LaTeX is so unbelievably stupid at times, and it's your fault if you do the 'recommended' thing and then end up having to change over 100 things because one person decided something ridiculous was 'best practice'

clauswilke.com/blog/2015/10/02/bibtex

My team at Google has an opening. Anyone interested in designing/testing superconducting qubits and/or investigating the microscopic origin of flux noise, charge noise, or quasiparticles, inquire within.

@Obliv I would assume that whoever "assured" that it's a fair coin, is a liar.

and then I'd call heads because if the coin really is fair, then the call doesn't matter, and if the person who made the assurance of fairness is a liar, then heads is a better call.

@ManasDogra see physics.stackexchange.com/q/142159/50583

@DIRAC1930 what do you mean, radiative corrections are just specific terms in the propagator (which is a 1->1 scattering amplitude), and the Bethe-Salpeter equation relates this propagator to the bound states

the focus on scattering is not some narrowing of the scope of QFT, instead we have come to realize more and more that all the physics is indeed in the n-point functions (cf. various statements like Wightman reconstruction theorems - in almost every formal setting when the n-point functions of two QFTs agree, then the two theories are the same)

@ACuriousMind Yes but on top of the full QED Lagrangian (with interaction operator), people used to add an external classical four-potential as an extra perturbation through $A_\mu^{ex.}$ through $\overline{\psi }\gamma^\mu A_\mu^{ex.} \psi$

@DIRAC1930 that's just the background field method in modern parlance

So this background field could be for example the Coloumb potential?

what is Q(x,p) is it a special hermitian operator

i mean why its writen as a funtion of x and p

@DIRAC1930 it's just any classical background, so sure, you can choose the Coulomb potential

@PrateekMourya I assume they just meant any self-adjoint operator

I feel like processes such as the Lamb shift etc. should be at least schematically laid out in modern QFT courses

i.e. hermitian?

I've seen this a few times; the people writing this just haven't thought very carefully about what the $Q(x,p)$ is supposed to mean, they're just thinking that usually the operators we're looking at are stuff like $p^2 + V(x)$, i.e. given as simple "functions" in terms of $x$ and $p$

so...what does it mean exactly

nothing, there is no special significance here, if it bothers you just assume they had written $Q$ instead of $Q(x,p)$ :P

What are all the things one need to know before learning theory of elasticity.

alright!

@DIRAC1930 I feel like I got such a schematic idea from my QFT course; are you sure you're not just complaining that your QFT course didn't include it?

For instance, the pretty "modern" book by Weinberg discusses bound states in external fields in chapter 14

Perhaps

I don't have a copy around to check but I think Itzykon & Zuber should also at least discuss Bethe-Salpeter, if not specifically the Lamb shift

L&L 4 has around 100 pages just on radiative corrections

as in radiative corrections to processes with a background field

so?

Landau and Lifshitz often put their focus on different aspects than the mainstream, that's part of why these books generally produce rather polarizing opinions :P

Hmm I'm not too sure, most of the books from around the time roughly after Dyson focus the majority of the time on more physical aspects

you also have to consider that before the rise of collider experiments stuff like the Lamb shift was the kind of thing QED was invented to explain

so of course you'll see less mention of the n-point functions in terms of scattering amplitudes in the early material

Going to try and do my first proper QFT calculation calculating the lifetime of Positronium

When looking at the theoretical derivation of Hawking radiation, if we declare that the field is 0 at the singularity of black hole, the solution to the Klein Gordon equation in schwartzchild spacetime becomes e^{-ikv} - e^{-ik\beta(U-2R)}. Would it be correct to say that e^{-ikv} is the field that enters the black hole, and e^{-ik\beta(U-2R)} is the field that leaves the black hole?

I’ve been searching for a while for a source that clarifies this but I can’t find any

So it seems like the SYK model isn't a completely quantum theory since you have this disorder averaging. Is there an active field of research trying to find more low dimensional models or is it something that is too difficult?

The Mario movie was actually pretty good

What is up with game adaptations not being terrible nowadays

They're learning from their mistakes, I guess.

As all good problem solvers should :P

(?)

