The h Bar

10:03 PM

Good evening

Hey since we have strong enough microscopes these days to 'see' atoms (though the images are recreated using something like electrons), can we see a wavelength of light?

@acuriousmind it's a great evening actually. Soo beautiful outside

i can't imagine what it's like in germany..

ACuriousMind

It's cold and cloudy here

Obliv

well on a good day anyway

ACuriousMind

@Obliv What do you mean "see a wavelength of light"?

@Obliv I don't think good days in Germany are any different from good days in any other temperate climate

There are some pretty regions, though

0celo7

@ACuriousMind Wow, I really thought you had died.

@ACuriousMind I finally understand why the action on a G bundle is on the right.

ACuriousMind

10:08 PM

I'm simply busy, my schedule will normalize next week

0celo7

@ACuriousMind Do you want to hear about my insight?

@ACuriousMind I take that as a no.

Oh well.

ACuriousMind

If it is the thing about the fundamental vector fields and the anti-homomorphism, you already typed out once, no need to repeat it. I had the impression that it's simply a convention, too: You can fix it to be a homomorphism by switching the definition of the Lie bracket of vector fields

@0celo7 Let me type, dammit :P

Obliv

@acuriousmind actually I take it back. I have to keep thinking about light for a while then I'll ask some questions

0celo7

@ACuriousMind You don't want to switch the definition of Lie brackets of vector fields

Although that's a convention too

You'd flip so many signs

ACuriousMind

Sure, but it's a convention. I'm 99% that flipping the sign doesn't destroy anything.

0celo7

10:14 PM

Sure

Let's look though.

@ACuriousMind I think you have to flip all the signs in Morse theory.

ACuriousMind

You could also flip the Lie bracket on the algebra

0celo7

The basic object in Morse theory is a Lie bracket.

(The Hessian)

@ACuriousMind Ok, well it's hard to know if you've seen my rants if you don't aknowledge them :P

acknowledge

I can't spell that word to save my life

@ACuriousMind I say it was a great insight because it's said incorrectly in many places.

Wiki, for instance.

ACuriousMind

@0celo7 I don't think e.g. the MO thread I already linked to you when you first asked this question way back is wrong. You've just found another thing where it would conflict with a traditional convention.

0celo7

@ACuriousMind No, I mean the homomorphism thing.

ACuriousMind

Oh, what does wiki state?

0celo7

10:21 PM

Fundamental vector field

In the study of mathematics and especially differential geometry, fundamental vector fields are an instrument that describes the infinitesimal behaviour of a smooth Lie group action on a smooth manifold. Such vector fields find important applications in the study of Lie theory, symplectic geometry, and the study of Hamiltonian group actions. == Motivation == Important to applications in mathematics and physics is the notion of a flow on a manifold. In particular, if M {\displaystyle M} is a smooth manifold and X ...

"The map {\displaystyle {\mathfrak {g}}\to \Gamma (TM),X\mapsto X^{\#}} {\mathfrak g}\to \Gamma (TM),X\mapsto X^{\#} can then be shown to be a Lie algebra homomorphism.[5]"

Problem: their actions are on the left.

With the standard convention for brackets, it's only right actions that give homomorphisms.

ACuriousMind

Well, have you checked that the citation [5] does use conventions such that the statement is wrong ;)

0celo7

@ACuriousMind Also, Wolf's Spaces of Constant Curvature leaves it as an exercise to verify that the left action leads to a homomorphism. So of course little 0celo7 is very confused he keeps getting the wrong sign.

ACuriousMind

I'm willing to believe that many people might have gotten it wrong, though

0celo7

Bishop-Crittenden do the same thing.

user218912

@0celo7 why are you calling yourself little 0celo7?

0celo7

10:24 PM

@ACuriousMind No, I will do that next time I am bored.

Because I am little?

@ACuriousMind The one book that actually gets it right is the category theoretical book by Kolar et al.

Which I own for unknown reasons.

They specifically state you get an antihomomorphism for left actions.

ACuriousMind

Heh

0celo7

Truth be told, that book is a superb reference for basic geometry.

oops I don't remember the formula for trapezoidal area

Obliv

@acuriousmind do you know what a matter field is

user218912

any good animes right now?

0celo7

@IceLord How do you have time to watch animes

ACuriousMind

10:33 PM

@Obliv It's a quantum field whose associated particles make up matter (e.g. electron fields, quark fields,...).

user218912

@0celo7 Idk

Obliv

@acuriousmind how do you define matter

ACuriousMind

@Obliv Stuff we can touch.

