The h Bar

Silly Goose

02:15

does anyone know why ocneanu's rigidity is not published in a paper by ocneanu? it is referenced, e.g., here in reference 16

Slereah

06:33

@imbAF they are both generators, yes

Tobias Fünke

-11

Q: the unsolvable equation I've created?

Imagine you're standing at the edge of the ocean, staring out at the vast, unpredictable waters. You’ve always been told the ocean is calm, structured by natural laws we’ve studied for centuries, controlled by patterns we can predict. But in front of you, something is shifting. There’s a ripple. ...

I finally understand quantum mechanics!

the equation at the very end made me understand it

John Rennie

06:49

-11. That made me wonder what the most downvoted (non-deleted) question is. It turns out if you sort by score -11 isn't even on the first page.

satan 29

07:30

@JohnRennie whats the most no. of downvotes then?

John Rennie

On the site home page you can sort the questions by "Score" (you need to click the More link to get this). Then jump to the last page.

Some of the questions on that page aren't that bad. I think people sometimes get carried away with downvoting.

-28

Q: Do we need Maxwell's Equations since they fail to account for an experimental fact at least in one occasion?

This question is an outgrowth of What is the difference between electric potential, potential difference (PD), voltage and electromotive force (EMF)? , where @sb1 mentioned Faraday's law. However, Faraday's law as part of Maxwell's equations cannot account for the voltage measured between the rim...

electromagnetism maxwell-equations

This seems like a reasonable (if not great) question. I would guess people downvoted because they saw it as a crank post suggesting Maxwell was wrong.

qwerty

people have such knee jerk reactions

makes me disappointed in humans

and we all love jumping on a bandwagon to show we're on the "right" team

PM 2Ring

08:23

I'd agree, but I don't want to jump on the bandwagon.

How about some complex numerology?

-6

Q: i am the universe, wouldn’t you agree?

The imaginary unit i or the square root of negative one is fundamental. Tachyons are the singularities of general relativity where space becomes time. It is the time evolution operator in the Schrödinger, Dirac, and Ehrenfest equations, supports Gödel's incompleteness theorem, and Euler’s Identit...

singularities measurement-problem grand-unification tachyon

Ryder Rude

Carrie is a good horror movie

PM 2Ring

When people talk like they've discovered The Secret Of The Universe it's hard to know if they're merely misguided, or if they have a serious mental health issue.

I've known a few people who had psychotic breakdowns who started talking like that. One of them was a well-educated guy who'd been a friend for several years. He became very charismatic. He even attracted a few disciples...

He got treatment, and made a full recovery. But it was quite surreal while it was happening.

Ryder Rude

@PM2Ring interesting

@PM2Ring yes

qwerty

@PM2Ring I'm glad to hear. it's very sad when it happens

Ryder Rude

@PM2Ring it seems like this person has studied all this stuff but still has this gibberish opinion

maybe they recently had a mental breakdown after having studied it

or maybe they werent sober when they posted it

people shouldn't be giving final answers about the secrets of the universe

i don't think it is wrong to talk about big ideas but it should be done in a speculative manner

PM 2Ring

08:45

@qwerty So when I see stuff online from people who might be starting a psychotic episode, it's frustrating. I wish there were some way to get them to professional help, but I don't know how you do that.

One thing I tried years ago was to say "Have you told your theory to a friend or acquaintance who's been to university? How about your doctor?" But I don't know if that ever worked...

@RyderRude It's impossible to know from just one post. And even with an ongoing dialogue, it's difficult to diagnose mental health issues online, even for trained professionals.

And of course people without significant mental health issues can have weird theories that they strongly defend.

Ryder Rude

reading it again, i sort of relate to the very last sentence of the post, as i too have had that idea before (that the metric field is related to consciousness, as it determines reference frames)

so it seems like there is a genuine train of thought behind this post

but I can't make sense of the rest

@PM2Ring yes

the problem is that we can only see the post. we can't see the train of thought. maybe they didn't bother to write out the complete idea

PM 2Ring

Quantum mysticism can be very alluring. And actually knowing some mathematics & QM doesn't make you immune. ;) Hey, maybe there is some great quantum mystic revelation hiding in plain sight, waiting to be discovered. But I know that trying to find it is a dangerous path...

Even contemplating mathematical infinities can be dangerous. Just ask Cantor and Gödel.

