The h Bar

sigh...turbulence sucks

Danu

12:52 AM

^ agreed

Oddly, I'm trying to model it

But I need it to be divergence free

And it's not working

6:21 AM

@ACuriousMind if enough people downvote they hit the low quality question block

(hint hint)

12:09 PM

Today in the exam I panicked and invented a new Feynman diagram

Pretend that that's a scalar particle.

Only one interaction, the black dot

Yes, it's completely wrong. The wording in the question confused me and for some reason i felt that this was necessary to be able to answer the question. Even though I knew it's wrong

hooray for exam panics

1:05 PM

@ManishEarth No, it isn't. It's called a tadpole, and could exist, e.g. in $\phi^3$-theory

2:13 PM

@DavidZ I have already cast a downvote on everything they posted when it appeared, but, as always, someone appears to be voting this stuff up...

3:41 PM

I've got a physics question inspired by fictional science. Looking for a quick answer. Ask in here, or on the main site?

@Keen If it is fictional science, it is probably off-topic on main.

@ACuriousMind It's a 'here's a magic device that doesn't exist, it outputs this' scenario where 'this' is what i'm asking about and is non-fictional.

and i'm looking for a back of the envelope sort of answer, not a rigorous one, since it's not a serious problem

@Keen Then...can't you skip the magic device and ask us about that only? If it is really non-fictional (meaning it can occur without the magic device), then you don't need that part at all, right?

Okay, here's my question, so I can get specific advice on it:

I'm discussing Iron Man's repulsors over on sci-fi. They're essentially a magic device that consumes energy and outputs thrust. There doesn't seem to be any fuel tanks, no fuel consumed. Ergo, magic device. Possible explanation is that it does consume a small amount of fuel, if it accelerates particles to exceedingly large velocities.

But shooting out individual particles to produce enough thrust to lift a human still seems outlandish to me.

and said particles, what would the impact be on the environment around the thrust? It intuitively feels like it would just cause destruction.

Uhh, Iron Man is powered by his arc reactor

3:51 PM

@KyleKanos Yeah, I'm ignoring the energy this mess would consume.

I'm basically saying that the "fuel tank" issue is moot

@Keen But the repulsors could just be turbines. You don't need fuel to get thrust if you have energy available

@KyleKanos How so? I'm never got enough into physics to understand thrust.

@ACuriousMind In the comics they can produce thrust in a vacuum.

And blowing air is how VTOL works

@Keen Uh, what?

That makes zero sense

The "fuel" comes from the arc reactor

@KyleKanos You can generate thrust via electricity and no reaction mass?

@KyleKanos Well, I mean I understand Newton's Laws, but I don't know the specific equations used when analyzing thrust or rockets.

3:54 PM

Hm, if they are working in a vacuum, they indeed need to generate the thrust by expelling matter he has "on board".

@Keen Are you talking sci-fi or real-world?

@Keen But your question is then essentially "How fast would X particles need to be to generate thrust Y", which is off-topic as homework-like

@KyleKanos Uhh, real world, I think. Essentially the issue is that his suit lacks fuel tanks.

@Keen Again, fuel is moot because his suit is powered by the arc reactor....what part of that sentence confuses you?

@ACuriousMind My main question is the second issue of 'say you're spitting out particles at Z velocity, what side effects would that be', since I assume that Z would be some huge value.

@KyleKanos ~~fuel~~ -> reaction mass

3:57 PM

@Keen Fuel = arc reactor

What part of that is unclear?

That ACuriousMind said that it needed to expel matter.

According to Marvel (not science/scientists), it expels neutrons

Neutron generator

Neutron generators are neutron source devices which contain compact linear accelerators and that produce neutrons by fusing isotopes of hydrogen together. The fusion reactions take place in these devices by accelerating either deuterium, tritium, or a mixture of these two isotopes into a metal hydride target which also contains deuterium, tritium or a mixture of these isotopes. Fusion of deuterium atoms (D + D) results in the formation of a He-3 ion and a neutron with a kinetic energy of approximately 2.5 MeV. Fusion of a deuterium and a tritium atom (D + T) results in the formation of a He-4 ion...

heh ... i was compared to Ron Maimon... in a not so nice way

So table-top devices + "advances in human technology in sci-fi" = Iron Man repulsors

@SabreTooth wat

Where?

