The h Bar - 2016-10-01 (page 5 of 6)

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6:02 PM

One reason I am bad at indices is because seeing only one component at a time caused me to lost track on where I am in the set

@0celo7 you are described in one comment to Lubos' blog as "innocent young cheer and playfull student called Ocelot7" (second non-indented comment here). Thought you might want to know your fame has spread beyond this chatroom ;) (Also, anyone not in the mood for insults probably shouldn't follow that link)

I stumbled across that more or less randomly while looking for something else entirely

@ACuriousMind Wow. Clicked, read half a sentence. Lol'd.

Am I a character in his charming slavic folk tale

@ACuriousMind LOL

@ACuriousMind what?

6:08 PM

@0celo7 What exactly is unclear about what I said

@ACuriousMind The comments in the post linked near the beginning of that blog entry need some clean up.

Man, being able to see deleted posts is really interesting.

I take it back

@DanielSank They probably do. I can't do anything other than you about that, though, so flag one or two of them

Motl is a genius

People post some ridiculous stuff.

6:09 PM

I can't find that comment.

@ACuriousMind I did. Also, be a mod already so you can clean stuff up.

Hence unclear

@Slereah May not want to post that in chat. "Be nice" and all that.

Note "may".

That is just a quotation

I know.

6:10 PM

I do not necessarily share that opinion

May not.

(Although I do)

-_-

@0celo7 Scroll down to the comments. There's a thread started by a "Bernd Schimmelpfennig" and then below that Dilaton mentions you

Ah, second nonindented comment.

I thought the month ban was over a holocaust joke?

6:12 PM

I honestly can't remember what it was for, if I ever actually knew in the first place

In my mind I attribute it to David Z being jealous of me.

Makes me feel good.

@DanielSank Sooooooooooooon. ::laughs evilly::

@ACuriousMind Very good.

Note that although you will wear the moderator hat, you're still my court jester.

Don't get too big for your britches.

The dungeon has a vacancy, and I'm still the sock-puppet king of this site.

I will add some bells to the moderator hat to be sufficiently jester-like

"It's so bizarre why so many people want to be actively and constantly threatened, annoyed, and controlled by similar stupid scumbags. He's clearly abusing all the rules and people's compassion etc. all the time for something that is demonstrably not the purpose of the server and that is not bringing anything good to anyone - except for himself while masturbating."

6:14 PM

@ACuriousMind Ah, very good indeed.

Well he has my vote for mod

Wait ACM is now a mod officially?

No.

@0celo7 No, the election starts only in 2 days, silly

I'm making what I think is a reasonably safe assumption.

6:16 PM

PSE elections are a microcosm of the 2016 US elections

Explain.

@BernardMeurer Did you read knzhou's take on that in their rather hilarious nomination?

@ACuriousMind I'm not voting

Look at the comment's section under ACM's nomination, what a shit fest :p

Is JD the candidate for the Prohibition party in this analogy

6:17 PM

@ACuriousMind yeah, that made me laugh hard

@Slereah Lol, JD is the AfD of PSE

@BernardMeurer And that's after they've been sanitized by a CM twice.

Oh man

I'm looking at the third parties running

It's always a treat

@ACuriousMind A cm?

Nutrition Party

It fascinates me how so many users do not understand what being a mod means.

6:19 PM

@BernardMeurer Community Manager, the actual SE employees and not volunteer diamonds

Someone commented that ACM should be a mod because he's nice and answers questions.

Link to the shitfest?

I mean, I get that we want our mods to be nice and knowledgeable, but being a mod is different from being a 20k rep user because you have:

1. Unilateral close votes

2. Can deal with flags

@DanielSank Banning 0celo7. Although the mods have been leaving me alone lately.

3. Can handle other exceptional cases, which might e.g. require contacting users or the SE support team.

6:20 PM

physics.stackexchange.com/election/3

Unbelievable how I feel like almost nobody understands this.

@0celo7 I wonder if I can get a few more votes if I promise to change that :P

Interestingly, @ACuriousMind says in his nomination post that he's running because a good candidate should understand what mods do.