Shouldn't it be $\pi$ for spin 2? (cf last line in the pic)

Good nights physicists.

physics.stackexchange.com/questions/758702/… I am wondering about the comments on this post saying that symmetries have nothing to do with unitarity. But doesn’t Wigner’s theorem say that we can represent any symmetry (as defined by say Weinberg QFT vol. 1) as a unitary operator

@SillyGoose "Unitarity" is a weird word because in physics it often specifically refers to unitarity of time evolution/the S-matrix and not "the property of being unitary" for any operator

i.e. given standard physics parlance that question is asking whether spacetime symmetries have something to do with unitarity of time-evolution which is a bit strange

of course, if it is not using "unitarity" in this specific sense, then of course Wigner's theorem is the answer

(personally the physics use of "unitarity" annoys me deeply and I wish people were just more explicit about what they mean: You sometimes get people saying "oh, sure, because of unitarity!" and they mean some very specific property of the S-matrix derived from unitarity of time-evolution)

ah i see

did not know about that usage of unitarity :0

hmm. i wonder whta LM stands for

local-measurement-equivalent?

I've been meaning to ask this... I don't understand what is meant by centralizer here. In Dummit and Foote, a centralizer of a group $G$ wrt to a subset $S \subset G$, denoted $C_G(S)$ requires $S$ to be a subset of $G$.

In this paper $S$ is certainly not a subset of $G$. Or, I feel like I must be misunderstanding what $S$ is.

yes actually @Semiclassical :)

neat

@bolbteppa the Latex aspect which drives me nuts: the fact that \det is built in but \tr is not

So to my understanding $G$ is some $n$ tensor product of $SU(2)$, okay that is nice. Then, $S$ is the set of all reduced density operators of a given density operator $\rho$. Any $\sigma \in S$ is certainly not generically unitary. Hence $S \not\subset G$ generically. So, I don't see how $C_G(S)$ makes sense as a concept.

@SillyGoose I think they mean to take the centralizer within the group of all operators on your Hilbert space, then intersect that with the unitaries

oh

i would not have thought of that heh

Hm is it wrong to say that the meat of entanglement is correlation. I wrote an answer on a question asking concerning entanglement, and I made such a claim that entanglement is more about correlation between measurements on two (or more) systems rather than "what this composite state represents: is it one entity, two, etc". Since someone downvoted it, I am doubting my claim now :P

@SillyGoose it's not merely correlation, it's non-classical correlation

systems can correlate classically, but there is no classical entanglement

hm what is the difference between non- and classical correlation? I've read through some of the existing stack questions asking this question to no avail :P.

@SillyGoose one simple answer is correlations that violate Bell's inequality

what does violating bell's inequality translate into physically?

my too-short-to-be-useful answer is that Bell's inequality comes about from that fact that $(X+Y+Z)^2 \geq 1$ if $X,Y,Z$ must have definite $\pm 1$ values

@SillyGoose uhhhhhhhh

what do you mean by "physically"

Well, to my understanding a classical correlation is something like "put a red ball and blue ball into a drawer. close the drawer. pick a ball at random and check its color. if you pick another ball at random it must be the ball you didn't pick." There is a physical visual to go along with the concept "classical correlation"

What is the analogous physical visual for non-classical correlation

Bell's inequality is a bound on the ways in which variables in a "classical" (local and realist) theory are correlated. In particular, in such a classical theory, when you have a state of a system composed of two subsystems, you can always assign a definite state to each of the subsystems

@SillyGoose again, my pretty silly answer (but not entirely silly!) would be: if you walk in one direction for 1 meter, turn 120 degrees CW and walk another meter, then turn 120 degrees again and walk another meter, you're back where you started

entangled states are states that generically violate Bell's inequality, and they are also (by definition of entanglement) the states for which such an assignment of definite states of the subsystems are impossible

the longer spiel i'd give is to analogize a classical correlation to a so-called raffle model

@SillyGoose I think the reason people fight so much about interpretations of quantum mechanics is precisely because there is no such visual for truly quantum correlation

what is the raffle-model :0

it's boutique, first off. don't expect to see it elsewhere

wait so it sounds like non-classical correlation is just correlation that comes about from a state being entangled (or can be described in such a way)?