4:02 PM

@ACuriousMind yeah, had no idea I was considered to be like that - I mean, I know I can be annoying at times (anxiety disorder can do that), and I admit I can be abrupt at times, but surely, I am not that rude

but thats cool, I will focus my attention away from that site

(may even delete my membership from there)

Again: where?

in one of my other sites - don't want to cause a fuss, just thought that you guys would get a giggle that Maimon's online ghost still lingers

haven't been back there, except to write an apology to the members of that site if I had indeed been an ass

@SabreTooth @KyleKanos: Your profile pictures look eerily similar when they are so small (slightly different coloring, but still...)

What ever happened to the eye? ;)

got punched

I've been thinking about changing my profile picture

4:12 PM

too much like mine? ;P

Naw, just getting bored with SN 1987A

that happens... at some point, we all get bored with SN 1987A

mine is a result of a little experiment of mine

Yeah, it's not really doing anything interesting right now

Give it a hundred years though

I'll write that in my diary

but, just to reassure me, I am not as bad as Ron am I?

If i have been, I apologise

I never really knew Ron, so I cannot say

4:26 PM

(yes, the comparison has got me strangely and unexpectedly rattled)

@SabreTooth I don't recall you ever being memorably rude or...argumentative here

@ACuriousMind yeah, that was meant not so much for you, but for other people who might not be doing the same

I have been trying to figure out if and when I may have been

@SabreTooth you haven't been suspended multiple times, have you?

no, not at all

4:29 PM

then you're fine ;-)

well, I am hopin gI have not been rude to people

@Keen asking for the impact on the surrounding environment would be off topic, I think, but I'm not sure. You could give it a try. It might get closed.

Asking something like how many particles per second you'd need, or so on... it would be off topic to just ask us to calculate it outright, but if you tried to calculate it yourself and were confused by something in the process, that would be on topic.

But anyway... let me note that jetpacks exist. Iron Man's repulsors are basically tiny jetpacks. (When he's flying and not shooting, at least)

where can I buy one?

@SabreTooth trust me, there is no way you could be as rude as certain former members of this site without having gotten multiple mod messages about it.

@SabreTooth a jetpack? I actually have no idea

@DavidZ Eh, getting to the Neutron radiation Wikipedia article was sufficient for me to see how bad spraying things with neutrons could get.

I'm good.

4:35 PM

Fair enough.

@DavidZ hmmm a mystery then...

@Keen honestly for questions like that, it usually winds up better to keep them on Science Fiction & Fantasy and then post a link to them here in chat.

@DavidZ it wold seriously help shorten my commute to work .... I walk 5km with both knees on the verge of collapse

I don't know if you have a physics tag there, but if you do, we could add the feed to this chat room's notifications

@DavidZ Enough of the SF&F community is allergic to anything that smells like reality that I doubt that would be beneficial.

4:38 PM

haha

I find the world building SE amusing; they seemingly go there for advice on how to construct realistic words, and then they abandon all of physics in their answers!

@JamalS:

2 days ago, by David Z

It irks me in a professional sense to see all these questions on Worldbuilding asking whether something can be done in a somewhat scientifically realistic manner, and then all the answers tossing most of established physics out the window :-P

Funny that you and David come up with different conclusions based on identical facts!

@JamalS I think it is less realistic worlds, but more consistent worlds they're after

@KyleKanos Well, I don't know that they're inconsistent conclusions

@DavidZ I think irk and amusing are incongruent, but that could be me

4:44 PM

I used to be a member there

@KyleKanos maybe it takes a certain tolerance for non-rigorousness or something, but I think you can have both at the same time

@DavidZ: I wouldn't say I'm tolerant of non-rigorousness. For example, I absolutely detest the Big Bang Theory (a classmate told me I should watch it) because of the way physics and physicists are portrayed.

@JamalS it sounds like you don't like it for reasons having little to do with its lack of scientific rigor

@JamalS yes, this! I get tired of being likened to Sheldon (I work in a rural high school with a PhD in Physics = circus freak here...hate it)

For example, there's a clip on YouTube where they're at a physics competition, and the last problem is a scattering amplitude, but it's tree-level. The main character is 'struggling' to compute it, yet he's a theoretical physicist, and should notice the delta function in the integral makes it trivial.

4:55 PM

The problem with TBBT

@ACuriousMind Except that this paper was on KG/Dirac/vanilla QFT only :P

@ManishEarth I'm not sure what you mean by that - $\phi^3$ theory is perfectly fine in "vanilla" QFT, it's even renormalizable (even superrenormalizable, I think).

