« first day (2496 days earlier)      last day (2418 days later) » 
00:00 - 17:0017:00 - 00:00

12:19 AM
@ACuriousMind I require the name of that sitcom.
 
@DanielSank It (sadly) doesn't exist (yet).
 
12:35 AM
@ACuriousMind instead of a PhD you should make a new Seinfeld
 
@0ßelö7 You know Seinfeld? I see at least some pop culture got to you ;)
 
@ACuriousMind The only things I don't know are old movies and Simpsons
 
I don't like the Simpsons either.
 
They're yellow - can't be good.
What are they hiding?
 
Yeah, and their voices are annoying.
 
12:49 AM
I think I'm more attuned to the zeitgeist than anyone here, barring @BalarkaSen
 
(semi-removed)
Welp, 7 days and counting...
Welp whelp
Have you @0ßelö7 seen the movie called "The Raging Bull"?
 
 
1 hour later…
2:17 AM
@BalarkaSen TITLE: Group Actions on Large Scale Spaces II
We will prove the following theorem:
Let X be a discrete bounded geometry metric space and let $\mathcal{X}$ be the large scale structure induced by the metric on X. Let G be an amenable group that acts on X by coarse equivalences. If $(X,\mathcal{X})$ has property A, then $(X,\mathcal{X}_G)$ has property A, where $\mathcal{X}_G$ is obtained from $X$ by trivializing the action of $G$.
 
Uh, cool story
I don't know what that means though
 
Me neither
the last talk was pretty vague
I asked you for geometric group theory references and didn't get a response
 
Oh, did you?
I guess I didn't see the ping
 
Aug 30 at 20:08, by 0ßelö7
@BalarkaSen what the hell is geometric group theory and how do I learn it
the talk went off the rails, something about Novikov conjecture and KK theory
the K theory of a K theory?
something like that
 
I don't know that flavor of coarse geometry. For geometric group theory; I haven't really tried to read it thoroughly but I have heard Kapovich's notes are good
Alternately you could dive headlong into Gromov's "Hyperbolic Groups" paper but he's unreadable
 
2:30 AM
unreadable by your standards or mine?
 
mine. I have never been able to read a single thing by Gromov
or at least whatever I tried to read
 
what's the issue?
 
Exposition, I guess. It's all sorts of ideas clamming into each other hither and tither.
I'm sure you can find better references that what I mentioned by just googling though
 
meh, my reading list is out of control
 
pick 5 essentials out of your reading list and burninate the rest
Like, really
 
2:35 AM
@BalarkaSen I was commissioned for some talks at the new geometric analysis seminar
one on my research, another on regularity theory/Laplacians
so there's two books on my reading list
 
gr8
 
@BalarkaSen do books for class count?
 
probably not
 
ok, currently reading chapter 5 of "Classical Fourier Analysis"
that's for a class though, maybe
@BalarkaSen Ok I'm not actively reading 5 books.
But there are 10 books that are waiting for me
@ACuriousMind Jeez, how many things are you subscribed to? I just got an email with 50 papers
 
3:19 AM
@vzn I can't do anything before the 26th of Sep!
 
3:45 AM
o/
 
 
2 hours later…
5:28 AM
@skullpatrol o\
 
 
2 hours later…
7:03 AM
Weapon idea from one of my chemistry collegue:
Alcuberrie cannon:
Given that we knew there's this horizon problem with the warp bubble, and the divergence of energy at the boundary of the bubble which result in the energy to be released to the surroundings when the drive stopped at its destination
, we can perhaps exploit this fact by propelling e.g. a lead bullet sized object in a warp bubble, then disengage the warp bubble to deal massive destruction at the target
But now, with this idea in mind, things will get interesting. Since the alcuberrie drive effectively allow FTL travel, it means such scheme will allow the realisation of FTL bullets, so the dynamics of what happens when a warp bubble collide with some matter energy will be important. Perhaps spacetime warps just right to preserve causality
 
 
1 hour later…
Anonymous
8:26 AM
I need some advice on this: I need some details and clarifications regarding the methods used in this paper. Also, I want to know about the recent developments they have made in this area (if any). Would it be better to mail just mail one of the authors and wait for a reply and if they don't mail the next author. Or, should I mail all of them at once? Or, should I send out blind carbon copy mails?
 
