to speak of a CSCO - the notion of CSCO only makes sense for concrete operators on a concrete space
20:57
could you give an example of a space were the claim is valid ?
In order not to overthink this, I would simply modify the statement to say "$J^2$ and $J_z$ may not build a complete set..."
@imbAF did you not already do that with the hydrogen atom and I agreed?
@ACuriousMind Aha so my guess was right
@ACuriousMind This is more general I guess, so better claim
well, you are wrong in saying $f(r)$ is "responsible for $L^2$ and $L_z$ not commuting" - they still commute, the set just isn't complete
sorry
but the example of the hydrogen atom - or really, any particle moving in 3 dimensions - is still correct
20:59
Sorry, they do commute but they do not build a CSCO
I expressed myself wrong
There's one more thing that I want to address. But I will try a couple of times to understand it myself
@ACuriousMind why did you emphasize 3D movement? Ofc there are no 2D movements in a H-like atom, but we hypothesize a 2D system where a particle is involved
@imbAF if the particle is moving in less (or more) than 3 dimensions, angular momentum works differently; there is no $\vec J$ in the general case
this isn't something quantum, just a quirk of how rotations in different dimensions work; don't worry about it if you haven't run into it
I haven't encountered something of that sort. But hopefully I will
as a first step, you can recognize that the cross product of vectors exists only in three dimensions, so the definition $L = r\times p$ is not possible in dimensions other than three
Yes of course the mathematics will change
@imbAF I would have to point out that ALL basic quantum examples that you have seen in the textbook are 1D except for the H atom, so if it is not applicable to the H atom, why would the textbook point out that $J^2,J_z$ alone do not manage to completely pin down the $\mathcal H$ eigenstates?
21:07
I just thought that that wouldn't be a factor in the physical behavior. But of course I am not sure and don't know the case considered
can you, like, PLEASE focus on learning the basic quantum theory as it is and leave the complications to later?
@naturallyInconsistent Quite logical thinking. But, how could I know, what other systems behave like that
I thought the H atom would, because I know it. And my guess was correct
@naturallyInconsistent I am lol. The notes made a claim, without proof or nothing. That's why I asked
That H atom alone behaves like that is already sufficient evidence that the stated observation that the author gave is correct.
But we haven't spoken about the H atom yet
or any other 3D system
@naturallyInconsistent It's "correct" to someone who knows what the author is after; to someone putting every sentence under a microscope it is not
21:12
Am I wrong for doing that?
@imbAF look at your beloved-to-hate Griffiths's contents page. There is only one 3D thing being covered, and that is the H atom.
And besides it doesn't require to put the sentence under the microscope in this case. They made a claim, yet no reason why that holds. It doesn't require to much to grab one;s attention
@ACuriousMind it is the putting under microscope that is ouchies
@imbAF Yes and no; of course it is important to question the things you read, but as I have also advised others here:
Very often in the course of learning it is the right thing to do to just put something one doesn't understand into a box like "ask/think about later" and first see where the text goes with this and what it does with it instead of stopping dead in one's tracks and refusing to continue until every last possible worry is resolved
many questions will naturally go away when you see how something is used or built upon
and those that remain you can then ask more sharply with more context
I mean, if you have any intention of going into QFT later, the attitude of putting everything under a microscope will become fatal. After all, we don't really know how to get a mathematically pinned down QFT, nor how to teach QFT to something near that level of goodness, and so if you go after every single statement, you will not be able to progress at all.
21:16
Then it get's tricky to estimate when it's worth asking something and when it's not. I guess I will refrain myself from asking then
@imbAF I'm not trying to get you to not ask questions - but rather to try as much as possible to first finish the chapter, if not the book, to then ask the question with full context
at least in my personal experience, a statement that appeared cryptic to me when it first appeared becomes much clearer when you see what the text is going to use it for
Yeah you are right
I guess my OCD is the problem xD
Like, what if the next section of your text just applies that statement to the hydrogen atom? Then your question of "is the hydrogen atom an example of this?" would have just answered itself
@ACuriousMind +1
21:40
What is the most generic scenario in which an equation which holds within correlator brackets also hold as an opeator equation? E.g. When is $\langle \phi(x_1) \phi(x_2) \rangle =0 \implies \phi(x_1) \phi(x_2) = 0$ true?
If this is too broad to ask, I am looking for some particular examples... E.g. It is given in a note that this is valid for unitary CFTs. When else is it valid?
@Sanjana with respect to what state are these brackets?
@Sanjana 2d CFT or general?
@ACuriousMind The interacting vacuum of the theory
@ACuriousMind For general unitary CFTs
$\partial f$ is a null state: a primary as well as a descendant, that's why.
I have also seen in some other contexts where this happens. So I was wondering whether there's some theorem or something... you know...
If this this this happens in an interacting QFT, then you can forget about the correlator brackets.
Ah, it's just because conformal symmetry makes the two-point function of a field with itself $\frac{C}{\lvert x_1 - x_2\rvert^{2\Delta}}$, where $\Delta$ is the scaling dimension. So this being zero means $\Delta = 0$, but the field with scaling dimension zero is the zero field.
Note that your initial question was weaker than the claim in the screenshot: It's really $\phi = 0$ as an operator equation that follows here
@ACuriousMind Neat. But, where does unitarity play a role in this?
no idea, I hate when physicists use that word because they never mean the same thing twice :P
21:53
@ACuriousMind Got you. Yeah I remember some other conversations where you pointed this out
@ACuriousMind I mean this must be a proof sketch cz $\partial f$ is not a scalar although it looks like so.
I'm not sure what you mean
The notation makes it look like $\partial f$ is a scalar but it is not. Sorry for not giving that info earlier. You used the two point correlator of two scalars to arrive at the conclusion. You could guess that $\partial f$ is not a scalar when they say that they have to use spin $l$ correlators for the job.
supremely bad notation in that case :P
it's probably still an argument about how the conformal symmetry restricts the possible form of two-point functions, but I don't know it off the top of my head
@ACuriousMind Yeah that's why I said it is still a proof sketch...
(and I'm not going to research this anytime soon since tomorrow I depart for a week of heavy metal)
22:05
Wowww... Happy journey!
 
1 hour later…
23:35
@naturallyInconsistent what are some good books for linear algebra, which offer a good theoretical explanation and that it relates the mathematical concepts with ones in quantum mechanics
23:52
5
Q: How does metallicity change the mass-radius and mass-luminosity relations in main-sequence stars?

TerranAmbassadorI'm building a star cluster for my space opera setting and I'd like to include some highly-metallic stars as anomalies and resources. Wikipedia defines stellar metallicity as the fraction of a star's composition that isn't hydrogen or helium (i.e. if a star is 98% percent hydrogen and helium and ...

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