because a state being entangled is equivlaent to not being able to define definite states to your subsystems

it's not exactly equivalent (you can have entangled states that don't violate the inequality) but it is the case that all states that violate the inequality are entangled in some sense

oh

here's the most basic elements of a Bell experiment. you have Alice and Bob, each of whom have access to a subsystem, and each can make one measurement of type a,b,c (one measurement per state prepared)

bleh i guess i got to learn about bell's inequality

oh perfectt here we go heh

each outcome is either +1 or -1. if Alice and Bob pick the same measurement type in a given round, then they'll definitely get opposite results. But if they pick different types, they may or may not get the same result.

oh. i should've labelled the measurement types as X,Y,Z.

oh well

wait so is the extra condition about having to measure the same measurement type all that distinguishes classical and non-classical correlation?

not at all

but i do need it to make the scenario simpler

if you want a more physical model for this, take the initial state of the system to be a singlet state

Alice and Bob each get one of the two particles, and each can measure the spin of their particle along three different (not necessarily orthogonal!) axes

if they choose to measure along the same spin directions, then they'll definitely get opposite results (since the singlet state has zero total spin)

but if Alice measures along the +z direction and Bob measures along the +x direction, then there's no correlation: all four possible pairs of outcomes are equally likely

ok?

yes :)

okay. since they definitely get opposite results if they measure the same thing, we can proceed from Alice's perspective as follows. She measures one of her spin components, and guesses that Bob will obtain the opposite of that component.

if Bob chooses to measure along the same component as her, she'll definitely be right. if not, there's some probability she'll be wrong

wait so is alice just guessing "spin up" or "spin down" without respect to the axis of Bob's measurement

since it seems like if Bob measures along an arbitrary axis there's llke a 0% chance that Alice guesses correctly if Alice is also specifying the axis

she doesn't know what axis Bob is using, no, but we're assuming there's only three possible choices of axis

ohoh sorry i missed that

yeah. this is the bit about there being three measurement settings (I said types earlier, but settings is better)

now, suppose that both Alice and Bob's measurement outcomes already existed before the measurement was actually made.

let me grab a picture i'll need

So is this supposition imagining our quantum situation as classical? As in the values of these measurements are at most most unknown only because of a lack of information and not because of an inherent statistical nature of the (quantum) theory?

right

that's a possible 'ticket' in the raffle

if Alice gets the LHS and measures along the b-direction, she gets +1. if Bob gets the other side, he'll also get -1 if he measures in the b-direction

of course, that ticket is boring

but there's not that many tickets to actually consider: on the face of it there's 8, one for each possible choice of signs on the LHS. but i'm assuming Alice and Bob will get each side equally often, so four is sufficient.

the claim is that, if you're assuming definite values, then what you're doing is equivalent to drawing a ticket, distributing it between Alice/Bob, and having them measure the result

right

at this point, you can start making statistics like: what is the expected value of Alice's outcome for the a-direction? it's zero, because she's just as likely to get the LHS of any ticket as the RHS).

more interesting is to consider correlators, i.e., products of her outcome and Bob's. but if we say we're using her 'guessed' value, that amounts to products between Bob's outcomes

let me get a better ticket for this

do you have a ticket generator sitting somewhere for moments exactly like this XD

better than that: i have a paper i'm screenshotting

which would be theft, except that i'm a coauthor :P

Ohh i see

Haha

i'm not the one who drew these, but i did help design them

anyways. So now suppose that Alice measures along the b-direction while Bob measures along the a-direction. then if we use Alice's guess, we'll end up with (+1)(-1)=-1 regardless of which side of the ticket we used

so we only really need to pay attention to the LHS of the ticket to figure out what product we'll get

Right

a * b gives -1, a * c=+1, b * c=-1 being the only possible outcomes for this ticket

in particular, if Alice and Bob measure the same direction then the product is 1 regardless of what ticket they received

now, they only ever measure two of the three observables. but if we're imagining that they have definite values, then that doesn't really matter: there should be some fact of the matter as far as the third observable. that's the upshot of the raffle model

as such, one can make the following observation: what possible values are there for the sum of the outcomes on a given side of the ticket?

if you think about that for a moment, it has to be a+b+c=+3,+1,-1,-3. those are the only four possibilities if a,b,c have to be +/- 1

in particular, this means that (a+b+c)^2 is always at least 1

if you expand that out, that's a^2+b^2+c^2+2ab+2ac+2bc >=1. since a^2=b^2=c^2=1 for every ticket, we can simplify this to ab+ac+bc >=(1-3)/2=-1