Anonymous
I wouldn't want to come across as a desperate and pestering mailer. Moreover, they might not be willing to reveal some of the developments, but that's okay. :P
 
Anonymous
It seems the authors are from UW Washington...
 
9:10 AM
@Blue easy: email the corresponding author
they're usually easy to identify because their email address is on the paper
if there's more than one corresponding author, send a single email to all the addresses on the paper, so that they know the others have also received it.
 
 
2 hours later…
Anonymous
10:50 AM
@EmilioPisanty Thanks a lot. I'll do that =)
 
11:03 AM
@ACuriousMind How is this unclear?
it's a shit post, but it's quite clear what the core question is
 
@EmilioPisanty That's a link to this room ;)
 
user84215
Can you tell me which moderator has removed my pin on one of my posts?
 
What was the post? I'm probably the main depinner hereabouts.
 
user84215
the first post in the physics workshops room
 
11:08 AM
Link?
 
You mean a post in a different chat room? If so it wan't me since I only have power over posts in this room.
 
@EmilioPisanty I agree the main question is probably how Voyager took photos, but pretty much in every sentence of that question I find myself wondering: What is this sentence referring to? What old post? What low light? What images exactly? Why would the speed make them blurry if the target is far away?
I think it would do us a disservice if we communicated that such a post passes as "clear" here.
 
user84215
@ACuriousMind Can you tell me why I can not pin my message?
 
@ACuriousMind I think we're served well enough by piling downvotes on the thing, but OK
 
11:12 AM
@MathematicsAminPhysics you can't pin it here because you aren't a room owner here.
I can't see the room owners in the Physics Workshops room, but it looks like you're a room owner there so you should be able to pin it there.
 
@MathematicsAminPhysics I haven't been to your room at all and moderation action on stars/pins is one of the few things that are not logged (at least not in a place I can see), so I have no idea who might have acted on that message
 
user84215
I am the room owner, and I can not pin my message.
 
Have you happened to previously unstar/remove pin them? I recall in one of my rooms after I remove the star, somehow the system thinks I have removed the star and thus the message cannot be pinned anymore
 
user84215
Maybe
 
@JohnRennie "Can Superconducting Cosmic Strings Piercing Seed Black Holes Generate
Supermassive Black Holes in the Early Universe?"
I think vixra has leaked into arxiv
 
11:17 AM
:-)
 
user84215
Now what should I do?
 
@MathematicsAminPhysics I've no idea why you can't pin posts in your own room. Maybe send a mail to the SE team?
 
Back in that time, my message is short and incoherent enough, thus I just retype and star/pin the new one. But if the original is really important, I guess you might need to ask the SE team about how exactly this works, or whether this is a bug
 
Hm, I can pin messages more than once but that might be a mod power.
 
@0ßelö7 That's a really weird title, but breifly skimming, at least maths layout-wise, definitely not vixra material
as for physics-wise, I lack the background to comment
Johnrennie, try pin, unpin and repin a test message in this chat and see if you can repin it after you unpin/remove star it
 
11:22 AM
Pinned ...
Unpinned ...
Ahhhh I can't pin it again.
 
user84215
Yes, I think this is also my problem
 
What does the error said. Is it something along the lines of "you already unstar this message"?
 
In that case, the system is working as intended.
 
user84215
My error is "You have already voted, but the voting has been cleared by a moderator"
 
@ACuriousMind so is American politics
Doesn't mean it's not broken
 
11:26 AM
@ACuriousMind why is it a good idea to stop chat owners from multipin a message?
 
@MathematicsAminPhysics yes, that's the message I get too. You'll need to repost the message as a new post then star and pin the new post.
 
