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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)
Once interaction starts, there is no more notion of particle states
makes sense
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 $\sum f_k a_k + \bar f_k a^\dagger_k$
Fock space smock space
Jim
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.
15:03
But you can't sum solutions if it's non-linear, though
The only number operator I need is my set fingers
so in the Fock space, the field essentially become a linear combination of the normal modes of harmonic oscillators?
basically, yes
It's a practical math thing because it's easy to solve
I recall in today's spring school the professors says QFT is basically modelling things with an uncountably infinite number of harmonic oscillators
kinda yeah
that is what the $k$ is
it's a real index
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)?
@Secret None
That's why it's not really a useful analogy.
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?
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
@Secret $a$ and $a^\dagger$ are operators. $r$ is not an operator in QFT. You are trying to compare two very different things
that too indeed
wait, position is not an operator?
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
15:28
@Secret No, QFT does not have a naive position operator.
The wavefunction isn't $\Psi(r)$
In non-rel QM, you just have $a = x + \mathrm{i}p$ for the haromonic oscillator, as Slereah says.
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
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?
Basically you do averages over a volume
like $\phi_\Omega = \frac 1 {V(\Omega)} \int_\Omega \phi dx $
Then you recover $[\pi_\Omega, \phi_\Omega] = -i\hbar \frac{1}{V}$
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?
It helps yeah
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
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?
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
Jim
ah, okay then
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
Whenever I see dmcKee, I always think of D B Cooper.
Jim
Jim
why?
The dark glasses.
Jim
Jim
Yeah, I can kinda see it
Once you get that D B Cooper in your head it will never go away.
Jim
Jim
15:55
I think you underestimate my power to be distracted
Wanna talk about inflation and why it's superfluous?
This is worth a read.
@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
Jim
I don't know if it's superfluous. But it is a little fluous
@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)
ELI5 why there is no number operator in an interacting theory
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)
@0celo7 Why should there be? I.e. how would you even begin to define it, lacking the mode expansion of the free fields?
@ACuriousMind count!
@0celo7 Count what?
Particles
Jim
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.
16:07
@0celo7 Define particle in an interacting theory, then
Jim
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
@ACuriousMind little thing that bounces around
It's pretty obvious
@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
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
@0celo7 No one says the experimentalist can't.
16:13
Well the number operator counts
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.
This is why I'm not going into theory. The other reason is hippy hair.
@0celo7 You also should realize that "interacting theory" means in the middle of a high-energy collision, not "stuff lying about in a box".
Put the collision in a box and slow time
Then count
Stop trolling :P
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
"Put the collision in a box and slow time" is not a legitimate physics procedure, is it?
Jim
Jim
@0celo7 hippy hair?
@ACuriousMind not with that attitude
@Jim Context somewhere around this:
20 hours ago, by ACuriousMind
Yep, we're all filthy hippies over here.
Jim
Jim
@ACuriousMind gotcha
@ACuriousMind maybe not now
But in theory
Jim
Jim
16:20
I thought you were staying away from theory?
Well
In theory as in "in the future"
Jim
Jim
Okay, from now on I will replace all usages of "In theory" with "In the future"
Yes
Do that
Jim
Jim
In the future, it could work
Note that ACM has still not answered my question
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
obe
In theory I am finished this problem set. (in the future).
DO THE PROBLEM SET
obe
obe
D:
It's already overdue and I told you how to do it
obe
obe
I know I have to sit down and do it.
16:25
I did them while playing BF4
You should be able to do them now
obe
obe
Right god.
Jim
Jim
All hail the Right God
obe
obe
Who can do QFT problem sets while playing BF4.
Jim
Jim
ghostbusters?
obe
obe
Was that a reference?
Jim
Jim
16:28
no?
The random pauses in my talking where when I was concentrating on not dying
obe
obe
Lol I remember that.
Jim
Jim
While I sympathize with the political plight of Canadian scientists, this user name exhausts and annoys me
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
Jim
@0celo7 are you describing yourself?
16:33
@Jim please explain why that's exhausting and annoying
does anyone have the Steward badge?
Jim
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
What?
How is it going to take CW 97 years then?
Jim
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
@0celo7 That's not hard to figure out. :P
16:36
So he never does close votes?
Jim
Jim
@0celo7 1000 votes. He got 30 in 3 years. 970 to go at a rate of 10 per year
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 $\langle A_k|\hat{A}=A_k \langle A_k|$.
He made it sound like Steward is for Late Answer reviews.
@NeuroFuzzy yes
@0celo7 It's for 1000 reviews of a single type
Jim
Jim
you can get a steward badge in each of the different review categories
16:38
Ok, so he'd take 97 years for doing Late Answer reviews?
