The h Bar

Secret

15:00

I am in aust, and aust has opposite seasons:
One interesting thing I found is that there is something in common between equlibrium chemistry and the freeze out process in big bang nucleosynthesis

@Slereah Is it because once you start having things scattering and interacting, the particle number is no longer conserved thus it is no longer meaningful to talk about particle numbers? (I am not famialr with Fock space yet, but I will eventually)

Slereah

Once interaction starts, there is no more notion of particle states

Secret

makes sense

Slereah

the whole $a a^\dagger$ business only works because you are doing a Fourier transform of the field

and then you can say that the field is $\sum f_k a_k + \bar f_k a^\dagger_k$

0celo7

Fock space smock space

Jim

Oh, well if you're in Australia, then that's different. Forgive my assumption. More than 50% of people live in the northern hemisphere, so it's a good starting place.

Slereah

15:03

But you can't sum solutions if it's non-linear, though

0celo7

The only number operator I need is my set fingers

Secret

so in the Fock space, the field essentially become a linear combination of the normal modes of harmonic oscillators?

Slereah

basically, yes

It's a practical math thing because it's easy to solve

Secret

I recall in today's spring school the professors says QFT is basically modelling things with an uncountably infinite number of harmonic oscillators

Slereah

kinda yeah

that is what the $k$ is

it's a real index

Secret

15:07

and he showed us this video to illustrate in QFT how a wave and a field arises from the collective but independent motion of the oscillators
https://www.youtube.com/watch?v=yVkdfJ9PkRQ

so, as he narrates, at some point of the collective motion, you get one wave split into two waves, which corresponds to how one particle decays into two other particles

But then I am wondering...

What's the quantum field theory analogue of the chaotic motion shown in this pendulum analogy (e.g at 1:04)?

ACuriousMind

@Secret None

That's why it's not really a useful analogy.

Secret

15:25

Is it correct to say that when we use ladder operators $\hat{a}$ and $\hat{a}^\dagger$, we are describing the states in the system using these as variables instead of the usual $\mathbf{r}$ and $\mathbf{p}$ ?

I.e. while I know the expressions for the ladder operators, I am curious on what kind of property or symmetry it exploits such that it can describe states very nicely in terms of using these operators to move up or down the states in the quantum system?

Is using ladder operators simialr to perform some kind of coordinate transformation, which is why it makes the maths nicer?

Slereah

as said, the ladder operators are $a \approx x + ip$

or in QFT, $a \approx \phi + i \pi$

up to whatever factors, sums, integrals and whatnot

It is basically just a change of basis for the Hilbert space

ACuriousMind

@Secret $a$ and $a^\dagger$ are operators. $r$ is not an operator in QFT. You are trying to compare two very different things

Slereah

that too indeed

Secret

wait, position is not an operator?

Slereah

it's a basis of the Hilbert space, not the phase space

Not in QFT

There are no point-like objects in QFT

Only fields

ACuriousMind

15:28

@Secret No, QFT does not have a naive position operator.

Slereah

The wavefunction isn't $\Psi(r)$

ACuriousMind

In non-rel QM, you just have $a = x + \mathrm{i}p$ for the haromonic oscillator, as Slereah says.

Slereah

It is $\Psi[\phi(x)]$

Instead of having the particle at a position at a moment t, you have the field in a particular configuration $\phi$ at a moment t

You can recover QM from QFT, by the way

Secret

cannot see how to recover QM form QFT from the top of my head, but i guess it has something to do with the configuration of the field taking certain values, or going back to non relativistic systems?

Slereah

Basically you do averages over a volume

like $\phi_\Omega = \frac 1 {V(\Omega)} \int_\Omega \phi dx $

Then you recover $[\pi_\Omega, \phi_\Omega] = -i\hbar \frac{1}{V}$

Secret

15:39

so in a small volume , the average configuration of the field in there can be considered a point like entity, that's why you recover the particles, and the commutation relation per unit volume?

