The h Bar

Anonymous

7:29 PM

Can the transition matrix property shown here for a single qubit be shown to be equivalent to any of the Bell's inequalities ?

Semiclassical

@blue yep. like I said in CC (but forgot to ping) I think it's equivalent to the maximal violation of the CHSH inequality

Anonymous

@Semiclassical What's "maximal violation" though ? I know the statement of CHSH as given on wiki: en.m.wikipedia.org/wiki/CHSH_inequality

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

hello people im seeing a weird edit got approved. the person added 2 good words and modified "there" by "their" but I can't seem to make sense of "there" nor "their" in that context. im not english native speaker, please have a look at physics.stackexchange.com/questions/410585/… edit of the answer of The Sympathizer (which by the way I think provides the best answer)

Semiclassical

It's the Tsirelson bound.

i.e. "The mathematical formalism of quantum mechanics predicts a maximum value for S of 2√2 (Tsirelson's bound),[4] which is greater than 2, and CHSH violations are therefore predicted by the theory of quantum mechanics."

from that wiki page

Anonymous

Ah, interesting

Anonymous

7:35 PM

Not exactly sure how to translate that to the transition matrix formalism though. Gotta read a bit more about it I guess

Semiclassical

I can try to explain it (and would actually like to try---it's something I"m trying to better understand myself)

Anonymous

@Semiclassical Please go ahead!

Anonymous

(Whenever you are free)

Semiclassical

mmkay. So here's the setup. At a source, you entangle two electrons in a singlet state and send one off to Alice and the other to Bob.

Anonymous

Okaies

Semiclassical

7:37 PM

Actually, would this be better in CC?

i guess not :)

heather

...and hello over here.

Anonymous

It's fine here :)

Mr. Xcoder

\o guys

Novalium Company

@Blue I think I finally understood about boiling.

Semiclassical

Alice and Bob each have two Stern-Gerlach devices. The orientations of the devices won't change throughout the experiment, but they don't have to be related to each other i.e. Alice's two devices can and will have different orientations than Bob's.

Anonymous

7:41 PM

So far so good

Semiclassical

I'll specify the operators for these observables as $A_1=\vec{S}_A\cdot \hat{a}_1$, $A_2=\vec{S}_A\cdot \hat{a}_2$, $B_1=\vec{S}_B\cdot \hat{b}_1$, $B_2=\vec{S}_B\cdot \hat{b}_2$

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

there's a dude who makes thousands of edits to his own posts. this makes his answer pop up in the front . he's been told not to do that yet it repeats it. 28 edits in less than 24 hours to an asnwer lol

Semiclassical

where $\vec{S}_A$ is the spin of Alice's particle vs. $\vec{S}_B$ for Bob's particle. (So really it's stuff like $S_x\otimes I$)

heather

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP raise a custom flag and let the mods know.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

can someone tell me whether the last sentence is illogical as I believe it is? "Once the star's core is predominantly iron/nickel, the processes of fusing which have powered it, for all its life, simply stop. They can't fuse to more stable elements, so they just don't fuse."

he claims that fusion stops because this would not lead to more stable elements

I do no think logic is involved in this deduction, am I wrong?

Semiclassical

7:45 PM

I imagine it's more gradual than simply 'stopping' but otherwise it seems sensible enough. fusion proceeds until it's not energetically favorable to do so

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

exactly, it has to do with energy. not about stability

heather

fusion can still occur, the star just loses energy from it, as i understand...obviously it slows down though.

Semiclassical

for simplicity, I'll take the outcomes to be $\pm 1$ rather than $\pm \hbar/2$ (so units where $\hbar=2$, w/e)

eh. I imagine 'stable' just means 'lower energy' here?

Anonymous

@Semiclassical Fair enough

Semiclassical

I'm not an astrophysicist tho

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

7:46 PM

no he means as stable as won't decay

Novalium Company

@Blue You studied psychology or something right?

Semiclassical

So now we can start asking about the statistics that the observers would see. Obviously it'll depend on the measurement settings, but there are some basic facts.

First, $\langle A_1\rangle =\langle A_2\rangle = \langle B_1\rangle = \langle B_2\rangle = 0$.

Jalapeno Nachos

Does anyone here do data-driven physics?

