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12:55 AM
@Sanjana hm where does the fussing occur?
 
1:24 AM
@Sanjana have you tried greiner?
@MoreAnonymous Amazon is quite known for its toxic workplace that even veterans would cry at.
@SillyGoose It is that T matrix's definition Equation (6.10) again
@ClaudioMenchinelli Could you convert $\hat z{}^2$ into trigonometric functions, then convert that to $\hat L_{\pm}$ and $\hat L_z$ operators? Before converting that to $\hat J$ operators. The radial parts obviously drop out and you can use tables of generalised Laguerre polynomial integrals that is actually worked out. In fact, it is often derived without doing any actual integration.
@ManasDogra You have the answer now, but I would have put it this way: You have a $\sum_nf_n$ where $n=\frac L{2\pi}k$. The crucial insight is to note that $\Delta n=1$ is to be implicitly multiplied to the original sum, so that it is $\sum_nf_n\Delta n=\sum_kf_k\frac{2\pi}L\Delta k$ so that when you send $\sum_k\Delta k\quad\to\quad\int\mathrm dk$ you still have an extra factor $\frac{2\pi}L$
@MoreAnonymous You have not at all understood what @lucabtz was trying to tell you. If your stationary rock had infinite mass (and pick it to be the origin), then it is a Schwarzchild black hole of infinite size; this means your entire universe is the inside of this black hole and you cannot even speak of measuring any motion. You are not making any meaningful physical point.
@lucabtz Would it be slightly better if it were Ramen please?
 
2:02 AM
how can a boson's spin degree of freedom be described as a two level system?
 
 
2 hours later…
3:43 AM
:65694795 that is a very famous mathematical trickery that explicitly violates spin-stats theorem but gives the correct results. It is not meant to be physically meaningful, only mathematically correct and convenient
 
 
1 hour later…
5:07 AM
@naturallyInconsistent Is there a book by Greiner that discuss scattering?
 
@Sanjana no. However, I have read many coverage of the partial wave analysis, particularly the Coulomb wave phase shift formalism of quantum scattering, and if there is any series that ought to have it, Greiner's many books should have a coverage.
and his coverage of stuff are so good
 
Yeah...True.
But, I had already gone through those stuff, but in a different manner. My issue was with Sakurai's particular way of doing it.
In the way I learnt it, the $\varepsilon$ regularization Sakurai is using came out naturally while trying to regularize a pole in a Green's function integral that came out while solving Lippmann-Schwinger eqn. Here, he shows that regularization should also be done in the "definition" of T matrix (which is equivalent to other more standard definitions, which he derives later)
 
@naturallyInconsistent Thanks. It is much clear now.
 
5:24 AM
@Sanjana errm, so the problem is?
 
@naturallyInconsistent Why are we regularizing the $t_0 \to -\infty$ and not the $t \to \infty$ limit? And also, Sakurai discuss about the fact that $\varepsilon t \to 0$ carefully although the limit on $\varepsilon$ is to be taken before the limit on $t$ is taken, so why discuss this at all? Practically speaking, the $t \to \infty$ turns out to be trivial anyway...
Again the way I did it, the first question had the simple answer that depending on the b.c.s we had a $+$ or $-$ infront of the $i \varepsilon$. Here the answer should be the same, but I don't see that directly in the way he discusses this. I mean, I don't see the motivation, right there.
 
@Sanjana Again, earlier miao miao told you that miao miao cannot divine what it is that Sakurai is thinking about, but meowthinks that as long as you see it done for one side, it is kinda obvious what we can do to treat both sides.
 
@naturallyInconsistent Yeah... Indeed, Sakurai has a section on it, where he says that it can be done and it corresponds to the other b.c...
@naturallyInconsistent Hmm. That's why I was asking if the Goose wondered about this too and solved it ^_^
2
 
5:41 AM
meowth. More likely for you to play with Sakurai that way.
 
I feel somehow although Sakurai, Goldstein, Jackson --- these books are "standard", some sections out of it are (for good!) left out in most lecture notes.
E.g. I had a friend whose supervisor mocked him because he didn't learn Schumann resonances which is discussed in Jackson. So according to the supervisor, the topic is standard by definition because the book is! Is that reasonable given that not all grad courses on E&M discuss Schumann resonances, ig?
 
I just think that is inevitable. Lecture notes, by definition, are meant to be done within the constraints of lecture timeframe, and so necessarily miss stuff. They are thus not meant to be comparable with proper textbooks.
Even though miao miao always manages to have too much time left in lecture class, because miao miao thought things through and made the lecturing flow smoothly, there will still be quite a bit of stuff that miao miao would leave out of class time, simply because it makes no sense doing them in class.
 