Slereah

It helps yeah

Secret

Still I am no very familar with the expression $-i\hbar\frac{1}{V}$ although I have seen simailr things like $-i\hbar\nabla$, $i\epsilon\hat{L}_k$, so I felt I like I am oversimplfying something in my above picture

NB (off topic): Because I am such a visual learner in steroids, I have word versions of the pictures I am drawing. For example the above statement is actually the same as an illustration if one use it to draw a picture

This is how I tried to gain some physical intuition by essentially reducing eveyr problem into a geometry problem, along with doing more problems in textbooks an…

Jim

you are a visual learner in steroids? Is that some kind of expression I don't know about or do you take steroids to learn?

Secret

on steroids, I mean
I am abusing a common idiom or adjective to describe that I am really passionate about turnign things into pictures to better understand them, by essentially treating every concept as a combination of a manifold with each point on it having some kind of property (e.g. visualising the paramter space of a PV diagram by watching how the thermodynamic system expand and heat in a way that one can see some kind of patterns between the volume and pressure)

Jim

ah, okay then

Secret

15:51

So basically, since 2015, I often attempt to visualise configuration space and phase space by trying to understand what our eyes are seeing as the system take some trajectory in phase space

and the 4 spatial dimension visualisation exercise I learnt form another forum since 2011 really help on that

John Duffield

Whenever I see dmcKee, I always think of D B Cooper.

Jim

why?

John Duffield

The dark glasses.

Jim

Yeah, I can kinda see it

John Duffield

Once you get that D B Cooper in your head it will never go away.

Jim

15:55

I think you underestimate my power to be distracted

John Duffield

Wanna talk about inflation and why it's superfluous?

This is worth a read.

Secret

@ACuriousMind @Slereah
I learnt from today's class that the vacuum state is defined as $\hat{a}(\mathbf{k}|0\rangle)=0, \forall k$ whic has zero momentum and total energy. The professors said that it is an important state when things start interacting

And also in the complex field example of the harmonic oscillator in QFT, you have two different lowering ladder operators that can anihilate the ground state into the vacuum state

Is this vacuum state important in virtual particle related phenomenon such as casimir effect?

Jim

I don't know if it's superfluous. But it is a little fluous

ACuriousMind

@Secret I don't understand the question.

It's the vacuum state of the free theory

The vacuum state of the interacting theory will be different

Why do you think it has anything to do with the Casimir effect?

(Well, in a sense it has, because the Casimir effect arises since the finite volume theory between the plates shifts the energy of the ground state compared to the infinite theory outside)

0celo7

ELI5 why there is no number operator in an interacting theory

Secret

16:03

@ACuriousMind
Probably influenced by popsci magazines often saying how virtual particle pairs pop in and out of existence thus the vacuum is not actually empty, and that gives a pressure. And the casmir effect if I recall, the region between the plates exclude some of the modes thus the pressure outside the region pushses the plates together

Given now I finalyl have some beginning understanding of QFT, I think I should update myself on what is the rigoruos version of the popsci's "virtual particles pop in and out of existence in the vacuum"

(you can show me the maths if necesary, I am not afraid of the algebra as long I can udnerstand the physucal meaning of the important terms in the expressions)

ACuriousMind

@0celo7 Why should there be? I.e. how would you even begin to define it, lacking the mode expansion of the free fields?

0celo7

@ACuriousMind count!

ACuriousMind

@0celo7 Count what?

0celo7

Particles

Jim

@JohnDuffield It was a good read. But nothing that says inflation is superfluous. As I told my former supervisor when we were working on a paper, inflation was designed and implemented to take arbitrary and unknown initial conditions and result in an observable universe that produces the same observables we see today. Given that, it isn't surprising that no matter what we put into it, it usually gives us the desired outcome.

ACuriousMind

16:07

@0celo7 Define particle in an interacting theory, then

Jim

Obviously that would open up a can of worms of ambiguity and fine tuning of the initial conditions

Inflation does exactly what we wanted it to. It's not inflation's fault we didn't know what we really wanted from it

0celo7

@ACuriousMind little thing that bounces around

It's pretty obvious

ACuriousMind

@Secret The Casimir effect is just a finite-size effect: The momentum modes become discretized in the direction perpendicular to the plates. After an ugly bit of regularization and getting rid of infinities, you can calculate the ground state energy per unit area. It depends on the distance between the plates because the allowed momentum modes depend on it, and so (classically) the derivative of the energy with respect to that distance gives a force in that direction.

