The h Bar

More Anonymous

04:23

@ACuriousMind this is what I was trying to talk about that day. Hopefully its better articulated:

https://philosophy.stackexchange.com/questions/96175/describing-the-universe-using-symbols

Silly Goose

07:12

1) taylor expanding a function f(x) and 2) approximating its derivative to determine f(x - h) up to first order where h is some constant should be equivalent, or am I going wrong here?

ACuriousMind

09:02

@SillyGoose What do you mean by "approximating its derivative"?

it does sound as if you just mean a Taylor expansion to first order, but given that you have to ask this question you must have some other definition of that operation in mind

Mr. Feynman

@ACuriousMind now, that's a good way to start the day

ACuriousMind

@MoreAnonymous I do not understand at all how your "third" position is supposed to be an alternative to the first two, they do not even seem to answer the same question. You seem to conflate two different things: The idea that symbols can be explained in terms of other symbols, and the idea that a physical theory can have a more reductive/fundamental theory from which it is derived

Symbols are not physical theories, there is no obvious notion of one set of symbols being "more fundamental" than another. You can explain the symbols of one language in terms of other language: This does not mean one language is more fundamental than another

Mr. Feynman

That webcomic also reminds me of the question of the other day (the one where you posted Wigner's essay). I think that the only meaningful question about using math would be "why does our math work?* - that is, do we have the key or one of the keys? I don't think that Physics without Math (as OP suggested) would even be Physics

At the end of the day Math is a language. How do you explain things if you can't communicate?

ACuriousMind

09:24

Yes, that is one of the points Wigner makes: Mathematics is essentially constructed to be able to express a specific kind of complicated argument

The "miracle" is that this seems to be sufficient: Physics needs "only" mathematical arguments, and not like 10 different frameworks for 10 different kinds of arguments

Slereah

Amazing that the field that was build out of trying to describe the real world can be used to do so

Mr. Feynman

09:46

@ACuriousMind Yes, that's the amazing part. Some mathematical concepts were indeed invented to explain specific things but it turns out they work for a big variety of phenomena

@Slereah Well, it is indeed. Who tells you the world must follow any kind of logic in the first place? :P

Slereah

@Mr.Feynman Probably Plato or something

ACuriousMind

mathematics is only the shadows on the cave wall, to glimpse actual reality we must invent hypermath

Mr. Feynman

What are hyperspinors?

By the way, I've started reading some category theory and that really feels like hypermath :P

More Anonymous

10:12

@ACuriousMind strictly speaking the third is not an alternative to the first 2. But it's saying it doesn't matter ...

@ACuriousMind yes they are representation of the theory

I don't remember making the claim on set of symbols are more fundamental than the other

The third is saying more specifically one does not have access to knowing position 1 or 2 ....

Relativisticcucumber

in sakurai, he writes $\vert N \vert ^2 \int dp' e^{\frac{ip'(x'-x'')}{\hbar}} = 2 \pi \hbar \vert N \vert ^2 \delta (x' - x'')$, but i am confused because i thought this integral does not converge. am i missing something?

Mr. Feynman

@Relativisticcucumber That is an integral representation of the Dirac Delta distribution

Relativisticcucumber

:o

Mr. Feynman

It is not convergent in the ordinary sense, in fact $\delta$ is not a function

Relativisticcucumber

very interesting. okay thank you

Mr. Feynman

10:26

If you prefer, you can think of it in terms of Fourier transforms

ACuriousMind

@Relativisticcucumber The Fourier transform of $\delta(x)$ is the constant function, and that's simple to show by definition of $\delta$: $\int \delta(x)\mathrm{e}^{\mathrm{i}px}\mathrm{d}x = \mathrm{e}^0 = 1$. So since the Fourier transform is its own inverse, that integral must be equal to $\delta(x'-x'')$, right? ;)

At least, this is all the explanation you'll get for what's happening here at usual physics levels of rigor

Mr. Feynman

::tempered distributions knock the door

Relativisticcucumber

bahhh thanks

ACuriousMind

I think the better way to make this make sense without going into the full distribution theory is to think about nascent $\delta$s

Mr. Feynman

Do you mean Dirac's original definition?

Relativisticcucumber

10:32

wait is @Mr.Feynman the old feynman?

