The h Bar

12:00 AM

I think the story they're telling goes like this (having glanced at another of their papers which they cite): If I make a position measurement at time t and find the particle at position q, then the wavefunction afterwards will be a Gaussian wavepacket centered at q

(I'm ignoring the role of the potential b/c that's another headache to deal with)

bolbteppa

My sense is all you need is the Schrodinger equation to arrive at the path integral as it's Green function, though they seem to do something different, partially similar at the end, I wonder why lemma 7.1 doesn't apply to orthodox qm

Semiclassical

And the nice thing about such a Gaussian wavepacket is that you'll be able to find a bohmian trajectory between q at time t and an arbitrary q' at time t+dt

(typically that's not true, since bohmian trajectories don't cross and therefore the later location at time t' would be determined. but the gaussian wavepacket is a bit weird.)

so in that case it'll make sense to consider the evolution from q at time t to q' at time t+dt from the Bohmian path integral. (i sorta find it weird to call it a path integral--their terminology, not mine)

and then maybe that story makes sense? still pretty weird.

Cows

I guess I will be using the word "oracle" quite a bit today lolz

Semiclassical

i think one has to allow some weirdness, though. Otherwise the Bohmian instance that "velocity and position aren't independent" has no way of making sense given that the propagator is supposed to allow arbitrary initial and final spacetime points.

so something sneaky has to happen

@bolbteppa I think it does apply, it's just that you'd never write it

there's no particular impetus within orthodox QM to care about the wavefunction modulus along different points of a Bohmian trajectory

bolbteppa

Basically I am hoping to come across a reason in Bohm why one wants to use wave functions

Semiclassical

12:16 AM

I think one account would simply be: Same reasons as orthodox QM does.

bolbteppa

Well, my sense is that's actually what people do, and my sense is that's the death-knell for Bohm, keeping my mind open and keeping an eye out, thinking about it etc

and now maybe thinking Bohm was doing something different to this, which could potentially be more justifiable, not sure, interesting to randomly dip into

Semiclassical

Well, I don't think Bohm denies the orthodox formalism. You still need the wavefunction in order for particles to know what trajectories to follow.

It's just that you insist that there's still meaningful trajectories.

bolbteppa

Yeah I am not sure at all, and it's made a lot worse by the fact this is dipping into the whole interpretation business which one can easily go down rabbit holes with

Semiclassical

It's not "X instead of Y" but "X plus Y"

Cows

Ok let me wipe clean what I think I know for a second and ask some questions

so

given some n-bit binary sequence . . .

say

1111 1110 1101 1100 1011 1010

..

Semiclassical

12:22 AM

the trouble is that the X here (namely, the trajectories) isn't intended to alter the empirical output

with the implication being that "standard QM + trajectories" and "standard QM" give the same measurable predictions

Cows

actually let us do just 10

Semiclassical

which raises the question of why bother with the + at all

bolbteppa

I'm only really finding a new appreciation for the standard way of introducing quantum fields as the limit of n Hamiltons equations in the limit as n goes to infinity coupled with the replacement of PB's by commutators and how it reproduces the occupation number formalism, I liked going the other way of normal particle QM to the results of fields without fields

Cows

we can say something in particular like {1,0} characterizes some state Q

we don't have to know the physical state but can abstract it

Then an oracle

(an abstract machine)

then slaps this into something desired

bolbteppa

I tried to break the field perspective and ended up appreciating it more, trying to do that with Bohm, but the difference is a lot of people claim Bohm suffers huge flaws so will be interesting to see what happens