More generally, it might be used for any fermion field.

Obliv

@0celo7 it's not like scholars don't have hobbies. probably a lot less time for them depending on the person though

0celo7

My hobby is scholaring.

Obliv

10:35 PM

@acuriousmind but isn't the form of touch just an interaction through an EM field

0celo7

He was joking about that.

heather

@0celo7, trapezoidal area is $\frac{b_1 + b_2}{2}\times h$

ACuriousMind

@Obliv Ultimately yes. What's your point?

Obliv

@iceL what do you wanna watch I have seen some stuff u might enjoy and havent watched

heather

Could anyone help answer this question?

user218912

10:37 PM

@Obliv idm

user218912

the best anime i've seen is terror in resonance if you want an idea of what I like.

Obliv

@acuriousmind well, that's where I'm confused. If everything interacts through the EM field (everything observable), then why do we consider things 'neutrally' charged

@iceL wow my friend once showed me that a while ago but we only saw 2 eps together. he really liked it. Okay, this is not similar but reaaally good. have you seen steins gate;?

ACuriousMind

@heather I don't know the first thing about quantum computing, but isn't a gate ultimately just a linear operator applied to the input state? What about applying that operator to your different inputs can you not do?

user218912

@Obliv yes like 3 years ago xD

ACuriousMind

@Obliv Because something that contains an equal amount of negative and positive charges is overall neutral? I don't understand what confuses you.

Obliv

10:40 PM

wait no

ACuriousMind

Neutrinos are definitely not "stuff we can touch" :P

user218912

@ACuriousMind do you ever understand what confuses someone? lol

0celo7

@ACuriousMind Let $(X,d)$ be a $\sigma$-compact metric space, $\mathscr F\subset C_E(X)$ a family of continuous functions with $E$ Banach. If $\mathscr F$ is equicontinuous at each point of $X$ and $\mathscr F(x)$ is bounded in $E$ for each $x\in X$, then any sequence of functions in $\mathscr F$ has a subsequence converging uniformly on compact sets.

HELP

It's like A-A but WORSE

ACuriousMind

I can't help with that

0celo7

you took a functional analysis class, what do you mean?

heather

10:41 PM

@ACuriousMind, I am unsure how to incorporate the control qubit into the calculation, if that makes sense - with a normal NOT gate, I just multiply the matrix and the input vector, which is easy even if it is a superposition, but I can't just do that with the extra qubit.

ACuriousMind

@IceLord Yes - everytime I answer a question not with a counter-question, but with an answer :P

user218912

@ACuriousMind hmm

ACuriousMind

@0celo7 I never saw a "$\sigma$-compact space" as far as I recall, for one.

0celo7

@ACuriousMind Think manifolds

user218912

@Obliv so do you have any recommendations?

0celo7

10:42 PM

If you want A-A on manifolds this is what you have to do

ACuriousMind

I didn't do functional analyisis on maniflds

Obliv

@acuriousmind Well, okay. How do we detect neutrinos, then? They interact with 'ordinary' matter through the weak field, right?

0celo7

ahhh

Obliv

@iceL I have seen a lot of good stuff but nothing similar to zankyou comes to mind right now. I'll think about it at home

0celo7

The hint is: use a diagonal argument

heather

10:43 PM

@Obliv, en.wikipedia.org/wiki/Neutrino_detector

Obliv

or if i do think of something i'll let you know

heather

might assist; it has information about neutrino detectors

0celo7

@ACuriousMind Ok but can you help with this? "If $(f_n)$ is a sequence in $C_E^b(X)$ and $f_n\to f_0$ uniformly on $X$, then $\mathscr F=\{f_0,f_1,\dotsc,f_n,\dotsc\}$ has the property that $\mathscr F(X)$ is precompact in $E$."

I think the claim is wrong.

ACuriousMind

@heather It appears to me that the matrix for the CNOT is explicitly given at Wikipedia, for example. It's not clear from your question what exactly the problem in applying it is.

0celo7

For take $X$ to be the closed unit ball of $E$, which is noncompact if $E$ is infinite-dim. Then take $f_n=\mathrm{id}_X$, so $\mathscr F=\{\mathrm {id}_X\}$, and $\mathscr F(X)=\mathrm {id}_X(X)=X$, which is closed but not compact, hence not precompact.