Mr. Feynman

Oh, just functions with bumps

@Relativisticcucumber I am :P

Relativisticcucumber

:oo

i have been gone too long

ACuriousMind

@Mr.Feynman I mean pick (e.g. the Gaußians $f_\epsilon$ whose width goes to 0 as $\epsilon\to 0$) functions $f_\epsilon$ that "converge to $\delta$ in the sense that $\lim_{\epsilon\to 0}\int f_\epsilon(x)g(x)\mathrm{d}x = g(0)$ and show that the Fourier transforms $F[f_\epsilon]$ converge to 1

Mr. Feynman

Oh, that's a useful representation too

Actually I'm a bit triggered we call these "representations" :P

ACuriousMind

the $f_\epsilon$ are often called "nascent deltas" and for Gaußians this is very simple because Fourier transform of a Gaußian with width $\epsilon$ is just one with width $\epsilon^{-1}$

Relativisticcucumber

10:34

what is the og definition by dirac? @Mr.Feynman

Mr. Feynman

@Relativisticcucumber In his QM book he defined $\begin{cases} \delta(x)=0\quad\text{if}\quad x\neq0\\ \int\limits_{-\infty}^{+\infty}\delta(x)dx=1\end{cases}$

ACuriousMind

so as $\epsilon\to 0$ the nascent deltas shrink to the infinitely thin spike we imagine the $\delta$ as, and their Fourier transforms become wider and wider until they become a flat line

@Mr.Feynman I really dislike that definition because it makes no sense :P

Mr. Feynman

Who says I like it? :P

Slereah

Dirac's original use apparently

Although he wasn't the first one to use it

Mr. Feynman

Although the second property can be really taken as a definition if you define $\int_{-\infty}^{+\infty}\delta(x)f(x)dx=\langle \delta, f\rangle$ and take the first property to mean $\langle \delta, f \rangle = f(0)$ (I'm kinda clutching at straws here :P)

ACuriousMind

10:41

at least he says it should be considered as a limit of functions, not a function

I can live with that

because $\lim_{\epsilon \to 0} f_\epsilon(x) = 0$ for $x\neq 0$ and $\lim_{\epsilon\to 0}\int f_\epsilon(x)\mathrm{d}x = 1$ does define the $\delta$

Mr. Feynman

Personally, I like more the definition without mentioning limits because I accept it as sloppy languge to mean what I know from distribution theory

ACuriousMind

the problem with distribution theory is that you have to make sense of what the heck people mean by $\delta(x)$

distributions don't have values at points

the limit approach makes sense of this much easier

the problem with the limit approach is that it might tempt you to do things that aren't allowed, like mutliplying two $\delta$s

Mr. Feynman

@ACuriousMind I accept it as only making sense under the integral sign, it being a bad way to write $\delta(f)$

ACuriousMind

yeah but you'll find plenty of places where people write it outside the integral sign :P

Mr. Feynman

Fermi's golden rule intensifies

I've heard horrifying arguments about Fermi's golden rule written symbolically in terms of $delta$ and the probability being infinite, so I know where your pain is coming from

Each handwave is like a katana cutting a limb

Slereah

10:53

I can probably accept some fake definition of $\delta(x)$

But then how do you write $\delta'(x)$

DIRAC1930

This en.wikipedia.org/wiki/Real_representation doesn't seem to be very helpful because I can construct the map $j$ by complex conjugation of the vector space for any representation.

ACuriousMind

@DIRAC1930 no, you can't

for one, complex conjugation is a basis-dependent operation, there is no "the complex conjugation"

for the other, complex conjugation w.r.t. to a given basis is a $G$-equivariant map if and only if the matrix representation of $G$ in that basis consists entirely of real matrices

Mr. Feynman

@Slereah $\lim_{h\to0}\frac{\delta(x+h)-\delta(x)}{h}$

That must have hurt ACM :P

I need to find the force to read about Poincaré algebra in Weinberg, I'm sure it contains a lot of useful stuff, yet I'm scared of that book

ACuriousMind

11:09

Weinberg is a very nice book to not read as your first QFT book

it is extremely valuable to read when you already know how QFT is usually done

Mr. Feynman

This is just a chapter about Lorentz/Poincaré and symmetries so I don't think it will hurt much

The second chapter

My primary reference is Maggiore's book if you know it

DIRAC1930

Hmm so cc is defined for the fund rep of $SU(2)$ with matrices $S$ as $S^* = C S C^{-1}$. If I change the rep of $S$, to for example an equivalent rep through $S\rightarrow S' = GSG^{-1}$, cc of $S'$ will be $GC S C^{-1}G^{-1}$ so I would need to find the matrices $X$ such that $X G S G^{-1} X^{-1}= GC S C^{-1}G^{-1}$

Is this right

Mr. Feynman

Oh well and Peskin too but I'll start real QFT during the second semester

I have a question, the answer might be straightforward. If I wanted to define the rotation group abstractly without mentioning matrices, how would I characterize the composition law and its topological properties?