 
3 hours later…
8:28 AM
@naturallyInconsistent I can add a field to the geometry side for a cancellation
 
@MoreAnonymous no you cannot, and you have no understanding what it is you are talking about there.
 
I could flesh out some calculations if you wanted me to
 
8:47 AM
You should. Before you continue with more of this.
 
@naturallyInconsistent is this a response to me? sorry I can't see what the response tag is
 
@naturallyInconsistent i love ramen
do they make good ramen there in Taiwan?
 
@SillyGoose correct, you deleted the comment, that's why. It is to the Schwinger spin boson trickery
 
oh
but how can that be correct if it violates spin-statistics
 
@lucabtz obviously. Japan is too close by. And you can even find much less salty ramen here.
 
8:56 AM
@naturallyInconsistent i was in taiwan yesterday for several hours :D
 
@SillyGoose It is explicitly a mathematical trick and explicitly stated to violate spin-stats, i.e. it cannot be a correct physical thing.
@SillyGoose did ya have fun? Miao miao definitely did. Brought someone home miehehehe
 
oh
@naturallyInconsistent XD i had a boba milk tea it was pretty good
 
@SillyGoose there are other mathematical tricks that violate spin-statistics you will find
 
hm but does that just mean that this toy model is not realistic
 
@naturallyInconsistent my fiance is from the philippines and ive been there several times. the ramen is great there too
 
8:58 AM
it is manifestly not realistic, by virtue of violating spin stats
@lucabtz om nom nom~
 
how strange...
i don't get how this will work out to be a model of anything but i will see i guess :P
 
@SillyGoose but it is not? There are plenty of such accounting-like tricks that are manifestly unphysical yet quickly generates correct answers.
 
because it presumes a 2D representation of $\mathfrak{su}(2)$ that is not the unique (up to isomorphism) irrep whose representation space elements certainly transform as spinors
but i mean mathematically nothing of the like can exist
 
e.g. the entirety of using levi-civita tensor in Cartesian, cross products, etc.
 
but i mean if this generates correct results, is it not evidence that representation theory is not a good way to organize quasi-particles
 
9:01 AM
@SillyGoose yes; the whole point is that it quickly generates the correct stuff to populate any matrix you want, but it is ultimately just trickery. It is never asserted to model spins correctly.
 
why so many people do not use the latex package physics
tbf i wish it was included on this site as well
\frac{\partial f}{\partial x} vs \pdv{f}{x}
 
@naturallyInconsistent hm can u elaborate on these
 
@naturallyInconsistent i love a-sha -- is this popular there
@lucabtz omg im going to cebu this wkd
 
@SillyGoose All of those are actually supposed to be redone in terms of, say, Hodge dual.
@Relativisticcucumber not sure what ya mean; there are too many possibilities.
@lucabtz not very impressed by it; miao miao uses the COOL package
 
@Relativisticcucumber ive never been to cebu
@naturallyInconsistent whats that
@Relativisticcucumber you should try lechon baboy, they say it is very good in cebu
its a roasted pig
 
9:12 AM
@lucabtz look it up~~
 
wait i must be missing smth @SillyGoose @naturallyInconsistent because it seems atoms, e.g. Rb87 are bosonic and if we look at the "ground state" of Rb87, it seems there are two states for the lowest energy (i guess two comes from zeeman splitting or smth) so why cant these be the two states in discussion?
@naturallyInconsistent in the states a-sha is a famous ramen and its supposedly the most famous ramen in taiwan but ive always wondered if thats actually true
 
@naturallyInconsistent ctan.org/pkg/cool ctan is down
 
@Relativisticcucumber the energy degree of freedom lives in the infinite dimensional factor of Hilbert space. at least that is how i see it. in the discussion in the text, it is presumed that all of the infinite (i.e. spatial) degrees of freedom between the two spin levels are the same
 
@SillyGoose im not sure what you mean
you just want a two level system, right? so you just need a system where only two levels are practically accessible, no?
regarding the spin
 
@Relativisticcucumber The wacky fowl is talking about a trick that does not correctly capture every aspect of spins.
 
9:17 AM
in chapter five, the two levels are presumed to be two spin levels. i.e. to disregard all degrees of freedom that are not spin.
 
yes
im referring to fig. 4.1
i think i dont understand what ur saying
 
@Relativisticcucumber at least miao miao doesn't know of it. But it might well be for home-cooking?
@lucabtz OH NOES!!!! CTAN can never be down!!!
 