@0celo7 Well, you're following the rules lately. Interesting coincidence.

@ACuriousMind Change what?

Greetings and salutations, @JohnRennie.

6:21 PM

@DanielSank it's because JD hasn't been as annoying lately

@0celo7 The "leaving you alone" part. That was a joke, in case I need to make that clear.

@ACuriousMind I'm on mobile.

Evening all

Not my duty to help you overcome your technical limitations!

its 2PM, not evening.

@ACuriousMind as a mod I expect you to fix mobile chat

6:23 PM

@DanielSank To be fair, I think the questionnaire shows that the other candidates at least also have a reasonable idea of what being a moderator means

@ACuriousMind Yes but the voters seem not to.

I agree

Or, well, at least the ones who leave comments seem often not to

We can't know what all the voters who don't comment think

I made that meta post with all the links to try to help, but I don't think anyone followed the important one.

@ACuriousMind True dat.

@ACuriousMind They gave me a free curry wurst for being a loyal customer!!!!

uh...that's good, I guess?

6:26 PM

Yes it is.

Can anyone here modify pinned messages?

Editing messages after 2 minutes requires a full mod, I think

@Loong

dmckee seems to be active, though, so you might ask him

@0celo7 ?

6:28 PM

@ACuriousMind When I made that post with useful links for the election I hadn't realized there was already a "canonical" meta post for the election.

There's your full mod

@Loong can you help us out?

@DanielSank Yeah, that's because the canonical meta post is completely useless :P

All the links I put up should be in the other post, and my post should be deleted. I don't want to do it now though, because the pinned star in chat points to the wrong place.

@DanielSank which one?

@ACuriousMind Right, but I should have edited it.

6:29 PM

@DanielSank Which pinned star points there?

I see three pins, and neither points to your link collection

"Due to increased traffic..."

Ahhhh

Didn't click on that because I assumed it would point to the election page itself :D

Which was rather silly in hindsight

@ACuriousMind So you see the problem?

Yes

The situation is ok as is now, but there's one more meta post than we need.

This bothers me.

user116211

6:31 PM

WTH!!

user116211

@MAFIA36790 I deleted my answer — Valter Moretti 5 mins ago

@DanielSank Although we could fix it by just deleting the useless canonical post, no?

@MAFIA36790 Yeah.

Saw that.

@ACuriousMind Yep.

We surely could do that!

user116211

@DanielSank No, no....wait, lemme give him ACM's response's permalink.

Seems easier than to modify the starred message and transfer all the links over

6:32 PM

@MAFIA36790 Huh?

@ACuriousMind Yes.

I totally agree.

Do we need a mod for that?

Yes, 20k powers cannot delete positively scored non-closed questions.

Ok, yo @dmckee @DavidZ any of y'all around?

user116211

@Secret Well, the generalised definition of Cartesian product of Indexed Family of sets is more beneficial when the index-set is infinite and actually equivalent to the conventional definition we are generally accustomed to.

@DanielSank Yeah. Sorta.

The post linked in the Motl blog post mentioned above makes me think of the inimitable Professor van Manderpootz. And not in a good way.

It is, which is why I need to be more comfortable with working with indices and components of some large object, rather than trying to work with the whole thing all at once (because for infinite set, the object is so large and irregular that no diagrams can faithfully represent them)

Btw, currently on the finite set section, thus halfway through Ch. 1

6:36 PM

@dmckee See here please.

Note the comment.

Ah. Taken care of. Good thought.

@dmckee Thanks.

PSA: Anyone interested in the mod election: please read this FAQ on what it means to be a moderator.

@Secret Frankly his tendency to keep throwing in every concept he has every thought of means that even when he has a correct answer to the core question he almost always writes total nonsense elsewhere in the same post.

2

@DanielSank It might actually be better to post that in the election chat room

@ACuriousMind Will do.

6:40 PM

Although that one has been rather silent lately

@dmckee Whom are you talking about?