What I mean is: if I consider it (isomorphic to) a matrix group, then it's trivial to derive the properties, but is there any prior characterization that can be made without this?

ACuriousMind

@DIRAC1930 no, but this is indeed a common problem with how physicists talk about complex conjugation

there is no "complex conjugation", there is only complex conjugation with respect to a basis, or there is a map $V\to \bar{V}$ to the conjugate vector space

note that saying "the fundamental rep of $\mathrm{SU}(2)$" does not choose a basis of that rep

so you can't talk about complex conjugation

see this for an explicit demonstration of why complex conjugation is basis-dependent

DIRAC1930

11:55

Ah okay I think I understand

So in QM, if I have a basis $\psi$ for a vector space $V$, complex conjugation of a vector $\Psi$ in $V$ would be $*: \Psi = c_n \psi^n \rightarrow c_n^* \psi^n$?

ACuriousMind

yes

DIRAC1930

Why do people seem to complex conjugate $\psi$ then?

DIRAC1930

12:14

Hmm so how do I show that the fundamental rep of $SU(2)$ doesn't have a real structure?

More Anonymous

I don't think I'm making this conflation. I don't see myself as making the claim some language is more fundamental than the other.

The post simply states there are either an infinite regress of symbols or it can be stopped at some point.

The third position is saying more specifically one does not have access to knowing position if there will be infinite regress or a fundamental symbols and it doesnt matter for describing phenomena

Pinging you a 2nd time about this (cause I reread the first and it came across as jumbled)

DIRAC1930

12:32

Is there an example I can try. Does $U(1)$ have a real structure of the real line or something?

ACuriousMind

@DIRAC1930 In general this is a somewhat hard problem (the only general way is to compute the Frobenius-Schur indicator), but for the fundamental of $\mathrm{SU}(2)$ there's a direct way: There is a map $\mathbb{C}^2\to\mathbb{H}$ via the matrix representation of quaternions and this turns the fundamental rep of $\mathrm{SU}(2)$ into a rep on $\mathbb{H}$. This shows the fundamental rep is quaternionic and hence cannot be real.

@DIRAC1930 no complex irrep of U(1) is real nor quaternionic - the Frobenius-Schur indicator is $\int_0^{2\pi} \mathrm{e}^{2\mathrm{i}n\phi}\mathrm{d}\phi = 0$ for all $n\in\mathbb{Z}$.

DIRAC1930

I'm still confused about this $j:V\rightarrow V$. Say if I have the fund irrep of $SU(2)$ with the Pauli matrices as my basis. Why isn't $\xi_\alpha \rightarrow \xi^*_\alpha$ where $\xi_\alpha$ are the coefficients of a spinor one of these $j$ maps?

ACuriousMind

(although this is simpler to see by noting that there is no continuous group homomorphism $\mathrm{U}(1)\to \mathbb{R}$ except the trivial one by looking at $x^n = 1$)

@DIRAC1930 note the condition that the map should be equivariant

complex conjugation absolutely is an anti-linear map $j$ with $j^2 = +1$

but it won't be equivariant w.r.t. to the SU(2)

DIRAC1930

12:48

Ah so this statement isn't true for my case: 'That is, applying a symmetry transformation and then computing the function produces the same result as computing the function and then applying the transformation.' here en.wikipedia.org/wiki/Equivariant_map

ACuriousMind

you should more concretely look at en.wikipedia.org/wiki/…

DIRAC1930

$S \psi^* \neq (S \psi)^*$

or something

ACuriousMind

the statement is just "complex conjugation doesn't commute with SU(2) multiplication"

and if you think about what complex conjugation commuting with a matrix multiplication means, then you should be able to figure out this happens if and only if the matrix has only real entries

which is what I meant here:

2 hours ago, by ACuriousMind

for the other, complex conjugation w.r.t. to a given basis is a $G$-equivariant map if and only if the matrix representation of $G$ in that basis consists entirely of real matrices

DIRAC1930

So if $S$ is real $S\psi^* = (S \psi)^*$ since $S=S^*$

ACuriousMind

@MoreAnonymous I really don't know what your point is or what it has to do with quantum field theory. Yes, in a situation where there are two possible options, we can take the position that it doesn't matter which of them is true

this statement seems so entirely obvious to me that I have no idea what you're actually asking or why we would need the notion of renormalization to come up with it

