The h Bar - 2015-06-09

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3:47 AM

@ChrisWhite Thanks :)

4:40 AM

The Philosophy of the Large Hadron Collider / guardian

> There have been many tedious discussions about the value of philosophy for modern science. I find it much more interesting to ask if and in what way modern science can advance philosophy

@DanielSank hi re QM pushed-to-the-limits, wondering if you or anyone at google/ ucsb/ elsewhere noticed/ have any opinion on these new developments? nature.com/news/quantum-physics-what-is-really-real-1.17585

5 hours later…

9:30 AM

user image

pretty [pretty crowded] ising model graphs!

3 hours later…

12:12 PM

Hi guys, I have just been forwarded this paper by a friend of mine physics.stackexchange.com I am not currently affiliated with an institution and I won't pay for the right to read research that should be public. I have asked a mate to send it to me. In the meantime, I wanted to ask if anyone has read it and generally what's peoples thoughts on the experiment?

To me I think this is likely to be a CERN "Neutrinos breaking the speed of light" experimental timekeeping error or something involving entanglement that they have not considered [I don't know the details yet]. Any thoughts?

nature.com/nphys/journal/vaop/ncurrent/full/nphys3343.html

3 hours later…

3:26 PM

@Killercam I've only read the abstract, but it seems it's "just" (not to diminish the experimental achievement here) a demonstration of a prediction of quantum mechanics, namely the behaviour of a photon in a delayed choice experiment. What troubles you about it?

4:04 PM

@NeuroFuzzy so you are working with ising models now? what ref are you using?

Well, from what I have read, every time the two grates were in place, the helium atom passed through, on many paths in many forms, just like a wave. But whenever the second grate was not present, the atom invariably passed through the first grate like a particle. The thing that I am worried about is, the second grate’s very existence in the path was random. And what’s more, it hadn’t happened yet.

They allow the particle through the first grate before the randomly generated second gate it either there or not there, and it was as if the helium particle “knew” whether there would be a second…

2 hours later…

5:57 PM

@ACuriousMind Do you have access to BLT?

@0celo7 Yes.

@ACuriousMind They try to explain some group theory, but it still seems like magic.

Page 219, to be exact. Equation (8.79) is the key to most of my current problems.

I just have no clue how to get it.

I'm not sure if the Clebsch-Gordan comment is supposed to be a hint, or if they're saying it will be derived in chapter 11.

I have no idea what that equation is supposed to tell me...what is $A$? Why do the $\gamma$ have $p$ indices?

@ACuriousMind $A$ is defined on the following page. $\gamma^{[p]}$ is the antisym product of $p$ gamma matrices.

Ah

6:03 PM

There should probably be a $\mathrm{T}$ on $\psi$ in the definition of $A$.

Well, it sounds to me as if they promise here to derive it in chapter 11

No wise insights on tensor products of spinor reps?

The big issue I have is why we have even for same chirality and odd for opposite. Furthermore, why are there duality conditions on the four-form?

I can reconcile the duality condition by counting degrees of freedom, but the other one has me completely stumped.

I have honestly no idea what the heck they are doing there, I can't help you

:/

Another data point for "physicists can't explain group theory".

4

6:15 PM

How the heck did you do superstrings then?

@0celo7 Lol...we "did" superstrings by the lecturer telling us "This is the spectrum, and we can do this and that and this...", but I never saw a proof (and all my technical questions were mainly answered with "One can show this, but we don't have the time"). Basically, we didn't do superstrings, we were just told a bunch of nice stories :P

Rats

Alas, I have no time to go on a hunt to understand what they are doing the, I have to prove stupid statements about Sobolev spaces :/

That's your fault for taking functional analysis

You should have taken "String theory 2, now with extra group theory proofs"

@0celo7 I wanted to see operator theory and maybe some nice things about Banach spaces in general or something like that, not prove dozens of weird statements about Sobolev spaces without any indication why these statements would be interesting at all. :(

@0celo7 I would have, if such a thing existed.

6:24 PM

What are the odds that the authors don't know this either

Weigand tells me to look at Polchinski, and Polchinski is like: "magnets" and boom there are the answers.