On that note...

Mar 25, 2016 at 19:58, by ACuriousMind

Reminds me of this quote: "Ludwig Boltzmann, who spent much of his life studying statistical mechanics, died in 1906, by his own hand. Paul Ehrenfest, carrying on the work, died similarly in 1933. Now it is our turn to study statistical mechanics. Perhaps it will be wise to approach the subject cautiously."

Ryder Rude

@PM2Ring lol. it depends on how someone approaches it. one can, in principle, be objective and propose concrete models. i recently came across a paper from Chalmers which gave a concrete model

this paper arxiv.org/abs/2105.02314

@PM2Ring yes. thinking about infinities and Godel is more dangerous than thinking about QM

one has to be careful

i think about all this stuff but I am too careful to be committed to any claim :)

Tobias Fünke

09:03

@PM2Ring this. I so often have to suppress my urge to recommend seeking help.

@PM2Ring yes. I also had a friend who then developed serious mental health issues, and started talking like that...

Ryder Rude

if u think about a subject day and night, u might go crazy

cuz the subconscious mind becomes a mess

Tobias Fünke

@PM2Ring but Cantor had mental health issues not because of his theories but because of the response from the (mathematical) society, no? Or am I wrong here?

PM 2Ring

@RyderRude I have 2 main issues with "consciousness causes collapse" theories. 1. We don't know what consciousness is well enough to plug it into a mathematical physics theory. 2. WTF happened before conscious life evolved? No collapse for billions of years? Or do we resort to panpsychism?

Ryder Rude

@PM2Ring Chalmers' idea is a concrete model which addresses these criticisms. i have linked the paper...

PM 2Ring

@TobiasFünke Ok. Cantor got persecuted by segments of the mathematical community, particularly Kronecker & his allies. But I suspect he was already in a vulnerable state, verging on being manic, from intensive work on infinity, combined with the stress of presenting this stuff to a hostile community.

And Gödel was fairly sane for most of his career. He only got weird & paranoid towards the end of his life.

FWIW, one of his last conversations was a brief phone call with mathematician / scifi author Rudy Rucker.

@RyderRude I'll read it later.

Ryder Rude

09:19

@PM2Ring i just like to keep these ideas on my radar. i don't endorse any of these (and Chalmers doesn't either)

the paper just says that this is at least as serious of an idea as the other approaches like MWI, hidden variables, etc

we have barely made any progress on the measurement problem. we shouldn't prematurely get rid of ideas

PM 2Ring

Fair enough. And it can certainly be a great basis for a scifi story. :)

Ryder Rude

yes. maybe Greg Egan has written on this stuff

PM 2Ring

Eg,

Quarantine (Egan novel)

Quarantine is a 1992 hard science fiction novel by Greg Egan. Within a detective fiction framework, the novel explores the consequences of the Copenhagen interpretation of quantum mechanics (or rather of its consciousness causes collapse variant), which Egan acknowledges was chosen more for its entertainment value than for its likelihood of being correct. == Plot summary == The novel is set in the near future (2034–2080), after the Solar System has been surrounded by an impenetrable shield known as the Bubble, presumably by an extraterrestrial civilization for unknown reasons. The Bubble permits...

As a novel, it has a few flaws. But it's still worth reading, IMHO. But bear in mind that Egan is not a big believer in happy endings...

Ryder Rude

> Within a detective fiction framework, the novel explores the consequences of the Copenhagen interpretation of quantum mechanics (or rather of its consciousness causes collapse variant), which Egan acknowledges was chosen more for its entertainment value than for its likelihood of being correct.

@PM2Ring oh

i watched Carrie yesterday and it doesn't have a happy ending either

it is about school bullying and cult

PM 2Ring

09:57

After all that doom & gloom, here's a bluegrass version of Teardrop, from The Lovell Sisters

Teardrop, The Lovell Sisters

►

I can't decide whether I prefer that version, or the Punch Brothers with Sarah Jarosz...