Cows

12:26 AM

let me think for a . . ${\displaystyle f:\{0,1\}^{n}\rightarrow \{0,1\}}$

I am not quite sure how

classically they get

actually let me think for a bit before I ask

btw I am trying to figure out the classical . . . . 2^(n-1) + 1 thing

Deutsch–Jozsa algorithm

The Deutsch–Jozsa algorithm is a quantum algorithm, proposed by David Deutsch and Richard Jozsa in 1992 with improvements by Richard Cleve, Artur Ekert, Chiara Macchiavello, and Michele Mosca in 1998. Although of little practical use, it is one of the first examples of a quantum algorithm that is exponentially faster than any possible deterministic classical algorithm. It is also a deterministic algorithm, meaning that it always produces an answer, and that answer is always correct. == Problem statement == In the Deutsch-Jozsa problem, we are given a black box quantum computer known as an oracle...

let me just get a paper hehe

actually kind of interested in the classical part first

bolbteppa

Page 5 goes into the dB Vs. B thing a bit

Semiclassical

yeah

I'm not actually sure how historically valid that is, I'll note

bolbteppa

It vaguely sync's up with what we said before about Bohm looking at classical limits

Semiclassical

right

it's more the dB side I'm not sure about

dB was known for the so-called 'method of double solution' which is different

Sir Cumference

1:10 AM

@Slereah Sigh, I keep forgetting what that link is and end up downloading the same pic of Emilio.

I think my memory should get checked...

Fduck I did it again...

user351417

1:39 AM

@SirCumference Apparently it's there on github as well.

Cows

ok just spoke with a friend and go the stuff taken care of

so basically the statement of the problem constraints us enough to be able to narrow tdown the evals to $2^{(n - 1)} + 1$

the error and probability are naturally computed from there

So this is the classical case

Now scaling up reduces the classical efficiency

. . . . . . . enter Hadamard et al

Convo with Mike just now over hangouts was quite helpful hehe :*D

Not so random question. . . . . are people paying for "Quantum recommendation systems" now? Can one start making moola selling these? Just curious. . . .

DanielSank

2:06 AM

@heather I look forward to your contributions :-D

heather

@DanielSank oh dear...pretty sure you don't want my crappy code breaking everything ;)

Cows

3:36 AM

In a conference call with two of my friends talking about software and hardware . . code . . . quantum computine algos etc :D

Captain Bohemian

3:59 AM

have never paid attention to code

I only try to study the principles underlying quantum calculation.

Cows

4:23 AM

omg just realized why counting states is important , while talking about potatoes

Physics Meta

4:48 AM

-1

Q: What is the best/easiest way to write math symbols on this site?

For those with an Android phone, what is your favorite app, or whatever, for displaying math symbols on this site?

discussion

vzn

@Semiclassical gaussian wavepackets always sounded a lot to me like solitons but never saw a ref that stated that...

Semiclassical

they're not quite the same, since gaussian wavepackets by their nature expand over time

a soliton, by nature, has to be able to keep its shape over time; a gaussian wavepacket won't do that.

the basic issue (from what I remember) is that the wavepacket is expanding in free space subject to a linear ordinary differential equation, and because of that its expansion is basically inevitable

where you do see solitons is in nonlinear differential equations, e.g. the KdV equation

vzn

@Semiclassical yes, understood (was just thinking along those lines), but figure the math is very similar and there seems to be a continuum such that a gaussian wavepacket that spreads very slowly is like a soliton as a limiting case.

Semiclassical

well, they are wave equations. so some similarity is definitely there

but for that key component of stability in the absence of an external force, you really do need a nonlinear wave equation

vzn

5:07 AM

btw am very excited about the vinante ultracold cantilevers experiment/ finding. think it is right along the lines of an idea you hypothesized/ voiced in here once, that there could be a subquantum regime that is just "very subtle to measure" wrt current technology. some )( luck for me, reddit didnt hate it. :) o_O journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.110401 reddit.com/r/quantum/comments/8zpzqa/… ps thx much @Secret

@Cows lol oracles are one of the weirder aspects of the theory & there are some really funky yet deep ideas surrounding them. some of the most celebrated results are in the area eg Todas thm. see also en.wikipedia.org/wiki/Magical_thinking

Eulb

5:58 AM

howdy

@BalarkaSen o/

Balarka Sen

Hey!!

