I do think it's debatable whether Weyl spinors in themselves are large enough a topic to warrant their own tag and could not just be tagged 'spinors', though
@heather Oh, it just has one question? I'd just swap out the tag for 'spinors' then on that question and let the system remove the weyl-spinor tag on its own.
@Mostafa suspended for saying "JEEhad" too much and trying to ask why it was a problem, Shog9 suspended him, and he got suspended maybe...a month ago? (@Bernardo is that last right?)
@heather Affine Lie algebras are technically Lie algebras, but they are not what people imagine under a "typical" example of a Lie algebra. That tag carries more information than 'Weyl spinor' did, I'd just leave it alone
perhaps that's not a sign of being too smart, but of being not at all socially aware, which is the opposite of one kind of "smart" (don't know anything about Ron Maimon)
besides, isn't he still around on Physics Overflow...?
You: Why did he get suspended? Me: I don't think it's useful to discuss that here, he can discuss it with me if he wishes. Y'all: must've been "JEEhad" - clearly this is a dark magic word Also y'all: there's objectively nothing wrong with saying "JEEhad"; since we've now established that was the reason, we can declare the suspension bogus. Me: ...
@heather I have zero interest in whether y'all use or do not use "JEEhad" to refer to your disagreements over some schoolwork.
@BernardoMeurer I'm damn tired of folks here taking what should be a calm discussion over some little disagreement or policy, and egging folks on until they're all worked up over it.
You left out a lot of context; I'm tempted to think you're just cherry-picking bits of that conversation to make others look bad while ignoring your own behavior and how it contributed to the situation.
Here's something I would consider a really critical bit of context:
(for casual readers: "...
@Shog9 and gave no real answer. i didn't make anything up (that's kind of offensive, honestly) - i made a fairly logical conclusion that many other people in this chat made.
@Mostafa suspended for saying "JEEhad" too much and trying to ask why it was a problem, Shog9 suspended him, and he got suspended maybe...a month ago? (@Bernardo is that last right?)
@Shog9 well, first off, i'm not the only one who says that, and second off, it's basically the logical conclusion - he said something you didn't like, he disagreed with your assessment, he discussed it with you, and it ended in him being banned. i read the darn transcript, and was there.
@heather It's misinformation when you say that he was suspended "for" that - it's a classic post hoc ergo propter hoc. Lengthy suspensions are rarely based on one specific incident, but on a pattern of behaviour. And we do not want to publicly drag out these patterns - effectively "shaming" them far beyond the length of their suspension - unless the suspended user themselves asks for that.
@BernardoMeurer I pretty much expected that, but point is, if he wants to have this discussion, Shog already invited him to the Tavern. There's no point in having that conversation here, and without him.
@Shog9 really? That was the most unclear statement ever.
okay, not ever.
but really, it was unclear.
> You have an unfortunate tendency to pour gasoline on a fire when you see one; if you can avoid that, it'd make handling the little disagreements that crop up in the h-bar a lot easier.
so first, you call it a "little disagreement" - so not really a large incident.
second, this "pour[ing] gasoline on a fire" - what, was that discussing it with the mods in the room at the time and trying to understand why it was a problem?
> not helping resolve the situation, getting in the way of folks that were.
as I read the transcript, 0celo was talking with the mods trying to understand the problem. if that's not "helping resolve the situation" i dunno what is.
@BernardoMeurer Immerse bonded areas in warm, soapy water. Peel or roll skin apart; a spatula or teaspoon handle or even a pencil will help. Remove cured adhesive with warm, soapy water (may take several applications). Fingernail polish remover with an acetone base has also been successful for removal of cured adhesive from skin. (from here )
so...i still stand by my original comment and i'm not really sure how "misinformation" comes into this, and i'm also not sure how trying to understand your reasoning is the same as being "addicted to drama" @Shog9
@heather Ok, I'll make this really simple & put all the relevant information in one place:
@0celouvskyopoulo7, you're one of the most-flagged, most-suspended users on this entire chat. Like, top-5 all-time. Even calculated as a ratio of suspensions per message posted, you're near the top of the list.