@Secret Presumably because the star board is intended as a "recent highlights" thing, not a "best of all time" thingy. If there is something you want to tell users permanently, stick it in the room description, not the star board.
 
user84215
Thanks
 
Right, make sense
 
11:45 AM
I know I've asked this before, but
wasn't there a user with this username before?
 
user84215
I want to send a question on the Meta and advertise the physics workshops room. I hope this is not off-topic.
 
this one's marked as Member for 10 days
 
Two "delete my account", sounds suspicious
 
@Secret har har. I obviously looked already.
 
@MathematicsAminPhysics not on the Physics meta please
 
11:47 AM
@MathematicsAminPhysics you can submit an advert on the community-ads thread
 
user84215
@JohnRennie Questions about chat rooms are off-topic?
 
yeah, they are not related to the main, afaik
 
user84215
@EmilioPisanty How can I do that?
 
@MathematicsAminPhysics the physics meta is for questions about how the physics SE works. That's all.
 
11:48 AM
12
Q: Community Promotion Ads - 2017

Grace NoteIt is a bit late into this new year, being that we're already in the second month, but we are now cycling the Community Promotion Ads for 2017! What are Community Promotion Ads? Community Promotion Ads are community-vetted advertisements that will show up on the main site, in the right sidebar....

post an answer that's up to the specifications
if it gets score 6 or bigger it will be displayed on the main site
... which of course means that you need six or more people to agree that it's worth advertising
@Countto10 Thank you for your effort. You did more than you had to. — Adam Feb 28 at 19:21
^ also that?
 
user84215
Thanks.
 
It occurs to me that if you placed them very, very far apart, such that more of the universe was between them than was outside them, wouldn't they be pushed apart for the same reasons? Could this be what causes the universe to expand? I do appreciate that this is a thought experiment only, but the universe is not a "container" in the sense of anything we observe on Earth. I am not sure that you can say more of the universe is on one side of the plate than the other. — user146020 Mar 15 at 21:55
user 146020 was countto10
but why is he constantly deleting his account?
 
Also possibly user108787?
 
(One reason I can locate his comments is because 1) Search bar = countto10, and then trace back and 2) I really don't like that emotion induced when I look at his comments
 
11:55 AM
@ACuriousMind & mods, can you confirm that?
 
He strikes me as ... snarky
 
and is there a way to tell whether it's the same person serially deleting and recreating an account with the same name?
That strikes me as vaguely off w.r.t. the rules of the game
 
I know nothing about the under the hood stuff of user accounts, Acuriousmind and other mods may be a better person to ask
but presumably each account has an IP address?
 
@Secret I'm not sure that's available to mods
but given that this is the (third?) time this happens, maybe it's worth pinging the community team to see if there's something weird going on?
 
sounds like a good idea
 
12:27 PM
@EmilioPisanty Deleting and recreating your account is not against the rules unless you're circumventing some sort of system restriction by that (like question rate-limiting or suspensions)
 
Countto10's comment may be snarky (at least how I felt), but i don't recall him running into any trouble with the community nor have a bad reputation
People deleting accounts and recreating a new one is also not unheard of. They do it for a variety of reasons
 
12:49 PM
fwiw, Countto10 is a very nice/good user. They delete their account because they do not like their rep getting too high (for whatever reason) and crazily enough, if they didn't, they'd probably over 100k now.
and i sincerely doubt anything weird is going on @EmilioPisanty/@Secret
 
Though this is a valid reason, why will someone worry about getting high rep?
 
::shrugs:: I don't know, each to their own.
 
What is up with Count to 10?
 
@heather I doubt they'd be anywhere near 100k, but yes, they'd have a lot.
 
@ACuriousMind well, I think most times they delete an account when it reaches ~10k , and I know they've deleted their account a lot, so. But you're right, maybe not 100k.
 
12:56 PM
What happens when someone get 10K rep?
 