How's it going

Eulb

struggle lol

but magically still in the game

Sir Cumference

Howdy :)

Balarka Sen

Indeed?

Eulb

did you listen to the other album i sent you? it's some post-metal-rock-otherstuff thing

indeed?

Sir Cumference

6:01 AM

@Eulb You're Canadian where did you pick that up from

Eulb

you i think

Sir Cumference

Oh. Cool

Eulb

i say it to friends irl now too

look what you've done to me

Sir Cumference

I'm a bad influence. :(

Eulb

sure are partner ;)

lmao

Sir Cumference

6:02 AM

Well then again I only started saying it because I saw someone online say it

Balarka Sen

@Eulb I was travelling all day yesterday (moving to college, about a thousand miles from home), didn't get around to it. Let me grab my headphones.

Eulb

oh you're moving in already?

Balarka Sen

Yeah.

Eulb

that's awesome dude take your time

Sir Cumference

Oh, is your college in India?

Eulb

6:03 AM

being alone you'll find that albums become even closer friends LOL

Balarka Sen

Thanks. Haha yeah I'm planning to live on math and music

@SirCumference Yep.

Sir Cumference

Neat, physics or math major I presume?

Balarka Sen

Math major :)

Eulb

it's balarka, take a guess @SirCumference

Sir Cumference

@Eulb Well he does hang out with us. Then again most people here probably aren't doing physics as a career...

Balarka Sen

6:05 AM

I have at least one physics course each semester starting from next semester.

Eulb

when does your semester start?

Balarka Sen

In a week I think

Tomorrow I verify documents for admission

Eulb

woah time flies

Sir Cumference

I have to wonder what your first math course is gonna be. It seems like you know all the basics

Eulb

your TAs are gonna be in for a wild ride lol

Balarka Sen

6:07 AM

Eh, revision is good. I have been pacing through mathematics so fast that it's becoming a bit droll

I want to prove things rather than being an encyclopedia

So I'll take my undergrad seriously

Sir Cumference

Good idea. Not to sound scary, but missing out on info in college is a lot less forgiving compared to high school :/

Eulb

i don't think you know how much balarka knows lol

Balarka Sen

First semester is Calculus I, Algebra I, Probability I and Computer Science I (ugh), apparently.

Eulb

i'm afraid to ask him for help even sometimes

Balarka Sen

Fortunately there's exactly two CS courses; first semester this year and second semester next year

Sir Cumference

6:10 AM

@Eulb Well yeah :P But it's just a fair warning. My shakiness with first year classical mechanics haunted me for a few years

Balarka Sen

@SirCumference It's true!

@Eulb Don't be

Sir Cumference

If I had missed a single concept in Calc III, I'd probably be more screwed

Balarka Sen

If you ask questions enough you'll see I give dumb answers a lot

Calc III being multivariable?

Sir Cumference

@BalarkaSen Yeah

Balarka Sen

That shit's useful

I am glad I learnt it from Ted

Sir Cumference

6:11 AM

If there's one subject I'm glad to have an intuition for, it's that

Eulb

@SirCumference who would win. a physics and astronomy sophomore at a top university with years of learning experience or a lil wedge boi with a block

Sir Cumference

@Eulb J-junior...

Balarka Sen

@Sir Yeah I imagine Physics students know more multivariable analysis than Math students

Eulb

not yet 3rd year hasn't started yet

Sir Cumference

@Eulb I still have nightmares about rolling without slipping

3

Eulb

6:12 AM

@BalarkaSen not analysis but computational vector calc yes.

Balarka Sen

I think even that is already advantageous

Sir Cumference

@Eulb Eh, an intuition is extremely helpful in making sense of the physics.

bolbteppa

Everybody has nightmares about rolling without slipping :(

Sir Cumference

Otherwise physics becomes memorization, and my memory is pretty terrible

bolbteppa

Just try to do a moment of inertia calculation without a convenient choice of coordinates :(

Eulb

6:14 AM

@BalarkaSen i was kidding, i've mainly been talking to my friends when i go to the library to study so i don't have many questions anymore.

bolbteppa

Impossibleee

Sir Cumference

@bolbteppa The bottom of a cart wheel is being pulled by a string. Which direction does the wheel go?