Anonymous

@NovaliumCompany A bit busy here (trying to comprehend what SemiC is saying) ;) Let's talk tomorrow ? (Fwiw - not formally)

Novalium Company

@Blue Ok sorry, see you tommorow. :)

Semiclassical

7:49 PM

That follows from the fact that if you have no knowledge of the second particle, the singlet state $|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle$ will give up and down in equal probability. (the singlet state has zero angular momentum and is thus rotationally symmetric, so it doesn't matter which quantization axis I pick)

Anonymous

Right, agreed

Semiclassical

Second, the outcomes are all $\pm 1$ so $\langle A_1^2\rangle=\langle A_2^2\rangle = \langle B_1^2\rangle = \langle B_2^2\rangle =1$

Anonymous

Mhhm

Semiclassical

So if we could think of these as random variables in the usual sense, then they'd have zero mean and unit variance. That makes life substantially simpler.

What remains, of course, are the correlations $\langle A_k B_l\rangle$.

Anonymous

@Semiclassical That's the confusing part, yeah

Semiclassical

7:54 PM

So to measure those you'd have Alice and Bob pick their desired axis, measure a bunch of singlet state pairs, and see how often they got the same result vs. opposite results

Anonymous

True

Semiclassical

Now, if you had a generic state, this could be a bit of a pain to compute. But in the singlet case things aren't bad.

In particular, if you take the axes to be identical, then we know the results will be opposite (A measures up, so B measures down etc)

Anonymous

@Semiclassical Yes!

Semiclassical

so therefore in that case you'd have $\langle A_k B_l\rangle =-1$

Anonymous

Right

Semiclassical

7:56 PM

by contrast, if the axes are orthogonal, then you shouldn't be able to learn anything about the component of one from the other

(That's a bit slapdash but it can be shown more carefully.)

So in that case $\langle A_k B_l\rangle =0$.

Anonymous

Makes sense, yeah

Anonymous

They could even be at any other angle but I guess we don't worry about that case here....?

Semiclassical

actually, that's where I'm going

Anonymous

Ah, ok

Semiclassical

In the generic case, you can write $b_l$ as a linear combination of $a_k$ and an axis perpendicular to both i.e. $b_l = a_k \cos\theta +c\sin \theta$

where $\theta$ is the angle between these two unit vectors

Anonymous

7:59 PM

Fair

Semiclassical

and then the linearity of expectation guarantees that $\langle A_k B_l\rangle = (-1)\cos\theta+(0)\sin \theta=-\cos\theta$

(Write out the definitions of $A_k$ and $B_l$ in terms of that dot product with spins to verify this.)

Or, in coordinate independent terms, $\langle A_k B_l\rangle = -a_k\cdot b_l$

Anonymous

Interesting....I see

Semiclassical

That's an exercise in at least one edition of Griffiths, I should note

It really shouldn't be too surprising a result, mind. when $\theta=0,\pi/2,\pi$, you get $\langle A_k B_l\rangle =-1,0,1$ respectively

Anonymous

I think I've come across that one before, yeah :)

Anonymous

Go on

Semiclassical

8:02 PM

with the last one being the fact that, if Alice's axis is opposite Bob's, then they'll always come out as spin up since they've got opposite settings and they're measuring the singlet

Anonymous

Right, got that part

Semiclassical

So now we know how we'll predict the correlations we'll see. To create a more unified framework for this, we can arrange it as a two-by-two matrix $C_{kl}=\langle A_k B_l\rangle= -a_k\cdot b_l$

Now for me to ponder what notation to use...hmm. Let me define $U$, $V$ as 3-by-2 matrices whose columns are $a_1,a_2$ and $b_1,b_2$ respectively

What's handy is that this lets us write $C=-U^\top V$.

Anonymous

@Semiclassical Umm, what are the other two elements ?

Anonymous

It's 3*2 right ?

Semiclassical

You're misreading. remember, those are 3-vectors

which I'm intending as column vectors

so $U:=(a_1\,a_2)$ and $V:=(b_1\,b_2)$

e.g. $a_1$ is a 3-by-1 column vector

Anonymous

8:11 PM

Oh, okay okay....gotcha :P

Semiclassical

kk

before proceeding, I"ll note that while this formulation might seem pretty special

it's actually quite generic, as Tsirelson's own work proved. (basically, if you have any quantum state with operators that pairwise commute and have appropriate spectrum then you'll get some factorization like this)

the relevant paper is on his personal website here: tau.ac.il/~tsirel/download/qbell80.html

(full disclosure: I only partially understand the main theorem)

Anonymous

I'm not aware of his work, but interesting. I'll see. Thanks!