1:11 PM
hi :-)
 
hello @AccidentalFourierTransform
 
how are you doing?
back to high school already?
 
pretty good, how about you?
yep, back to high school
long weekend though, so i get to relax (labor day)
 
im good too, just came back from a week at the beach with my family
 
nice
 
1:15 PM
yeah, I have the nicest of the tannings :-P
 
@ACuriousMind I feel it does have an impact on accountability, particularly if that user participates further in discussions on meta
 
@EmilioPisanty maybe so, but that's certainly not this user's intention (at least as far as I've experienced with them).
 
do we know anything about countto10? I sometimes feel s/he is a kid, sometimes I feel s/he is an old person...
Emilio doesn't like cto10 to delete their account because of the rep loss every time that happens
 
nope, i think they have enough that the rep loss doesn't happen.
 
well in this case, it does happen
every time s/he deletes his/her account, I notice it :-P
 
Anonymous
1:22 PM
@AccidentalFourierTransform You're back! We were missing your -+++ memes :P
 
-5 or -10
@Blue oh hai
I dont have anything at the moment, but I'll ping you when I post a new one :-P
 
Anonymous
15 hours ago, by ACuriousMind
@Blue Probably gone into hiding to craft better +--- memes.
 
I would appreciate it if someone could explain-to-OP/mediate/step-in/vote-to-reopen/vote-to-close here.
2
 
welp, i can't, because i disagree with your closure.
It's a good physics question. Put some kind of mod disclaimer banner on it if you will, but it's on-topic, and a good question.
 
Anonymous
@heather You can surely vote to reopen in that case (?)
 
1:28 PM
@Blue yeah, I did.
 
@AccidentalFourierTransform don't you dare
 
blasphemy!
you love my maymays
 
Rule #1: Don't dare a mod @0ßelö7
 
Anonymous
@skullpatrol What?
 
Anonymous
When did he dare a mod? lol
 
Anonymous
1:32 PM
AFT is not a mod...but well, he would be the coolest mod ever :P The meme mod
 
Is @AccidentalFourierTransform a mod?
I dare you to ban me
 
@AccidentalFourierTransform I don't particularly care about any such rep loss
But if it's the same person, I'd like to know
 
@Blue if I were a mod I would spam the mod chat with memes 24/7
 
Oops @0ßelö7 my mistake
 
1:34 PM
and I would delete 0celo7's account
 
@AccidentalFourierTransform Do it
I dare you
 
@EmilioPisanty prove it. Delete your account and create a new one.
3
 
@AccidentalFourierTransform har har
@Qmechanic question looks OK to me
 
Me too. @Qmechanic
 
Anonymous
1:37 PM
@AccidentalFourierTransform That sounds fun. Nominate yourself for the next election and I'll vote for you. ;)
 
I will nominate myself, win, and delete my account
wonder what would happen
 
@AccidentalFourierTransform The runner-up candidate would be called to serve as mod.
 
what if they deleted their account too
and the rest of the turtles too?
 
Highly unlikely.
(given human nature :-)
 
I shall need help from other trolls then
 
Anonymous
1:46 PM
Re-election would happen :P
 
Re-re-...-re-election.
 
Anonymous
What if everybody except the current mods deletes their profile?
 
Anonymous
Oh, then no election...lol
 
@ACuriousMind What if someone deletes their account and then someone shows up to claim it?
How do you determine if it's legitimate
 
Congrats @heather on your CSed challenge question
 
Anonymous
1:58 PM
@0ßelö7 That's not gonna happen. If if they claim it they aren't going to get it back (most probably)
 
@0ßelö7 There is no way to claim deleted accounts, deleted accounts are unrecoverable.
 
^yep, I tried.
 
@skullpatrol thanks =)
 
Good job.
+1
 
Sid
Heh, our Maths test got postponed today(Or rather, we postponed it.) Our Maths Professor looked pissed.
 
2:06 PM
Good job
or bad? (depending on the topic :P)
 
Anonymous
@Sid Why?
 
Sid
Basically, we weren't prepared. Parties and all, meant no one really studied. So, we asked(read: demanded) for a week's delay. Then, the Bio teacher rushed in and said he wanted to take the class. So, test postponed
But, damn, the Maths professor seemed pissed off
 
Anonymous
@Sid Not a good sign....:)
 
Sid
And we have to submit a Math assignment tomorrow.
Better do this one..
 