I honestly don't know why, but it is annoyingly common to be referred to another source only to find out they also just state the result essentially without proof.

That's not a particular property of string theory texts.

@ACuriousMind Hmmmmmm. Is it perhaps possible, maybe, that if $\psi$ and $\lambda$ are opposite chirality, that $\bar\psi \gamma_{\mu\nu}\lambda=0$?

Poor Mary Star, they are telling her to stop asking so many questions in the math room :(

@0celo7 I really don't know anything about higher-dimensional spinors, so, well, I guess it's possible :P

@ACuriousMind If they have opposite chirality, the standard bilinear needn't be zero, right?

i.e. $\bar\psi\lambda=0$ is not true just because of opposite chirality?

6:30 PM

@0celo7 Mh...this always confuses me, but aren't the chiral subspaces orthogonal?

God, I don't know anything about spinors, it seems

@ACuriousMind Fuck if I know.

My first day of freedom and this book pisses me off more than high school ever did.

I think I spent my first day of freedom sleeping. :D

Let $\lambda$ have $+1$ chirality. Then $\bar\psi \lambda=\bar\psi\gamma_{d+1}\lambda=\psi^\dagger \gamma^0\gamma_{d+1}\lambda=-\psi^\dagger \gamma_{d+1}\gamma^0\lambda$.

Now suppose $\psi$ is Majorana.

Amen for sleep :D

Then $\bar\psi\lambda=-\psi^\mathrm{T}\gamma_{d+1}\gamma^0\lambda=-(\gamma_{d+1}\psi)‌^\mathrm{T}\gamma^0\lambda$ if we take a rep where $\gamma_{d+1}$ is symmetric.

@ACuriousMind See it????

$\psi$ has to have opposite chirality for consistency.

mfw that's wrong

Fuck my life.

GDI what is this

6:40 PM

...I know neither what mfw nor what GDI means.

my face when, god damn it

GDI these old people, mfw they are the smart ones

^^example sentence

Get off my lawn.

You don't even have a lawn :P

Get off the tarmaced street in front of my house where only hints of repressed nature are visible, then.

4

I'll stand on my multi-acre lawn over here and use a megaphone

6:44 PM

You'll have to buy a bigger megaphone, my hearing aid's battery is running low.

lol

@ACuriousMind Seriously, what the hell? Did I show above that the Dirac product of two spinors is only nonzero if they have opposite chiralities??

semi-interesting

full boring.

the philosophy of mathematics is cool though, and sort of non-trivial.

From an applied science point of view, yes.

@Icosahedron like ur face

6:55 PM

@0celo7 that's pretty spot on.

@ACuriousMind I give up. Time to grind some more in Skyrim.

I have all mage skills at 100.

I need a lot in light armor and blocking.

@ACuriousMind I've taken some nice naps.

I have an awards thing tonight, not looking forward to that.

@0celo7 this is me right now.

@0celo7 because you didn't win any awards?

@Icosahedron No

It's an hour of bullshit speeches, then an hour of clapping, then home.

Nothing gained.

Everything lost.

@Icosahedron You don't get invited unless you're getting something.

Did you write your speech?

Students don't give speeches

They'll have some teachers, some graduated people maybe

I'd refuse to give a speech if they asked me, which they wouldn't because I'm notorious.

7:05 PM

For giving bad speeches?

For not having anything good to say about anyone, ever.

I'm pretty sure admin hates me.

Exception: Zee.

Literally everyone who writes string theory books has no clue about group theory pedagogy. This is crazy.

Person: Do you have any friends online?

0celo7: Only a flaming loser that helps me understand things I read.

Ok, that makes me sounds like a sociopath.

@ACuriousMind You're not a loser <3

7:11 PM

@0celo7 This is contradictory to the point where it cannot be refuted, though it's true.

@0celo7 This too.

@Icosahedron You can't confirm nor deny that.

You read a few pages.

@0celo7 I read more than 50.

That's a few.

I can infer from what I've read.

@ACuriousMind After reading Polchinski very carefully, I have concluded that what I was doing was in principle correct, but I'm missing a single minus that changes the whole answer.

