The h Bar

7:00 PM

Because sometimes you overlooked something in your definition and it doesn't have the right properties.

@0celo7 Yes, we did.

My point about QFT meandering between classical and quantum is well illustrated by the fact that most books don't really talk about the Dyson equation :p

0celo7

ok so what about the affine map part

do we have something like $c-b=(t(c-a)-t(b-a),\psi_{t(c-a)}(c-a)-\psi_{t(b-a)}(b-a))$

Slereah

Every derivation I find is from the path integral

ACuriousMind

@0celo7 Yes!

Slereah

I know it is the easiest to do but I want Heisenberg!

0celo7

7:01 PM

well what then!

ACuriousMind

@Slereah I know no other

Slereah

Heisenberg. "You're Goddamn Right!" Breaking Bad Season 5 Walter White

►

0celo7

I know the first part is just $t(c-b)$, which makes sense

Slereah

Let's see the original paper

Hm

1949

ACuriousMind

@0celo7 Use that $t$ is a linear map. Now, we see that we have one problem left: We must define the $\psi_t$ in such a way that they respect the structure.

Slereah

7:03 PM

Path integrals were 2 years old

Could be done without them

0celo7

@ACuriousMind yes

so $\psi_{t(c-a)}(c-a)-\psi_{t(b-a)}(b-a)$

Danu

@ChrisWhite Do you think that's not right?

I think we can just restrict to 2 dimensions, and then it's fairly intuitive to me.

Slereah

"On the green functions of quantized fields"

ACuriousMind

Hmmm

Slereah

seems to be my thing!

Altho

Typewriter equations

I weep

ACuriousMind

7:06 PM

@0celo7: I may have backed us into a dead end

Let me think for a bit, though

0celo7

deer lord

Lord Venison, pls save @ACuriousMind's hard work!

Danu

@Slereah Schwartz' new book does, quite extensively in fact.

Slereah

Oh boy

Time to buy it legally

0celo7

Oh girl

Slereah

Quick, to the buying books legally website!

0celo7

7:09 PM

Amazon

Slereah

Sure, why not

0celo7

where else would you buy books

Slereah

what is Schwartz's book?

0celo7

QFT Noire

Slereah

an intriguing quantum mystery

ACuriousMind

7:10 PM

Ha!

Slereah

It all started when some hysterical broad walked into my office

ACuriousMind

@0celo7: We have a freedom left in the defintion of the $\psi_{t_0}$.

0celo7

@ACuriousMind was that in response to my pun

or the problem

ACuriousMind

We can choose which point in $t^{-1}(t_0)$ to map to the origin of $\mathbb{R}^3$.

@0celo7 Both :)

Slereah

schwartzqft.com

this book?

0celo7

7:12 PM

@ACuriousMind yes

@Slereah i think so

Slereah

Let's go with that

So what are the things that are like

0celo7

buy two copies and send me one

Slereah

Basis of Heisenberg QFT

Is it 1) Hilbert space shit 2) Lorentz invariance 3) $\hat H \vert \psi \rangle = i\partial_t \vert \psi \rangle$

Can I do all the QFT in Heisenberg with just that

At least for free fields

0celo7

expletive!

there are children in here

Slereah

Are you still 17?

Grow up man

I want to say swears

0celo7

7:16 PM

10 days

Slereah

Oh and $[\phi, \pi] = i\hbar \delta(x-y)$

Or the same but with vectors

Depending if I do equal time or covariant

0celo7

@ACuriousMind still thinking?

ACuriousMind

I'm thinking how to formally write it down.

0celo7

do you stroke your hair when thinking?

ACuriousMind

My beard.

0celo7

7:18 PM

I mean, it must serve some purpose

then why the long hair

I rub my stubble when thinking

Danu

@Slereah QFT and the Standard Model or something like that

Slereah

"Quantum mechanics involves two distinct sets of hypotheses - the general mathematical scheme of linear operators and state vectors with its associated probability interpretation and the commutation relations and equations of motion for specific dynamical systems. It is the latter aspect that we wish to develop, by substituting a single quantum dynamical principle for the conventional array of assumptions based on classical hamiltonian dynamics and the correspondence principle."