Sarah Jarosz crushes Massive Attack's "Teardrop" with Punch Brothers

►

Something a little more recent: Sierra Hull - "Boom" (On The Beach)

Sierra Hull - "Boom" (On The Beach) - Live in Mexico

►

Sierra Hull - "What Do You Say"

Sierra Hull rippin' fast bluegrass! "What Do You Say" Grey Fox 2023

►

Sierra Hull- Shine

Sierra Hull- Shine (Collective Soul)

►

Sierra Hull & Sarah Jarosz when they were still teenagers. "Old Dangerfield"

Sierra Hull & Sarah Jarosz "Old Dangerfield" at Grey Fox BGF

►

Ryder Rude

10:23

@PM2Ring cute song

@PM2Ring nice

thanks for the recommendations :)

PM 2Ring

Sarah Jarosz: "Jealous Moon"

►

I'll finish off with another one from the Lovell sisters, aka Larkin Poe. "Sledgehammer"

Peter Gabriel "Sledgehammer" (Larkin Poe Cover)

►

PM 2Ring

10:50

Our founder dropped by briefly a few hours ago

in Tavern on the Meta on Meta Stack Exchange Chat, 7 hours ago, by Jeff Atwood

worthless internet points are underrated

Ryder Rude

11:04

@PM2Ring soothing

@PM2Ring nice

Silly Goose

13:47

Finally starting to learn about topological phases :D

perhaps first though i should learn about effective lagrangians and renormalization group

Ryder Rude

15:23

what do u think is the relevance of Godel's first incompleteness theorem in physics

there are theorems which are physical in nature instead of mathematical. e.g. Heinsenberg's uncertainty principle

what if there are physical theorems that we can't prove?

DIRAC1930

@RyderRude I don't think this will make much of a difference since if it agrees with experiment, then that is most likely enough

Ryder Rude

@DIRAC1930 oh

Imagined we lived in a world where Heinsenberg's uncertainty principle is true but we can't prove it

then we can perform an experiment where we measure $\sigma _x \sigma _p$ and it is always less than h bar, but we can never get this result out of our theory @DIRAC1930

DIRAC1930

But we would still have the equation though wouldn't we?

Plus, theres not many things you can prove from first principles in physics

Like how would one prove something like classical mechanics

Ryder Rude

we would have the inequality by doing experiments, but we wouldn't have it from first principles

@DIRAC1930 i think our models are supposed to capture all measurable aspects of a system?

@DIRAC1930 from QM?

DIRAC1930

15:38

Okay yes I suppose so, but then how would one prove QM?

Ryder Rude

i think QM is a set of four postulates. these are assumptions. we r not supposed to prove these four things

but we should be able to get all other measurable aspects of the system using these four principles?

DIRAC1930

Yeah I suppose so

Ryder Rude

right

now, the math we use in physics is very advanced. so incompleteness theorem applies to at least the math aspects of physcs

which means there are theorems about the math of physics that can't be captured using any number of axioms

but I don't think these theorems are of physical relevance

but there is some result about the undecidability of spectral gap. it may be related

Ryder Rude

17:30

there r some papers about this arxiv.org/abs/physics/0612253

ACuriousMind

@JohnRennie It's not really comparing apples-to-apples to try to compare the vote counts on posts from 2011 (or generally high view counts) with much newer posts - especially not if you notice that that question has several other highly voted questions in the "linked" section. This is unusual for highly downvoted posts (most are simply bad and never get linked anywhere and don't get traffic after they're closed).

Sorting by "most downvoted" is less sorting for the worst questions and more sorting for somewhat-bad questions that for one reason or another drive much more traffic than the average post.

The question also has 18 upvotes, more upvotes than most posts that never receive a single downvote get

it's not as easy as saying it's dogpiling - high exposure means high vote counts, in both directions

@qwerty the above also cc to you - beware of judging absolute vote counts without setting it in relation to traffic or the individual up/downvote counts. +18/-46 is not a clear "bandwagon", it's closer to a 1:2 split in votes. With less exposure the question would have something like -4 total with +2/-6 votes, which no one would think particularly noteworthy

imbAF

20:06

I am reading my notes on group theory/ representation theory etc. taken during this semester, and I want to read more about the formula in the red box. Is there a name for it? How does it come to be? Any information or link where I can read more about it will be enough. Additionally what is $(L^{\alpha\beta})^\mu_\nu$

DIRAC1930

20:18

This is usually done in the intro part to any book on QFT

imbAF

what?