@0celo7 Stop trolling :P

0celo7

I'm not trolling

I want an intuitive explanation

I can chug through the math, I even understand Wald's algebraic approach

But that leaves me no reason why an experimentalist can't just count the little things in the box

ACuriousMind

@0celo7 No one says the experimentalist can't.

0celo7

16:13

Well the number operator counts

ACuriousMind

That does not mean that, on the theoretical level, a number operator (or even the notion of particle state) has to exist for interacting theories.

0celo7

This is why I'm not going into theory. The other reason is hippy hair.

ACuriousMind

@0celo7 You also should realize that "interacting theory" means in the middle of a high-energy collision, not "stuff lying about in a box".

0celo7

Put the collision in a box and slow time

Then count

ACuriousMind

Stop trolling :P

0celo7

16:16

And I did realize that

Wtf

I'm not trolling

You're just giving JD ammunition

These are legitimate physics questions

I'm not a physician

ACuriousMind

"Put the collision in a box and slow time" is not a legitimate physics procedure, is it?

Jim

@0celo7 hippy hair?

@ACuriousMind not with that attitude

ACuriousMind

@Jim Context somewhere around this:

20 hours ago, by ACuriousMind

Yep, we're all filthy hippies over here.

Jim

@ACuriousMind gotcha

0celo7

@ACuriousMind maybe not now

But in theory

Jim

16:20

I thought you were staying away from theory?

0celo7

Well

In theory as in "in the future"

Jim

Okay, from now on I will replace all usages of "In theory" with "In the future"

0celo7

Yes

Do that

Jim

In the future, it could work

0celo7

Note that ACM has still not answered my question

Secret

16:23

(This is a (possibly silly) crossover question, creating by mixing 0celo7's question x chemistry x QFTstuff learnt in spring school) Is some notion of transition state in chemistry applicable to describe interactions between particles?

obe

In theory I am finished this problem set. (in the future).

0celo7

DO THE PROBLEM SET

obe

D:

0celo7

It's already overdue and I told you how to do it

obe

I know I have to sit down and do it.

0celo7

16:25

I did them while playing BF4

You should be able to do them now

obe

Right god.

Jim

All hail the Right God

obe

Who can do QFT problem sets while playing BF4.

Jim

ghostbusters?

obe

Was that a reference?

Jim

16:28

no?

0celo7

The random pauses in my talking where when I was concentrating on not dying

obe

Lol I remember that.

Jim

While I sympathize with the political plight of Canadian scientists, this user name exhausts and annoys me

0celo7

it's easy to think about math when walking to the firefight

but pretty tough to shoot and talk about math

@Jim ELI(not Canadian)

Jim

@0celo7 are you describing yourself?

0celo7

16:33

@Jim please explain why that's exhausting and annoying

does anyone have the Steward badge?

ACuriousMind

@0celo7 Yes. It has been awarded 33 times.

Jim

@0celo7 NSERC is our government body that gives grant money to scientists etc. Canadian government has been cutting science budget for years and as a result nobody has money for new students, research etc. This person is overly upset about it and is advertising that everywhere in a way that does nothing but shout "I'm angry about this" to people that aren't affected. They aren't doing anything about it, just telling uninvolved people

0celo7

What?

How is it going to take CW 97 years then?

Jim

It's annoying because all of us have to make do with NSERC cuts. He chose to leave. But he acts like it's a political protest

ACuriousMind

@0celo7 That's not hard to figure out. :P

0celo7

16:36

So he never does close votes?

Jim

@0celo7 1000 votes. He got 30 in 3 years. 970 to go at a rate of 10 per year

NeuroFuzzy

Okay I'm dumb, can someone fix my brain please? So if you have a hermitian operator $\hat{A}$ with eigenvalues $A_k$, you should have $\langle A_k|\hat{A}=A_k \langle A_k|$.

0celo7

He made it sound like Steward is for Late Answer reviews.

@NeuroFuzzy yes

ACuriousMind

@0celo7 It's for 1000 reviews of a single type

Jim

you can get a steward badge in each of the different review categories

0celo7

16:38

Ok, so he'd take 97 years for doing Late Answer reviews?

But certainly he has other reviews?