@DIRAC1930 yes

DIRAC1930

12:54

So for the case $D_+ \xi \otimes \xi^* D_-$, we have $j$ given by Hermitian conjugation. $\xi \otimes \xi^*$ is Hermitian, therefore $D_+ (\xi \otimes \xi^*)^\dagger D_- = (D_+ \xi \otimes \xi^* D_-)^\dagger$ if $D_+^\dagger = D_-$

Slereah

I don't think renormalization has anything to do with the notion of encoding physics in symbols rly

there's nothing about it that would fundamentally change how it works

ACuriousMind

I completely agree

DIRAC1930

I think the case for $SU(2)$ may have something to do with the epsilon tensor. i'm going to try and work it out

My above case was for the $(1/2,1/2)$ rep btw

Slereah

Renormalization is just one of those things where there is a layer of more abstract entities to deal with in between the actual measurable results

but that's true of most of physics

if you want a lot of such ramblings about symbols and physics you can read Carnap or something

Is all of Libgen down lately?

DIRAC1930

13:22

@ACuriousMind Does my thing for $(1/2,1/2)$ seem right?

ACuriousMind

@DIRAC1930 I don't really follow your notation, but it is true that $(1/2, 1/2) = (1/2,0)\otimes (0,1/2)$ is real essentially because $(1/2,0)$ and $(0,1/2)$ are conjugate representations.

DIRAC1930

Hmm so $\xi^* D_- = D_-^T \xi^*$. From my condition, $D_-^\dagger = D_+$ or $D_-T = D_+^*$ therefore $D_-^T \xi^*=D_+^* \xi^*$

i.e. $D_+ \xi \xi^* D_-$ in your notation is $(D_+ \oplus D_+^* )(\xi \oplus \xi^*)$

Slereah

13:52

Carnap still uses notation from the Principia Mathematica

Gross

"Facts are particular events. “This morning in the laboratory, I sent an electric current through a wire coil with an iron body inside it, and I found that the iron body became magnetic.” That is a fact unless, of course, I deceived myself in some way. However, if I was sober, if it was not too foggy in the room, and if no one has tinkered secretly with the apparatus to play a joke on me, then I may state as a
factual observation that this morning that sequence of events occurred."

Pretty big if

More Anonymous

@Slereah let's say I want to describe an ideal gas equation. The reason I don't have to worry about the actual shape of the potential when describing this is simply cause it can be renormalized away

*the actual potential involved in a collision

Slereah

I'm not an expert in thermodynamics but I don't think that change the prospect of symbolic expression in physics much

Some informations are relevant or not in physics

regardless of renormalization

More Anonymous

I disagree if i had 0 potential then my molecules would go through each other and never reach thermal equilibrium

Slereah

I'm sure if the potential was sufficiently weird it may also change physics :p

More Anonymous

No it just needs to be extremely steep (or renormalizable) :p

Slereah

14:06

But the point is more that you have the objects of your theory, and the statements regarding measurements are mapped from those objects but not bijectively

Some informations in your theory may be dropped out by that process

More Anonymous

@Slereah yes that's what I was arguing for in position 3

Slereah

But why focus on renormalization so much

That is true of many things

The color of a ball will not affect how it falls

More Anonymous

I mean if assume qft is a fundamental theory

And it happens to be renormalizable

Then this is a tenable position imo

Slereah

Pretty big if

More Anonymous

Most phsyicsits would roll with it tbf

Slereah

14:10

Also you should beware of the difference between your theory and how you arrive at your theory

the renormalization process will essentially take the limit in a class of theories

More Anonymous

@Slereah yup

Slereah

The actual theory could arguably be the resulting observables from that process

Rather than the original Lagrangian you took

More Anonymous

@Slereah true ... But it would be indifferent to my measuring apparatus

Slereah

Yes, but then again, this isn't related to renormalization specifically

More Anonymous

5 mins ago, by More Anonymous

I mean if assume qft is a fundamental theory

Which is renormalizable

ACuriousMind

14:17

Yes, renormalization, like many other limiting procedures, forgets about some information. This is not a particularly deep statement, and I still have no idea how it is supposed to be related to symbols or infinite regress.

More Anonymous

@ACuriousMind if i want to describe phenomena in the universe. And assume qft is sufficient then irrespective of there being infinite regress or some fundamental description. Then since qft is renormalizable some information will never enter the picture

Slereah

That's bc those are not real informations

Might as well be worrying about gauge degrees of freedom

More Anonymous

So I say we can stop at that cause irrespective of whatever the truth is: i don't need it to describe the universe

@Slereah what do you mean by real information's?

