The h Bar

Mike Miller

16:01

"Is the path integral formulation of QM just a mathematical tool?" Well, it's not a mathematical tool, so... :)

(Isthere a way to shut that feed off?)

ACuriousMind

@Secret You have a strange conception of "no physical meaning". The wavefunction/quantum state is not measureable (and that is the sense in which it has "no physical meaning"), but it encodes everything you can possibly know about the state of a quantum object. So...you're surprised that you can get things with "physical meaning" from the thing that is explicitiy supposed to contain all relevant information?!

@MikeMiller Hehe

@MikeMiller Nope

Mike Miller

Drat.

Secret

(I felt I am risk falling into quantum interpretation hence philosopy discussion...)

Sometimes I felt really frustrated that despite we get a lot of useful results from quantum mechanics (and how the maths worked so well), I still yet to understand what, fundamentally, is a state (other than this "stuff" has the property that it can superimpose like a wave)

But ... well... .....

IMO, even electric fields are more stragithforward to wrap one's head around than this

this is really the core concept I am always confused by quantum mechanics

When I look at the frank condon factor $\langle\psi'\lvert\psi\rangle$ and the transition moment integral $\langle\psi'\lvert\mu\rvert\psi\rangle$in quantum chemistry (and dig through every step of the derivation)

, I always thought that I am starting to get quantum mechanics, it's all about how much the waveufnctions between states overlap each other, which is what controls the probability of a transition, but the above discussion reveals that my thinking is all wrong and I just have to accept that I cannot interpret it this way

typo, corrected version: it's all about how much the waveufnctions between states overlap each other, and the role of measurements (caused by physical things the environment such as electric dipole moment) perturb the wavefunction thus changing the overlap hence changing the probability of the transition (and by extension, possible photochemical pathways)

but since a wavefunction is not measurable, this is not the correct way to understand it, and I am left with putting the quantum states in some percuiler way and then integrating it somehow gives me the probability of transition

Secret

16:21

It always feels like there's a conceptual gap that prevent me from understanding quantum mechanics, unlike when I learn general relativity where most outcomes can be explained concisely in a short paragraph

or even in classical mechanics, thermodynamics, statistical mechanics and electromagnetism

DanielSank

Is there some way I can just give some rep to another user?

@Huy helped me understand something this morning and I'd like to "say thanks".

0celo7

Did you finally get a result for the delta thing

DanielSank

@ACuriousMind If you want, that's really what I'm trying to figure out.

0celo7

You should just put it here for future reference

DanielSank

@0celo7 No, still working on it.

@ACuriousMind This whole thing started with the following question: Suppose we have a probability distribution $P(x,y) = (1 / 2\pi r)\delta(x^2 + y^2 - r^2)$.

What is the marginal distribution for just $x$?

This problem can be done by parametrizing the distribution by $\theta$ instead of $x$ and $y$, but I really would like to understand how to do it "without thinking".

In other words, I want to get enough of a handle on the math that I can just sort of write down the answer.

Secret

16:26

@ACuriousMind
To me, entanglement, tunneling, supoerposition and various other things that people often quoted as counterintuitive in quantum mechanics is pretty straightforward once you see the maths (because it is basically linear algebra)

The only really nonintuitive thing to me in quantum mechanics, is that in some sense, I felt I don't really understand what actually is a quantum state, other than it can superimpose like a wave, and is nonlocal like a field

and that is where my major confusion about quantum mechanics lies

DanielSank

Anyway, the meaning of $\delta(x^2 + y^2 - r^2)$ pretty clearly means that the probability density's support is a 1D manifold sitting in the $xy$ plane.

@Secret Here's how I think of it: a quantum state is just a thing which represents the information available relative to a particular subset of the universe.

A common subset is "the experimenter".

@0celo7 I already posted it on Math.SE

0

Q: Integrate $\delta(f(x))$, where $f:\mathbb{R}^N\rightarrow \mathbb{R}$, over one variable

Let $f: \mathbb{R}^N \rightarrow \mathbb{R}$ such that the solutions of the equation $f(x)=0$ form an $N-1$ dimensional manifold $M$. Consider a probability distribution $P: \mathbb{R}^N \rightarrow \mathbb{R}$ which is uniformly distributed over $M$. We could write this as $P(x) = \delta(f(x))$ ...

Someone actually down-voted it!