Eulb

before i was all alone lol so i bugged you too much

but i am still kind of behind atm

Balarka Sen

@Eulb Ahhh nice

bolbteppa

Don't make me cry

Eulb

6:15 AM

@BalarkaSen are you in the same school as @Blue?

Balarka Sen

Nah

Sir Cumference

@bolbteppa I figure as my math intuition gets better, my physics intuition gets better though. I never had a real appreciation for energy, until I learned about dynamical systems and had some kind of epiphany about potential energy...

Balarka Sen

But there is a connection as Blue is in a school that's in the state where my home is

Eulb

say what

@BalarkaSen oh right

that's why i felt the need to ask

where are you then?

1000 km is pretty far

miles*

Balarka Sen

ISI Bangalore

bolbteppa

6:18 AM

My big problem is being able to do all of those baby physics problems without thinking

I have some mental image of Nirvana baby physics problems

Eulb

@BalarkaSen if you're so far away from your family why didn't you just study abroad?

Balarka Sen

Figured this is the optimizing solution for the university problem maximizing for a better mathematical syllabus and minimizing for hassle proportional to distance

Eulb

fair

Sir Cumference

6:37 AM

@Eulb Forgot, are you a (rising) junior?

Eulb

yeah

but i'll probably finish in 2021 not 2020 because i screwed up a bit by changing around my program a few times.

Sir Cumference

Might be in the same boat, depending on whether my schedule matches reality

Eulb

i deleted reality from my friend's list but i added him back again recently

Sir Cumference

:(

Balarka Sen

6:53 AM

Relatable

Slereah

7:15 AM

mornin

bolbteppa

Sleep is for those who understand qft :(

Replace the position operator $\hat{q}_i(t)$ by $\hat{q}_{\mathbf{x}}(t) = \hat{\varphi}(\mathbf{x},t) = \hat{\varphi}(x)$ (and PB's by commutators) to get a quantum field $\hat{\varphi}(x)$. Take $n$ quantum fields $\hat{\varphi}_r(x)$, $r = 1,\dots,n$. Take the matrix elements of $\hat{\varphi}_r(x)$ in a frame $x$ in going from $|\Psi_p\rangle$ to $|\Psi_q \rangle$ to be $\langle \Psi_p|\hat{\varphi}_r(x)|\Psi_q\rangle$. The matrix elements of the same operator in another frame $x' = ax + b$ with $|\Psi_p \rangle = U(a,b)|\Psi_a \rangle$, $U$ a unitary rep of Poincare, are $\langle \Psi_…

Balarka Sen

sleep is for nerds and those who understand qft are also nerds

3

bolbteppa

varphi is the best

Slereah

@bolbteppa why, do you want to use

The HILBERT BUNDLE?

bolbteppa

It's starting to make sense to use the HILBERT BUNDLE :(

Slereah

7:24 AM

Only one crazy man uses the Hilbert bundle

arxiv.org/pdf/quant-ph/0004041.pdf

Balarka Sen

What is that cover art

bolbteppa

There's a big complication in what I posted, it's like you're doing a unitary transformation of an operator not on a vector space but on a fiber bundle, $$\langle \Phi_b'|\hat{\varphi}_r(x')|\Phi_q'\rangle = \langle \Phi_b| U^{-1}(a,b) \hat{\varphi}_r(x') U(a,b)|\Phi_q \rangle = S_{rs}(a) \langle \Phi_p | \hat{\varphi}_s(x) | \Phi_q \rangle$$
gives
$$U^{-1}(a,b) \hat{\varphi}_r(x') U(a,b) = S_{rs}(a) \hat{\varphi}_s(x) $$
which is more than just a unitary change of basis of a linear operator into another linear operator, $U^{-1} T U = T'$.