@Relativisticcucumber oh i think chapter 5 is pursuing a different goal from chapter 4 i guess
 
? what is actually wrong w what im saying tho
 
chapter 5 is only considering spin degrees of freedom. so energy has no role
 
9:20 AM
4.2 is titled spin degrees of freedom
which is the section 4.1 the figure is in
 
but 4.2 includes in its analysis, e.g., orbital angular momentum which is a spatial degree of freedom
it also is including spin degrees of freedom to prior analysis it seems. because the punchline seems to be (4.6) and etc. which starts to talk about the creation and annihilation operators labeled not only by $x$ but also by $\sigma$, the spin d.o.f.
 
so then i dont really understand
im not v satisfied w "this is just a random math trick that is wrong"
 
i think i figured out something which maybe justifies it a little
 
i was reffing 4.2 bc the excerpt says to ref that section so i thought it should be relevant but ig not
@SillyGoose can u explain
 
what book is this?
 
9:23 AM
quantum atom optics
 
i am trying to airdrop to my laptop but its not working one sec :P
 
wait to clarify
ur q is
what r the two levels in discussion
right?
so what ab like a two component bec then
maybe im just rambling stupid stuff
 
@Relativisticcucumber i agree with you that you can have a two level system in which the two levels are occupied by bosons
i dont know if this is the case here though, but in general you can
 
yeah i think so too and i think those two levels can be described solely by spin
i thought thats what a two component bec is but i might be wrong
 
9:28 AM
I think this is not a "usual" representation of $\mathfrak{su}(2)$. In particular, the commutation relation has a factor of $2$ which eliminates actually the spinorial nature of the representation
 
@SillyGoose i dont like to read handwriting :P
 
@Relativisticcucumber this idk about
 
@SillyGoose idk what any of this is saying
 
@Relativisticcucumber it is more like how can a boson have only two spin-z states
 
@SillyGoose the two isnt really a huge deal, you can kill it redefining the $S$ operators rescaling them
 
9:30 AM
But the fixed factor in the commutation relation is what gives rise to the spinorial nature in the usual representation theory of angular momentum
 
@SillyGoose it isnt litteral spin, its just that two states systems are clearly bound to Pauli matrices
 
in particular, in usual theory of angular momentum, the commutator is fixed to $[S_i, S_j] = i\hbar\epsilon_{ijk}S_k$, with a factor of $1$. this ensures that rotations in $3D$ space perform a full rotation as you take the parameter from $0$ to $2\pi$
 
@SillyGoose but this spin is not angular momentum, it is not spin at all
 
@Relativisticcucumber
 
why cant you trap atoms of two spin types? then the two spin types are the two levels
 
9:33 AM
the hilbert space is $\mathbb{C}^2$ so clearly $SU(2)$ plays a role here, but it isnt related to angular momentum
 
@lucabtz hm but this seems to violate my definition of angular momentum
 
@SillyGoose most likely your definition is wrong
 
@lucabtz what do u mean oh no
 
@Relativisticcucumber it is normal to use pauli matrices to speak about two-level systems, but they do not represent the angular momentum
you model a two level system on $\mathbb{C}^2$, write the hamiltonian as $H = \frac{\Delta E}{2} \sigma_3$
 
what im confused ab is why we r trying to show that a boson has two spin states
 
9:35 AM
i take a state of definite angular momentum to be a state that transforms in an irrep of $\mathfrak{su}(2)$
 
with $\Delta E$ the energy gap between the two levels
 
instead of assuming we have two different types of states that an atom can occupy
 
@SillyGoose you can have irreps of su(2) which have nothing to do with angular momentum
 
hm so what is a definition of angular momentum
 
say an irrep of su(2) stong isospin is not related to angular momentum
 
9:36 AM
okay that is sensible
 
it is the same here
this is not su(2) rotations, is some other su(2)
 
@Relativisticcucumber my bone to pick...
 
so i think the name given in this text is just an utter misnomer lol
 
@SillyGoose ? wait what happened XD im so lost
 
9:37 AM
@SillyGoose you can think of the bosons as spinless in the su(2) rotations (trivial representation)
 
well @lucabtz is saying this is not a two level spin system at all. it has nothing to do with spin. at least that is my understanding
@lucabtz wait what do you mean by this
 
@SillyGoose i dont think that is true
 
@SillyGoose you call it spin generally even though it isnt related to spin sometimes
@SillyGoose you can think that the total hilbert space is $\mathbb{C}^2 \otimes S$ and $S$ is your usual representation of the su(2) rotations
im just saying take $S = {0}$ to be the zero dimensional space
so the bosons have spin 0
otherwise you may have degeneracy unless you add some spin dependent terms to the hamiltonian
 
@SillyGoose what does this mean
 
hm but the defined schwinger boson operators do not act trivially on the bosonic states
@Relativisticcucumber my bone to pick about the naming
 
9:40 AM
total hilbert space of a single boson im speaking about
 
@SillyGoose what?
 