ACuriousMind

@DanielSank Yep. But that heavily relies on $0$ being a regular point of $f(x,y) = x^2+y^2-r^2$. So I don't think you can hope to write down the generic $\delta(f(x,y))$. What you're is having some $N\subset M$ and then extending a density on $N$ to a density on $M$ by just saying "whenever integrating over $S\subset M$, just go to the original density on $N$ and integrate over $N\cap S$.

DanielSank

I've been getting a "lot" of down votes lately...

ACuriousMind

I'm not convinced writing this as a $\delta$ is really the "proper" way to think about it.

I'm not convinced it isn't either :D

DanielSank

@ACuriousMind Give me a better one.

@ACuriousMind First of all, I really don't care about smoothness assumptions. I can usually tell when something is irregular enough that I have to think twice before using formulae derived under heavy assumptions.

So regarding the regularness issue: I don't care :-D

Huy

16:33

@DanielSank np

DanielSank

@ACuriousMind Have you heard of the coarea formula?

It seems deeply related.

ACuriousMind

16:45

@DanielSank Okay, lets say you want that $\delta(f(x))$ is such that $\int_M g(x)\delta(f(x)) = \int_{f^{-1}(0)}g(x)$. Now, $\int_{f^{-1}(0)}g(x) = \int_\mathbb{R} \left(\int_{f^{-1}(t)}g(x) \right)\delta(t)\mathrm{d}t$. But now you can't apply the coarea formula because there's that $\delta(t)$ inside the integral.

DanielSank

What's $M$?

ACuriousMind

@DanielSank Whatever the space is that $f$ is defined on, it's the $\Omega$ from the coarea article.

DanielSank

Ok, let's write $\Omega$ to keep things straight.

Are we assuming $f:\Omega \rightarrow \mathbb{R}$?

ACuriousMind

Yes

And now you want that integrating against $\Omega$ evaluates functions on the set of $x$ with $f(x) = 0$, right?

DanielSank

There's something weird here.

What's the domain of $g$?

Oh, I see... $g:\Omega \rightarrow \mathbb{R}$ as well.

ACuriousMind

16:50

@DanielSank Also $M$. The "weird" part is that I suppressed the measures, the integral $\int_{f^{-1}(0)}$ is with that Hausdorff measure.

Hm...let's compare this to the case in zero dimensions

DanielSank

"Hausdorff measure" $\rightarrow$ "Volume element"?

ACuriousMind

Then my formula there gives $\int_\mathbb{R} g(x)\delta(f(x)) = \sum_{f(x) = 0} g(x)$.

DanielSank

Which is wrong.

There needs to be a $1/\left \lvert f'(x) \right \rvert$.

That's actually pretty easy to prove.

ACuriousMind

@DanielSank Almost, it is normalized differently

@DanielSank Yes, so that's not the definition of $\delta(f(x))$ we want.

DanielSank

@ACuriousMind Correct.

@ACuriousMind Oh.

Earlier today @Huy helped me remember something I had forgotten: If we have a linear map $T: \mathbb{R}^k \rightarrow \mathbb{R}^n$ with $n>k$, then the volume of the $k$-dimensional parallelogram you get by acting $T$ on the set of unit vectors in $\mathbb{R}^k$ is $\sqrt{\text{det}(T^t T)}$. I think this will help us (because I think that's how you prove the coarea formula).

Huy

16:56

that's an unfortunate name for the linear map

DanielSank

@Huy Why?

$T = \text{transform}$ in my little brain.

Huy

because $T^T T$

ACuriousMind

Well...the "easiest" thing to do here would be to exhibit $\delta(f(x))$ as the derivative/Jacobian of another "function", then we could apply the coarea formula directly :D

DanielSank

@ACuriousMind ...go on...

Mike Miller

@Huy of course this just means $e^{T \ln T} T$

Huy

16:58

that looks like a smiley face

DanielSank

@ACuriousMind perhaps this chat post by Daniel Fischer will help.

Huy

$T \ln T$

DanielSank

I am having trouble comprehending it.

Mike Miller

not seeing it

crying eyes and a long nose or something?

Huy

yea

OK looks like a person

Mike Miller

16:59

i think that's a stretch.

Huy

.________________.

DanielSank

^Yarrrrr the white whale!

@ACuriousMind Part of my problem is that I don't know how to relate what math people call $d\sigma(x)$ to something I can compute.

0celo7

What is this coarea formula

DanielSank

Coarea formula

In the mathematical field of geometric measure theory, the coarea formula expresses the integral of a function over an open set in Euclidean space in terms of the integral of the level sets of another function. A special case is Fubini's theorem, which says under suitable hypotheses that the integral of a function over the region enclosed by a rectangular box can be written as the iterated integral over the level sets of the coordinate functions. Another special case is integration in spherical coordinates, in which the integral of a function on Rn is related to the integral of the function over...