Since $\hat{q}_{\mathbf{x}}(t) = \hat{\varphi}(\mathbf{x},t)$ is just a position operator at each point of space-time, it seems like a quantum field is a linear representation of the Poincare group, but it's more complicated, thinking bundles fix this

Those notes are for QM only, seems to be a gigantic difference in QM vs qft bundles

But they look cool

Slereah

it is interesting

bolbteppa

7:41 AM

So it makes sense to me that you have spacetime $M$ as your base space, and you have a Hilbert space at each point $x$ of $M$, but I think the Hilbert spaces are representations of the Poincare group at each point, since $|\Phi_a'\rangle = U(a,b) |\Phi_a \rangle$ has to hold. It vaguely makes sense you want a collection of $n$ operators $\hat{\varphi}_r$ defined at each point of $M$, $\hat{\varphi}_r(x)$,

but I think the operators have to be a linear representation of the Lorentz (not Poincare) group, but then there is this insane matrix element law of transformation
$$\langle \Psi_p'|\hat{\varphi}_r(x')|\Psi_q'\rangle = S_{rs}(a) \langle \Psi_p|\hat{\varphi}_s(x)|\Psi_q\rangle$$
which I can't put language to, e.g. the 'structure function law of transformation in a bundle' or something

Slereah

I'm not sure it's a great idea

you may recall that quantum fields are operator-valued distributions

You may not be able to define an operator as a point

In fact what's his name claims that any relativistic theory doing such a thing would necessarily be the trivial theory

zero dimensional Hilbert space

2

Q: Theorem of QFT as an operator-valued function theory

From Axioms of relativistic quantum field theory, I find the following theorem : We conclude this section with the following result of Wightman which demonstrates that in QFT it is necessary to consider operator-valued distributions instead of operator-valued mappings : Proposition 8.1...

quantum-field-theory mathematical-physics

^cf here

Plz remember that $\hat{\varphi}(x)$ is a physicist lie

$\hat{\varphi}[f]$ only

bolbteppa

Gawd

Whatever modification you do to the physicist lie to get $\hat{\varphi}(f)$ from $\hat{\varphi}(x)$ you would probably just do to the fiber bundle understanding of operators on a manifold and the transformation between matrix elements

I am 100% sure you can describe the lie in terms of bundles anyway, but fleshing it out is crazy, let alone that transformation law

Slereah

7:59 AM

Well the set of test functions does form a vector space

you could use this as a base space, I guess

but usually people use QFT as a sheaf

not a bundle

bolbteppa

Ignoring the point $x$ and pretending we just have $n$ operators $\hat{\varphi}_r$ in a single vector space, it seems to be saying $U^{-1} \hat{\varphi}_r U = S_{rs} \hat{\varphi}_s$

Balarka Sen

infinity stack you mean

bolbteppa

8:23 AM

I don't think the qft lie is in sending $n$ Heisenberg equations for $\hat{q}_i(t)$, $i = 1,\dots,n$ to $\infty$ to send $\hat{q}_i(t) \to \hat{q}_{\mathbf{x}}(t) = \hat{\varphi}(\mathbf{x},t)$ is it

arxiv.org/pdf/1010.3892.pdf

Page 7 mixes bundles with distributions, looks crazy

It even references the notes you gave in that section, or same author anyway

Slereah

it is the same author yes

As I said it's not really a popular topic

bolbteppa

8:39 AM

I don't see how else you're going to explain
$$\langle \Psi_p'|\hat{\varphi}_r(x')|\Psi_q'\rangle = S_{rs}(a) \langle \Psi_p|\hat{\varphi}_s(x)|\Psi_q\rangle, x' = ax + b, a \in O(1,3)$$
properly