ACuriousMind

@MoreAnonymous But what does this have to do with QFT or renormalization specifically? If I have a description that is "sufficient" (for whatever we mean by that), then I don't have to care about whether or not my description is "fundamental" or not.

That's just...how science works

More Anonymous

@ACuriousMind ah I'm claiming qft is fundamental and there is no phenomena which can't be described by it .

ACuriousMind

14:23

you can claim that

what does that have to do with symbols or infinite regress?

More Anonymous

Well cause then I don't go into infinite regress or fundamental symbols. Since both these scenarios are inaccessible

ACuriousMind

I could make the same claim about classical mechanics

More Anonymous

@ACuriousMind yes but no one makes the claim classical mechanics is fundamental

ACuriousMind

@MoreAnonymous You're saying QFT is fundamental, you can't then just turn around and claim you have evaded the problem of either having a fundamental theory or infinite regress!

More Anonymous

They are inaccessible because that information turns out to be irrelevant to my measuring apparatus

ACuriousMind

14:25

and again, the infnite-regress-or-not in terms of a tower of more/less fundamental theories doesn't really have anything to do with symbols

@MoreAnonymous ...so what?

you're still saying there's a fundamental theory

you're just saying some of its details are in practice inaccessible to us

More Anonymous

@ACuriousMind yup

ACuriousMind

that's not novel or specific to QFT at all

Slereah

@MoreAnonymous discovers the notion of observables

More Anonymous

So once I can describe all phenomena in the universe

ACuriousMind

that's, like, the whole idea of statistical mechanics

there's some fundamental microscopic theory but we can't access all its details so we construct a theory that operates on macroscopic observables

again, you're just describing how science works

Slereah

14:27

This isn't even a stat mech scenario here

Because he's not even talking about informations that is physical

More Anonymous

@Slereah what do you mean by information that is physical?

@ACuriousMind i don't claim to have done something new. Just interpreted existing work

Slereah

although I guess since he's talking about the shape of potentials here I guess that could be in this case

but yeah that's essentially just "specific experiments do not access all informations"

More Anonymous

For stat mech this claim would not have sufficed cause I can still measure the position of a particle from say density.

Ryan Unger

I was plagiarizing one of my own SE answers for a class I'm teaching tomorrow and I found a mistake

whoops

More Anonymous

@RyanUnger curious ... Which one?

Ryan Unger

14:31

actually maybe it's correct, I don't think I actually write a statement

so it's not even wrong at worst

Slereah

For shame

Ryan Unger

10

A: Stokes theorem in Lorentzian manifolds

This is version two of my proof. The OP discovered a sign error in my first attempt that revealed my argument to be circular. The correct proof is below. Not surprisingly, this has to do with the signature of the spacetime metric not being positive definite. Furthermore, this issue is very subtl...

I think it is correct as stated actually

More Anonymous

@Slereah the claim is deeper if u say qft is fundamental: "no experiment has access"

Slereah

Access to what

also if that is an accurate statement, I would be very suspiscious of saying QFT is the fundamental theory :p

Ryan Unger

the funny thing is that if the boundary consists of two homologous spacelike hypersurfaces, then the orientations I use in that post are actually incorrect

so that produces an extra minus sign

Slereah

14:33

If it predicts properties that no experiment can measure, I would very much question if those properties are real

Ryan Unger

which is perfectly consistent

More Anonymous

Brb

Slereah

This is why I am not too fond of the disdain that some physicists have for philosophy

It just makes physicists do bad philosophy instead :p

More Anonymous

@Slereah atleast i try to get a second opinion rather than say this must be it ... ://

Slereah

strangebeautiful.com/lmu/readings/…

More Anonymous

14:45

@Slereah i was thinking of something like the length scale in the ising model

8

A: I'm missing the point of renormalization in QFT

Here's some complementary perspective to the excellent answer by AccidentalFourierTransform. This turned out very long, but this is a huge topic which can't be entirely summarized in one answer. An important point I want to make is that renormalization is conceptually independent of either the qu...

Slereah

Some Carnap on observables

You can have theories which have properties that cannot be measured

That is true

But they can theoretically be rewritten without those properties

also plenty of physical theories are written with more information than necessary, in general

More Anonymous

I mean yes but I wouldn't say they don't exist rather we dont have or need access to them

Slereah

Mostly because if you write theories with the most barebones theoretical objects, those tend to be very difficult to work with

@MoreAnonymous Why assume that they exist?