@Jim ohhhhhh

Ok

ACuriousMind

You are seriously overanalyzing that joke :P

Jim

@ACuriousMind maybe you're underanalyzing it

how would you know unless you analyze it more?

0celo7

At least I don't overanalyze QFT

just count the little bastards, there's your number operator

Jim

@0celo7 Count the exact number of jelly beans in this jar without opening it up. That's where the number operator shines

NeuroFuzzy

[hit enter on accident] and $\hat{A}|A_k\rangle=A_k | A_k\rangle$. So then if $\hat{B}$ is any operator, $\langle A_k|[A,B] | A_k \rangle=A_k \langle A_k|\hat{B}| A_k\rangle-A_k \langle A_k|\hat{B}| A_k\rangle=0$...

0celo7

16:41

@Jim easy

shink down to nano size and slip in

then count

Jim

that's cheating

ACuriousMind

@NeuroFuzzy That is correct

0celo7

@Jim that's engineering

ACuriousMind

The expectation value of a commutator vanishes if the state is an eigenvalue of one of the two observables.

Jim

but in the future, it could work

0celo7

16:42

yup

in theory, anything is possible

we just need to wait until we can change the laws of physics

some smart guy will figure that out eventually

NeuroFuzzy

@ACuriousMind Is there functional analysis weirdness if $\langle p'|[\hat{p},\hat{x}] |p'\rangle=-i \hbar \langle p'|p'\rangle$?

ACuriousMind

@NeuroFuzzy Yes

NeuroFuzzy

OH

okay then

Secret

$\hat{H}_{ren}=\hat{H}-\langle 0| \hat{H} | 0\rangle = \int d^3\mathbf{k} \omega_{\mathbf{k}\hat{a}^\dagger(\mathbf{k})\hat{a}(\mathbf{k})+\frac{1}{2}\delta^3(0)-+\frac{1}{2}\delta^3(0)}=\int d^3\mathbf{k} \omega_{\mathbf{k}\hat{a}^\dagger(\mathbf{k})\hat{a}(\mathbf{k})}$

We are also being introduced the concept of renormalisation, which one professor said it can be interpreted as after summing up all the oscillators, (thus the ZPE of them will add up to give a diverging term), and then pull this diverging ZPE like term down so it form the baseline of measuring the energy and we then set t…

ACuriousMind

Because, technically, the momentum operator's eigenstates do not lie inside the Hilbert space

0celo7

16:45

Slereah does not know any GR

He doesn't even know what the Kasner metric is

Secret

who is expert in GR here?

Jim

define "expert"

NeuroFuzzy

@ACuriousMind Thanks! Hehe i just spent half an hour trying to figure stuff out. I'm trying to go over most of my undergraduate QM for my class.

@Jim uses a subset of big enough words to convince other people who use the same set of big words that your big words are good.

0celo7

@Secret ACM

Jim

@NeuroFuzzy Not I, in that case. I usually try to avoid the big words and focus on the ideas

Secret

16:49

@Jim
let's see...
Know about black holes, all machinery in GR that is used to formulate it), spaceimte manifold and all interesting details of various matrices such as Friedman (forgot) robertson walker, schazchildopen problems in cosmology, and most importantly open problems and attempts at unifying GR and QFT

ACuriousMind

@0celo7 @Secret Just to keep me from a length GR post: He's kidding.

John Duffield

@Jim : I say inflation is superfluous.

Jim

@Secret Damn, I narrowly missed that one too.

@JohnDuffield Well, I know you have rational reasoning for that opinion. And given the state of cosmology currently, I can't really argue with you. I don't agree, but it's a valid opinion that I can't argue with

Well, I can. Just not sincerely

John Duffield

@Jim : the rational reasoning comes from understanding GR as described by Einstein, then applying that to black holes, then flipping things round for the universe as a whole. The evolution of the universe over time is a bit like pulling away from a black hole in space.

Just not sincerely? OK, I'll save my breath then.

0celo7

@ACuriousMind you can explain math better than anyone here

in that sense, you do know a lot of GR

especially stuff like HE, which is just geometry + energy conditions

Secret

17:00

The professor then elaborated on that the reason why the total energy is not globally defined on the spacetime manifold is because e.g. at one point and the neighbourhood on the manifold, the total energy may be E_1, but at another point, it becomes E_2

He then say that QFT still works at small enough regions on the manifold where the spacetime is pretty much flat, hence the gravity is weak and thus there isn't too much interaction between gravity and the energy density of fields present there

0celo7

jesus

Slereah

And I am home

0celo7

how do you type that fast

@Secret matrices?