DIRAC1930

All the stuff you wrote down

imbAF

I posted what I did in my intro

I am asking for where can I look for further info about this

without the need to fully dive in in representation/group theory (atm)

ACuriousMind

@imbAF It's just an explicit form for the generators of the Lorentz algebra (here denoted $L^{\alpha\beta}$) in the fundamental representation. It doesn't have a name and there isn't a lot more to say about it - it's just what you find if you look at the literal 4-by-4 matrices SO(1,3) acting in the usual way on Minkowski space.

and indeed it should occur in most better intros to QFT, e.g. it's eq. (3.22)/(3.23) in Weigand's notes

imbAF

I plan on starting with WeigandQFT and Mark Srednicki if I ever need to have additional info if Weingand doesn't provide or suggests

But two questions

DIRAC1930

20:29

@imbAF I would learn the essentials of representation theory for like SO(3) first

then SO(1,3) will be very analogous

imbAF

Any suggestions regarding group and representation theory ?

@ACuriousMind what do you mean with: " it's just what you find if you look at the literal 4-by-4 matrices SO(1,3) acting in the usual way on Minkowski space." ?

Slereah

Hall?

link.springer.com/book/10.1007/978-3-319-13467-3

ACuriousMind

@imbAF You can just look at the matrices of the Lorentz group and you can write down an explicit formula for them (e.g. in terms of a rotation and a boost). Then you can write down an explicit formula for their algebra. The algebra is a vector space spanned by the matrices $L^{\alpha\beta}$

imbAF

Also, in my notes, we consider the commutation between different generators $L^{i}=\frac 1 2 \epsilon_{ijk}L^{jk}$ and $L^{0i}$. And we see that two rotations give a rotation, or two boosts give a rotation. But what is the idea behind considering commutations between generators? Just simply consider this mathematical procedure out of the blue, or is there some logic behind it?

DIRAC1930

This is the theory of Lie algebras, Lie groups and their representations

imbAF

20:34

Does it make sense to ask why we consider the commutation or is illogical ?

ACuriousMind

@imbAF The commutator $[A,B]$ tells you how $A$ infinitesimally transforms under $\mathrm{e}^{\mathrm{i}B}$ (or vice versa). You should be familiar with this notion from basic QM.

imbAF

In QM, the interpretation I gave to the commutator was such that, in case it is zero, I can at the same time measure the physical quantities represented with the two operators (hermitians) whose commutator I am considering

ACuriousMind

if you want to learn this on a mathematical level and not just the physics "just expand to first order", you have to learn Lie theory properly, as people are saying

imbAF

Or that, measuring one quantity won't affect the other

@ACuriousMind Most definitely will start with it, this weekend. Finally the semester is over and vorelsungsfreizeit begins

Ryder Rude

@imbAF this is different

this statement is not about measurements but about lie groups

@ACuriousMind i mean this statement

DIRAC1930

20:39

I would actually start with SU(2)

imbAF

I see. I mean, I would expect that the commutation in physics, between operators and in math, between generators would be different in intepretation

ACuriousMind

@imbAF it's all just linear algebra

imbAF

@DIRAC1930 Would it be enough to start with the book you linked, go through the entire book?

Ryder Rude

for matrix lie groups, what ACurisouMind means is that $e^{-At} B e^{At}$ differentiated at 0 is $[B,A]$

ACuriousMind

if you understood the relationship between the Heisenberg equation of motion and the time evolution operator being $\mathrm{e}^{\mathrm{i}Ht}$, you would understand my statement above

Ryder Rude

20:42

u should be familiar with the first expression as the transformation generated by $A$

DIRAC1930

en.wikipedia.org/wiki/Translation_operator_(quantum_mechanics) Have a look at the section "Momentum as generator of translations" if you are unfamiliar with generators first

imbAF

@RyderRude Isn't this the Baker–Campbell–Hausdorff formula ?

Ryder Rude

@imbAF this is simpler than that. it just follows from differentiation

but the BCH formula is related, yes

ACuriousMind

@imbAF This is BCH to first order, yes. But that's what BCH tells you, too - the commutator is the infinitesimal/first-order version of the conjugation by the exponential!