You can add as many fake properties that are not measurable in a theory if you want

You could add infinitely many fields in your QFT that don't interact with the other

More Anonymous

@Slereah i mean i understand renormalization as a limiting process. Surely I am limiting the a variable that "exists"

Slereah

Are you?

Sometimes that can be true, but sometimes renormalization is just going through hypothetical theories that are not true

Do you think the complex degrees of freedom exist when you solve the oscillating pendulum in the complex plane

More Anonymous

14:53

I see ... Either way I hope what i was saying earlier doesn't come of as full of balony ... I got

@Slereah ofcourse not

*i got to go :(

@Slereah also this was mean >_<

Slereah

😈

Big mistake to wonder if things are real imo

You should walk around in a haze never sure if anything is real

also drop acid

More Anonymous

@Slereah this sounds more loaded than what I've been going on about ?

Slereah

You see some theory with certain properties that cannot be measured and assume that those properties are real

Bad idea imo

"There is a temptation at times to think that the set of rules provides a means for defining theoretical terms, whereas just the opposite is really true. A theoretical term can never be explicitly defined on the basis of observable terms, although sometimes an observable can be defined in theoretical terms."

A wise man

"If a child does not know what an elephant is, we can tell him it is a huge animal with big ears and a long trunk. We can show him a picture of an elephant. It serves admirably to define an elephant in observable terms that a child can understand.

By analogy, there is a temptation to believe that, when a scientist introduces theoretical terms, he should also be able to define them in familiar terms. But this is not possible. There is no way a physicist can show us a picture of electricity in the way he can show his child a picture of an elephant."

More Anonymous

15:09

Yea I disagree with this wisdom ://

Also where is this from?

the quotations?

Slereah

the link I gave earlier

More Anonymous

@Slereah here .. for thecuriousmind :P

DIRAC1930

15:34

I think maybe the hermitian matrix is $\xi^\alpha \epsilon_{\beta \delta} \xi^{*\delta}$ not $\xi \otimes \xi^*$ as I had earlier

So really I should have $\imath \xi \otimes \sigma_2 \xi^*$ or something

Silly Goose

16:03

@ACuriousMind by approximating the derivative, i mean:

$[f(x) - f(x - \delta x)]/ \delta x \approx \frac{\partial}{\partial x}f(x)$ if $\delta x$ is small.

ACuriousMind

@SillyGoose and how is that related to Taylor series?

Silly Goose

It's in the context of this derivation in Sakurai: physics.stackexchange.com/questions/419473/…

The accepted answer uses taylor expansion to achieve the result, but I arrived at the same result using approximate derivative as mentioned above

i think one of the comments mentions it is not a taylor series since you aren't expanding around a particular point but what seems to be a variable. so the result of taylor expanding in this context is to get a function of taylor expansions of a function as a function of the point of expansion?

but my question is more about if the two methods (derivative approx) and taylor expanding) are equivalent.

ACuriousMind

ah, I see

so we have $f(x) = f(y) + f'(y)(x-y) + \mathcal{O}(y^2)$ as the Taylor expansion around $y$

if you write $y = x - \Delta x$, then this becomes $f'(x-\Delta x) + \mathcal{O}(\Delta x) = \frac{f(x) - f(x-\Delta x)}{\Delta x}$

Taking $\Delta x\to 0$, you get $f'(x) = \lim_{\Delta x\to 0} \frac{f(x) - f(x-\Delta x)}{\Delta x}$.

so yes, this is equivalent

Silly Goose

yay okay thank you

similar question... instead of performing a change of variables to get from the 2nd line to the 3rd line, we can just apply the translation operator to the bra? Use dual correspondence to figure out what the ket looks like then correspond it back into its bra?

hm well maybe not. i don't think the translation operator can be moved around with impunity

Relativisticcucumber

16:27

if the propagator has an integral formulation for a definition, why do we need the Feynman path integral approach to solve for it? seems much more complex, but im not sure if im misunderstanding something

Slereah

Which integral formulation did you have in mind?

Relativisticcucumber

something like $K(x'',t;x',t_0)=\sum \langle x'' \vert a' \rangle \langle a' \vert x' \rangle e^{\frac{-iE_{a'}(t-t_0)}{\hbar}}$

ACuriousMind

@Relativisticcucumber you never "need" the path integral :P

it's not intended to be an amazing computational tools or anything, think about it more like Lagrangian and Hamiltonian formulations: You don't need the Hamiltonian formalism if you have the Lagrangian one, but some things are much easier to express in one of these formalisms than the other

likewise, some things are nicer in the operator formalism, others in the path integral formalism

Slereah

@Relativisticcucumber That one is already solved if you have the energies :p

It is easy to compute if everything is solved indeed!