Semiclassical

np

Anyways. So far we've attacked the problem from the following direction: If we know the settings for Alice and Bob, what correlation matrix $C$ should we expect to see?

However, what's pertinent for the Tsirelson bound is the reverse question: If someone gives me a correlation matrix, can I deduce whether it was obtained from a setting like this?

Anonymous

@Semiclassical Right, but one thing to note here is that we're dealing with the TM for a 2-qubit system whereas the answer I had linked was for a single qubit TM. Anyhow I guess we'll come to that

Semiclassical

eh, I'm not actually sure you can do a Bell inequality with one qubit, if only because there's no notion of separability

you need to be able to say "Alice measures something in one place and Bob measures something else in a different place" in order to test whether local hidden variables are plausible

Anonymous

8:18 PM

@Semiclassical Yeah, thats exactly what I was confused about. @heather's answer says that we can explain why the Hadarmard state is different from a classical coin using a Bell inequality

Anonymous

However, the other answer I linked had a TM logic based on quantum interference

Semiclassical

hmm, nice

I'll have to look more at that later.

the case I know about, I should admit, is entirely in this "two spatial domains" case

Anonymous

Yeah, I sort of understand where you're getting at with this one (two spatial domains)

Anonymous

I think this deserves a Q&A thread on the main site :)

Semiclassical

yeah, maybe

Mithrandir24601

8:24 PM

@Blue Slightly different, but you can do a thing sending 2 photons into a beamsplitter (equivalent to Hadamard) and do a Hong-Ou-Mandel (HOM) dip, which is an intrinsicly quantum, non-classical thing

Semiclassical

momentarily distracted, back

Anonymous

@Mithrandir24601 Haven't heard of it before, but I'll check, thanks

Mithrandir24601

(as an aside, I met the very same Jeff Ou about half a year ago :) )

Hong–Ou–Mandel effect

The Hong–Ou–Mandel effect is a two-photon interference effect in quantum optics which was demonstrated in 1987 by three physicists from the University of Rochester: Chung Ki Hong, Zhe Yu Ou and Leonard Mandel. The effect occurs when two identical single-photon waves enter a 1:1 beam splitter, one in each input port. When the photons are identical, they will extinguish each other. If they become more distinguishable, the probability of detection will increase. In this way the interferometer can accurately measure bandwidth, path lengths and timing. == Quantum-mechanical description == ==...

Semiclassical

So now we want to start doing actual bounds, and here I'm at my shakiest

Anonymous

@Semiclassical Lol, same

Mithrandir24601

8:29 PM

@Semiclassical Yeah, an actual CHSH is impossible with a single qubit to my knowledge. Don't know if there's an equivalent thing for 1 qubit or not...

Semiclassical

I sorta doubt it. it'd be like testing nonlocality without actually having observers which are spatially separate

you can test other quantum stuff, I'm sure, but it seems unlikely you'd be in a position to test that

Mithrandir24601

Or actually... I wonder if you could do some sort of weak interaction/measurement scheme and look at the averages? You have to look at averages anyway

Semiclassical

possibly, but i think you'd have to worry about loopholes a lot more

Mithrandir24601

@Semiclassical but you can split a single photon

enumaris

I decided to check out all the MTG cards I could still find yesterday...stuff I've had since high school...I thought they'd be worth some money now, but actually they've all depreciated in value...T_T

Mithrandir24601

8:31 PM

OK, that sounds wrong, but I'm referring to splitting the wavefunction

Semiclassical

now, if we want to compute the CHSH statistic, it's convenient to introduce another matrix...and I am finding myself running out of symbols. hmmm

Mithrandir24601

@Semiclassical This is true

David Z

@enumaris sorry to hear that :( that's how it goes most of the time

Semiclassical

I'd use $U$ since the matrix I"m about to do is unitary but I already used that...hrm

enumaris

Serra avatar was 20 bucks when I was in highschool, now it's 14...

Semiclassical

8:33 PM

I'll use $S$ for summing, I guess

enumaris

I found a City of Brass from the Arabian Nights set which would be 300 bucks...and I was excited...but then I found out I had the chronicles version...only 15 bucks T_T

the whole collection is probably <$100 LOL

Semiclassical

$H=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}$

enumaris

it does bring back memories tho

so at least it has some sentimental value

Mithrandir24601

@Semiclassical Try using $H$

David Z

Yeah, I find it better to focus on the sentimental value rather than the monetary value.