I love you Dyson

you can tell he wrote this on a typewriter because he writes quantum states as $\vert \alpha )$

Wait no

He just writes like that to be a dick!

I take it back I hate you Dyson

wait fuck what even is that notation

$(\alpha'\vert)$

What's going on

Oh wait I think that's supposed to be an expectation value

$(\alpha'\vert) = (\Psi(\alpha'), \Psi)$

Oh wait $\Psi(\alpha')$ is the eigenvector of some operator

I DON'T KNOW

I guess it's the probability of being in state $\Psi(\alpha')$

People should be forced to use the same notation

under pain of torture

ACuriousMind

7:38 PM

Okay, this was silly

@0celo7: I'm sorry

We don't need all those $\psi_t$ :/

Since $t : \mathbb{R}^4 \to \mathbb{R}$ is a map between vector spaces, we have a projection $\pi : \mathbb{R}^4 \to \ker(t)$ (by rank-nullity, essentially)

Now define $\phi(b) = (t(b-a),\psi_0(\pi(b-a)))$.

And this now is an affine map since $t,\psi_0,\pi$ are all linear.

I'm sorry for doing so much unnecessary stuff, but that gave you good example of how crooked the paths are by which one usually finds the solution to such exercises ;)

Slereah

heheh

From the wikipedia article :

"There are not many books that treat the Schwinger–Dyson equations."

that is a problem with physics books, they always want you to follow a little historical path

It is very rarely "Here's a bunch of axioms now we do things"

No you have to start QFT from 1905

ACuriousMind

7:56 PM

@Slereah I find that very annoying, too

Slereah

well it's fine, but it would be nice if some did it otherwise

ACuriousMind

It would be nice to have QM and QFT in a fashion that's not shaped by the historical baggage, but by logical stringency, yes.

Slereah

Here's a rule for a book : You can never use classical physics

You get some quantum axioms and you deal with it

ACuriousMind

Lol, someone over at AskUbuntu flagged the bot that posts warnings about spam to chat as spam.

Slereah

Hm

Finding a Heisenberg derivation of the Schwinger Dyson equation is not trivial

The original paper doesn't use path integrals but at some points, there's a scalar and he says "Let's call this the action now"

For no explained reason

ACuriousMind

8:05 PM

That doesn't sound right :D

Slereah

Like there is some hermitian operator $W_{\alpha\beta}$ and he just says that $W_{\alpha\beta} = \int_{\sigma(\alpha)}^{\sigma(\beta)} \mathcal{L} dx$

But he does not really relate those to the Hamiltonian or anything

Although

Maybe

the old timey notation doesn't help to read it fast

What's that website that lists papers per topics?

I forget the name

ACuriousMind

@Slereah arXiv? :P

Slereah

Nah

Like

0celo7

great

Arnold is downstairs

Slereah

There's a page about a topic

0celo7

8:09 PM

but the mail room is closed

Slereah

and different sections of various aspects of the topic

and links to papers

0celo7

@ACuriousMind never apologize

ACuriousMind

@Slereah Wikipedia? :P

I have no clue what site you are talking about

Slereah

Hm

Fuck what is it again

I end up there once in a while but I never remember the name

0celo7

Wald does not take a historical path

but then again I think his chapter 4 is garbage

(that's the chapter where he talks about the field equations)

Slereah

8:12 PM

Ahah!

phy.olemiss.edu/~luca/Topics/list.html

That's the one

0celo7

ole miss?

TIL they do anything other than football there

Slereah

phy.olemiss.edu/~luca/Topics/qft/green.html

mite b cool

0celo7

@ACuriousMind ah

my book is downstairs but I can't get it :(((((

Slereah

by the way is there a huge difference between equal time commutation relations and covariant commutation relations

0celo7

@ACuriousMind How on Earth (get it) are you supposed to do the problem on page 12 without knowing conservation of energy or the work-energy theorem?