Slereah

why are you applying an operator rotating the different scalar fields for a physical rotation

bolbteppa

Not sure what you mean, but that equation apparently has to hold in general for any qft, seems like this might be how they realized group reps show up in qft btw

Slereah

What is $_r$

bolbteppa

If you take $n$ independent quantum fields $\hat{\varphi}_r(x)$ with $r = 1,\dots,n$, each of which satisfy the Heisenberg equations $\partial_t \hat{\varphi}_r(x) = i [\hat{H},\hat{\varphi}_r(x)]$, $\partial_t \hat{\pi}_r(x) = i [\hat{H},\hat{\pi}_r(x)]$, and commutation relations between $\hat{\pi}_r$ and $\hat{\varphi}_s$, the matrix elements of these $\hat{\varphi}_r(x)$ operators have to satisfy
$$\langle \Psi_p'|\hat{\varphi}_r(x')|\Psi_q'\rangle = S_{rs}(a) \langle \Psi_p|\hat{\varphi}_s(x)|\Psi_q\rangle, x' = ax + b, a \in O(1,3)$$

Slereah

Why do different quantum fields mix together when the space is physically rotated, though?

bolbteppa

8:54 AM

B&J motivate it by saying in going from a classical field theory to a quantum field theory, the Lagrangian being a relativistic scalar is no longer enough to guarantee relativistic invariance in a theory, you need to put conditions on the operators they became, which you do by examining their matrix elements and ensuring a transformation as above holds between different frames

Slereah

It's a physical rotation, not a gauge rotation

bolbteppa

This transformation law implies
$$\langle \Psi_p'|\hat{\varphi}_r(x')|\Psi_q'\rangle = S_{rs}(a) \langle \Psi_p|\hat{\varphi}_s(x)|\Psi_q\rangle = S_{rs}(a) \langle \Psi_p'|U(a,b)\hat{\varphi}_s(x)U^{-1}(a,b)|\Psi_q'\rangle $$
which gives
$$U(a,b)\hat{\varphi}_s(x)U^{-1}(a,b) = S_{rs}^{-1}(a) \hat{\varphi}_r(ax+b) $$
which is the usual transformation law for quantum fields

(For some reason they reverse the indices on $S^{-1}$ to $S_{sr}^{-1}$)

Slereah

What is B&J

bolbteppa

Bjorken and Drell

amazon.com/Relativistic-Quantum-Fields-James-Bjorken/dp/…

I haven't seen it explained this way in any other book, they usually just tell you to do it :p

Slereah

Drell doesn't start with a J >:|

bolbteppa

9:06 AM

Oh man

James Bjorken

James Daniel "BJ" Bjorken (born 1934) is an American theoretical physicist. He was a Putnam Fellow in 1954, received a BS in physics from MIT in 1956, and obtained his PhD from Stanford University in 1959. He was a visiting scholar at the Institute for Advanced Study in the fall of 1962. Bjorken is Emeritus Professor at the Stanford Linear Accelerator Center, and was a member of the Theory Department of the Fermi National Accelerator Laboratory (1979–1989). He was awarded the Dirac Medal of the ICTP in 2004; and, in 2015, the Wolf Prize in Physics and the EPS High energy and particle physics...

BJ :(

Slereah

BJ is for William Joseph Blascowicz

bolbteppa

The backwards indices thing is concerning too

Slereah

What page is it?

bolbteppa

20 and 21

They want to know whether the Noether procedure on a classical Lagrangian will transfer into the quantum domain which is a cool way to motivate this

Balarka Sen

@ACuriousMind is our BJ <3

(that was unnecessarily sexual)

Slereah

9:14 AM

Errr, the chapter "Lorentz covariance of the Dirac equation"?

bolbteppa

Volume 2 sorry!