Anonymous
In solid state physics, a particle's effective mass (often denoted m*) is the mass that it seems to have when responding to forces, or the mass that it seems to have when a mass with other identical particles in a thermal distribution. One of the results from the band theory of solids is that the movement of particles in a periodic potential, over long distances larger than the lattice spacing, can be very different from their motion in a vacuum. The effective mass is a quantity that is used to simplify band structures by modeling the behavior of a free particle with that mass. For some purposes...
 
Anonymous
2:15 PM
The wikipedia article doesn't state how how calculate $m_x*$ when $m_x$ is given. They are just defining it. Any idea how to find effective mass along an axis?
 
Anonymous
Suppose the E-k diagram is given and we take the parabolic or ellipsoidal approximation
 
solid state physics is not even wrong
 
Anonymous
@AccidentalFourierTransform I wholeheartedly agree XD
 
Anonymous
They use hand-wavy QM
 
Anonymous
Anyway, do you know the answer? :d
 
2:18 PM
well, $(m_x^*)^{-1},(m_y^*)^{-1},(m_z^*)^{-1}$ are the eigenvalues of $E_{,ij}(\boldsymbol k_0)$, right?
 
Anonymous
@AccidentalFourierTransform What is $E_{,ij}(\boldsymbol k_0)$ ?
 
Anonymous
What does that comma stand for?
 
the Hessian of $E=E(\boldsymbol k)$ evaluated at the equilibrium position, $\nabla E(\boldsymbol k_0)\equiv 0$
the matrix of second derivatives
the comma stands for a derivative
$E_{,ij}=\frac{\partial}{\partial k^i}\frac{\partial}{\partial k^j}E(\boldsymbol k)$
 
Anonymous
I don't know what Hessian is. Perhaps you could simplify the explanation by taking 1D instead of 3D ?
 
ok
you are given a function $E=E(k)$
 
Anonymous
2:21 PM
@AccidentalFourierTransform Okay
 
you calculate its minimum by solving $E'(k_0)=0$
 
Anonymous
@AccidentalFourierTransform Yes, agreed
 
then you calculate one more derivative, and evaluate it at $k_0$
you get $E''(k_0)$
 
Anonymous
Alright
 
finally, $m^*=1/E''(k_0)$
 
Anonymous
2:23 PM
@AccidentalFourierTransform Ooooo....gotcha!
 
Anonymous
Thanks
 
in three dimensions, you need $k_0^1,k_0^2,k_0^3$, and you find those by solving $E_{,1}=E_{,2}=E_{,3}=0$
where the comma denotes a derivative with respect to $k^1,k^2,k^3$
three equations, three unknowns
you then calculate the matrix $E_{,ij}$, which means that you take all possible second derivatives
and arrange them as a matrix
$$\begin{pmatrix}E_{,11}&E_{,12}&E_{,13}\\E_{,21}&E_{,22}&E_{,23}\\E_{,31}&E_{,32}‌​&E_{,33}\end{pmatrix}$$
 
Anonymous
Generally are they equal for the three orthogonal axes? I mean is usually $m_x*=m_y*=m_z*$ ? Or do we have to calculate them separately? And what would be the reason that they are not equal along the 3 axes (if so) ?
 
where for example $E_{1,2}=\frac{\partial^2 E}{\partial k^1\partial k^2}$
you then evaluate that matrix at $\boldsymbol k=\boldsymbol k_0$ which you calculated before
finally, you diagonalise that matrix
the inverse of the eigenvalues are $m_x,m_y,m_z$
in general, $m_x\neq m_y\neq m_z$
they are only equal if the material is isotropic or homogeneous, I dont know
 
vzn
@AccidentalFourierTransform hi whats new, you cited koopman-von neumann QM awhile back, have been meaning to ask, have you studied it any?
 