0celo7

I've never heard of it before today

DanielSank

17:03

Neither had I.

0celo7

@DanielSank I googled it earlier.

DanielSank

I think it's easier to understand if written $\int_\Omega f(x) dx = \int_{-\infty}^\infty \left( \int_{u^{-1}(t)} f(x)/|\nabla u(x)| dH_{n-1} \right) dt$.

Here $u:\Omega \rightarrow \mathbb{R}$.

and same for $f$.

$\Omega \subset \mathbb{R}^n$.

0celo7

dH?

DanielSank

so $u^{-1}(t)$ is a manifold (level set) of dimension $n-1$. $dH_{n-1}$ is some kind of measure on those $n-1$ dimensional manifolds.

The problem for me is that I don't know exactly what $dH_{n-1}$ means.

0celo7

I own no books that mention it. I have no books on my iPad that mention it.

DanielSank

17:08

@0celo7 unfortunately this is what Wikipedia writes.

It's the Hausdorff measure, whatever that is

0celo7

Maybe I need a book on geometric measure theory. After I take measure theory. After I take analysis.

DanielSank

Well yeah, or we can hope that @ACuriousMind knows wtf is going on.

0celo7

He probably does.

@DanielSank If you're doing QM for a living why don't you learn some measure theory, etc. so you can make use of mathy results like this?

DanielSank

@0celo7 What does any of this have to do with quantum mechanics?

I came to this question by trying to solve problems in a stochastic processes textbook.

0celo7

@DanielSank Nothing. I'm just surprised you don't know any measure theory.

DanielSank

17:17

I'm an experimentalist, remember?

I specialize in screwdrivers.

Secret

Hmm... if interpreted it that way.

Suppose we have a molecule in some state $\lvert\psi\rangle$

Then

$\lvert\psi\rangle$ is all the information of the molecule available relative to its immediate environment

Now an external electric field $E$ result in charge separation in the molecule

Thus

$\hat{\mu}\lvert\psi\rangle$ redistribute the information available relative to the immediate environment (by filtering out some information already present or supplying new information from the electric field (thus information is conserved)) so that a new (possibly nonlocal) pattern of information…

Sounds reasonable to me...?

typo: electronic should be rovibronic

DanielSank

o_O

What's $\hat{\mu}$?

Secret

transition dipole moment operator

in the above interpretation, it is a particular way on how information is rearrange and redistributed by the electric field

In the standard (mainstream) understanding, the transition dipole operator is tied to the dipole moment of the molecule, which then governs the likinhood of a transition as it emit or absorb a photon

vzn

17:40

@Secret new research asserts the wavefn is real, by a leading australian qm lab, but it is not yet disseminated into the widespread community, more details in my blog

FenderLesPaul

Isn't there "new research" like that every other month?

and it always get debunked somehow

vzn

17:58

@FenderLesPaul you younguns think science is just what is in textbooks and what was devised/ discovered decades ago. there is new science being built right now...

FenderLesPaul

I don't think I think that...

vzn

Secret, some of your confusion/ perplexity might be addressed in seeing science/ qm as (admittedly against conventional wisdom) still a work in progress, and surely you know your own questions cut to the heart of classic ongoing questions about QM...

agreed it is tricky to separate signal from noise in new claims/ research, but thats exactly the nature of science... agreed there are a lot of flimflam claims in physics but it is possibly/ presumably not unique among science branches in this regard.

no_choice99

18:24

hi people I've got a question about black holes and planets and stuff like that. IS the astronomy or physics website the better suited?

no_choice99

18:40

nevermind I posted in the astronomy website (although it's beta)

DanielSank

18:58

@vzn Until someone shows something that the standard qm theory doesn't predict, it's all just a matter of interpretation.

Please summarize this new theory about the wave function being real, whatever that means.

Slereah

19:25

Guess who got a copy of Grimm and Jaffar for Christmas B)

Crom and Jaffa Cakes

Glib and Jar

Grub and Jello

Groot and John

Gringo and Knave

0celo7

19:48

@Slereah wtf who gives you math books for christmas

also, do you even know enough FA to begin reading that

FenderLesPaul

I didn't get shit

stomps floor in defiance

0celo7

@FenderLesPaul I'm getting an AMS membership

that's about it

FenderLesPaul

What's AMS

0celo7

Amer. Math. Soc.