@0celo7 FOLLOW THE ARROWS

MOBILE DAMMIT

THEN GET OFF MOBILE

^

6:43 PM

MY LAPTOP IS AT HOME AND IM NOT

@ACuriousMind done

@ACuriousMind Getting in my car...I wasn't planning on going home so early

@DanielSank Saw it; very good

Did you know you can write the lowering operator like this:

$$a = \frac{u}{2 u_\text{zpf}} + \frac{v}{2 v_\text{zpf}}$$

where zpf means "zero point fluctuation"?

$u$ and $v$ are the two degrees of freedom, e.g. position and momentum.

@DanielSank Is that the standard deviation of the respective operator in the ground state?

6:48 PM

@ACuriousMind Yes.

$\sigma_u$ if you want.

I always ask when "fluctuation" is mentioned because people tend to mean different things, and sometimes nothing precise at all by it

@ACuriousMind Yes. Totally agree.

I should probably switch my notes to write $\sigma_{u/v}$.

Then I didn't realize the prefactors worked out that neatly

Yeah, it's pretty sweet.

How to use that in e.g. the quantum harmonic oscillator, what are the values of the zpf variables?

6:51 PM

I like this form because you get everything nice and dimensionless, yet you retain the physical meaning of the scale you used to make things dimensionless.

@Secret Very good question!

Write the Hamiltonian like this:

$$H = \frac{1}{2} \alpha u^2 + \frac{1}{2} \beta v^2$$

where $[u,v] = i \gamma$.

Then $$u_\text{zpf} = \sqrt{\frac{\gamma}{2}} \left( \frac{\beta}{\alpha} \right)^{1/4}$$

@DanielSank I like this form because I can actually remember it. I usually just wave my hand and say the creation operator is "some linear combination" and look the form up when needed

@ACuriousMind Yep.

If I have not mistaken, you converted the usual hamiltonian into phase space thus we are in the quantum phase space of the harmonic oscillator?

is that how it works?

It's far better than that dumb-ass formula we find in books like Griffiths with the square root of mass and spring constant and god knows what else.

@Secret The usual Hamiltonian already lives on phase space. What do you mean?

@DanielSank My thoughts exactly

@Secret Daniel has just chosen to rename the prefactors, nothing else has happened.

6:55 PM

Ah I see, just realised

@ACuriousMind Right. I write the SHO Hamiltonian completely generically, work out all the useful relations from there, and then just plug in the right constants for my problem at hand.

I've found this surprisingly useful in my life, so much so that it's an appendix in my thesis.

Dangit. Obviously, I forgot an $i$ in that equation for $a$.

For ladder operators, I tend to search for them by trying to factorise the quadratic operator expression. using the usual (a-ib)(a+ib) and then the commutator will take care of the cross terms

I recall my QM teacher said finding ladder operator is a great skill for systems all the way up to QFT level

@DanielSank Well, yeah. All harmonic oscillators are formally the same up to their energy in classical mechanics and their unit of energy in quantum mechanics.

I'm embarrassed by how long it took me to recognize the implications of that, but I did get it eventually.

@dmckee Yes. I'm not proposing anything groundbreaking... just good notation.

Look, here's Wikipedia's unbelievably terrible way to write the lowering operator:

The above $\alpha,\beta$ notation also make it easy to see where the major and minor axes of the ellipse is in the phase space

7:00 PM

$$a = \sqrt{\frac{m \omega}{2 \hbar}} \left( x + \frac{i}{m \omega} p \right)$$

That is so unhelpful it hurts.

I have absolutely zero intuition for what any of this means.

@Secret Yes.

You know what... instead of typing all of my thoughts in this chat, here's a link to my writeup:

github.com/DanielSank/theory/tree/master/Quantum/…

Back in my QM class, I deal with this guy as follows:
$a=@ (x+iKp)$

(I have a tendency to use spirals as symbols)

Clone the repo (note that there's a submodule), and just build QuantumOscillator.tex.

What are zpf variables?

7:16 PM

@ACuriousMind

How can one show that a cofinite set is "big" in the non-principal ultrafilter

I'm not sure how that follows

I mean $N$ is cofinite and it is big

But beyond that

Not sure how to show it

If I pick a singleton, by the axioms either {x} is big or N - {x} is big

Oh wait

I get it

It's the freeness condition of the ultrafilter

"U contains no finite subsets of J"

Does ACM know PhD set theory?