Ryder Rude

@imbAF but again, the key thing is not the mathematical manipulation. u should understand the meaning of the operation $e^{-At} B e^{At}$

u must have encountered in in the Heisenberg picture of QM

this is why i say it is not related to measurements

imbAF

20:47

This discussion just makes it even more clear the supperficial level of understanding I have, for even the most basic stuff, while I am doing QFT. And I don't get it, because it's not that I don't question everything new I learn. So I am stunned that I don't understand anything

I don't know what can i do more

Ryder Rude

have u studied the Heisenberg picture? @imbAF

imbAF

I guess I am not that aware, otherwise I would have asked myself at some point, even in physics, why would the commutator of two hermitian operators, be an indicator that we could measure both quantities at the same time. I just took that as a fact and didn't question it.....

@RyderRude Idk what you mean with study it. We have used it in certain cases. We have made the transition from Schroedinger to Heisenberg or even interaction picture

Ryder Rude

oh

imbAF

But just making a transition, I don't think it constitutes studying it

and having a fundamental understanding

DIRAC1930

It was like this for me dw I had never really come across any of this stuff formally so I had to learn at least the basics in my spare time

Ryder Rude

20:51

it is okay. i think ur course is structured differently

imbAF

Would learning representation and group theory

help me understand stuff like

"why would the commutator of two hermitian operators, be an indicator that we could measure both quantities at the same time. " ?

DIRAC1930

There is a book by Peter Woit that I liked

I think he has free lecture notes online

math.columbia.edu/~woit/QM/qmbook.pdf

Ryder Rude

@imbAF this is linear algebra. it is unrelated to group theory....

ACuriousMind

@imbAF You should at some point have learned that commuting operators have simultaneous eigenbases

that statement is actually not related to the commutator/exponential thing above

imbAF

@ACuriousMind joint eigenbasis I know that

or simultaneous is another thing, different than joint one?

ACuriousMind

20:54

no

that simultaneous/joint eigenbasis is the reason you can measure them "at the same time"

Ryder Rude

u already have the "most fundamental" understanding of this statement

imbAF

Yes, but why would the commutation as a process (if I can call it as such)

allow me to "see" that

ACuriousMind

I don't understand the question

imbAF

Or it doesn't and just knowing that they have a joint basis

is actually enough to let me know about the measurement at the same time

Ryder Rude

starting with the commutation =0, prove that they have a joint basis @imbAF

@imbAF yes. The joint basis is what u really need. the commutation is just sufficient to guarantee the joint basis

Allie

20:56

because if the commutator is 0 and you make a measurement of A it can collapse to an eigenstate that is also an eigenstate of B, so that you can determine both without imprecision

In general if the commutator is not 0 and you make a measurement of A, the eigenstate of A that it collapses to will be a superposition of eigenstates of B, meaning there is uncertainty in wheat measurement of B you will get

and then if you try to measure B right after, it collapses to an eigenstate of B that now is a superposition of A's eigenstates. so you cant measure A and B simultaneously

imbAF

Yes I am very familiar with the collapse and with whether one can make measurements at the same time or without affecting a succesive measurement

depends on whether the operators have a joint basis or not

Allie

then i have no clue what your question is

srry

imbAF

I mostly wanted to understand how the commutator is an indicator of all that

Allie

because the commutator being 0 implies there is a common eigenbasis

Ryder Rude

u have to do the proof @imbAF

imbAF

21:00

To me, if I have to give my opinion

Ryder Rude

do it for finite dimensional Hermitian matrices

imbAF

simply, if two operators have joint basis, and you consider their commutator, it is zero. So that's how one relates commutation result with whether simoultanteous measurement is possible

Allie

yup

imbAF

Just a shortcut, where instead of checking the eigenstates of the operators, you just check the commutation

Allie

although with what you're asking its kind of the reverse statement

imbAF

21:01

Ah ok

Allie

the commutator being 0 implies a common eigenbasis

i agree with RR you should write the proof, its quick and simple and cute

assume the commutator is 0 and show that there is a common eigenbasis

imbAF

Ok

@Allie This is actually good to do, cuz I have never done it

Allie

yes

you can also think of it as $AB$ is measuring B then measuring A, and $BA$ is measuring A then measuring B. if these are the same, then the measurements did not interfere and so you can determine them simultaneously

imbAF

@DIRAC1930 Thanks for this

It looks nice

@Allie I have done an exercise where I calculate the probability of measuring a_n and later b_n for A and B that commute. Then I calculated the probability of measuring the reverse. And the result was the same

But I guess this one is different than

proving what RR says

Allie

its pretty much the same idea

ok my turn to ask a question

i read somewhere recently that in classical mech, the lagrangian formulation/action has no real physical signifiance beyond being a useful way to formulate newton's laws. but in quantum mechanics it does end up being significant. true? exaggerated? outright false?