But if he Lagragian in question had no $\phi^3$ term, then yes, the diagram is nonsense :P

5:10 PM

@ACuriousMind no, it had one

it also had mass terms

well, thank you all, I feel little bit better and a bit more self assured than before about the ron maimon comparison bestowed upon me

doesn't this violate momentum stuff?

@ManishEarth No, it's a perfectly valid contribution to the VEV of $\phi$.

hmm

Ah#

Sry

5:13 PM

?

As a scattering diagramm, it is nonsense, since, as you say, momentum is not conserved

@ACuriousMind yeah

It is perfectly valid if you are interested in $\langle \Omega \rvert \phi \lvert \Omega \rangle$, though

Feynman diagrams do more than just compute scattering :)

Right :p

I wanted the Wick's theorem term

term out of $\langle \Omega \rvert ::\phi^3:: \lvert \Omega \rangle

@ManishEarth Well, everyone understand a slightly different thing by "Wick's theorem". But your diagram is exactly how one would graphically represent $\langle \phi(x)\phi(y)\phi(y) \rangle$.

5:25 PM

@ACuriousMind x,y?

@ManishEarth spacetime points

We seem to be using a different notation,

okay, we aren't

but there's only one spacetime point?

$\phi$ is not an operator - $\phi$ is an operator-valued distribution. To get an operator you can take the expectation value w.r.t. $\lvert \Omega \rangle$ of, you have to take $\phi(x)$ at some point $x$ in Minkowski space

So, $\langle \phi^3 \rangle$ is not unambiguous notation. It is a different thing if the arguments of the three $\phi$ are allowed to be different or not

Hm, but we always gave each $\phi$ and $\psi$ different spacetime args only if they were from a different interaction

Interaction take place at a point, that is correct. So, if you are interested in $\phi(x)^3$, then its expectation value is zero, since there's no possible diagram - you cannot contract an odd number of things in a way such that all are contracted

5:31 PM

@ACuriousMind yeah

But still, the tadpole would be the graphical representation of contracting two of the fields and leaving the third alone

Yep

It's been too long since I've dealt with these diagrams...I'm a bit rusty

heh

Or rather, I now take them for granted and don't think too much about where the rules for them actually came from

5:34 PM

@ACuriousMind: Out of curiosity, what are you working on at the moment?

@JamalS Physics-wise: String theory and Higgs mechanism/beyond the Standard Model, math-wise: algebraic geometry

@ACuriousMind: What string theory text do you predominantly work from?

I like Witten's two volumes the most

@JamalS Not found one I like working with yet...the course I'm taking is surprisingly slow, so don't need a book for that. I've only briefly looked into Polchinski, and I'm not sure what about him it is I don't like, but I...don't.

@ACuriousMind: You're not the only one, I don't like his text either.

@ACuriousMind: If you haven't looked at Green, Schwarz and Witten's Superstring theory, you should.

Becker, Becker and Schwarz's String Theory and M-Theory: A Modern Introduction is also a good book, but they deal with topics quite quickly, and there is some handwaving.

@JamalS I planned to look around for another book to read over Christmas break (an intention probably doomed to failure, as always when I intend to read books over the holidays...), anyway, so thanks for the tips :)

5:50 PM

Interesting: The front page of the users tab is filled up with 10k+ users

It was not that way when I first started, or when I hit 10k

6:13 PM

@KyleKanos Quite a few people hit 10k in the last few weeks

@ACuriousMind Yep, I imagine there's a way to look at the number of users in certain rep brackets as a function of time using the Data Explorer, but I don't (yet) understand SQL

Me neither

The only things that are real are Fig Newtons. Everything else is a Fig Newton of your imagination. — Hot Licks 1 hour ago

Am I missing a pun or other joke here?

It looks as if it is supposed to be a play on "figment of imagination", but I don't see how you get from figment to Fig Newton

Also, WTF did I just reject?

@ACuriousMind That's clearly vandalism

@ACuriousMind I think that's what it is supposed to be

It also conflicts with the author's original intent :D The rejection messages are just great, aren't they? :)

It is also no improvement whatsoever.

The only thing it is not is attempt to reply

6:33 PM

6 more reviews on the First Posts before I get a shiny new badge

9:58 PM

Hello there

I have a question about fundamental quantum mechanics

could I have a discussion with someone about it_

?

@TheQuantumPhysicist Depends. What do you want to know/discuss?