2:27 PM
probably isotropic
 
Anonymous
@AccidentalFourierTransform Oh. Makes sense :)
 
@vzn nope, I just know it exists
 
vzn
@AccidentalFourierTransform have you heard of any nice refs on it delving into it more deeply? maybe have heard it alluded to occasionally...
 
Anonymous
@AccidentalFourierTransform Do you know any source which states the values/graphs of $E_1,E_2,E_3$ for Si and GaAs semiconductors?
 
Anonymous
I just wanted to see how much different they are along different axes
 
2:30 PM
@vzn no, sorry. A mathematician friend of mine told me about it. I have read the wikipedia page only, nothing else
@Blue any book on solid state physics should contain graphs like that
 
Anonymous
@AccidentalFourierTransform My book gives only 1D graphs :P
 
Anonymous
Could you name any book which has it?
 
buy a better book :P
 
Anonymous
@AccidentalFourierTransform Like?
 
but there is a reason for that: 3D plots are cumbersome as fck
with those symmetry points and stuff
:: shudders ::
 
vzn
2:32 PM
reading wikipedia it (K-vN) does not req that observables commute. so maybe what am looking for is a classical system where observables do not commute in the koopman-von neumann theory.
 
Anonymous
 
@Blue e.g. Solid State Physics, by Ashcroft and Mermin?
 
Anonymous
One more question...I can't understand why the heavy hole and light hole and split-off bands separate
 
Anonymous
I searched the net...but found nothing useful
 
Anonymous
Most of them state something like spin or smh
 
Anonymous
2:35 PM
@AccidentalFourierTransform Thanks...I'll check
 
thats the book we used in class
it is supposed to be good
 
Anonymous
I'm using the one by Neamen
 
Anonymous
It's way too simplified at places
 
the authors have a couple of important theorems so they know what they are doing
 
Anonymous
Wow...that book is costly...I'll probably download the pdf
 
2:37 PM
you should download it even if it was free
 
Anonymous
@AccidentalFourierTransform BTW did you look at my last question ? ^ :)
 
Anonymous
@AccidentalFourierTransform libgen :P
 
@Blue there is no question anywhere :-/
 
Anonymous
3 mins ago, by Blue
One more question...I can't understand why the heavy hole and light hole and split-off bands separate
 
Anonymous
A question without the question mark :P
 
Anonymous
2:39 PM
 
Anonymous
What does "light-hole" and "heavy hole" and "split off" mean anyway?
 
Anonymous
I guess they are referring to effective mass
 
Anonymous
But...how?
 
¯\_(ツ)_/¯
beats me
 
Anonymous
Same here ;_;
 
2:42 PM
light hole=light mass, heavy hole=heavy mass
$m^*_l\ll m^*_h$
 
Anonymous
@AccidentalFourierTransform Okay. But why would some holes have heavy mass and some would have lighter mass along the same axis?
 
Anonymous
Also, what's that split-off for?
 
good old memes
 
Anonymous
1
A: Why do we have heavy and light hole bands in semiconductors?

Jon CusterTo delve more deeply into the origin of the various bands, you should go look at the literature where these bands are calculated. The classic reference for silicon and germanium is Energy-Band Structure of Germanium and Silicon: the k.p Method. Since this is still fairly early in band structure c...

 
Anonymous
I hate answers which only contain the link to some paper...
 
Anonymous
3:10 PM
@AccidentalFourierTransform Any idea what $(l,l_z)$ means here physics.stackexchange.com/questions/229328/…
 
Anonymous
ActualIy suppose the above derivation suffices. It seems that the introduction of the spin orbit effect is what leads to these two bands being different, which is related to how only total angular momentum $J = L + S$ is conserved due to spin orbit containing an LS product type of term. This means we have $j = 3/2$ and $j = 1/2$ states (the latter being the split off band), and the former being of two types; $(l,l_z) = (3/2,\pm3/2)$ and $(3/2,\pm1/2)$. Why exactly the $\pm$ is irrelevant for the energy is not clear to me entirely, but $l_z = \pm3/2$ is the heavy band and the other the light. — user129412 Jan 14 '16 at 14:47
 
Anonymous
$l$ is angular momentum..I know that
 
Anonymous
What what does the notation $(l,l_z)$ mean?
 