ACuriousMind

21:07

a very strange statement :P

the comparison makes no sense in the first place because quantum mechanics is usually formulated in terms of Lagrangian or Hamiltonian mechanics, there's no "Newtonian QM"

perhaps that's what this statement is trying to get at but I would rate it very poorly if that's what it's trying to say :P

Allie

the lagrangian formulation would be the path integral formulation, right?

ACuriousMind

not necessarily

even in the operator formalism you usually start QFT from a Lagrangian, not a Hamiltonian

not that it really matters because the two (Lagrangian and Hamiltonian) formalisms are classically equivalent (if you take proper care of gauge symmetries)

Tobias Fünke

hello

Ryder Rude

hi

@Allie i have a post about this

Tobias Fünke

@imbAF In finite dimensions, two hermitian (actually normal) operators commute if and only if their spectral projectors commute. So what you obtained is totally fine

imbAF

21:14

@DIRAC1930 between Brian C. Hall book and Peter Woit's one. which would you suggest for me to start with?

@TobiasFünke spectral projectors ?

Tobias Fünke

In particular, this implies that the product of two projectors is a projector again, and if you think about it, this is exactly what it means for a joint probability to exist

Ryder Rude

i cant find it

Tobias Fünke

$A=\sum\limits_{a\in \sigma(A)}aP_a$, where $\sigma(A)$ is the spectrum of $A$ and $P_a$ projects on the eigenspace corresponding to the eigenvalue $a\in\sigma(A)$.

imbAF

I see

Tobias Fünke

for a clear statement, a short proof and proper interpretation, see e.g. Isham's QM book

Ryder Rude

21:17

this post physics.stackexchange.com/a/741889/156987 @Allie

User198

In the path integral approach.

A particle that is going from A to B is going through all the paths possible (i.e. it is in a superposition of all the paths), but when we measure it, it chooses one of those paths?

ACuriousMind

@User198 no :P

User198

xD

Ryder Rude

@User198 this is pop sci

ACuriousMind

or rather, using the mathematical formalism does not necessarily entail taking any position about "what the particle does"

it's just a mathematical fact that the path integral correctly computes the transition probabilites

Ryder Rude

21:19

although some many Worlders would agree with this

Tobias Fünke

@User198 this is very "anthropomorphic"

ACuriousMind

the path integral formalism, just like the operator formalism, does not inherently carry with it any such interpretation at all

Tobias Fünke

I second what ACM says.

DIRAC1930

I found Halls book very terse. Woit's book has a ton of insights not found anywhere else. He also does the non-rel case very indepth in ways that are not found anywhere else

User198

Ok

"In calculating the probability amplitude for a single particle to go from one space-time coordinate to another. The path integral assigns to all these amplitudes equal weight but varying phase, or argument of the complex number. Contributions from paths wildly different from the classical trajectory may be suppressed by interference. "

From wiki

Allie

21:26

meow

User198

Now, every path has its appropriate action value. And for systems whose action is much bigger than the value of the planck's constant (classical systems) all the paths supress, only the path that has stationary action remains.

But for quantum systems, where action has the value simmilar to planck's constant, there isn't only 1 path (like in classical mech) that will be chosen 100% of the time right?

ACuriousMind

you should look at the actual formulae, not vague natural-language descriptions

the path integral is not actually about "choosing paths"

Ryder Rude

See Carrie from Brian De Palma

User198

@ACuriousMind Ok will do thanks

Ryder Rude

it is a horror movie

User198

21:34

Its about menstruation.

I don't see how that will help me understand the path integral approach?

Ryder Rude

lol

i didn't mean u specifically

and it is about school bullying and cult practices

@User198 are you reading any philosophy ideas these days

User198

@RyderRude I was watching a video about how Penrose says that consciousness can not be computable.

And that it is beyond maths and physics.

Ryder Rude

i am familiar with his ideas

User198

Or sth like that.

Ryder Rude

Penrose says that a computation machine couldn't possibly have proven Godel's theorem

User198

21:40

Cool

Ryder Rude

which i and a lot of other people dont agree with cuz Godel's proof is a proof like any other proof

User198

If I have an electron in free space, can its wavefunction be constant in that space? i.e. that probability of finding it anywhere is the same?