@Slereah enlighten us

Slereah

8:21 PM

$[\phi(\vec x), \pi(\vec y)] = i\hbar \delta(\vec x - \vec y)$ versus $[\phi(x), \pi(y)] = i\hbar \delta(x - y)$

Apparently both can be used

ACuriousMind

@Slereah You get only the equal-time commutation relations in the Heisenberg formalism.

Slereah

What is the covariant one used for?

0celo7

Jidenna - Classic Man ft. Roman GianArthur

►

ACuriousMind

@Slereah Oh, I think mainly when you want to make the Lorentz covariance of the approach manifest somewhere

Slereah

but which one does that, is what I wonder :p

Hm

Really some troubles to find a proof without path integrals

ACuriousMind

8:27 PM

@Slereah No one does that, I've never seen an useful application of them except that their vanishing for space-like separation is a causality statement.

Slereah

>no one does that

Fat chance!

like nobody is going to make a random paper about some quantum formalism

0celo7

lol

he's got you there

ACuriousMind

Alright, I may be underestimating your ability to find weird shit :D

0celo7

ok

how do you find escape velocity without using work-energy theorem or conservation of energy

Slereah

do the kinematics, see when it never comes back down?

changing direction would have a change of sign of the speed

find if the height has a maximum

so second derivative > 0

or < I dunno

inspirehep.net/record/9122/citations?ln=fr

Well, let's see who quoted that fucking paper

0celo7

8:36 PM

He's finally lost it

Jebus

that's scary!

Slereah

Hm

Not sure I will manage to find it

Maybe I should follow Schwinger's derivation

And then prove that the action is indeed the action made from the Hamiltonian

Basically the approach seems to be that for some quantity $\langle \psi \vert \hat \phi(x) \hat \xi(x) \vert \psi \rangle$, you can apply the operator $\delta$ to transform from one representation to another unitarily equivalent one

Hm

Maybe the trick would be to use the time evolution operator as a unitary transformation

It is the most obvious one

and it already has the hamiltonian in it

0celo7

seriously, how do you do the escape velocity problem

I have no clue how to do that kinematics problem...do you want to integrate Newton's equation or something?

Slereah

@0celo7 Just solve the Newton equation for the radial component with gravity?

$\frac{dp_r}{dt} = G\frac{mM}{r^2}$

When p(t) is 0, you have reached the apex

So... $\ddot r = G\frac{M}{r^2}$

0celo7

forgive my ignorance of second order equations

how do you solve that

Slereah

Yeah I dunno either

Hm

Let's see

$r^2 \ddot r = C$

Free fall

In Newtonian physics, free fall is any motion of a body where its weight is the only force acting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it and moves along a geodesic. The present article only concerns itself with free fall in the Newtonian domain. An object in the technical sense of free fall may not necessarily be falling down in the usual sense of the term. An object moving upwards would not normally be considered to be falling, but if it is subject to the force of gravity only, it is said...

Apparently it's complicated

0celo7

8:52 PM

how can free fall be complicated

Slereah

Well usually it's written as $\vec F = m\vec g$

2

Q: Free-fall according to Newton's gravitation law

Most analysis of free-fall assume that bodies fall with constant acceleration. If however one analyses free-fall according to Newton's gravitation law, one is lead to a differential equation which I don't know how to solve. The differential equation one is lead to[1] is $\frac{d^2x}{dt^2}=-\fr...

So to answer you I'm not so sure solving the system is the best solution :p

0celo7

so how does Arnold expect the reader to solve this problem

Slereah

Ask Arnold, perhaps

ACuriousMind

@0celo7 Integrate the equation once as in the math.SE post

0celo7

Ah

thanks

er

maybe I should work it out

ACuriousMind

8:59 PM

I think you can determine the escape velocity from that one now without solving the differential equation

0celo7

just set $v=0$, right?