I just can't wrap my head around non-determinism

I can't find a mathematical model that will generate determinism for determined wave-content

generate non-determinism, sorry

Is that topic OK for you?

A "mathematical model" with non-deterministic measurement outcomes is easy

Please go ahead

Take an electron, and care only about its spin

There are two possible measurement outcomes - up or down

So, QM says the space of states of that electrons is spanned by an "up" vector and a "down" vector

10:03 PM

Btw, it'll shock you, but I'm a PhD in particle physics and atomic physics...

Now, if the state of the electron is (up + down) instead of merely up or down, you have equal probability to measure up or down spin.

My question is a little more fundamental

I thought so ;)

^_^

Could I please state the issue?

Sure

10:05 PM

OK. Looking at the mathematical problem of uncertainty. The simplest form is that if you would have a wave-packet, then the frequency of the wave is not determined unless you have an infinite measurement in the time domain

isn't that right?

In other words: The Fourier transform of an infinite sinusoid in the time domain is a sharp peak in frequency domain

Or infinitely long sinusoid

@TheQuantumPhysicist Mhhh...isn't the Fourier conjugate to time energy?

Ah

frequency or energy, up to an \hbar

Yeah, I forgot about the hbar

OK :)

So we agree on my statement?

@TheQuantumPhysicist Yes, under the caveat that time is not an operator in QM, so the uncertainty relation does not hold naively for it

I.e. it holds as an abstract property of Fourier transforms, but not as a property of QM operators

10:09 PM

Well, we could simply build an experiment and measure for some time. If our measurement in time is infinitely long, then energy is very well defined (apart from experimental systematics, etc... I mean a perfect experiment)

Q: What is $\Delta t$ in the time-energy uncertainty principle?

13

In non-relativistic QM, the $\Delta E$ in the time-energy uncertainty principle is the limiting standard deviation of the set of energy measurements of $n$ identically prepared systems as $n$ goes to infinity. What does the $\Delta t$ mean, since $t$ is not even an observable?

I'm pointing to the transition from classical to quantum mechanics. So I'm trying to derive that non-determinism is inevitible

Yes exactly. From that thread, you're showing that the width of your spectrum peak has to be infinitely small by having an infinitely long measurement. Right?

So the width is simply $\sigma$

@TheQuantumPhysicist the width of the spectrum peak is $\sigma_H$ in the notation of joshphysics there

...isn't it?

Yes, right.

So are we OK until now?

Shall I continue or should we discuss further to clarify concepts?

@TheQuantumPhysicist Not yet. $\Delta t$ measures "how quickly" the observables in a system change. Why then, would a long measurement imply a big $\Delta t$?

10:17 PM

Let's assume we have a stable system, which doesn't change over time (let's say we cool it enough, for example, that it takes the age of the universe to change its state)

Then the energy peak can be made arbitrarily narrow, yes

Perfect! So an infinitely long measurement will just keep narrowing the energy peak to zero.

Now the reason for the non-determinism in energy is that we cannot measure time infinitely long. If we could do that, then voila, we have a 100% determined energy

Now this follows exactly when talking about measuring a signal. If you want to determine the frequency of a signal, you have to measure for infinitely long time and count the zero crossings. More zero crossings -> better precision in frequency.

I fear you are describing experimental uncertainty, not fundamental uncertainty here

What's the difference? We have already established that if we could measure an observable for infinitely long time, then we have it with infinite precision. Right?

The point of the QM uncertainty is that the inherent statistical variances of the operators obey it, not that the statistical variances of our measurements obey it (though the latter follows from the former)

You are thinking of "energy" as a well-defined property of the system, but that is only the case when it is in an energy eigenstate

Now, by saying that the system doesn't change over time, you have essentially restricted yourself to an energy eigenstate, making the statement that it's energy can be precisely determined true

10:25 PM

So QM is non-determined because observables can change without us being able to control them?

In an energy eigenstate, the expectation value of operators doesn't change over time, so you divide indeed by zero in $\Delta t$, yielding precise energy

Well, I'm thinking of energy as the frequency of the wave-packet. Isn't that the correct definition?

Energy is the expectation value of the Hamiltonian

Well, when we calculate expectation values, we're not anymore in the wave-function. We move to our classical picture. I'm thinking of the system before taking an expectation value

In my head: The system before the expectation value is simply a wave!

Okay. I dislike the wave picture, but it works here:

A wave-packet is a superposition of different Fourier modes, right?