You can explain away the odd misunderstood comment here and there, but this is long past that point. Either your preferred mode of operation here is intentional trolling, or you simply have zero interest in not causing problems and thus put no effort into avoiding issues even when they're pointed out to you, repeatedly.
So here's what's gonna happen, what has already started happening starting, oh, on about 6 months ago, and what you should expect in the future: if you're involved in a discussion and it goes south faster because of your involvement, you're gonna lose access to chat for increasingly long periods of time. If you get flagged, same deal. You have years of past experience here that by all appearances you haven't learned from - either learn from 'em now, or go somewhere else.
@Mithrandir24601 I think so, yeah, but it's not really named in this text, they do mention that...So we get density matrix $\tilde{\rho} = \frac{A_n \rho A_{n}^{\dagger}}{p_n}$
@ACuriousMind I don;t know the rigorous one but for this purpose it is just the operator that define the change on initial density operator $\rho$ to $\tilde{\rho}$. It could be taken as a definition I guess, there is a motivation which precedes it...but not a direct proof...just a motivation.
@Mithrandir24601 From the decomposition theorem we can write these measurment operators $A_n = U_n \mathcal{P_n}$
@ACuriousMind How to I get latex to render on my chat?
@Mithrandir24601 We can then write the density matrix after the measurment as $\tilde{\rho_n} = U_n(\frac{\mathcal{P}_n \rho \mathcal{P}_n}{p_n})U^{\dagger}_n$
@JohnDoe This might be a bit of a roundabout way, but a Kraus operator is equivalent to a quantum channel. Acting on said channel by a unitary gives you the same channel, hence acting on a Kraus operator by a Unitary is the same Kraus operator (Kraus operators, by the way, aren't unique)
@ACuriousMind That's a direct quote, not my quote, I suspect it means that the eigenvalues don't change, hence ...Am I right about the unitary operator?
@BernardoMeurer Do you have any experience unlocking a modem that is locked to an operator? I have a Huawei USB 4G modem that is locked to some mobile operator but I want to use it with another sim card from another operator.
@JohnDoe Well, sure the unitary doesn't change eigenvalues. But I really mean that it's not very clear what "extracting information by measurement" is supposed to mean specifically.
@Mostafa No, all I know about anything remotely like this is that unlocking locked iPhones involves something called an IMEI code and a fee to a third party service
Give me the exact model of the modem and I can try and find something
Yeah they should be normally unlocked using the IMEI code (some websites give the unlock code given the IMEI) But this model uses a different algorithm.
@BernardoMeurer There are many websites like that. They only get some information that the tool generates (hashes, etc.) and send you the unlock code after you pay.
@JohnDoe i.e. $\xi\left(\rho\right) = \sum_{n, m, l} U_{n, m}\mathcal{P}_m\rho \mathcal{P}_l^{\dagger} U_{n, l}^* = \sum_n\mathcal{P}_n\rho\mathcal{P}_n^{\dagger} = \sum_n p_n \bar{\rho}$, so it's the same channel, so acting on a channel by any unitary doesn't change the channel, so there's nothing you can learn from that
it wasn't really that bad, surprisingly - i was expecting it to have you know facts off the top of your head; instead it gave you passages to read. it was more thinking/reading than science.
@Mithrandir24601 So in a sense the channel $\xi\left(\rho\right) = \sum_n A_n\rho A_n^{\dagger}$ is averaging the outcomes for a particular measurement? Since $\sum_n p_n \tilde{\rho}_n = \sum_{n} A_n \rho A^{\dagger}_n$.