Suppose Alice gets the LHS of the ticket and measures b, while Bob gets the RHS and measures a. Alice gets -1 and thus guesses +1, while Bob gets -1. So the product is (+1)(-1)=-1

if they got the opposite sides of the ticket, then Alice would get +1 and guess -1, while Bob gets +1. so the product is now (-1)(+1)=-1

the signs of their outcomes have flipped, and thus the product is unchanged

and we could read that product directly off the LHS of the ticket: the a and b results have opposite signs, therefore the product is -1

hm wait so how is the product = 1 if A and B measure the same direction. wouldn't it still be (A's measure) * (A's guess) = -1?

suppose they both measure in the a direction. Alice gets LHS -> Alice measures +1 -> Alice guesses -1. Bob gets RHS -> Bob measures -1

so (-1)(-1)=+1

if Alice got the RHS, then Alice would end up guessing +1 and agreeing with Bob's +1

so we are considering the product (A's guess) * (B's measure)?

right

oh wait have we been considering that product all along I see I think I misunderstood that bit

doing this via text is tough

another way to frame the arithmetic is that, at random, you're given a set of three +/-1 values labelled a,b,c

haha yeah i could also try following along via the paper if the paper was on this

and what you get to measure is the product of two of the values

you can measure the same value twice, in which case you get a^2=b^2=c^2=1

but if you don't, then you could get ab=+/-1 depending on which set of values you were given

okay I see so $\sum_i x_i^2$ is invariant w.r.t. choice of ticket, namely it $= 1$

right

by contrast, $(\sum_i x_i)^2$ isn't invariant. but it can't be smaller than 1

if you expand that out, you get $a^2+b^2+c^2+2ab+2bc+2ac=2(ab+bc+ac)+3\geq 1$

okay got it

so $ab+bc+ac\geq -1$

you of course can't measure $ab+bc+ac$ directly

right

but the supposition is that you can measure $ab$ and average to get $\overline{ab}$ and so forth

and thus $\overline{ab}+\overline{bc}+\overline{ac}\geq -1$

there are, naturally, assumptions built into that. we're assuming that which ticket Alice/Bob get doesn't depend on what they plan to measure

is this like a fact that if any such value of a random variable is greater than $r$ the expectation of that random variable must be greater than $r$? I guess it makes sense because by definition of an expectation the expectation is at least greater than the smallest value

yeah

err actually im not really sure about the continuous case but at least it is clear in the discrete case

okay I follow

nah, that's exactly it. if $X\geq 0$ with probability 1 then $\overline{X}\geq 0$

following on my assumptions bit: if the person handing out the tickets knows what Alice and Bob plan to measure after handing off a given ticket, they could pick the ticket in order to make things look very weird

for instance, they could arrange it so that you get $ab=-1$ when you're looking at that pair, and similarly for the other two products

but that would require the equivalent of a,b,c all having opposite signs, which is nonsense: $a=-b=-c\implies b=c$

anyways. my claim at this point is that $\overline{ab}+\overline{bc}+\overline{ac}\geq -1$ is nothing but Bell's inequality, phrased in terms of expectation values

wait so this invisible hand giving out tickets in a weird way is representative of a quantum correlation? in that it is producing results that don't make sense with a classical correlation model?

@SillyGoose it's actually worse than that: the invisible hand can cook up correlations which even QM can't do

oh i see i saw an answer on stack which talked about that you can make correlations that qm cannot do, so this is a generalization of quantum correlations to just non-classical correlations in the literal meaning of the word: every correlation that is not classical

right.

as an indication of that, note that $\overline{(a+b+c)^2}=3+2(\overline{ab}+\overline{bc}+\overline{ac})$

but the square of any quantity should never be negative

so we should still have $\overline{ab}+\overline{bc}+\overline{ac}\geq -3/2$, which is violated if all the averages were -1