Semiclassical

8:35 PM

works for me

Mithrandir24601

nooooooo! :P

:)

enumaris

I want to make a deck and play with it...but I think I've lost too many cards to actually be able to construct a working deck lol

Semiclassical

The reason it's handy to introduce this (symmetric, unitary) matrix is that we can express the CHSH statistic as $\text{tr}(CH)=\sum_{kl} C_{kl}H_{kl}=\frac{1}{\sqrt{2}}(C_{11}+C_{12}+C_{21}-C_{22})$

enumaris

I also don't want to buy a new deck cus too much has changed in the last 15 years...

>.>

Semiclassical

which is a valid CHSH statistic. (there's actually more than one you can pick, owing to the symmetries of the system, but you can only maximally violate one of them at a time I think?)

So our goal is now to find an appropriate bound on $\text{tr}(CH)$. there are a few that can be formulated.

one is to note that, since the outcomes are all +- products, the pairwise correlations are all between -1 and 1. that gives the upper bound $C_{11}+C_{12}+C_{21}-C_{22}<4$.

This, however, is not a very useful bound---it assumes that we can pick these four correlations independently.

enumaris

8:42 PM

anybody here play MTG?

Semiclassical

To obtain a more useful bound, I'll take a page from how one proves Cauchy-Schwarz and consider the quantity $\text{tr}[(UH-V)^\top (UH-V)]=\sum_{kl}(UH- V)_{kl}^2$ which is nonnegative and only vanishes when $V=UH$ identically.

David Z

@enumaris This would actually be a good discussion to have in the Board and Card Games room

enumaris

oo

is that room active?

David Z

Moderately so

Go have a look

enumaris

Will do

and is that just a suggestion or is that a mod-ly - "don't discuss that here as it's off topic" thing?

Semiclassical

8:47 PM

(gotta have it as $UH$ since $U$ is 3-by-2 and $H$ is 2-by-2, lol)

on the other hand, the linearity of trace lets us rewrite that to $\text{tr}(H^\top U^\top U H)+\text{tr}(V^\top V)-2\,\text{tr}(H^\top UV)=\text{tr}(U^\top U)+\text{tr}(V^\top V)-2\,\text{tr}(CH)$

enumaris

@Semiclassical feels like you've lost your audience..

Semiclassical

eh, my audience is partly just me

moreover, $U^\top U=(a_1^\top\, a_2^\top)\begin{pmatrix} a_1 \\ a_2\end{pmatrix} = a_1^\top a_1 +a_2^\top a_2=1+1=2$

Mithrandir24601

@enumaris I'm still here! Only I'm eating at the minute

enumaris

XD

Semiclassical

and similarly $V^\top V=2$

David Z

8:52 PM

@enumaris Just a suggestion

enumaris

@DavidZ cool :D

Semiclassical

except that what I just wrote is total nonsense

enumaris

btw

I notice if I tab complete a name, spaces are truncated

does that affect the notification on the other end?

David Z

@enumaris Yeah, it makes sure that only the person you want to ping actually gets pinged

Anonymous

@enumaris Not really :P I'm still noticing but things are going a bit over my head now, partly because I'm very sleepy :P Will catch up on what SemiC wrote, tomorrow

Semiclassical

8:54 PM

$\text{tr}(U^\top U) = \text{tr}[\begin{pmatrix}a_1^\top \\ a_2^\top \end{pmatrix}(a_1\,a_2)] = a_1^\top a_1+a_2^\top a_2 = 2$

there

and similar the trace of $V^\top V$ is $2$.

enumaris

@DavidZ so you see a ping even though there's a missing space?

@Blue it's been over my head since statement 1

Semiclassical

So that lower bound can be rearranged to $2\,\text{tr}(CH)\leq 2+2$

i.e. $C_{11}+C_{12}+C_{21}-C_{22}\leq 4/\sqrt{2}=2\sqrt{2}$

David Z

@enumaris Yeah. For example when you write just @David it pings everyone in the room whose display name starts with "David".