Everyone does except you

Even @Secret knows

@0celo7 ACM knows EVERYTHING

2

@0celo7 As stated in the transcript, $u_\text{zpf}$ just means the standard deviation of the $u$ operator for the ground state of the harmonic oscillator.

@Sanya I dunno about that.

@ACuriousMind how does the Ramsey decay time depend on the magnitude of 1/f frequency noise in a two level system?

The exponent will be accepted as an answer.

@dmckee Ah, yes, agreed.

@DanielSank ...what even is this?

7:28 PM

@0celo7 unclear what you're asking

I hear you have some fancy new formalism for creation/annihilation operators?

@0celo7 put $u \rightarrow x$, $v \rightarrow p$, $\gamma \rightarrow \hbar$, $\alpha \rightarrow k$ and $\beta \rightarrow m \omega^2$ and see if it makes more sense.

yes I know what the symbols mean

I have to find the relevant chat posts though

@0celo7 Ok so what's the problem?

@0celo7 They're not too far up. Kindly read the chat log for a minute before asking questions like "...what even is this?".

@MAFIA36790 and other editors: Note the difference between the spec. ref. tag and the res. recom. tag.

May 4 '15 at 5:27, by David Z

@KyleKanos rule of thumb: specific-reference has exactly one correct answer, resource-recommendation has an open-ended list of correct answers

7:30 PM

@DanielSank I'm stupid

I can't find it

user116211

@Qmechanic I was just about to ask you ;))

@DanielSank But...why?

@0celo7 Read the chat log.

What is the Hamiltonian in terms of $X$ and $Y$

@DanielSank I am reading, and it's not making sense

@0celo7 I posted a link to my github repo with my entire writeup.

Could you try reading it?

Do you know how to use git?

7:32 PM

I don't know how github works...

desktop.github.com

Download and install that.

@DanielSank I couldn't even tell you what a Ramsey decay time is off-hand :)

^ Point proven.

ACM doesn't know everything.

He knows a lot of the things.

@ACuriousMind How hairy is my back?

Too hairy.

7:36 PM

what do the thingies $a$ and $a^\dagger$ mean?

@0celo7 You'll learn in quantum mechanics.

Ladder operator

In linear algebra (and its application to quantum mechanics), a raising or lowering operator (collectively known as ladder operators) is an operator that increases or decreases the eigenvalue of another operator. In quantum mechanics, the raising operator is sometimes called the creation operator, and the lowering operator the annihilation operator. Well-known applications of ladder operators in quantum mechanics are in the formalisms of the quantum harmonic oscillator and angular momentum. == Terminology == There is some confusion regarding the relationship between the raising and lowering ladder...

@0celo7 They're called 'a' and 'assassinated a'.

@DanielSank (a) I'm reasonably familiar with quantum mechanics before taking the class (b) we already covered them in class

What do they mean

Why this strange linear combination of $x$ and $p$?

Why did Dirac (or whoever) write it down like that?

The mean the creation and destruction of a quanta in a mode. The creation operator adds one to the count in a mode and the destruction operator reduces the count by one (or has no effect on the ground state).

That doesn't answer how Dirac dreamt them up.

7:44 PM

By wanting an operator that goes from $\lvert n\rangle$ to $\lvert n\pm 1\rangle$. Since you only have $x$ and $p$ in the algebra of operators, the first thing to try is a linear combination of them. Exploiting the commutation relations of $x$ and $p$ with $H$, you can arrive at the correct coefficients.

I think Dirac just did it the classical way of rewriting the Hamiltonian

@ACuriousMind Before we have $N$, what is $|n\rangle$?

@0celo7 Eigenstate of the Hamiltonian with the n-th lowest energy

$a^2 + b^2 = (a + ib) (a-ib)$ is a classical formula

Note that $N$ and $H$ only differ by a constant

7:46 PM

@ACuriousMind yes I know

I guess they knew the eigenvalues were quantized by computing in the X basis?