10:28 PM

We try to determine some macroscopic observable from that wave, and that's why we have uncertainty!

Exactly

So it doesn't have a single, well-defined frequency to begin with!

It is the superposition of different frequencies

Well, it has a set of frequencies! Well defined frequencies

But, in general, these frequencies are infinitely many, i.e. $\mathbb{R}^+$

Yes, I accept that. However, if I have a mathematical wave-packet, I can determine the frequency components in this packet infinitely precisely; probably worst case with a least-squares fit!

But alllllllllll frequency components can be read from that wave! I can know exactly every component there

Right, given that you can measure the wavepacket

But the QM wave isn't accessible

10:31 PM

Exactly!

Great! Now we're at the point where I have a problem

So we see QM non-deterministic because we can't measure the packet!!!

This is what blows my head!

You shouldn't ascribe reality to the wave function in that way

It is just a mathematical "trick" to get the QM results. It possesses no physical reality, it is unclear to say what it would even mean to "measure" it

Well I understand that wave-functions are a model. But yet we claim that QM is non-determinstic due to the wave-nature of QM, which is not compatible with our macroscopic world, right?

This incompatibility can be seen in the wave picture with the concepts of infinite measurements (that we were talking about)

This is how I understand it!

Well, there's nothing really wrong with it. I think, if you find it enlightening/pedagogical, it's good

But then this means our world is deterministic!

Yes, the evolution of the wave-function/states is deterministic

10:38 PM

The fact that we can't access the wave is the reason for non-determinism! This is my conclusion!

You have just redeveloped a very down-to-earth version of the Many Worlds Interpretation

Really? Tell me how!

MWI rejects the idea of collapse, so there is a universal wave function, and it evolves as dictated by the Schrödinger equation, which is a deterministic evolution

OK? ... Why is it called many-worlds?

Because you have to explain why the "other parts" of the wave function aren't observable

10:41 PM

other parts?

If the wave function evolves, every state that can be reached by time evolution
will* occur with some probability in it. Take Schrödinger's cat, where the wave function contains both a living and a dead cat

After you observed the cat living, you will never come to state of the universe where the cat was found dead

Aha, I see what you mean.

Yet, since the wave function is universal and deterministic, there's still a state with a dead cat in there

Gotcha!

So we have no proof that our world is non-deterministic... it's just our ignorance and assumption at the same time!

@TheQuantumPhysicist Well, even if the wave function is universal and deterministic, the outcome of measurements is still not

Even given the complete wave function, you could never predict with certainty the measurement result of an observable it is not an eigenfunction of

10:45 PM

Why? We had arrived to the conclusion that all frequency components of the wave are fully determined intrinsically, apart from our inability to measure it!

If I have the wave-function, I can know all its frequency components!

A simple Fourier transform or least-squares fitting will tell me allllllll the contributing energies

or frequencies

Yes. But consider the following: I give you an electron, and I tell you its wavefunction is the position eigenfunction at $x_0$.

What will you measure for its momentum?

That depends on what you mean by measure

When we talk about momentum, we're defining a classical observable. A momentum (in the wave-form) is a spectrum!

Slam a target into it and measure the recoil, for example

It's not a single number anymore!

Okay, yes

10:48 PM

Actually that's the second part of the discussion, which includes interactions

If you consider only the probability distributions of the measurement results, then they are completely determined by the wave function

I still haven't modeled interactions in my head :)

We don't need to talk about probability at all! That's what I'm trying to do

But most people would still call that indeterministic, since the specific result of a single classical measurement isn't determined

We're just at the wave level... probability means that we already gave up and we accepted non-determinism

When you say "momentum is a spectrum" - what else are you saying than "This is the probability density for measuring momentum"?

10:49 PM

The classical measurement is not determined because we're talking about two incompatible systems

I'm saying that the evolution of the system in space cannot be determined by a single momentum. All contributing momenta have to be taken into account

Unlike a classical particle, for example

a single vector suffices!

Alright, you want to avoid saying "there's a probability distribution for the momentum", and so you prefer saying "there's several momenta in that wave"

Exactly!

When I say probabilities, I'm admitting that I don't want to bother with how this evolves. I still don't want to do that.

Let's move to the next point, probably it will be more helpful.

That's not what QM does when it talks about probabilites, either, by the way!

Well, from what I know, quantum mechanics uses all possible momenta to create a probability distribution

right?

My second part of the question, is about interactions.