Semiclassical

which is the Tsirelson bound

enumaris

I see

Semiclassical

8:56 PM

i.e. that's the largest possible value which that sum can take if you're using the singlet state in this manner

enumaris

ain't nobody in the other room

feelslonelyman

Semiclassical

Vto be precise, all I've proven is that it's bounded in this way. to actually achieve this bound, I need to ensure that the bound is saturated. but I noted that equality occurs when $V=UH$---or, in more suggestive terms, $$(b_1\, b_2) = (a_1\,a_2)\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}\implies b_1=\frac{a_1+a_2}{\sqrt{2}},\, b_2 = \frac{a_1-a_2}{\sqrt{2}}$$

which looks an awful lot like a rotation :)

enumaris

hard to decipher a rotation from raw TeX code

Semiclassical

then turn on mathjax

enumaris

can't...-.-

Mithrandir24601

9:02 PM

27

A: Any chance of MathJax in chat?

As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An altern...

enumaris

yeah I'm not allowed to install plugins lol

Semiclassical

additionally, we have to have $1=\|b_1\|^2 = \frac12 \|a_1+a_2\|^2 = \frac12(1+1+a_1\cdot a_2)\implies a_1\cdot a_2=0$

and then $b_1\cdot b_2 = \frac12(1-1)=0$

So in order to achieve this correlation matrix, we need the axes to be at right angles to each other. We might as well pick $$a_1=\hat{x}, \,a_2=\hat{y}\implies b_1=\frac{\hat{x}+\hat{y}}{\sqrt{2}},\, b_2 = \frac{\hat{x}-\hat{y}}{\sqrt{2}}$$

Mithrandir24601

@enumaris It's not a plugin - just bookmark the first link here, then click the bookmark

Semiclassical

which amounts to the axes of Bob being orthogonal to each other and at a 45 degree angle to those of Alice

So we actually can achieve this with the singlet state.

enumaris

how do I bookmark a link rather than a page...

Semiclassical

9:07 PM

click and drag it

heather

@Blue I wasn't saying you could do bell's inequality with one qubit; I was trying to say that the idea of bell's inequality and the lack of hidden variables is what makes quantum computing quantum, and therefore why qubits are different from classical bits.

Semiclassical

So we actually can achieve this with the singlet state, i.e. $C_{11}+C_{21}+C_{12}-C_{22}=2\sqrt{2}$ is possible with QM

what remains is to show that this is not achieveable with local hidden variables...

enumaris

oh figured it out

so do I have to click that link every time a new latex equation shows up or...$test$

oh guess not

nice

Semiclassical

it's handy yeah

enumaris

greato

Semiclassical

9:09 PM

@Blue Here's an argument that I've only tried recently and so I'm testing it on you.

We set up Alice and Bob's axes in the above manner. Suppose we take $(A_1,A_2,B_1,B_2)$ to be a discrete random vector taking values $\pm 1$

as noted earlier, these random variables have unit variance.

in addition, from the fact that the measurement axes for Alice are mutually perpendicular, we assume that we'd get $\langle A_1 A_2 \rangle =0$. (you can't actually test this experimentally, but I don't see an easy way around it so oh well)

similarly, $\langle B_1 B_2\rangle =0$.

hrm. that doesn't actually give me what I want. dangit.

enumaris

blink

Semiclassical

well, this was the part I was most tentative on so I'm not surprised it doesn't work yet

oh well

enumaris

keep at it, you'll get there :D

Anonymous

9:27 PM

@heather I clearly see you quote my comment-question based on the Hadamard state (which certainly is 1-qubit).

Anonymous

There are several differences - the way that measurement works (see the fourth paragraph), this whole superposition idea - but the defining difference (Mithrandir24601 pointed this out in chat, and I agree) is the violation of the Bell inequalities.

heather

yes. see my latest comment.

> I'm using the idea of Bell's inequality to point out that, firstly, measurement is fundamentally different in quantum mechanics, and secondly, that quantum mechanics is not deterministic. Both of these points mean that any quantum system, including a qubit, is going to be fundamentally different from any classical system

i quote you to set up the issue - what really is the difference from a probabilistic classical system and a quantum system?

Semiclassical

Where I’m aiming: suppose that you had a probability distribution of random (A1,A2,B1,B2) vectors

heather

i think part of the confusion is from the fact that i'm not really addressing your specific question, with the hadamard state, but the overall idea behind the question.