ACuriousMind

10:45 PM

@Obliv Yeah, and you detect them by that weak interaction (e.g. recall that they are emitted in beta-decay, so in principle the inverse reaction can happen)

heather

@ACuriousMind, yes, I have the matrix for the cNOT, but I am unclear in how to apply it. Normally, you simply multiply the matrix by the input qubit (represented by a vector or ket/bra notation) but in this case you have two vectors for two qubits, the control qubit and target qubit, and I don't know how to solve the problem in this case. If it is that unclear, I can edit to clarify that this is my problem and remove the specific examples...

ACuriousMind

@heather You don't have two input vector. You have one four-dimensional input vector $a\lvert 00\rangle + b\lvert 01\rangle + c\lvert 10\rangle + d\lvert 11\rangle$ to which you apply the 4D matrix given there.

heather

@ACuriousMind, oh, duh...why didn't I realize that?! Wow, that...okay. If you post that as the answer, I would accept it. Thanks so much!

ACuriousMind

@0celo7 That appears to be a correct counterexample

Obliv

@acuriousmind my problem is that , if I think of EM waves as the result of a breaking of symmetry, causing a change in the EM field, of charged particles, then shouldn't the weak force be caused by a breaking of symmetry of particles that have the properties necessary to affect the weak field?

And where does gravity come into play in this picture?

ACuriousMind

10:53 PM

@Obliv I don't know what you mean by EM waves being a result of "breaking of symmetry". Symmetry breaking has a very specific meaning in physics and this isn't it.

Obliv

@acuriousmind probably the wrong word. It's due to a charged particle accelerating, though, right?

I thought that had something to do with the symmetry of the vacuum (see above the very first messages I exchanged with rob)

ACuriousMind

@Obliv Yes, but that's a classical picture. We were talking about "interactions" and fields before.

Obliv

yeah and aren't interactions due to the acceleration of particles that exist in the field

ACuriousMind

I think you are confused because you are mixing classical and quantum field theoretic concepts freely, which will not lead to any meaningful statements about physics.

@Obliv No. The acceleration of particles does not typically feature in quantum field theoretic interactions.

Obliv

@acuriousmind The EM field interacts through the acceleration of charged particles. This is not in general to all fields?

ACuriousMind

11:01 PM

@Obliv Classically, an electromagnetic wave is emitted from an accelerated particle. But certainly, even two static charges "interact" by their electrostatic fields - they repel or attract. Even in the absence of acceleration you have electromagnetism.

I don't know where you got the idea that acceleration is somehow fundamental to interaction.

Obliv

everything in my brain is wrong

0celo7

@ACuriousMind I'm answering linear algebra questions on math chat

ACuriousMind

@0celo7 oO Who are you and what have you done to 0celo7? ;)

Is this the coming of 0celo8?

0celo7

no

0cel8

Obliv

ARGH I'm missing the point.

@acuriousmind Okay, so electrostatic fields may be considered as coherent states of the superposition of many travelling EM waves. In this way, acceleration is necessary for the interaction of particles in the EM field, right?

Danu

11:15 PM

> physics in the h bar

THE DAWN OF A NEW ERA

3

Obliv

to produce** the EM waves.

ACuriousMind

@Obliv I don't know what you mean by "necessary". Classically, an "interaction" is exerting a force on each other, which is the same as exerting accelerations, so that sounds like a tautology ("interaction is necessary for interaction"). Quantumly, an "interaction" is a highly complex time evolution where stuff goes in and other (or the same) stuff goes out and the in-between eludes our ability to describe it in natural language.

If you're saying that because I may represent a static electric field as the superposition of waves then that's not as deep as it sounds because you can Fourier transform most functions and therefore represent them as a "superposition of waves". That doesn't mean that those waves are a useful way to think about the situation.

Obliv

Well, an EM wave can be created by accelerating a charged particle. In this way, acceleration of the charged particle is required to produce an EM wave (which I considered an interaction with the EM field and the particle, I suppose this was a misuse of the word?). Then static electric fields exist in QFT? @acuriousmind

ACuriousMind

11:32 PM

Asking whether "static electric fields exist in QFT" is not as meaningful a question as it may sound. The electric field is an operator in QFT. What does it mean for it to be "static"?

