The h Bar

6:00 PM

Because that also looks like the zealous hope of someone sceptical of quantum physics :P

@Danu If you calculate the optical chirality it gives zero, and the conclusion from the questions here was that higher-order chiralities reduce to this for monochromatic fields.

DanielSank

@DavidZ go for it

Danu

@EmilioPisanty Right

Emilio Pisanty

@Danu The global symmetry in principle would permit different mirror versions in the far field for the two enantiomers.

However, they also need to be able to have a chiral local interaction with the field

And that's where the zilches give zero, so no luck.

As far as I can tell.

0celo7

@ACuriousMind In class, we showed that $Na^\dagger|n\rangle=(n+1)a^\dagger|n\rangle$. The prof then claimed that $a^\dagger|n\rangle=d|n+1\rangle$. Why?

David Z

6:02 PM

@DanielSank By the way, as far as your question, I think the biggest difference is that as a moderator, most of my time on the site is spent on "janitorial" work. My first stop when I come to the site is the moderator flag queue. It typically has about 10 flags, and I go through and handle the easy ones (~5), then think about the hard ones. Sometimes it takes 10-20 minutes to handle a difficult flag, not counting time spent waiting to confer with the other mods.

ACuriousMind

@0celo7 What's the definition of $\lvert n+1\rangle$?

DanielSank

@DavidZ Interesting. This feels more "different" than, say, people constantly pinging you in chat with whatever questions and comments?

0celo7

@ACuriousMind It's a state s.t. $N|n+1\rangle=(n+1)|n+1\rangle$. But how do we know there's just one?

ACuriousMind

@0celo7 Aha, I thought that that was your underlying question!

Wait while I locate the relevant Qmechanic post

David Z

@DanielSank You mean, is this the janitorial work a more important/relevant difference to how I would use the site as a 10k user, than the repeated pings? Yes, I'd say so

DanielSank

6:05 PM

@DavidZ ok

ACuriousMind

@0celo7 Here.

0celo7

@ACuriousMind uhhh, what

Secret

quantum guys: I still find it strange in about the interpretation that it is an uncertainty. From what I remember about the derivation, it is derived using cauchy schwartz and also terms like $\langle A^2 \rangle$ (which suggest variance), which when plugged into cauchy swartz gives the commutator term. But since a commutator [A,B] is an infintesimal unitary transformation of A by B (which act as a generator), then wouldn't the interpretation that the inequality is along the lines of the lower

Danu

@EmilioPisanty Invent allll the tensors :D

Secret

bound on how much A is being transformed by B?

David Z

6:07 PM

@DanielSank BTW the fact that I spend most of my time handling flags rather than posting questions and answers is a reflection of what I, personally, see as the best use of my time here - it's not an obligation of moderators in general.

Emilio Pisanty

@Danu No, my gut feeling is that if the simple tool says no then the simple tool is right.

ACuriousMind

@Secret Writing @quantum just pings QuantumBrick currently. Don't use @ unless you actually want to ping someone.

Emilio Pisanty

Here it's worth going up a single step in the ladder, because first-order contributions could conceivably cancel.

Secret

ok sorry about that

QuantumBrick

@ACuriousMind indeed.

Emilio Pisanty

6:08 PM

(e.g., take two co-propagating, counter-rotating fields at $\omega$ and $2\omega$ (another favourite of mine). Set the relative intensities right and the optical chirality / 000-zilch will vanish, but the field is still chiral.)

Secret

(cont.) so why there is no interpretation of a notion of a infitesimal transformation in HUP as hinted by the presence of the commutator?

Emilio Pisanty

but more than that is unlikely to give something new.

0celo7

@ACuriousMind So we just assume there's only one eigenstate?

DanielSank

@DavidZ That's puzzling.

Danu

@EmilioPisanty My feeling is TENSORS

0celo7

6:10 PM

are we quotienting out the various degenerate kets?