Semiclassical

Eg (+1,-1,-1,-1)

heather

9:32 PM

(or at least, attempting to address the question =)

Semiclassical

So when you measure the two spins, you’d pick a random such vector, look at two of the components, and discard the rest

The point is to show that the correlations I just obtained are not compatible with this formulation

That’s the goal. (For my own reference)

Anonymous

@heather I don't really know what to say. From my point of view the answer sounds like a collection of buzzwords - hollow from the inside. But I'll try to frame constructive suggestions in the comments, let's see. I appreciate your enthusiasm to help the OP though. And I think the answer can be improved several fold.

enumaris

by vector are you considering the full set of 4 numbers as 1 single vector?

heather

@Blue it is providing a description of a qubit. how is that buzzwordian? (which is not a word, but whatever.)

Semiclassical

Right. So pretending that, if you measure A1 and B1, that the values for A2 and B2 were determinate but not measured

heather

9:39 PM

it might not be formalized, but analogies aren't meant to be formal. it's meant to give a sketch of the ideas.

Semiclassical

Assume that and arrive at a contradiction

I know how to do that in one case but not this one, hmmmm

enumaris

So A1 and A2 forms a uncertainty pair like [L_x,L_y]?

Semiclassical

Right.

enumaris

so Bell's inequalities would involve things like $\langle A_1 B_1\rangle$ right?

I think the inequalities would say what those are based on some angle between $A_1$ and $B_1$?? It's been like 10 years since I've had to deal with Bell inequalities so...I'm super hazy on the subject lol

Semiclassical

Yeah, some linear combination of those. Earlier I defined $C_{kl}=\langle A_k B_l\rangle$

enumaris

9:44 PM

I see

welp that's about all I gathered :'D

Semiclassical

So when I refer to the CHSH statistic as C11+C12+C21-C22, I’m really looking for (upper) bounds on that

That’s basically the equivalent of the Bell inequality here

I think I know how to see it, hmmm

enumaris

LIke are you trying to rederive the Bell Inequality?

Semiclassical

More like I’m trying to find a cute way to show that the specific case I gave earlier is definitely not compatible with the assumptions of Bells inequality

Without having to derive the full inequality

It’d help if I had paper to write on lol

enumaris

hmmm

Semiclassical

I think I see it but It’ll have to wait until I can sit down with paper

enumaris

10:01 PM

sounds good

enumaris

10:29 PM

I need to find something to read...

bolbteppa

10:39 PM

This is always great reading: motls.blogspot.com/2007/11/exceptionally-simple-theory-of.html

enumaris

unfortunately I'm probably too ill versed in Lie theory to be able to understand the exceptionally simple theory of everything

bolbteppa

So am I, that's not the fun

is the fun

enumaris

hmmm

enumaris

10:54 PM

the paper is unreadable to me

bolbteppa

'My connection of everything = connection for gravity + weak force + strong force + electromagnetism + electron + neutrino + up-quark + down-quark + other-generations' = genius, simple

enumaris

what connection is that

like a Levi-Civita connection?

bolbteppa

It's the famous 'madness connection' where you add apples and oranges and unify all of physics as a result

'The author is not constrained by any old "conventions" and simply adds Grassmann fields together with ordinary numbers i.e. bosons with fermions, one-forms with spinors and scalars, neglecting any traces of dimensional analysis, too. He is just so skillful that he can add up not only apples and oranges but also fields of all kinds you could ever think of.

Every high school senior excited about physics should be able to see that the paper is just a long sequence of childish misunderstandings. I understood these things when I was 14.'

enumaris

hmmm...

bolbteppa

This thing was worldwide news

enumaris

11:00 PM

maybe my knowledge of connections is not good enough

I never actually finished Bishop and Crittenden...lol

bolbteppa

He called it a brst-extended connection, there are loads of blogs of people like Baez and others trying to figure out what the hell he meant

enumaris

hmmm

bolbteppa

'Just googled "BRST extended connection" and the only results I got point to Garret Lisi's papers/talks.

On the other hand, the term "super connection" is indeed well known and simply denotes a superfield expansion. However, in that case, the fermion fields are all multiplied by the corresponding power of the Grassmann variable \theta so that all the terms in the sum have the same spin dimension.