But certainly he has other reviews?
@Jim ohhhhhh
Ok
You are seriously overanalyzing that joke :P
Jim
Jim
@ACuriousMind maybe you're underanalyzing it
how would you know unless you analyze it more?
At least I don't overanalyze QFT
just count the little bastards, there's your number operator
Jim
Jim
@0celo7 Count the exact number of jelly beans in this jar without opening it up. That's where the number operator shines
[hit enter on accident] and $\hat{A}|A_k\rangle=A_k | A_k\rangle$. So then if $\hat{B}$ is any operator, $\langle A_k|[A,B] | A_k \rangle=A_k \langle A_k|\hat{B}| A_k\rangle-A_k \langle A_k|\hat{B}| A_k\rangle=0$...
16:41
@Jim easy
shink down to nano size and slip in
then count
Jim
Jim
that's cheating
@NeuroFuzzy That is correct
@Jim that's engineering
The expectation value of a commutator vanishes if the state is an eigenvalue of one of the two observables.
Jim
Jim
but in the future, it could work
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
@ACuriousMind Is there functional analysis weirdness if $\langle p'|[\hat{p},\hat{x}] |p'\rangle=-i \hbar \langle p'|p'\rangle$?
@NeuroFuzzy Yes
OH
okay then
$\hat{H}_{ren}=\hat{H}-\langle 0| \hat{H} | 0\rangle = \int d^3\mathbf{k} \omega_{\mathbf{k}\hat{a}^\dagger(\mathbf{k})\hat{a}(\mathbf{k})+\frac{1}{2}\delta^3(0)-+\frac{1}{2}\delta^3(0)}=\int d^3\mathbf{k} \omega_{\mathbf{k}\hat{a}^\dagger(\mathbf{k})\hat{a}(\mathbf{k})}$

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
Because, technically, the momentum operator's eigenstates do not lie inside the Hilbert space
16:45
Slereah does not know any GR
He doesn't even know what the Kasner metric is
who is expert in GR here?
Jim
Jim
define "expert"
@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.
@Secret ACM
Jim
Jim
@NeuroFuzzy Not I, in that case. I usually try to avoid the big words and focus on the ideas
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
@0celo7 @Secret Just to keep me from a length GR post: He's kidding.
@Jim : I say inflation is superfluous.
Jim
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
@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.
@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
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
jesus
And I am home
how do you type that fast
@Secret matrices?
I like to think this is my (future) math advisor
downvote him
he would probably not ask such an elementary question though
Jim
Jim
17:02
@0celo7 lots of math, but I noticed no EFT. That's one of the most important and math-heavy physics tools
@Jim what
I have no clue what you're talking about :O
Jim
Jim
@0celo7 effective field theory
what about it
Jim
Jim
They guy you linked doesn't have it in his top 4 tags
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
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
Jim
Jim
my mistake
if I make a math double major official, Jochen Denzler will be assigned as my faculty advisor
Jim
Jim
must be the same guy
No way it's a coincidence
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
Huy
17:24
I met the fastest lecturer I've ever had today
I feel violated
Basically, you got a lecture from Sanic ?
Huy
Huy
he was a french guy
I blame you
Huy
Huy
Tristan Rivière
No one likes the French
Jim
Jim
17:32
What about the French?
They've done nothing
@0celo7 :-(
Their greatest accomplishment is Canadia...a country that is literally a hat
Jim
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
Florida is a leg
Mexico is the middle leg ;)
Jim
Jim
17:36
Florida is the middle leg
And that's what everyone in America calls it
This has to be the stupidest conversation we've had here
Jim
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
You're a Canadian supremacist
Jim
Jim
Saying we look more like a shirt or a torso than a hat makes me a supremacist?
Yes, I bet you want a war too
And you think you'd win
Jim
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
Jim
Jim
^ that
How about the war of 1812 :p
Jim
Jim
exactly, Canada won that
The U.S. Was not a superpower back then
It's hardly comparable
Jim
Jim
17:52
comparable to what?
we aren't comparing anything
There's no way Canadia wins
Jim
Jim
wins what?
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
Jim
No they wouldn't
We already have our entire population lined up along the border
Nice and easy to shoot at
Bombers don't have to travel far
Jim
Jim
17:58
besides, invading Canada is like invading Russia. You can't win. Different reasons why, but same result
I guarantee there's a strategist somewhere in the Pentagon who has a file "Invasion Plan for Canadia" on his desk
100%
Why do you keep saying Canad i a ?
This isn't 1830/1940
We have cold weather gear
Jim
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
when were the other two

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