Yep, the HO states are some special sort of polynomials (Hermite? Lagrange? I can never remember) after all. Although @Slereah's way also sounds reasonable, I don't know what exactly the initial motivation was

Wtf Facebook, how do you know who my matlab TA is???

Hermite

@ACuriousMind If I read BBS will you be able to understand it now?

@ACuriousMind I think they're Hermite polynomials * a Gaussian

Hermite polynomials, I think

8:33 PM

@0celo7 Maybe

@ACuriousMind What is the point of reducing the structure group of a principal bundle?

I know one can reduce the structure group of $\mathcal{B}(M)$ to $\mathrm{O}(n)$, but once does not need this fancy Steenrod theorem to do that.

Hmmm ... minus forty rep on "user was removed". Who's been playing silly buggers?

@dmckee Huh?

I just lost forty worthless internet points because "user was removed" so the four votes he/she/it had cast for me in the last six months were canceled.

Not that it matters much or that it is all that rare, but it always makes me wonder what was going on.

Excepting the few times when I request that the CM team nuke a sock-puppet. Then I know what was going on.

@0celo7 The (non-)existence of a reduction (at least for an actual subgroup) generally answers the question whether the bundle is "truly" a $G$-bundle or "just" an $H\subset G$-bundle.

8:46 PM

@ACuriousMind Of course...so?

That, to me, is the point. If you want another answer you'll have to be more specific what you mean by "what's the point".

@ACuriousMind Ok, I don't have anything more specific.

@dmckee I dunno, but might it not just be some user who decided they want their account deleted? I don't think votes from accounts with low vote totals are preserved and transferred to the Community user

@0celo7 @ACuriousMind Actually, the raising/lowering operators don't just come from quantum mechanics.

It's all rather simpler than that.

I wonder who invented the $\wedge$ for forms. Bishop & Crittenden do not use it.

8:49 PM

@ACuriousMind Yeah. It could be voluntary deletion just as easily as a moderation action.

Maybe more easily. We've had spates of repeated sockpuppet activity, but they usually come from just one or two offenders. Most of the time it's pretty quiet on the cheating to enlarge my influence front.

Looked at one way it is a real tribute to the "most people are pretty decent most of the time" fact of life.

A simple harmonic oscillator has classical Hamilton's equations of motion $$\frac{\partial H}{\partial X} = \dot{Y}, \quad \frac{\partial H}{\partial Y} = -\dot{X} \, .$$

Now define $$ a \equiv X + i Y \, .$$

@DanielSank Not actually allowed in classical mechanics, since coordinates are real variables

rekt

@ACuriousMind Come as it may as a surprise to certain fanboys in this chatroom, you are incorrect ;)

@ACuriousMind Complexify phase space, that's fine since it's even-dim?

@DanielSank I'm no fanboy of @ACuriousMind

@HDE226868 HSM question: why do some books write $\phi$ for the emptyset?

8:55 PM

Then $$\dot{a} = \dot{X} + i \dot{Y} = i a$$ which is a fantastically nice and simple equation.

...because then $a(t) = a(0) \exp(i t)$.

@0celo7 No idea. I typically don't read history books about math.

The exponential function is an eigenvector of the derivative, whereas sine and cosine are eigenvectors of the derivative squared.

So with first order (i.e. Hamiltonian) equations of motion, $a$ is more convenient.

Now, you may think that convenience isn't convincing, but let's remember what argument I'm trying to convince you of in the first place.

I'm claim that Dirac and co. probably didn't think up $\hat{a}$ by wanting something that turns $|n\rangle$ into $|n+1 \rangle$.

Indeed, since $a$ exists in classical mechanics description of the harmonic oscillator, I bet Dirac and friends knew about it already.

Beyond that...

It's a fine argument

Once you deal with coupled oscillators, $a$ and $a^*$ are almost necessary in order to calculate anything.

In fact, if you write the equations for, say, two coupled harmonic oscillators with disparate resonance frequencies, you'll find that by writing stuff in terms of $a$ and $a^*$ you can easily determine that certain terms can be neglected.

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