ACuriousMind

@Secret Well, that is a valid interpretation, but it's the more impractical one. That $\delta A$ is an uncertainty is not an interpretation however, since its definition is just the definition of the standard defination of a random variable

Emilio Pisanty

@Danu lol

@Danu your feeling is wrong

your feeling should be SYMMETRY

I see, makes sense

As always :'(

@Danu =P

6:11 PM

@DanielSank how so? (also note my edit to the previous message)

ACuriousMind

@0celo7 The "nicer" assumption is that the representation is an irreducible representation of the ladder algebra. If you have degenerate kets it would be reducible, so the system would fall apart into two disjoint, non-interacting pieces, which we certainly don't expact for a single HO.

DanielSank

@DavidZ Ah.

Physics Meta

5

Q: In 2016 (pre-election) what activities are involved in being a Physics SE mod?

With a moderator election looming it would be useful to hear from our current moderators how much the role affects their use of the site. Some obvious questions would be how many hours a day does it require, and what are are the day to day tasks you need to do in that time. However it would also ...

discussion

0

Q: 2016 moderator election README

Election page Question pool for potential moderators Present (2016 September) moderators' thoughts on what it's like being a mod. Main Stack Exchange FAQ on what it means to be a moderator. This is important; in order to vote for a moderator we need to know how being a moderator differs from e.g...

^clickbaity title

Which?

6:14 PM

Why not just "General information concerning the 2016 moderator elections"

ACuriousMind

"read me"

Danu

Yours :P

DanielSank

That's standard notation...

Should I have not all-caps'd it?

Secret

Q2. Why measurement must sample from a spectrum of eigenvalues and project the state onto an eigenstate? Using the decoherence interpretation, why only eigenstates become entangled to the environment pointer states but no entangled states formed by a superposition state of the system entangled with one of the pointer states? That is, why entanglement with the environment must occur between eigenstates an not the superposition states?

ACuriousMind

@DanielSank Yes, because it's supposed to be clickbait. I don't think that's a bad thing.

Danu

6:14 PM

No "read me" in general, IMO.

DanielSank

@Danu A readme file is something you should read so other stuff makes more sense.

That's the point of that post.

ACuriousMind

We call these things "readme" for the same reason people call their virus executables "clickme" :P

DanielSank

@ACuriousMind True.

David Z

@DanielSank for people who are familiar with software, yes; outside that community, I'm not so sure.

Danu

@DanielSank I don't think it is a good title, because I personally for instance don't want/need to read that post.

Emilio Pisanty

6:15 PM

@DanielSank I'm with Danu on this one - standard in some conventions, but I don't really think it's one that applies

Danu

If you had just called it "general info concerning..." then I wouldn't have clicked

Now I clicked, for nothing.

DanielSank

So edit it.

Danu

I wanted to leave that to you, but sure

DanielSank

It's really frustrating that the software community has such good standards with regard to communication, but physics simply does not.

Danu

It has nothing to do with that, IMO. I understand what README means.

It's just not true that I need to read that post

So I don't want the title to say that I do

DanielSank

6:17 PM

@Danu omg

David Z

@Danu just edit it already :-P

Danu

@DavidZ I did :)

0celo7

@ACuriousMind Geodesic balls tonight :D

David Z

OK cool

Danu

@DanielSank It's really not a big deal, don't get me wrong.

DanielSank

6:17 PM

I don't like the new title XD

ACuriousMind

@Danu I don't need to read the readmes of most software either, the title doesn't mean "everyone needs to read this"

David Z

lol

DanielSank

It's not general information, it's a set of links.

@ACuriousMind Yes.

Listen to ACM. He is wise.

Danu

@DanielSank ...which collect general information

ACuriousMind

It's to say "if you are confused about what's going on with this thing, read its readme first", not "read the readme at all costs"

DanielSank

6:18 PM

^ That

Secret

In short, why measurement can only project to eigenstates, but not superposition states made from nondegenerate eigenstates?

Danu

@ACuriousMind Meh. I think the current title is more accurate and more useful, at least for me personally.

ACuriousMind

@DanielSank I notice an uptick in flattery lately. Are you implementing your election "winning" strategy? ;)

Danu

lel

Emilio Pisanty

@ACuriousMind Sure, that's what readme means to a programmer. That's much too subtle a reading for an item on the yellow box on the front page, though.

DanielSank

6:20 PM

@ACuriousMind Partially I'm trying to highlight how incredibly good you are at stating what I'm thinking. This serves two purposes.