ACuriousMind

@User198 only if it is confined to a box

it's a basic theorem that there is no uniform probability distribution on $\mathbb{R}$

User198

Because it has to be normalizable?

Tobias Fünke

+++

User198

21:42

On R it will only look like a smoothed and smoothed Gaussian as time passes?

ACuriousMind

not necessarily, it's a Gaußian if it starts as a Gaußian

User198

Okok

ACuriousMind

if you start with some other initial condition, it evolves into what the SE tells you that evolves into

User198

Ok thanks

Ok, now going back to the path integral approach; Is there a situation where all the paths a particle can go, are of equal probability?

Like it has many ways that it can travel from A to B, and they are all equaly probable.

Unlike in CM where, the path of stationary action is 100% probable, and all other are not.

ACuriousMind

no, there is also no uniform probability distribution on the space of functions

User198

21:47

Ok. Than why are some paths more probable for a particle?

ACuriousMind

the path integral really does not give you a probability distribution that would be easy to interpret in any literal sense

for instance, the space of differentiable paths has measure zero

even worse, the set of paths with finitely many kinks (non-differentiable points) has measure zero

so with probability one (if you want to interpret the Wiener measure as a literal probability measure for paths), the path has infinitely many kinks

User198

Ok, but a quantum particle chooses a path.

ACuriousMind

no

User198

It is in a superposition of paths?

ACuriousMind

even in the path integral formalism, QM is still about operators and expectation values

it's not about particles following definite paths

Ryder Rude

21:49

this is why i don't like calling path integral a "formalism"

Hilbert space and C* algebras are different "formalism"

ACuriousMind

it's simply that, mathematically, you can show that the theory of states and operators and expectation values is equivalent to computing the expectation values of those operators via the path integral

Ryder Rude

path integral is just a way to do time evolution and maybe calculate some expectations values

User198

So there is no "good" way to interpret why some paths have higher expectation values than other?

ACuriousMind

paths don't have expectation values

the question is ill-formed

User198

Ok ok I see

Thanks

Today I read about en.wikipedia.org/wiki/Classification_of_finite_simple_groups

Classification of finite simple groups

Did you know that "The proof consists of tens of thousands of pages in several hundred journal articles written by about 100 authors"

popularly called the enormous theorem

Really cool

Ryder Rude

21:54

some people say that this is a result u could also expect an alien civilisation to have

cuz it is an example of "discovered" math

User198

Cool

I read your answer now: physics.stackexchange.com/questions/741886/…

That is kind of my question.

Why would the EL equations be usefull in QFT (in path integral approach) if the principle of stationary action is a thing in classical mechancis, and not quantum mechanics

ACuriousMind

@User198 who claims the E-L equations are "useful" in the path integral approach?

Ryder Rude

i have written in that answer that the EL eqn of QFT does not correspond to a stationary action

User198

@ACuriousMind The OP of that question xD

ACuriousMind

As Qmechanic's answer points out, the quantum version of the E-L equations that still hold in the path integral formalism are the Schwinger-Dyson equations (because they are equations about expectation values)

Ryder Rude

21:57

cuz the "action" of that EL equation would be operator valued at best. no one even talks about that action

so it is meaningless to stationarise it

ACuriousMind

@User198 But that's not "in the path integral formalism"

Signor Feynman

After all this time I still don't understand if RM (which I've only heard of here) was not well seen here because of interactions with other users or because of crank ideas (I don't know the guy, just heard some rumors). Or both?

ACuriousMind

that's just in the operator formalism, where the operators obey the classical equations of motion as operator equations

Ryder Rude

@ACuriousMind my answer gives u another version of EL eqns in QM

Signor Feynman

Because every time I find a possibly useful answer of his without a source, I feel torn

ACuriousMind

21:59

@RyderRude no, that's just "as an operator equation" as usual, clumsily expressed

User198

@ACuriousMind Ok and can Schwinger-Dyson equations be derived by variying some action?

Ryder Rude

@ACuriousMind the OP's question is looking for my formulation of EL eqns. qmechanic mentions the Schwiger dyson equations as added trivia

Tobias Fünke

@SignorFeynman RM?

any german here watching ZDF at the moment? just out of curiosity

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