ACuriousMind

No, you want to get $v_0$!

You have to set "$x = \infty$"

Slereah

Oh yeah integrating once might work I think

you have to group it clever tho

0celo7

what

hmm, let me work this out

Slereah

Like $\frac{d(f(r) \dot r)}{dt} = r^2 \ddot r$

or something

"Multiply the equation by dx/dtdx/dt and integrate once"

apparently

0celo7

9:04 PM

sigh

sometimes I disappoint myself deeply

Slereah

heheh

So...

ACuriousMind

@0celo7 Found it out?

Slereah

$\ddot r \dot r = \frac{\dot r}{r^2}$

ACuriousMind

No, Slereah :P

Slereah

no?

0celo7

9:05 PM

I have $$\frac{1}{2}v^2=\frac{GM}{r}+C$$

now

er

Slereah

Now send r to infinity

Weeee

ACuriousMind

@Slereah You do $x\ddot{x} = \frac{-\dot{x}}{x}$.

0celo7

ok so $C=v_0^2/2$

ACuriousMind

Integrate by parts on the left, and logarithmically on the right.

Slereah

ah yes

0celo7

9:06 PM

then set $v=0$

boom

Slereah

Been a while since I've had to do integrals by hand :p

0celo7

perhaps, I get a negative where it should not be

@ACuriousMind I just used chain rule

ACuriousMind

@0celo7 I think that's what I mean by "logarithmically"

0celo7

$\mathrm{d}r=v\,\mathrm{d}t$

ACuriousMind

Well, that's a physicists' trick :P

0celo7

9:07 PM

tfw engineer

tfw less rigorous than a physicist

you're lucky I didn't plug that into MATLAB

Danu

When one says that $P_\mu:=-i\partial_\mu$ "generates translations", what is the precise statement? I only know that one typically sees it explained as "act with the exponential of it and you get translations" but I'd like a better understanding.

0celo7

@ACuriousMind So is $v^2/2=GM/r+C$ right? I get an imaginary escape velocity :O

Danu

The exponential map acting on $P_\mu$ yields translations? That's it?

ACuriousMind

@Danu Yes

Danu

I don't like this crappy "first order Taylor of the exponential" one always sees though

How do I go around that?

Also, how do I know that $P_\mu$, defined abstractly in terms of the Lie algebra, should be represented by $-i\partial_\mu$? :P

I would really like to have a separate course on the representation theory of the Lorentz & Poincare groups.

ACuriousMind

9:11 PM

@Danu The algebra of vector fields exponentiates to diffeomorphisms

The diffeomorphisms that come from the $\partial_\mu$ are precisely the translations in $\mu$-direction

Danu

@ACuriousMind You mean the algebra of left-invariant vector fields?

ACuriousMind

@Danu No, left-invariant under what?

Slereah

@Danu : $e^{a\partial_x} f(x) = \sum a^n \frac{\partial^n f(x)}{n!} $

Danu

Group action

ACuriousMind

Vector fields on a manifold, e.g. $\mathbb{R}^n$.

Slereah

9:12 PM

Hm

Danu

@Slereah I know that

Slereah

I forget how the rest of that goes

Danu

Take first order blabla

but I don't like that

Slereah

Well not first order

It works at all orders

Danu

@ACuriousMind I thought you referred to the Lie algebra

(which is the set of left-invariant vector fields)

It doesn't make sense to exponentiate anything else, does it?

ACuriousMind

9:12 PM

Okay, let me state my setting.

We have a manifold $M$ with an algebra of vector fields $V(M)$.

0celo7

>mathematician comes in

Danu

You mean $\Gamma(TM)$?