Relativistic Quantum Fields

That's volume 1

Which is like the Schrodinger picture to QM going straight off the Dirac equation

The chapter 1 in volume 2 starts by simply taking the Poisson bracket equations for $n$ particles, promoting them to quantum Heisenberg equations, taking the Heisenberg formalism as the pov, and then sending $n$ to $\infty$ to define a quantum field, and here they are asking how the Noether procedure going into the quantum realm, with this requirement on quantum field matrix elements being forced on them

Slereah

Guess I should read the whole chapter

Or do my job at work

Tough choice!

bolbteppa

idk but no other book made this point so clear before, shocking

Thinking of Srednicki etc they just tell you to follow this rule for fun

Slereah

It's the same with everything, rly

there's always one book

(or zero)

that makes the correct demonstration

bolbteppa

Yeah

Slereah

9:19 AM

and all the good ones are never in the same book

So you have to have 30 books on hand

bolbteppa

I haven't found a subject yet where that doesn't happen

Slereah

Oh wait, is $\phi_r$ supposed to be the components of a field with some rep of the Lorentz group?

ie a scalar with $r = 0$, a vector with $r = 0,1,2,3$ etc?

bolbteppa

No, on page 16 he says he just wants to generalize to several independent fields $\hat{\varphi}_r$

Slereah

Whaaat

I mean, I guess it could work if the rep in that case is like

$S_{rs} = \text{diag}(1,1,1...)$

For $n$ scalar fields???

I dunno

Let's read on

bolbteppa

Sure, that's what a complex scalar field is

I think

Slereah

9:23 AM

Phew

I was scared the whole world was insane for a bit

bolbteppa

On page 2 they say to think of Maxwell as a guide, so going to say $4$ fields is natural without knowing you'd end up with spinors etc

Slereah

I guess it's one of those...

Reducible representation of the Lorentz group

he's not demanding irreps

bolbteppa

That's what's cool, he's not even bringing up reps necessarily, the generality falls out of it

Until he writes the $S_{rs}(a)$ thing, no reps specified explicitly, even if they are there, which I think they are

Slereah

"Here $S_{rs}$ is a transformation matrix for the fields $\phi_r$ under the infinitesimal Lorentz transformations and differs from the unit matrix if the field is not a scalar field"

That makes more sense

bolbteppa

Yeah, but you can skip the classical section to page 20: "Going over now to quantum field theory"

Slereah

9:28 AM

"A general approach from an axiomatic viewpoint has been given by Lehmann, Symanzik and Zimmermann"

Poor saps

nobody remembers that one!

Oh wait

It's the LSZ guys

but then it's only for scattering!

bolbteppa

You would be forgiven for thinking qft was only scattering the way books explain it

In going from (11.65) to (11.67) there is a flip of the rs to sr which is upsetting, but apart from that it seems legit

Spooky, googled 11.67 B&D and your author referenced it in one of his papers arxiv.org/pdf/1203.6254.pdf

Or wait, $S_{rs}(a)^{-1} S_{sr}(a) = I$ so obviously the indices reverse :(

You know you are in the loop when you are googling specific equations from one book you are trying to generalize and find them specifically referenced in another paper the guy who generalizes them in another paper beyond comprehension

The paragraph around 11.63 motivating it is a tiny bit waffly, why are we enforcing Lorentz "covariance"

and why isn't 11.65 $S(a,b)$ instead of $S(a)$

Anonymous

11:16 AM

@BalarkaSen How's the place? Reached already?

Balarka Sen

I'm in a hotel now. I verify for documents for admission tomorrow.

Abhinav

@BalarkaSen@Blue@JohnRennie physics.stackexchange.com/questions/418404/…

Based on the answer I get that the there will be an opposite force acting on the spring which will be equal to the Force F on equilibrium but I don't understand the relation when there is no equilibrium

And from where this force comes

Anonymous

@BalarkaSen Nice! Roam around a bit today ;)

Anonymous

@BalarkaSen Do you get a room for yourself at ISI or is it a shared one?