I guarantee you've seen Bull. Amer. Math. Soc. in some bibliography somewhere

@FenderLesPaul do you still have Bredon?

FenderLesPaul

I do back at Cornell

0celo7

20:00

do you like it

Slereah

@0celo7 Well unlike you I have a real degree :V

Also I asked for it

0celo7

@Slereah you're also 10 years older...

Slereah

respect your elders you little scalliwag

0celo7

@Slereah uh, no

FenderLesPaul

I like Bredon yes

0celo7

20:09

@FenderLesPaul am I insane or is there no MyCopy available for it

FenderLesPaul

There is no MyCopy for it unfortunately

Nick

Happy Isaac Newton's Birthday. May all you physics people uncover the Grand Theory of Everything. Good luck, folks.

0celo7

@FenderLesPaul what's some crazy GR thing that no one knows about and like 1 person does

that uses some really obscure math

@Nick I already know it: GR.

Slereah

@0celo7 : General relativity with non-metric connection

0celo7

why would you do that

literally all of HE has to be thrown out then :/

Nick

20:14

@0celo7 Sprinkle some QM on your GR. Not tasty.

0celo7

dude literally no GR thing works any more

@Nick QM! Lol!

Slereah

Yeah pretty much

IIRC the source of the non-metricity is called

HYPER

MOMENTUUUM

0celo7

Proof?

FenderLesPaul

@0celo7 no idea

Slereah

that's not a theorem, that's a name

FenderLesPaul

20:16

Yau's proof of positive energy theorem maybe

because Witten's proof is so much easier

I'm guessing no one bothered to waste time on Yau's proof

0celo7

@FenderLesPaul oh fuck that man

Slereah

Metric-affine gravitation theory

In comparison with General Relativity, dynamic variables of metric-affine gravitation theory are both a pseudo-Riemannian metric and a general linear connection on a world manifold . Metric-affine gravitation theory has been suggested as a natural generalization of Einstein–Cartan theory of gravity with torsion where a linear connection obeys the condition that a covariant derivative of a metric equals zero. Metric-affine gravitation theory straightforwardly comes from gauge gravitation theory where a general linear connection plays the role of a gauge field. Let be the tangent bundle over...

0celo7

@FenderLesPaul So much? Fucking smart people...

Slereah

^this thing

0celo7

they're both equally terrible

Schoen-Yau's just take 150 pages

Nick

20:18

Alternatives to general relativity

Alternatives to general relativity are physical theories that attempt to describe the phenomena of gravitation in competition to Einstein's theory of general relativity. There have been many different attempts at constructing an ideal theory of gravity. These attempts can be split into four broad categories: Straightforward alternatives to general relativity (GR), such as the Cartan, Brans–Dicke and Rosen bimetric theories. Those that attempt to construct a quantized gravity theory such as loop quantum gravity. Those that attempt to unify gravity and other forces such as Kaluza–Klein. Those that...

There are just so many alternatives.

FenderLesPaul

@0celo7 relatively easier I should have said

not that it's easy in and of itself

FenderLesPaul

20:50

@0celo7 I told Wald I would love to work for him in grad school

he was like "kay"

0537

lol

Slereah

21:12

the correct theory is gnomes

Gnomes cause gravity

0537

lol you know pse still doesn't have the new profile design.

kevin Tah N.

It's Christmas eve guys!

Slereah

in what reference frame

kevin Tah N.

We should consider throwing a shindig on here

nice one @Slereah

lol

Slereah

A hootenanny

A hoedown

kevin Tah N.

21:34

lolz

Valentin Tihomirov

21:52

Why does law of Malus say that polarizers at 45 degree cause attenuation of 4x hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polcross.html#c2? Isn't Cos^2(45) = 1/2 rather than 1/4?

Valentin Tihomirov

23:23

It is some Bohm quantum mechanics says that at 30, 60, 120 degrees you will have half attenuation. I wonder, how is that possible that as you monotonically reduce correlation from 1 to 0, you will have the same value at 30 and 60 degrees?

0celo7

@kevinTahN. Proof?

Kaien Yang

23:41

hi everyone

is anyone on?

FenderLesPaul

no

Kaien Yang

darn

FenderLesPaul

sup?

Kaien Yang

hi

wait darn I'm a 7th grader

FenderLesPaul

I'm a Bolshevik.

Kaien Yang

23:44

bye

gtg

FenderLesPaul

bye

0537

what

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