0
Q: Homework-like questions

drake01I don't understand why one of my question was put on hold as it seemed homework like, I had got two out of the three multiple choices correct, but I could not figure out how to approach the third one, that's why I asked for some suggestions.

 
3:32 PM
@Blue $l_z$ is the angular momentum along the z direction
since $L^2$ commutes with $L_z$, we can find a basis common to both and label it by $l,l_z$
In solid state physics, a particle's effective mass (often denoted m*) is the mass that it seems to have when responding to forces, or the mass that it seems to have when a mass with other identical particles in a thermal distribution. One of the results from the band theory of solids is that the movement of particles in a periodic potential, over long distances larger than the lattice spacing, can be very different from their motion in a vacuum. The effective mass is a quantity that is used to simplify band structures by modeling the behavior of a free particle with that mass. For some purposes...
 
3:49 PM
i want to learn qm but i only know single variable calculus and linear algebra
would it be a bad idea to read a book and look online to understand the math that i do not know
and what book out there is the best for qm
 
4:06 PM
?
 
62
Q: What is a good introductory book on quantum mechanics?

PhaDaPhunkI'm really interested in quantum theory and would like to learn all that I can about it. I've followed a few tutorials and read a few books but none satisfied me completely. I'm looking for introductions for beginners which do not depend heavily on linear algebra or calculus, or which provide a s...

 
Anonymous
4:18 PM
@ııııııııııııııııııııııı I don't know what that means...$L^2L_{z}=L_{z}L^2$ ? Why should that be?
 
Anonymous
Also, I don't get why $j=3/2$ and $j=1/2$ only? Why not other possible values of total angular momentum?
 
@Blue Are you asking why they commute or what does that mean?
 
Anonymous
@PrathyushPoduval The former
 
@Blue Do you know what are the angular momentum operators? (If not, do you know the momentum operator?)
 
Anonymous
Huh...I found the proof in Griffiths just now!
 
4:27 PM
::bangs head against wall::
 
Anonymous
$[L^2,L_z]$
 
Yeah, and play around with the commutator relations
 
Anonymous
$=[L_x^2,L_z]+[L_y^2,L_z]+[L_z^2,L_z]$
 
Anonymous
Phew...
 
Anonymous
Got it
 
Anonymous
4:30 PM
@PrathyushPoduval umm, and do you know why $j=3/2$ and $1/2$ only?
 
Gas is soooooooo expensive
 
@Blue If you're asking about the quantisation of angular momentum, then it is done using the ladder operator technique (Its there in griffiths I think), If not, I don't know why it is only 3/2 and 1/2 (I have limited knowledge on semiconductors)
@0ßelö7 Use public transport then
 
user228700
@heather You...uhh, what happened, man? :-o
 
@heather ?
 
@Blue You did the quantum harmonic oscillator yet? It shows you the ladder technique without much pre-req knowledge
 
4:39 PM
@heather Programming? Writing a proof?
 
nope, still trying to figure out the mathematical representation for spdc
it's rather maddening
@0ßelö7 instead of thinking of gasoline, i immediately thought of lab materials =P
 
Anonymous
@PrathyushPoduval I know the ladder operator technique from griffiths...but I was concerned about semiconductors...thanks though!
 
@heather have you looked at nonlinear optics?
 
@EmilioPisanty this answer doesn't really deserve the 500 rep bounty :/
 
Anonymous
4:57 PM
@PrathyushPoduval I think they are dealing with stuff like Si. So they are getting limited values for $l$ and hence for $j$
 
@Mithrandir24601 yeah, i found a pdf of a book and i'm trying to figure it out.
 
Anonymous
$l$ could take values $0,1,2$
 
Anonymous
Ok...the OP mentions GaAs
 
00:00 - 17:0017:00 - 00:00

« first day (2496 days earlier)      last day (2418 days later) »