Balarka Sen

Instead I shall think about symmetry groups of tessellations

It's shared in first semester, and afterwards a single room I think

Who cares

It's all bullshit anyway. Not even sure why I am going to a university.

bolbteppa

11:26 AM

Do you want to be a mathematician

Balarka Sen

No

bolbteppa

Why are you investing so much time in math then

Anonymous

Ask @KaumudiH Her noctural roommate drove her crazy in the first year :P But well, shouldn't be a problem for you since you're nocturnal anyway :D

Balarka Sen

Because you stink @bolbteppa

Anonymous

@bolbteppa Pretty sure Balarka secretly wants to be a physicist :P

bolbteppa

11:28 AM

That's too strong I guess haha

Balarka Sen

@bolbteppa You're addicted to stink?

Unfortunate indeed

bolbteppa

Think of it as buying a load of time to study what you like, and exams as the necessary evil of ensuring you know at least the minimum they threw at you, even if it's done terribly

Balarka Sen

I'd engage in this more seriously (I appreciate that suggested point of view, actually, @bolbteppa - thank you) but I am in a very silly mood right now

glS

2:31 PM

0

Q: Does Bell's theorem imply a causal connection between the measurement outcomes?

It is a standard result that quantum mechanics does not allow for superluminal communication, which would seem to directly imply that the answer to this question is no. After all, Bell's result tells us about correlations, and we know that correlation does not imply causation. However, if we res...

quantum-mechanics quantum-information causality bells-inequality

@EmilioPisanty about the above question: after some thought maybe I found a better way to state (at least part of) the question, thought it strips away most of the physics. Suppose you have two random variables, $A$ and $B$, which are correlated, that is $p(a,b)\neq p(a)p(b)$. Suppose also that this correlation cannot be explained by a third variable, that is you cannot write $p(a,b)=\sum_c p(a,b|c)p(c)$ for any third random variable $C$ (...)

(...) does this imply that $A$ and $B$ must be causally connected? This, I believe, mathematically should mean that the marginals of $p(a,b)$ can be expressed as $p(a)=\sum_b p(a|b) p(b)$, where $p(a)=\sum_b p(a,b)$ and $p(b)=\sum_a p(a,b)$

(note that in the above I'm using a somewhat sloppy notation: the $p$ in $p(a,b)$ is clearly not the same function as $p(a)$ etc., but instead stand for probability of the given event)

mh, actually once I write this way, the answer seems trivial. The marginals can always be written that way, by the same definition of $p(a|b)$: $p(a|b)\equiv p(a,b)/p(b)$

Emilio Pisanty

2:54 PM

@glS why are you talking about causality without including the 'control' variables (normally $x$ and $y$)?

vzn

@glS there are some deep subtleties in bells thm & invite you to take the red pill & see how far down the rabbit hole goes :) physics.stackexchange.com/questions/402680/…

Emilio Pisanty

3:23 PM

SUB[59]: The submission numbers for my accepted OEIS submissions in chronological order

heh

gotta love sub[46]

John Rennie

Aren't the '0' to '9' characters all the same width in Helvetica?

Emilio Pisanty

@JohnRennie you use monospace to break any ties

then courier

glS

@EmilioPisanty well you can think of that reasoning as working in the case in which a single measurement setting has been chosen. But I think I understand now that my use of the term "causal relation" is meaningless, in that one can always write the marginal of a distribution as "being caused" by the other marginal

@EmilioPisanty which means that as long as $x, y$ are fixed, you can make the argument that $p(a,b)$ can always be thought of as "caused" by a hidden variable $\lambda$, but while true this is a useless observation. Rather, Bell's theorem is about the limitations imposed by the assumption of $A$ being only correlated with $X$ and $\Lambda$ and $B$ with $Y$ and $\Lambda$. I'll try to rewrite the question to make it more meaningful and then probably self-answer

Emilio Pisanty

3:38 PM

a'ight

glS

sorry, that was a bit of rubber ducking on my part =)

Emilio Pisanty

¯\ _(ツ)_/¯

glS

@vzn I don't really understand that question

Sir Cumference

Morning

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