@JohnDoe I'm not sure if 'averaging' is the right word, but you act with the measurement projector on the state/density matrix and get a sum of possible outcomes, weighted by the probability of getting that outcome. Neither the possible outcomes, or the probability of these outcomes changes when acted on by a unitary
@BernardoMeurer basically, yeah. my dad also says I'll be sure to get in and will be able to get more opportunities than in a bigger school, he says it's being a "big fish in a little pond"
@Mithrandir24601 Does it make sense that after measurement we can write $\tilde{\rho} = U_n(\frac{\mathcal{P}_n \rho \mathcal{P}_n}{p_n})U^{\dagger}_n$, where since they $U$ does not change the eigenvalues of $\rho$ is also does not change the eigenvalues of $\mathcal{P_n} \rho \mathcal{P}_n$ and hence does not play significant role?
@heather Parents can be like that... There is an argument with the whole big fish/little pond over little fish/big pond. But if you want to go to the likes of MIT as a grad, you're going to have to be a little fish someday. Also, they're probably going to have better teaching than a little pond, so you're going to be better at physics there than compared with said small pond. It'd be unbelievably tough though
@JohnDoe Hang on... I've just noticed something - what do you mean by $U_n$?
@Mithrandir24601 The unitary operator which corresponds to the measurement operator $A_n$ which we get from the polar decomposition $A_n = U_n \mathcal{P}_n$.
@Mithrandir24601 $n$ is the outcome of the measurement which corresponds to some measurement operator $A_n$ I think.
@Mithrandir24601 Well I think $U A U^{\dagger}$ is a change of basis with the same eigenvalues...
@Mithrandir24601 I think the idea is that the unitary operators hence does not change the uncertainty of the system just the states it is most likely to be found in.
@Mithrandir24601 Nah not at all. I don't have a good grasp on the subject yet, so I'm just feeling my way through. But it's interesting, starting to make a bit of sense...
@JohnDoe Sorry, I've just realised that I have been mistaking your labelling (my bad!), so the bit about the Kraus operators being the same as each other isn't (necessarily) true. It is true that it doesn't change the probabilities, 'just the states it is most likely to be found in' indeed :)
Hey, maybe someone can help me out as my search attempts have been failing me. I'm looking for a peer reviewed paper or something of the sort that tabulates the dielectric constant/relative permittivity of III-V semiconductors, specifically InAs and InSb. I'm interested in DC/low frequency values, and also low temperature. All I can find are websites that have tables with no citations.
@heather FWIW I think there might be some truth in there, but one has to be careful. If you want to get into grad school and do well there, it's important to (1) get good research experience as an undergrad and (2) take challenging classes. (And some other stuff) If you go to a top university for your bachelor's degree, assuming you graduate from that university, you're pretty much guaranteed to fulfill both of those requirements.
But if you go to an average university, it can be hard to tell in advance whether you'll have the chance to do so. I know there are major state schools where you can come out well prepared for a top grad program, e.g. Penn State does a decent job of this. I don't know about Iowa State.
Of course I will say that if you want to pursue a career in physics research, where you get your PhD does matter far more than where you get your bachelor's degree.
@DavidZ hmm, okay, i'll keep that in mind. Iowa State seems to have a decent number of research opportunities and a wide variety of classes. I could always do a few years there for the required classes/basic stuff found at any university and then transfer (if necessary).
I mean, I'd suggest identifying a few likely transfer "targets" - places you might want to transfer to - and doing a bit of research on what their requirements for transfer applicants are, as well as how the curriculum at Iowa State (or wherever you'd be going first) transitions into the curriculum at the transfer school. That way you'd know if, say, it's unlikely you'd make it in as a transfer student, or if you'd be coming in a year behind and have to delay your graduation.
That's all hypothetical, of course; the point is, transferring introduces additional complications to your education, and I wouldn't just assume that you can jump from one school to another and pick up right where you left off without doing a bit of checking to see that it's a plausible thing to do.
Otherwise you could find yourself stuck in your original school without having an option to transfer.
You left out a lot of context; I'm tempted to think you're just cherry-picking bits of that conversation to make others look bad while ignoring your own behavior and how it contributed to the situation.
Here's something I would consider a really critical bit of context:
(for casual readers: "...