1) It might make you want to run for mod, because a key element of a community leader is understanding what other people want/think and distilling those thoughts into useful works.

ACuriousMind

@EmilioPisanty I agree with that, but the reason @Danu gave was "I don't need to read this thing so it shouldn't be called readme". Which I guess proves your point about subtlety, but is not a good reason in itself ;)

DanielSank

2) It helps me rationalize to myself that if you are elected, my opinions on policy will be well received, etc.

Emilio Pisanty

@ACuriousMind Not sure I agree with Danu's reasons, but I agree that the old title is clickbaity

Not 100% behind the new one though

DanielSank

@EmilioPisanty I feel like software community standards are pervading our entire culture though.

Surely the idea of a readme is well understood.

I'm delighted by this discussion.

Emilio Pisanty

@DanielSank "our entire culture" is a bit strong

DanielSank

6:22 PM

Communication is important.

@EmilioPisanty Is it?

If I say "the system crashed" you know what I mean.

0celo7

that's a long way away from README

Danu

"entire culture" = hbar ;P

0celo7

@Danu I think we're representative of the population at large.

Emilio Pisanty

What about the many hundreds of 300+ rep users who use the site occasionally and only rarely go to meta or chat?

Danu

@0celo7 With you in here, we've even got Trump voters!

ACuriousMind

6:23 PM

@0celo7 Bwhahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha

DanielSank

@EmilioPisanty They've never seen a reaadme?

Emilio Pisanty

@DanielSank You betcha some of them haven't

0celo7

@Danu exactly

Emilio Pisanty

Sorry to break the news

0celo7

We've got like every segment of the population

Emilio Pisanty

6:24 PM

and if they did, they forgot about it

DanielSank

How about "2016 moderator election resources/links"

Emilio Pisanty

maybe they did read it

Secret

@danielsank @EmilioPisanty @ACuriousMind sorry some typo: Why measurement can only project to eigenstates (or for degenerate eigenstates, the subspace spanned by the degenerate eigenstates), but not superposition states made from nondegenerate eigenstates?

0celo7

Well, our resident homeless person Chris White isn't here any more...

but other than that

DanielSank

@0celo7 Yeah what's his deal?

0celo7

6:24 PM

Sorry, box-dweller.

not homeless

Emilio Pisanty

@DanielSank Better. I'd shy away from starting with a number

ACuriousMind

@DanielSank I'm sure some of them haven't.

Emilio Pisanty

particularly with the current font

0celo7

@DanielSank JD annoyed him; he left.

Emilio Pisanty

I hate the numbers in this font

DanielSank

6:25 PM

@0celo7 wow

0celo7

Deleted account.

DanielSank

Because of JD?

Emilio Pisanty

@DanielSank Not sure if that's why, but indeed the whole SE account is gone.

0celo7

We don't know the full story.

Emilio Pisanty

@0celo7 Is there any relevant chat transcript that's still public?

0celo7

6:25 PM

But something happened with JD, CW got suspended, then deleted his account.

Emilio Pisanty

Also CC @ACuriousMind

0celo7

@EmilioPisanty AFAIK, no.

Chris's comments were deleted, and everything else was behind closed doors.

Danu

@DanielSank Something like that is fine too

Emilio Pisanty

@0celo7 suspended or chat-banned?

0celo7

@EmilioPisanty Chat banned for a day.

ACuriousMind

6:26 PM

@EmilioPisanty There is no visible evidence left. CW made some Not Nice comments, was chat banned for a day, never returned and had his account deleted. We don't know anything more.

0celo7

Can we remove a user for being extremely disruptive, even if there's no direct evidence of trolling?

Emilio Pisanty

@ACuriousMind OK. Were you present at the time?

ACuriousMind

@EmilioPisanty At the time of the comments? Yes

David Z

Let me ask that we avoid speculation about what happened to Chris. It's not a punishment imposed for something that happened on this site, but I don't know any details beyond that, and AFAIK nobody else here does either.

Emilio Pisanty

@DavidZ Sure. I've been pondering whether to write directly (just to say sorry you're gone), and I wanted to brush up on the public knowledge.

0celo7

6:28 PM

9 page lab report

Still have to do analysis, problems, conclusion

when will it end??