Obliv

electric fields due to stationary charged particles ***

ACuriousMind

There are certainly states where the expectation value of the electric field does not fluctuate with time. But by now, I've totally lost track what question we are actually trying to resolve

Obliv

non-accelerating

***

ACuriousMind

"stationary particle" is a horrible thing to say because localizing a relativistic QFT particle is...difficult. You can derive Coulomb's law for the static electric field from QFT though.

Obliv

I'm trying to resolve this picture of what light is. If it's a wave that propagates through space and carries energy just as a mechanical wave would, I would like to know what its properties are as well as the field associated with it.

Slereah

11:34 PM

@ACuriousMind Commutes with the hamiltonian?

0celo7

rekt

Obliv

rob told me "The electromagnetic fields (and, while we're at it, the matter fields) are properties of the vacuum that arise when you demand that the vacuum be unchange under translations, rotations, and boosts." and we discussed more about it

so I thought EM waves were just disturbances in some field in which certain particles (charged) existed in

and they interacted with them (classical use of the word). They were then created due to an acceleration of charged particles. It made so much sense but now I don't know what to think. @acuriousmind

ACuriousMind

@Obliv Well, physics doesn't have a single, unified picture of things. In some situations, classical electromagnetism is useful, in some situations, the particle picture of photons is useful, in yet other situations, the picture of quantum electrodynamics is useful. All of these are models with a certain validity, and certain relationships to each other, there's nothing to "resolve".

@Obliv Classically, an EM wave is a disturbance created by charged particles accelerating. That's true. But you keep moving the goalposts: Sometimes you talk about EM waves, which are classical, then you talk about the weak force, which is fully quantum, then you talk about the EM field, which is much more general than a wave. I'm terribly confused as to what you want to actually know.

Obliv

@acuriousmind I'm not asking for a perfect model. I just want to know what causes them and what they do and what do they do it with

How could classical & quantum be that different that they give completely different ideas of what they are?

user218912

I changed my name to bloo (blue).

user218912

11:48 PM

(icelord btw)

ACuriousMind

How could the Bohr model and full quantum mechanics be that different that they give completely different ideas of what an electron does in a nucleus?

How could Newtonian gravity and general relativity be that different that they give completely different ideas of what very massive bodies do?

How could Newtonian mechanics and special relativity be that different that they give completely different ideas of whether or not there is a fastest speed?

Obliv

but in those cases there is a more experimentally accurate model

user218912

@0celo7 I'm gonna learn GR from carroll, starting right now.

0celo7

who the hell are you

user218912

fuck do we have to go through this every time I change my name?

0celo7

11:50 PM

flagging

ACuriousMind

@Obliv And yet special relativity won't help you understand a collision at ordinary speeds, and general relativity won't help you understand the Kepler problem

user218912

@0celo7 seriously?

0celo7

@ACuriousMind Of course GR will

GR helps with life in general

ACuriousMind

The "less accurate" models are often much simpler, and are tractable in situations where to "more experimentally accurate" models get hopelessly overcomplicated

If you want to be strict, then QED is of course "more accurate" than classical electrodynamics. That doesn't mean that QED helps you understand a capacitor, or a large antenna.

Obliv

@acuriousmind The thing is, light is massless in both classical & quantum right? They're both travelling at the same speed through empty space. (I think) they're both considered a particle and a wave. I just want to know what causes this wave. I want to also know what it interacts with. Are there different answers to these questions based on what model I'm using?

@bloo is that your favorite shade of blue

user218912

11:56 PM

@Obliv no.

user218912

my favourite shade of blue depends on how I feel.

user218912

it always changes.

user218912

for instance it already changed xD

Obliv

i thought I was going crazy lol

NeuroFuzzy

@bloo >:| stop messing with my brain!

Slereah

11:58 PM

you want to know something even worse?

We never measure real photons

All we ever measure are virtual photons

Since any photon that is detected is in the end just between two charge vertex

For very long range they get pretty close to on shell, tho

ACuriousMind

@Obliv It doesn't make sense to ask for the "mass" of light in classical electrodynamics, imo. It's a wave of electric and magnetic fields. What is the "mass" of a configuration of fields? Light is a wave in classical electrodynamics, and a certain quantum state in quantum mechanics that sometimes behaves more like classical particles than classical waves. It's not that simple.

@Slereah Don't make me go on another rant about people taking Feynman diagrams too literally

user218912

@Obliv I feel like I become more obsessed with blue as every day passes...

Slereah

@ACuriousMind Just integrate the momentum density and take its square?

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