@Relativisticcucumber i was complaining about the naming on the walk over
the use of "spinor"
 
im saying what is the difference between the first thing i said and the second
i dont care ab the word spinor -- i was under the impression that has nothing to do w this q?
 
it was a separate comment
 
i think they r the same tho
 
9:42 AM
well it is related to the question i asked
but it is just a separate comment
 
oh, i misunderstood since u liked it to respond to my comment
are we not trying to establish what the two states in discussion are?
imo that is not a rep theory question so im lost
i just wanna understand the q XD then i can just ask the author hes sitting next to me
 
that would answer some of it. i asked the question i was interested in answering though. and the question is representation theoretic
 
im not nitpicking ur q
i legit dont know what ur q is
 
at the beginning of chapter 5, he says consider $N$ two-level spin systems. These are bosons.
It is mathematically and physically impossible for a boson to have two levels of spin.
Via representation theory and spin-statistics theorem, respectively.
So what exactly is a two-level spin system for a boson
where in the question spin is interpreted as the spin part of angular momentum
 
ok i get it. i will ask
 
9:46 AM
@lucabtz hm i am not understanding
 
@SillyGoose i dont understand the problem
take a single boson
which can be in two levels $\left\lvert 0 \right \rangle$ and $\left\lvert 1 \right \rangle$
the first has energy $-\Delta E /2$ the other $\Delta E /2$
you have $\sigma_3 | 0 \rangle = - | 0 \rangle$ and $\sigma_3 |1 \rangle = |1\rangle$
so the Hamiltonian is $H = \frac{\Delta E}{2}\sigma_3$
this $\sigma_3$ is a Pauli matrix but has nothing to do with physical spin
people will speak about this as a qubit nowadays that quantum information is so popular
these two states are energy eigenstates NOT spin eigenstates, they have nothing to do with angular momentum
 
hm but the preceding chapter in this text is specifically about accounting for the angular momentum d.o.f. for BECs
because it starts with a heuristic derivation of BEC, then it accounts for spatial d.o.f. only, then weak interactions, then it accounts for angular momentum
 
@SillyGoose well i dont know the datails about the book, maybe this is not the case for the book
all im saying is that it makes perfect sense to put bosons in a two-level system
 
@lucabtz this i agree with
 
@SillyGoose then maybe I was misunderstanding you
 
9:54 AM
@lucabtz especially if the two levels are degenerate
@lucabtz The issue is that trying to treat the spin part of a bosonic system this way has no physical meaning except that it is convenient to do so
 
@naturallyInconsistent yeah if it is actually the spin dofs i agree
 
@naturallyInconsistent i think what i mentioned above is a sort of loophole. since we insert a factor of $2$ in the commutation relation, I think this will fix the generators such that rotations complete a full rotation as the parameter goes from $0$ to $2\pi$, mathematically eliminating the spinorial nature of the representation
 
@lucabtz and it explicitly said it is the Schwinger spin boson trickery in the screenshot that started this whole thing
@SillyGoose I dont think this is correct or whatever; the Schwinger spin boson trickery is explicitly violating spin stats and is only meant to quickly supply correct matrix elements. Don't try to read too much into it.
 
@naturallyInconsistent hm well it is mathematically true. i just checked
 
10:13 AM
@SillyGoose yes; as mentioned, it is a mathematical trick that is proved to give correct results but cannot be physically interpreted correctly because it fails spin-stats
 
10:40 AM
@SillyGoose hi. when we write the angular momentum commutation relations like this, $2\pi$ is no longer identified with a single full rotation physically. instead $2\pi$ is now two rotations, and $\pi$ is a single rotation.
basically, we're working with a rotation parameter that is twice the radian
so spinors still go to $-1$ under a $\pi$ rotation, while the other reps go to $1$
 
10:52 AM
e.g. the Pauli matrices go to -1 under a $\pi$ rotation
it's just a different parametrisation of the group becuz we have changed the generators
so the defining feature of spinors isnt that they go to -1 at 2pi (this depends on parametrisation convention)
the defining feature is that they go to -1 at whatever moment the other reps complete one full rotation
 
 
10 hours later…
9:18 PM
Wow, double digit silence?
99 hours later...
 

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