Jochen

16 1

I like to think this is my (future) math advisor

Slereah

downvote him

0celo7

he would probably not ask such an elementary question though

Jim

17:02

@0celo7 lots of math, but I noticed no EFT. That's one of the most important and math-heavy physics tools

0celo7

@Jim what

I have no clue what you're talking about :O

Jim

@0celo7 effective field theory

0celo7

what about it

Jim

They guy you linked doesn't have it in his top 4 tags

0celo7

what the hell are you talking about

the guy I linked happens to have the same first name as a prof I talked to

no clue if he's actually him

Jim

17:06

OH! Well, that is different. I thought you found a guys with lots of math in their top tags and that's why you linked him

0celo7

no

Jim

my mistake

0celo7

if I make a math double major official, Jochen Denzler will be assigned as my faculty advisor

Jim

must be the same guy

No way it's a coincidence

0celo7

he went to the same school as ACM for undergrad and did his PhD where @Huy is right now

that's the coincidence

It's crazy that the two Germans I've met went to UH

What's even crazier is that ACM won't give me any good Intel on the school

@ACuriousMind this is where you tell me that there's a secret underground death cult and that they were members

@skillpatrol @Rigor reserved ticket for the Georgia game

Huy

17:24

I met the fastest lecturer I've ever had today

I feel violated

Hippalectryon

Basically, you got a lecture from Sanic ?

Huy

he was a french guy

I blame you

Hippalectryon

Name ?

Huy

Tristan Rivière

0celo7

No one likes the French

Jim

17:32

What about the French?

0celo7

They've done nothing

Hippalectryon

@0celo7 :-(

0celo7

Their greatest accomplishment is Canadia...a country that is literally a hat

Jim

No, I mean don't the French like the French?

@0celo7 I'd say we're the torso

Admit it. I'f North america is a person. USA looks like a speedo. Mexico is the legs, and Canada is the torso. Visually, it's much more representative

Alaska is a shoulder or arm or something

Hudson's bay must be the neckline, I guess

0celo7

Florida is a leg

Mexico is the middle leg ;)

Jim

17:36

Florida is the middle leg

And that's what everyone in America calls it

0celo7

This has to be the stupidest conversation we've had here

Jim

It's on the list for sure. But I bet we've had stupider

It's in the stupid conversations that you can really learn who a person truly is

0celo7

You're a Canadian supremacist

Jim

Saying we look more like a shirt or a torso than a hat makes me a supremacist?

0celo7

Yes, I bet you want a war too

And you think you'd win

Jim

17:47

I don't think Canadians have ever actually wanted a war. But I know that, unlike how America claims, we've never actually lost a war. We are too young to be in many and we tend to side with the winners. So, no I don't want one, but yes, I think we and our allies would win

Hippalectryon

2

Jim

^ that

Slereah

How about the war of 1812 :p

Jim

exactly, Canada won that

0celo7

The U.S. Was not a superpower back then

It's hardly comparable

Jim

17:52

comparable to what?

we aren't comparing anything

0celo7

There's no way Canadia wins

Jim

wins what?

0celo7

Now the speed of the destructions depends on who wins in 2016

Shillary and Bernie are incompetent

Now if we get Trump, Cruz or Rubio, watch the fuck out

They'd burn your maple reserves and rek your economy in a heart beat

Jim

No they wouldn't

We already have our entire population lined up along the border

0celo7

Nice and easy to shoot at

Bombers don't have to travel far

Jim

17:58

besides, invading Canada is like invading Russia. You can't win. Different reasons why, but same result

0celo7

I guarantee there's a strategist somewhere in the Pentagon who has a file "Invasion Plan for Canadia" on his desk

100%

Hippalectryon

Why do you keep saying Canad i a ?

0celo7

This isn't 1830/1940

We have cold weather gear

Jim

Every other country would be like "You're invading Canada? What did they ever do to you? Pick on someone your own size!" And then come to help us and it'd be WW5

0celo7

when were the other two

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