So how do particles interact: An electron is flying to another electron. When those electrons meet, do they measure each other's expectation value and then decide to interact? Or do they interact in wave form?

Wave-particle duality. Always fun.

10:56 PM

Yep!

I prefer throwing it all out

How?

I wish I could! I'm crazy about understanding fundamental physics in a good way!

I prefer retreating to the abstract Hilbert spaces, and just saying that there is a time evolution operator that turns some states into other states.

Well, that's a little premature for me, because I want to understand whether our world is or is not nondeterministic intrinsically

I mean, it could easily be that our world is deterministic, but our ignorance makes us see it as non-deterministic!

"Interaction" is not really a term that enters here. Stuff evolves with time and becomes other stuff

@TheQuantumPhysicist You know Bell's theorem, right?

11:00 PM

Well, you have a Hamiltonian which takes care of the interaction

I know it by name and by conclusion, but not in detail. I know that it concludes that the superposition principle is correct.

Which doesn't negate the deterministic wave-form I'm trying to pose, does it?

Maybe a way of stating it. More precisely, it says that any theory that reproduces the (experimentally well-tested) results of the usual QM cannot be local and realistic

Where local essentially means "causal"/"no information transfer faster than light" and "realistic" means "all measurement results are predetermined".

I thought it's either local or realistic, but it doesn't negate both, does it?

Right, non-local realism and non-realistic localism is both allowed

Most people choose the latter, though

Yes, because they like relativity

except that quantum eraser experiments has shown that something faster than the speed of light is going on, if I got that correctly

Could you please tell me your point?

Are you trying to say that non-realistic localism = non-deterministic world?

Ah, yes, you said that it could easily be that our world is deterministic, and we only see it as non-de. How would you distinguish between a non-local deterministic and a local non-deterministic world?

If non-locality is allowed, everything could, in prinicple, influence everything else everywhere in an instant.

11:07 PM

Well, it just means that interactions can have infinite speed, which doesn't sound so crazy to me

You know the world isn't both. But if it is non-local, you cannot ever tell whether a result is random, or influenced by something infinitely far away

You're right

The distinction between predetermined and non-predetermined results becomes essentially meaningless, because you allow arbitrary influences

Well, it's just a way of saying that we don't like that because we can't measure that...

Determinism is really only a good concept if you believe in locality

If not, you cannot tell the difference between deterministic and non-deterministic. And concepts that cannot be empirically differentiated aren't science

11:10 PM

The last statement is very true. So it's a limitation of science that we're trying to avoid

It's not that it's impossible conceptually

In other words: I can't measure things that affect my system in the other side of the universe, and thus I claim that my system is local and hope that I'm right

No. But a statement about the world, of which you could never tell if it is false, it meaningless. And the statement "The world is deterministic" could never be disproven in a non-local world.

You're right. It cannot be disproven.

But it has, by Bell, been disproven in a local world. So the statement "the world is deterministic" is either false or meaningless

I see!

That's a very interesting conclusion

So non-determinism comes from Bell's theorem

because we want our world to be local, since all experiments support that, and since it would be meaningless to have a non-local world that nothing can be disproven with that regard

In a way, yes. At least, my confidence that the world is non-deterministic (if that is not a meaningless statement) stems from Bell's theorem (and the experiments that test it)

11:15 PM

So if we want to prove that this world is deterministic, we have to disprove relativity.

That's the first step, at least

Thank you very much. I enjoyed this little chat and learned a lot :)

@TheQuantumPhysicist That is a very dangerous statement ;)

You mean that relativity is very robust?

Yeah. I've never heard anyone doubting it whose arguments would have been even remotely convincing

And disproving relativity really attracts all sorts of crackpots

Well, Lorentz violation experiments are ongoing :)

I, myself, work in an experiment about CP violation

so all kind of crackpots already exist ^_^

Yeah, if we find a Lorentz violation, that would be very interesting.

11:23 PM

We'd be screwed actually... we'll need then 50000 years of measurements to build a model of the universe...

CP violation is not that bad, you can write down perfectly good QFTs that violate CP

ooh

Well, that wasn't there 50 years ago ;)

I keep seeing seminars, in which CPT and Lorentz are violated with QFT terms

@TheQuantumPhysicist I think the quote "We don't need to talk about probability at all! That's what I'm trying to do" is a misunderstanding of the fundamentals of quantum mechanics ala Copenhagen

Yes, it is always possible tomorrow, everything we believed until today is false. Not very likely though.