7:24 PM

Ah, sign errors. No derivation is complete without at least one.

@ACuriousMind This is very interesting.

I can't figure out where I went wrong though.

I think I understand Polchinski's explanation.

Or...not. This is quite depressing.

@ACuriousMind Hmm. $\bar\psi=\psi^\dagger\gamma^0$ must be wrong.

@0celo7 That's the definition.

@ACuriousMind Only in four dimensions I bet.

@0celo7 That...may be correct.

@ACuriousMind Or not.

7:38 PM

@0celo7 Yes, may implies that the opposite may also hold ;)

But if the definition is changed, they should state that somewhere. Oh, I forgot, they only should if they were the slightest bit pedagogical.

@ACuriousMind I think the secret lies in the transpose of $\gamma_{d+1}$.

I assumed it is symmetric, but who knows if that's true.

Eq. (8.75) in BLT

WTF I can't figure out how Polchinski gets that sign!

7:50 PM

I think the situation has finally come to an end

@ACuriousMind DON'T EVEN GET ME STARTED ON THIS JKL#P$#@PR&*(!!!

I take it you agree, then.

Lie algebra $\neq$ Lie group, damnit!!!

@Danu In a nice/acceptable or in a not-so-nice way?

@Danu That's not even the bad part.

@ACuriousMind Slightly unfortunate but I don't think any serious damage has been done.

Basically, it got really really really close to going too far :P

7:53 PM

Why the heck does this minus depend on $d/2$??

Literally no explanation given.

"It follows from the definition"

@Danu You're saying there was a supremum, but not a maximum?

@0celo7 The whole of math follows from the definitions :D

2

@ACuriousMind D:

@ACuriousMind I'm saying bounds were pushed, but not violated

That's...good, I guess.

@0celo7 You mistyped ":D". ;)

@ACuriousMind If you say so.

7:58 PM

@ACuriousMind It got too close to not address it right after

...so we did

@ACuriousMind I GOT POLCHINSKI'S ANSWER

Who knew $\sum_{k=1}^nk=\frac{1}{2}n(n+1)$ would be so useful!

@ACuriousMind Found the mistake. It turns out that $\bar\psi=\psi^\dagger\gamma^0$ is not the definition. It's only true when Majorana spinors exist!

That still means I proved two spinors with equal chirality are always orthogonal in dimensions where Majorana spinors exist. Not good.

What's the definition then? "Transforms in the conjugate representation" and we have to figure out whatever the hell that means in any particular dimension?

@ACuriousMind Forget that

Lol

Just read the first para under the bold title on page 216

wtf

I'm so confused

There they say it is $\gamma^0$

Back to square one!

8:16 PM

2 hours ago, by ACuriousMind

God, I don't know anything about spinors, it seems

^^

Forget the physics, the math is confusing!

Spinors ARE math

Spinors ARE confusing

idk what you mean by that

@0celo7 That they're really important in pure mathematics, too

@Danu I never said the opposite.

8:21 PM

@0celo7 I never said you did :D

Sir Michael Atiyah, What is a Spinor ?

I don't need that! I need to prove that everything I know about them isn't a lie!

@0celo7 Could your knowledge of spinors be summarized by the cake?

...because all may be lost ;)

@Danu I think I have shown that $\bar\psi\psi$ is identically zero in all even dimensions. It's probably all a lie.

There will be cake!

After you've disposed of your trusty companion spinor, that is.

@0celo7 Yeah... pretty certain about that being slightly incorrect :P

8:26 PM

@Danu What about $\psi^\mathrm{T}\gamma^0\psi=0$? Do you buy that one?

@0celo7 Most definitely not :P

:/

Halp me pls

@Danu You're my only hope

Oh, first suppose that $\psi$ is Weyl.

Do you believe it now?

No

These are the completely standard things that appear everywhere in QED

they're not zero :P

@Danu Ok, please find the flaw: Let $\gamma_{d+1}$ be the chirality operator and let $\gamma_{d+1}\psi=+1$. Note that $\{\gamma_{d+1},\gamma^0\}=0$.