Relativisticcucumber

16:43

i see - okay thanks

ACuriousMind

@MoreAnonymous You're still not saying anything specific to QFT (nor do I understand how this is supposed to relate to renormalization). E.g. Bohmian mechanics is also like "I promise the position of the particle really exists even when we're not looking we just can't ever determine it to precision". Ontologies that involve unobservable properties are not uncommon

they're just usually not well-liked, because mainstream natural science is/pretends to be falsificationist + following Occam's razor, and falsificationists who follow Occam's razor reject untestable claims about reality by principle.

Slereah

also a lot of informations involved in physical theories are fake rly

they are just around for convenience

DIRAC1930

So for the $3d$ irrep of $SU(2)$, using our Hermitian conjugation map $j$, I should be able to decompose the tensor product vector space into a direct sum and then I need to work out how the transformations act on this 4 dim real vector space and then I need to take the symmetric part to get the 3d irrep and then see how it relates to the fund rep of $SO(3)$?

Slereah

We like dealing with array of numbers but most things in physics aren't arrays of numbers

ACuriousMind

@Slereah well, there's no rule that you have to ascribe any ontological weight to any particular thing that appears in a physical theory

Slereah

16:53

@ACuriousMind Yeah but many people do

ACuriousMind

I know and it bothers me :P

Slereah

My hot take is that I don't believe in atoms

ACuriousMind

what does it mean to believe in atoms?

Slereah

Exactly

Like Carnap I only believe in elephants

and nothing else

ACuriousMind

elephants and turtles would be more traditional

only elephants is some radical stuff

Mr. Feynman

17:23

Elephants are homeomorphic to turtles though

ACuriousMind

the new Wikipedia layout is making me consider to make a Wiki account just to be able to customize it to the old layout

Hannah Vernon

I'm no physicist, as you can probably tell from my question, however "am I getting bigger because of the expansion of the universe?"

ACuriousMind

@HannahVernon no, see physics.stackexchange.com/q/2110/50583

Hannah Vernon

thanks!

More Anonymous

17:49

@ACuriousMind Okay so I've tried a few times to spell my view point and gotten nowhere. Perhaps a better starting point would be. Where specifically do you think am I falling short?

Also this was a pretty specific example I gave slereah

3 hours ago, by More Anonymous

8

A: I'm missing the point of renormalization in QFT

Here's some complementary perspective to the excellent answer by AccidentalFourierTransform. This turned out very long, but this is a huge topic which can't be entirely summarized in one answer. An important point I want to make is that renormalization is conceptually independent of either the qu...

About the length scale in the ising model

(Also its late so I may not be able to respond immediately)

ACuriousMind

@MoreAnonymous of what is that supposed to be an example?

Mr. Feynman

18:10

What does Wikipedia mean here talking about Galilei group? Preserving the metrics separately should account for $t=t'$ in Galilean transformations but what's different mathematically from preserving $g_{\mu\nu}=\mathrm{diag}(1,1,1,1)$ which leads to $\mathrm{SO}(4)$+translations? If what I'm asking is unclear, I mean what in more mathematical terms means to preserve the metric of $\mathbb{R}$ and $\mathbb{R}^3$ separately, like how would we write the condition in matrix form?

For the $\mathrm{SO}(4)$ case I've mentioned, it is of course $R^T R=I$

ACuriousMind

18:27

@Mr.Feynman if you just preserved the Euclidean metric on $\mathbb{R}^4$, then you would be doing some sort of Riemannian SR, not Galilean physics

in particular you're saying that time and space coordinates can mix, since some $\mathrm{SO}(4)$ transformations will not leave the time direction untouched

but Galilean physics says time is absolute

Mr. Feynman

@ACuriousMind Yes, I agree on all of that. It is definitely different. My question should have been more straightforward and ask: what is the condition a matrix has to satisfy to be in homogeneous Galilei group?

The Physics is clear

ACuriousMind

It's...not really a matrix

because the Galilean group acts on affine space, like the Poincaré group

you can't straightforwardly write down the "matrix" of something in the Poincaré group, either

note that translations are not linear operators on vector spaces

Mr. Feynman

I said "homogeneous" though. Isn't there something analogous to Lorentz group inside Galilei?