Secret

CW stuff

Emilio Pisanty

@Secret That is way old

0celo7

@Secret what?

Secret

cycle through all the letters of the alphabet to find all CW messages

ACuriousMind

@Secret that's basically the measurement problem, which I'm not going to discuss.

Bernard Meurer

6:30 PM

Okay guys

I really need some help with my maths

@ACuriousMind Are you busy?

Emilio Pisanty

@ACuriousMind @Secret pretty much my position too.

ACuriousMind

@BernardMeurer Only busy with the other conversations in this chat :P

So go ahead and ask

@ACuriousMind <3

why not ask me?

@0celo7 You too

6:32 PM

Too busy

got a lab report due in 2 hours

I really need like 3

Secret

@EmilioPisanty Well I am not exactly sure if my question is really the measurement problem, because I am not asking whether the state do collapse, but rather, why it collapse on eigenstates and never superposition states of nondegenerate eigenstates... Or is that still part of the measurement problem?

0celo7

I blame Kobayashi-nomizu

ACuriousMind

I forgot: Did we have any unresolved topic before the lengthy readme digression? :D

Bernard Meurer

We started today by defining an Inductive set as: $$X\subset \mathbb{R}, X\text{ is inductive iff }x\in X \Rightarrow x+1\in X$$

ACuriousMind

For the love of the almighty potato, use \text for words that are meant to be words

6

Emilio Pisanty

6:34 PM

@DavidZ Also: while there's good reason for laying off the speculation, it is undeniably a relevant aspect of the site's life. Chris's departure has a clear and definite negative impact on the site's ability to address a broad swathe of astrophysics. If (if) it was caused by other users being disruptive-while-keeping-the-formal-rules, then that is a cause for concern.

Bernard Meurer

@ACuriousMind thanks, didn't know how to do that :)

Okay, does that definition make sense?

ACuriousMind

@BernardMeurer Yes, although I have seen "$1\in X$" as an additional requirement to be called inductive

Emilio Pisanty

@BernardMeurer \text for longer stretches; \mathrm works but it kills any spaces.

David Z

@EmilioPisanty fair enough; I guess what I really want to ask is that everyone refrain from acting like they know what happened to Chris when they really don't.

Bernard Meurer

@ACuriousMind Seems reasonable, but let's go with the one the old man gave

vzn

6:36 PM

in theory salon, 24 hours ago, by vzn

sparkfun robot autonomous vehicle competition/ combat robotics, boulder 2016 / Tmachine blog

0celo7

@BernardMeurer what the hell

what kind of class is that?

Bernard Meurer

Continuing, we then proves that $\mathbb{R}, \mathbb{R}^{+},\mathbb{Q}$ are inductive

0celo7

PhD level logic? I've never even seen that

Bernard Meurer

@0celo7 Introduction to analysis

0celo7

What the hell?

Danu

6:37 PM

Get over it @0celo7

Obliv

@0celo7 can you show me your lab report i'm curious how the quality of our reports differ

0celo7

I've never even heard of these things

@Obliv Mine is literal trash

@Danu Get over what?

Bernard Meurer

Which is fine I got that

vzn

↑ for makers/ hackers/ "experimenters" etc like @Bernard :)

0celo7

I've never heard of these things

Emilio Pisanty

6:37 PM

@Secret It's still "why do projective measurements work like they do instead of otherwise", which is much too close to the MP for me to wade into at the moment.

Secret

ok then

Bernard Meurer

then he showed us something called (Literal translation) Principle of the finite induction (dunno if this is the name in english)

which says:

Obliv

@0celo7 give it to me...

vzn

@EmilioPisanty plz do contact him & let him know he is missed & ask/ urge him to rethink/ reconsider

Bernard Meurer

If $A$ is an inductive subset of $\mathbb{N}$, and $0\in A$, then $A=\mathbb{N}$

0celo7

6:39 PM

o.o

Is this actual math?

Obliv

what is a subconj.

ACuriousMind

@BernardMeurer "subconj."? Do you mean subset?