11:25 PM

Exactly, agreed ^_^

@bobby Copenhagen is an interpretation of the result. I understand it very well. But the thing was about other models of QM

If you read Landau you can clearly see the mathematical structure of quantum mechanics derived from Heisenberg's uncertainty principle + experiment. Heisenberg + measurement implies probability which implies born rule, it all comes from a one-line computation.

@bobby That mixes our ignorance. Please read the full discussion with @ACuriousMind

Good night! :)

I think it's nonsense tbh you are calling a wave function fundamental for no reason

@bobby It doesn't necessitate that the Born rule gives a probability, though, if we are nitpicky. It only necessitates that the frequency of occurence of the measurement results is given by the Born rule, not that there aren't unconsidered factors producing these deterministically. For that, you need Bell's theorem

You derive the wave function as a mathematical consequence of the probabilistic consequence of "expected value"

No I, & Landau, assume probability based on experiments. Probability implies the Born rule naturally, it's a 1-line computation

I know Landau states it as an axiom in section 2, but it follows from the basic statement of Heisenberg + the consideration of experiments

Heisenberg says "there's no concept of the path of a particle", but measurements show a probability distribution for position measurements. That's all you need, write out the expected value for position over n measurements, then re-write it as a dot product and then as a bilinear form. From this simple 1-line computation you unavoidably derive position as an operator action on a wave function, see page 1 of this books.google.ie/…

This assumes classical mechanics exists in the limiting case of course, but I see no reason to throw away CM and latch onto wave functions and axiomatics, it seems like a fruitless human-constructed dark alleyway

It is basically philosophy, and probably mis-guided :(

It seems like you guys don't know that simple 1-line computation, it basically shows how secondary concepts like wave functions and operators are

idk I hope you will convince me I'm wrong :)

11:37 PM

Either you or I do not understand what we respectively mean when we talk about probabilities and determinism

(Note, as an aside, that I dislike the wave picture myself)

You can certainly do the experiments, and then say that you can model the distributions of the outcomes by QM, i.e. Born rule + Hilbert spaces

But nothing here tells you if the fact that you cannot predict with certainty the result of any measurement is a consequence of you not knowing enough because of it being principally impossible

That's what Heisenberg's Uncertainty Principle says, it says "there's no concept of the path of a particle" which implies you can NEVER establish the mechanical state of a system, something specified by giving (x,v) & I.C.'s

It tells you it's impossible in principle to have the full information to do the predictions, then you use the concept of measurement to establish the notion of a "complete state", this is all in Landau

I so do not agree with taking the HUP as a starting point, but well, it's another equivalent way to QM

People don't like it because it requires you to admit classical mechanics exists, i.e. you use the limiting case of your theory to establish the rules of your theory. You can get all of QM by following this logic, it's so unbelievably natural and easy, non-ad-hoc

Nothing makes it better or worse than wave mechanics or matrix mechanics

@bobby There is a (mathematically) very beautiful way of getting quantum mechanics from classical mechanics, it is called geometric quantization. It, also, is non-ad-hoc to those versed a bit in Hamiltonian mechanics and group theory

You don't even need to assert the HUP there

You are basically doing the same thing with GQ, you are merely trying to deform commutativity (of measurements, only using operator language) into non-commutativity right? Then you use all these square-integrable assumptions, I mean it's basically the axiomatic approach to QM again

It's lovely, but lets not pretend it's some magical recipe :)

The only argument I know of against what I've said is something Gell-mann says, which is you can't apply this logic to the whole universe, so the whole consistent histories formalism was invented to allow you to do this. I do not know the details of it, but apparently it is basically Copenhagen ala Landau + some modifications to allow you to treat QM as fundamental

So my understanding is that a book like Shankar wants to give you 4 or 5 god-given axioms and tell you that you can derive CM and Heisenberg from these axioms so we can pretend QM is fundamental and feel philosophically special. But someone like Landau tells you that you can use CM + Heisenberg + experiment to derive QM too, plus those axioms, very naturally. I can only see it as philosophical or authoritarian reasoning that makes one take the Shankar approach over the Landau approach,

11:59 PM

@bobby No, geometric quantization certainly isn't the same axiomatic approach. See one of my all-time favourite Phys.SE posts and follow the nLab links to see how it is indeed a purely formal motivation that produces quantum mechanics out of the phase space formulation of classical mechanics.