Just do it in $d=4$

8:30 PM

First question: is it always possible to find a representation where $\gamma_{d+1}$ is symmetric?

You shouldn't get zero

@Danu I know, but I'm working on spinors in $2n$ dimensions and need to figure this out.

Just take n=2 for now

Fine, I know it is possible to diagonalize $\gamma_5$. Do you agree?

cf. Weinberg's first book, Sect. 5.4 to be exact.

Since it is symmetric, we have $\gamma_5=\gamma_5^\mathrm{T}$.

Sure

8:33 PM

Thus $\psi^\mathrm{T}\gamma_0\psi=\psi^\mathrm{T}\gamma^0\gamma_5\psi$.

Agreed?

No

Why?

$\psi$ has +1 chirality.

Now do you expect to multiply by $\gamma_5$ for free

You already agreed to $\psi=\gamma_5\psi$.

No

Of course not

8:36 PM

That's literally the definition of a Weyl spinor...

no

$P_R=(1+\gamma_5)/2$

That projects out the right Weyl part.

What exactly do you mean by a Weyl spinor

If $\psi$ is already Weyl, $P_R\psi=\psi$ implies $\gamma_5\psi=\psi$.

as far as I know, the standard definition is $\psi_L$ and $\psi_R$

which is $P_{R/L}\psi$ for the Dirac spinor $\psi$

8:38 PM

@Danu Does $P_R\psi=\psi$ not imply $\gamma_5\psi=\psi$.

These are not Dirac spinors, they have definite chirality.

Obviously not

@Danu The heck?

Please show how.

$(1+\gamma_5)/2\neq \gamma_5$

If the eigenvalue of $\gamma_5$ is 1, it sure is.

2/2=1

Hah right yeah

sure

(I'm done with my chess match now lol)

8:40 PM

So are we on the same page?

figuratively speaking, yes :P

Now anticommute the matrices in the middle. $\psi^T\gamma^0\psi=-\psi^T \gamma_5\gamma^0\psi=-(\gamma_5\psi)^T\gamma^0\psi=-\psi^T\gamma^0\psi$.

See my issue?

Yeah okay but that's actually okay

yeah

No problem

$\psi_R \gamma^0\psi_L$ is the only thing that should appear

I think

Normally

So how does a singlet appear in the decomposition of the tensor product of two Weyl spinors which have the same chirality?

Maybe we construct the bilinears using Majorana adjoints and not Dirac.

Let's see about this first. $\gamma^0P_R=\gamma^0(1+\gamma_5)=(1-\gamma_5)\gamma^0=P_L\gamma^0$

8:52 PM

Why $\ne0$?

Oh yeah I'm retarded :D

DELETE EVERYTHING!

lol

it's a mistake to ever "expand" the bar over the spinors :P

$\bar\psi_R\psi_R=\bar\psi_RP_R\psi_R=0$ simply

while $\bar\psi_L\psi_R=\bar\psi_LP_R\psi_R\neq 0$

@Danu So I was right, right?

Beats me how all this works then.

I'll choose to believe the book.

@0celo7 As long as you talk about Weyl spinors then their "naive contraction" is zero, yes. For sure this time!

9:02 PM

I got my carroll and shankar in the mail today.

Carroll yayyy :D I love it

But I don't think it was $300 well spent.

Lol

You should have just read HE

Why'd you waste all that monies

Because my laptop broke and I can't read when I'm not in my room anymore.

(desktop)

That sounds overpriced!

I love the internet :P

9:06 PM

@0celo7 HE is too elementary for me right now.

:D

@Icosahedron Oh shit, teach me your ways

Some things cannot be shared with others.

All I need to know is how to determine the tensor product of Weyl spinors

@Icosahedron Also why Shankar? That's something an engineer would read.

9:24 PM

@0celo7 Right.

Shankar's book is the foulest color in existence.

This needs to be rebinded.

@0celo7 What color is yours?

Mine looks like this.

how do you have money to rebind all this stuff

user image

10:01 PM

Yours is red!

How.

@0celo7 It costs $10.

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