Bernard Meurer

Yes

ACuriousMind

@0celo7 Yes, this is how a European first semester rolls :P

0celo7

@Obliv If in a parallel circuit, the currents add up to more than the current in the power supply, wtf went wrong

Bernard Meurer

6:40 PM

Proof: $$A\subset \mathbb{N};\mathbb{N}\subset A \Rightarrow \mathbb{N}=A$$

Secret

@ACuriousMind @EmilioPisanty Q3. Why do classical observables cannot be operators, is it because the classical state space is a set hence there is nothing it can act on? (thus we don't see something like a poisson bracket acting on some function like e.g. $\{A,B\}f(q,p)$)?

0celo7

Don't say measurement error

or human error

Obliv

that's the only possible option if you're not gonna give any detail about the procedure

Emilio Pisanty

@Secret I'm not sure why you think I'm the guy to answer your questions.

Bernard Meurer

b/c $\mathbb{N}$ is the intesection of all inductive subsets of $\mathbb{R}$ than contain 0

ACuriousMind

6:41 PM

@Secret Koopman-von Neumann mechanics makes your question non-sensical.

0celo7

Literally hooked up 3 resistors in parallel, measured the current in the power supply, measured the current around each resistor, added them up, got more than the power supply

like 1% more

Bernard Meurer

All of this makes a lot of sense to me

until this goof started showing "Proof by induction method"

0celo7

@ACuriousMind Crazy

Bernard Meurer

Which was like this (this will take a while):

0celo7

@BernardMeurer now knows more math than me

I don't get it

Obliv

6:42 PM

@0celo7 you sure you did the math right 8^)

0celo7

@Obliv Yes.

John Duffield

@DanielSank : no, not because of JD. He posted a comment calling for downvote collusion and banning, I said that's not very nice and flagged it, then he got some kind of chatban, 1 day I think, then he decided he wanted to leave PSE.

0celo7

Literally addition

Secret

> I'm a PhD student at the CDT in Controlled Quantum Dynamics at Imperial College London, and at the Max Born Institute in Berlin. My research is on strong-field and ultrafast physics, and I study tunnel ionization and high-harmonic generation.
=>This fit the description of a "quantum guy"

Obliv

incorrect.

0celo7

6:43 PM

@Obliv for the current, you add them

what is wrong with that?

ACuriousMind

@Secret May I point out that you never really interacted with Emilio in chat before and now just started pinging him with your questions?

Just because people may be qualified to answer your questions they still can get rather irritated if you ping them with those questions without any sort of prior rapport.

Bernard Meurer

$$P(n) = \sum\limits_{k=0}^{n}\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!};\forall_{n\in\mathbb{N}‌}$$

Emilio Pisanty

@Secret Not sure how to put this, but I'm going to pass on your question.

Obliv

@0celo7 physics.bu.edu/py106/notes/Circuits.html you have to get the equivalent resistance and then divide the voltage by this quantity and you'll get the current

not just addition

Secret

ACM: Actually we do interacted before today

0celo7

6:45 PM

The resistors adding inversely is equivalent to the current adding simply @Obliv

Secret

but nvm, that's all my quantum questions for today

Obliv

yea but instead of measuring the current maybe you should try this method

you'd only need to know the voltage & resistances

Bernard Meurer

Okay, fine, we then show that the solution set for this is $$A=\{n\in\mathbb{N}:P(n)\}$$

Obliv

maybe u made a mistake measuring the currents

Bernard Meurer

Does that make sense @ACuriousMind?

ACuriousMind

6:48 PM

??? Three messages over a time span of over a year don't even mean he remembers you. I certainly don't keep track of users I spoke that rarely, and I would be rather confused if they started pinging me with their questions. As a matter of fact, I've ignored some particularly obnoxious variants of these.

@BernardMeurer I think you may have messed up what exactly $P(n)$ there is supposed to be

What is meant to be going on is this: We have the proposition $P(n)$, which is true for a given $n$ if $\sum_{k=0}^n \frac{k}{(k+1)!} = 1-\frac{1}{(n+1)!}$ holds and false otherwise.