ACuriousMind

sure it's $\mathrm{SO}(3)$ :P

the rotations

just like $\mathrm{SO}(1,3)$ is the "rotation" part of the Poincaré group

Mr. Feynman

Mhh, sure that is a subgroup. Though that does not remove just translations

The Galilei group has dimension $10$ so without space and time translations we should be left with rotations as you say but also boosts

ACuriousMind

18:42

oh, right, there's an $\mathbb{R}^3\rtimes \mathrm{SO}(3)$

the "boosts" are the uniform motion transformations $v\in\mathbb{R}^3$ that act as $(t,x,y,z) \mapsto (t, x+v_xt, y + v_yt, z + v_zt)$

Mr. Feynman

I guess the problem is casting it in matrix form along with rotations. $\begin{pmatrix} 1 & 0 \\ -v& R\end{pmatrix}$ with $R$ rotation

ACuriousMind

I mean, that's not a problem, you just did it :P

Mr. Feynman

The question is: what condition does this matrix satisfy to belong such group in the same way $A\in\mathrm{SO}(4)$ needs to be orthogonal and with unit det?

I mean, I've constructed the matrix because I "happen" to know Galileo tranformations but I wouldn't have been able to proceed with the condition Wiki mentions

ACuriousMind

@Mr.Feynman what problem do you have in showing that this matrix preserves both $\mathrm{diag}(1,0,0,0)$ and $\mathrm{diag}(0,1,1,1)$?

conversely, showing that these are the only matrices that do so should be just a bunch of tedious but entirely uneventful equation solving

Mr. Feynman

@ACuriousMind sure, it's easy. I just wondered how I would have done without guessing the matrix (which I did already knowing the result :P)

ACuriousMind

18:57

actually it's not even that tedious

write $\begin{pmatrix} a & w^T \\ v & A \end{pmatrix}$ and the first condition is $a=1, w^T = 0$ and the second $A\in\mathrm{SO}(3)$

Mr. Feynman

Oh, I'm dumb I should have worked normally but imposing things for two different metric matrices, that's what I missed

So I had to impose two matrix conditions. For some reasons I was trying to impose only one...

Sorry for bothering for such a stupid thing

ACuriousMind

no worries

DIRAC1930

19:38

$\xi = (-X_0 + X_3,\, X_1 + \imath X_2)$

Therefore $\imath \xi \otimes \sigma_2 \xi^*$ will be a Hermitian matrix

DIRAC1930

20:15

The more I experiment with this stuff the more confusing the question of what a spinor is becomes

The construction in Cartan's book is the only one which aims to answer the question

He calls what is now defined in modern terms as $\epsilon_{\dot{\alpha} \dot{\beta}}\xi^{*\dot{\beta}}$, a 'Spinor of second kind' which he defines as $\imath C \psi^*$ where $C$ is essentially the metric tensor so they the same thing.

His Hermitian matrix comes from finding the equations for the 'isotropic direction' but I have no idea what an isotropic direction is

DIRAC1930

20:34

On second thoughts, maybe he noticed the mapping from $\mathbb{R}^3$ to the set of Hermititian matrices $X$ first and subsequently notices that $S X S^{-1}$ leaves the scalar product invariant ($\det X$). Perhaps then he realises that he can write $X$ as $\xi C \xi^\dagger$ therefore his transformations are essentially $S \xi C \xi^\dagger S^{-1}$ or $(S \oplus (S^{-1})^T) (\xi \oplus C \xi^*)$

For the $SU(2)$ case he has $S^{-1} = S^\dagger$ so his equations reduce to $(S \oplus S^*)(\xi \oplus C \xi^*)$

However for $SU(2)$ $S$ and $S^*$ are equivelant reps

however this is not the case when the above construction is used for $SO(1,3)$

@ACuriousMind What do you think?

I think maybe this is what people mean when they say that 'a spinor is the square root of a vector'

ACuriousMind

22:55

@DIRAC1930 There are various ways in which one can interpret "spinors are square roots of vectors", indeed I think that $(1/2,0)\otimes(0,1/2) \cong (1/2,1/2)$ is one of the more straightforward ones, "squaring" the spinor representation (in the complex sense where "squaring" is $z^\ast z$ as often as it is $z^2$) yields vectors

but there are various other possible interpretations of that phrase, e.g. when you look at vectors as a subspace of the Clifford algebra you can find that the action of the $\gamma$-matrices on these vectors is the square of their action on spinors

imo it's one of these things that sounds really profound but mostly is either too vague to be useful or just some neat detail

Mr. Feynman

23:46

SE is on maintenance and so is the chat

The site should be read-only but the chat seems to be working fine

ACuriousMind

uhhh, not yet

maintenance is announced for 2-5am UTC

it's 23:48 UTC right now

Mr. Feynman

Oh I misread

I'll be sleeping at that time

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