0celo7

>discussion turns from bona fide physics to math

Danu

this is the standard notation as far as I know

@0celo7 tfw representation theory of the Lorentz algebra and group is full-on physics

ACuriousMind

The finite-dimensional sub-algebra $\mathbb{R}^n$ generated by the constant vector fields $\partial_\mu$ exponentiates to the translations among the diffeomorphisms $\mathrm{Aut}(M)$ (this exponentiation is finding the integral curves).

@Danu Yeah, same thing.

Danu

9:15 PM

@ACuriousMind I don't really know what you mean by exponentiation---I've only seen it defined on the Lie algebra.

Do you mean just finding the integral curves?

I guess you do

ACuriousMind

Yes.

Danu

I never heard that referred to as exponentiation

ACuriousMind

It's "exponentiating" in the diffgeo sense of the exponential map from the tangent space at a point to a small area around a point

Danu

@ACuriousMind I've not seen exponentiation defined like that

only on the Lie algebra.

ACuriousMind

@Danu You've never seen the exponential map?

Danu

9:17 PM

@ACuriousMind Not the one from Riemannian geometry, no

I told you, right? My course on "Riemannian geometry" was a course on Chern-Weil theory instead

We covered geodesics in 2 hours in the last week.

ACuriousMind

Oh, okay

Danu

But I do recall, now that you mention it, some mention of an exponential map

And that it coincided with the one on the Lie algebra too

(when both are defined)

But yeah, not much I learned about that :(

ACuriousMind

Indeed it does.

Slereah

Oh my god

Fraktur

I hate fraktur so much

Danu

Really? I think $\mathfrak{g}$ and $\mathfrak{h}$ look awesome.

0celo7

9:20 PM

it's the best

Danu

and of course, those are the only ones anyone in their right mind ever uses ;)

Slereah

"However, since his theory is based on his own formalism of quantized fields, it is not immediately evident which connection exists between his results and those of conventional field theories. It is with this relationship that the present paper will deal."

That's right

Fuck Schwinger

journals.aps.org.sci-hub.club/pr/abstract/10.1103/…

^this seems to be what I want, mb

0celo7

well the escape velocity is imaginary

so Newton was wrong

yay

either that or I didn't integrate right

Danu

@0celo7 Better doubt Newton first :)

0celo7

99% chance I took the wrong sign in the universal gravity thing

Danu

9:23 PM

@ACuriousMind In any case, it's something I seem to know, just under a different name. Now, what were we talking about? :P

0celo7

oh my

I integrated in the wrong direction

looks like Newton was right after all

Danu

How do I prove that it exponentiates to the translations?

0celo7

Big Data - "Dangerous (feat. Joywave)" [Official Music Video]

►

ACuriousMind

@Danu Well, you have to find the diffeomorphism that is the exponent and observe it's a translation :P

So, you have to find the integral curves for the constant vector field $\partial_\mu$.

Danu

@ACuriousMind Right, that's pretty easymode

ACuriousMind

9:25 PM

The exponent just moves the points along these curves, so you just need to show that the curves just translate the point in the $\mu$ direction

Danu

In fact, it's an explicit example in my diff.geo. notes:

who needs Riemannian geometry? :P

Or at least, I think this example amounts to the same thing.

It's framed a bit differently

0celo7

I feel like ACM needs these bursts of intelligent discourse after fussing with me for 2 hours about some linear algebra bullcrap

Danu

^lol

ACuriousMind

@Danu No, that's not the same :P

Danu

He needs a break from you!

0celo7

9:27 PM

he got two days :P

ACuriousMind

@Danu That there is showing that the constant vector fields are the ones invariant under translations. We want to show that the integral curves of the constant vector fields are the translations.

Danu

@ACuriousMind All the words coincide, though :D (just missing "exponential")

@ACuriousMind Yeah, I know; but it's fairly close to being the inverse problem :P

ACuriousMind

That's correct :P

0celo7

in related news, I helped a grad student polish a tungsten piece for a fusion exeriment

Danu

@0celo7 BRAVO!

ACuriousMind

9:34 PM

lol