What Mr. Lisi is doing is complete nonsense as he simply adds fields of different spins. '

http://backreaction.blogspot.com/2007/11/theoretically-simple-exception-of.html

vzn

← lol, hasnt been ridiculued/ castigated/ vilified by LuMo for my ToE yet, shew! so far so good, must be on the right track o_O :P =D

enumaris

hmmm

bolbteppa

11:11 PM

@vzn did I not do that for him? :p

enumaris

my knowledge of a connection is that it connects fibers in a bundle together so it allows you to take differences of objects lying on different bundles.

in that vein, if you have different fiber bundles all connected to a manifold it seems at least plausible to connect all the different fibers together in some funky way...and maybe...he uses the addition (+) operator to denote such an operation...but I can easily imagine that different fiber bundles may not be "compatible" with each other and there may be no general way to connect fibers from different bundles.

hence my "my knowledge of connections is not good enough" comment

vzn

@bolbteppa not really my thing but try dirac sea + solitons instead :P sciencedirect.com/science/article/abs/pii/0375947487903939

enumaris

I need to be better at science to refute this

vzn

↑ (The calculation of Dirac sea effects on finite solitons) oh look heres another one... Dirac sea correction to the topological soliton mass inspirehep.net/record/28496?ln=en :)

bolbteppa

An Exceptionally Simple Theory of Everything

"An Exceptionally Simple Theory of Everything" is a physics preprint proposing a basis for a unified field theory, often referred to as "E8 Theory", which attempts to describe all known fundamental interactions in physics and to stand as a possible theory of everything. The paper was posted to the physics arXiv by Antony Garrett Lisi on November 6, 2007, and was not submitted to a peer-reviewed scientific journal. The title is a pun on the algebra used, the Lie algebra of the largest "simple", "exceptional" Lie group, E8. The paper's goal is to describe how the combined structure and dynamics of...

Basically some mathematicians cleaned up what he was trying to say, I am not sure if it makes sense tbh, but yeah this is a great example of needing to be better to refute nonsense

It certainly makes Lie theory scarier seeing how subgroups and reps just pop up

vzn

11:25 PM

hey youre + others always talking about the Standard Model™, arent baryons part of it? Baryons as Chiral Solitons link.springer.com/chapter/10.1007%2F978-3-540-49454-6_4

bolbteppa

@vzn the Dirac sea perspective seems to have a lot of validity to it

Again, it's simply nuts that this picture lets you understand charge renormalization before it magically arises in computations of scattering amplitudes

vzn

← thinks Dirac Sea just yet another aspect of the Fluid Paradigm :)

bolbteppa

@vzn your fluid stuff is a mess of classical elasticity theory and general relativity, it makes no sense to think that's fundamental

vzn

@bolbteppa so youre interested in renormalizing? did you see that 1st ref?

> A scalar field typical of those used to model 16O in quantum hadrodynamics is also studied, and it is shown that, when the effective potential is supplemented by the next term occurring in a derivative expansion, the renormalized shift in the energy of the Dirac sea is well approximated.

bolbteppa

11:40 PM

@vzn the Dirac sea has real problems for bosons

9

A: What was missing in Dirac's argument to come up with the modern interpretation of the positron?

Dirac's derivation of the existence of positrons that you described was a totally legitimate and solid argument and Dirac rightfully received a Nobel prize for this derivation. As you correctly say, the same "sea" argument depending on Pauli's exclusion principle isn't really working for bosons...

But there is a QFT fix I think, trying to make sense of it

QFT (and Strings by association) are just insane compared to other subjects

Semiclassical

I feel like a big reason why people talk so much about philosophy of QM is because it's just technical enough a subject that there are interesting things to say about it but not so much that it's overwhelming. QFT, on the other hand....

bolbteppa

Yeah totally

'The necessity of mass renormalization already occurs in classical electrodynamics; experiments on a free electron measure $m$, the parameter in the Lorentz force law, plus the inertia of the electron's self-field. For a classical electron of radius $\approx a$, the electromagnetic self-energy is $\approx \alpha/a$, and the observed mass is $\approx (m + \alpha/a) $. For a point charge, $a \to 0$ and the correction to the mass becomes infinite. This is true also in Dirac theory' what...?

Semiclassical

One tangential point, looking at that answer: In the stochastic electrodynamics that the fluid people tend to like (e.g. Bush) you have them wanting to take Zitterwebegung as something physical rather than merely formal.

which...meh

see the discussion on page 286 here: pdfs.semanticscholar.org/0a00/…

bolbteppa

11:57 PM

My understanding of Zitter is if you try to view the Dirac equation as a one-particle theory, Zitter arises unavoidably and shows anti-particles have to exist hence the theory is multi-particle, something the solutions also show

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