Bernard Meurer

and we can prove this to be true $\forall_{n\in\mathbb{N}}$ by proving: $$\begin{cases} 0\in A, & P(0)\text{is true} \\ A\text{ is inductive}, & \forall_{n} P(n) \Rightarrow P(n+1)\end{cases}$$

Obliv

@0celo7 I want to do cool derivations and use beyond basic algebra in my lab reports but the procedures rely on such simple theory it's not fair :(

Danu

@BernardMeurer You're using too few }'s in the subscripts

0celo7

@Obliv how?

it's a digital ammeter

ACuriousMind

Your set $A=\{n\in\mathbb{N}\mid P(n)\}$ is then the set of all numbers for which that equation is true. The idea is now to show that $A$ is an inductive set containing $0$, since then you've shown that $A=\mathbb{N}$ without checking the truth of $P(n)$ for every $n$ individually.

Secret

6:51 PM

ACM: Noted, I will try to gain some rapport for future cases like these

Bernard Meurer

@ACuriousMind yeah, I got it this far

this all makes sense to me

then comes his proof

which I copied from the board verbatim and will paste here

and that I did not understand one bit

Danu

So he shows it for 0, then shows that the case $n$ implies the $n+1$ case (hopefully)

Obliv

@0celo7 idk dude maybe the wires at the end of the circuit had less resistance.

ACuriousMind

@BernardMeurer Okay :) Then let me just point out that the $P(n) = $ further up was bad notation because $P(n)$ is a logical proposition which is not equal to the sum you wrote it is equal to

Danu

The base case (checking $P(0)$ in this case is usually easy)

@ACuriousMind He should've written brackets around the second quality

Bernard Meurer

6:54 PM

@ACuriousMind Bad notation on my part there, he denoted it differently but idk how to do it on LaTeX and was lazy :P

Danu

(I'm commenting for Bernard, not for you ACM ;) )

ACuriousMind

@BernardMeurer Okay. I just thought you were confused about what $P(n)$ is because that is a frequent source of confusion if not stated clearly.

Bernard Meurer

$$\sum\limits_{k=0}^{n}\frac{k}{(k+1)!}=\sum\limits_{k=0}^{n}\frac{k}{(k+1)!}+(n‌+1)/((n+2)!)=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}=1-\frac{1}{(n+2)}$$

Obliv

that's a beautiful line of latex there @bernard

Bernard Meurer

$$\sum\limits_{k=0}^{n}\frac{k}{(k+1)!}=\sum\limits_{k=0}^{n}\frac{k}{(k+1)!}+\fr‌ac{n+1}{(n+2)!}=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}=1-\frac{1}{(n+2)}
\sum\limits_{k=0}^{n}\frac{k}{(k+1)!}=\sum\limits_{k=0}^{n}\frac{k}{(k+1)!}+\fr‌ac{n+1}{(n+2)!}=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}=1-\frac{1}{(n+2)}
\sum\limits_{k=0}^{n}\frac{k}{(k+1)!}=\sum\limits_{k=0}^{n}\frac{k}{(k+1)!}+\fr‌ac{n+1}{(n+2)!}=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}=1-\frac{1}{(n+2)} $$

Obliv

6:58 PM

forgetting a { or } somewhere

ACuriousMind

@BernardMeurer you forgot the slashes before the "sum" so the \limits doesn't know what to do

Secret

Background for my trio of questions today: I have just finished reading the two Susskind books yestdy. I then became curious of the mathematical analogies and difference between classical and quantum (but unlike most, being slightly more familar with QM since I learn that first in uni, I tend to ask the opposite question on why classical behave that way, why classical does not behave like QM, why QM behave this way etc.). That way I can identify the quantum only things to better study about them

Danu

In other news, it's unnecessary to use \limits after \sum in display mode.

$$\sum_{hi}^{there}$$ works just fine

Lol ironic

Bernard Meurer

halp

ACuriousMind

xDD

Secret

6:59 PM

(In the photo I am going to post is a summary of the content of the 2 Susskind books)

Danu

@BernardMeurer Want me to edit for you?

(I can do it if you like)

Obliv

$\sum\limits_{k=0}^{n}\frac{k}{(k+1)!}=\sum\limits_{k=0}^{n}\frac{k}{(k+1)!}+\fr‌‌ac{n+1}{(n+2)!}=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}=1-\frac{1}{(n+2)}$

Transcript for