QM will still satisfy that inequality

so Bell's inequality and related work is really all about probability theory

that's our line on it, more or less

that said, it's still genuinely weird

because while you can have the equivalent of $\overline{ab}+\overline{bc}+\overline{ac}=-3/2$ in classical probability

the way that would ordinarily arise is that $a,b,c$ can take on values other than $\pm 1$

so it's not just that the correlations are weird, it's that they're weird in the context of "the only possible outcomes are definitely $\pm 1$"

do you know of a (non-textbook sized ;)) resource that runs from classical to quantum to non-classical correlations in the language of probability theory?

not really, no

it's somewhat of a research direction of ours

also so this inequality is for a very particular situation, is there like a generalized inequality for a general situation?

oh interesting

depends. there's a number of ways one can generalize it

i imagine it can get quite nasty quite quickly :0

one way which we haven't attempted to do so is: what if the +/- outcomes are not balanced?

that's a big omission, b/c if you think about it that assumption is awfully strong

it holds only b/c we're assuming that we have a pure singlet state

but in practice that's not going to hold. you're going to have some mixed state instead

so this is the assumption of "perfect correlation" like if I measure + in the z direction, Bob will measure - in the z direction?

right

you can still do interesting computations in that setting. the singlet state still defines what correlations are possible

:0 how cool in what area does your research lie? is it mathematics?

but what it doesn't tell you is what marginals are possible (i.e., what Alice gets on average for +z). that's much harder and i've stayed away from the quantum marginals problem

@SillyGoose i guess this would be "foundations of physics"

i'm sorta in limbo as far as that goes

not a very stable situation, to put it simply

ooh are you a PI?

nah

i have my PhD, but my formal job title since then has been "teaching specialist"

aka "you graduated but you still want to work here, so back to the teaching mines with you"

oh I see it's cool that you still get to research what interests you :D

well, this semester we were trying to run an intro course that outlines our approach to teaching QM

with...mixed results

what is your approach to teaching QM?

I am trying to find unis in the states to apply to with quantum info groups that also do quantum foundations :P

well, in the context of what we were talking about it's trying to use this raffle model and such concepts to introduce non-physics students to QM and Bell's inequality

there's been some substantial challenges with that, some of which we could have anticipated better and other stuff which we probably wouldn't have realized until we actually did it

QM seems like a subject incredibly hard to teach someone with no background in linear or prob :P

for instance, we knew that Dirac notation and Hilbert space in general would be a challenge. but we probably didn't design the schedule well enough to avoid that being an uphill battle

@SillyGoose you aren't wrong

one thing we did insist on is not using any calculus

so no Schrodinger equation period

you can't avoid teaching some amount of linear algebra on that approach tho

Did you do like ladder operators and so on / more algebra based QM then?

nah. more just "everything is finite-dimensional and therefore like doing spin"

ahh i see

hence basically just linear algebra

which doesn't trivialize it ofc

people can learn how to solve boundary value problems in ENM ;)

lol

as a reminder, this was aimed at non-physics students

this is not the kind of course which would replace a traditional intro-QM course for physicists/chemists/engineers

how was the linear algebra taught? was it more concrete like a "vector is a n-tuple..." or was it also introduced in a more abstract way like "a vector is a symbol that lives in a vector space...can be represented in a basis..."

closer to the former

in truth tho i think our current approach to it wasn't the best

we needed to spend more time on the 'classical' side of things

oh interesting

for instance, one way to introduce vectors is: suppose you have two random variables X,Y. X is your independent variable while Y=aX+b is dependent

suppose you take a bunch of samples of X,Y. you can then form all your samples of X to get a 'sample vector' $\vec{x}$, and similarly a sample vector $\vec{y}$

i have a dream of one day making a physics course intended to define things like system and state and so on independent of a physical theory and then from there introduce like quantum theory. so one doesn't draw their intuition from a particular theory of physics but from the framework through which physics operates. but of course this is probably impossible XD.

what's cute is that, if you want to do regression analysis

then geometrically that amounts to projecting $\vec{y}$ onto $\vec{x}$

(not quite that simple, i'm leaving out an annoying part of the story)

hehe

in particular, you can think of random variables as being vectors in a Hilbert space

where variables are orthogonal when they're uncorrelated

and then regression is just "project the response variables onto the subspace generated by the explanatory variables"

the smaller the remaining uncorrelated component, the smaller the MSE

so basically turning regression analysis into a geometric story about Hilbert space

in terms of course design, tho, we probably did that later when we should have done it eaerlier