The h Bar

DIRAC1930

15:26

Are there any examples of Godel's incompleteness theorem?

SnoopyKid

15:45

Hello, which one can obliterate bones to very fine ashes or powder (not vaporized into gaseous state as in nuclear explosions) that are totally unrecognizable as human bones in less than 10 seconds? The temperature of cremators or conventional explosives such as TNT or dynamites?

Signor Feynman

That doesn't sound suspicious at all

4

ACuriousMind

@DIRAC1930 What do you mean by "examples"? First of all, there are two incompleteness theorems, and since they are general statements about formal systems, every (sufficiently powerful) formal system is an example.

@SnoopyKid what kind of serial killer question is this

Signor Feynman

@ACuriousMind I'm telling you, this must be the effect of one of those links of Slereah's

SnoopyKid

@ACuriousMind no I'm just curious. You should be suspicious only if I am a legit engineer lol 🤣🤣

Allie

16:33

hi besties

Tobias Fünke

hello

Allie

hi tobias!!!!

i accepted the offer to nyu

Tobias Fünke

nice :)

do you have to move or do you live around NYC already?

Allie

i live about an hour away, accessible by train

i know, surprising for America :P

but yeah i will be moving to NYC

i have a question about phononssss

in a phonon, is it the entire crystal that's vibrating?

Ryder Rude

@DIRAC1930 hi. The continuum hypothesis is independent from ZFC

Tobias Fünke

16:39

yes, a phonon is a vibration mode of the crystal

Allie

okay, cool

Tobias Fünke

@Allie nice :) I've never been there, though. Maybe I will visit it at some point

Ryder Rude

@DIRAC1930 the statement "every natural number is either even or add" is independent from Robinson's arithmetic (which is less powerful than Peano arithmetic)

Allie

so its like a normal mode basically, where each mode is independent of the others

@TobiasFünke if you do let me know. I'll be there for the next 5 years...

Ryder Rude

the fact that independent statements exist is not surprising at all. e.g. The statement "a group is abelian" is independent from group theory axioms @DIRAC1930

what makes Godel's theorem suprising for arithmetic is that there is nothing u can do to complete the theory of arithmetic

Tobias Fünke

16:40

@Allie yeah, there are in principle interactions, but as a starting point you treat phonons as a bunch of modes which are not interacting

@Allie hehe

Ryder Rude

while u can complete the theory of groups

Allie

okay, cool.

my next question thoughhh

so during the inelastic scattering of a photon off a crystal

where a phonon is created as a result

Ryder Rude

also see this math.stackexchange.com/a/2046406 for why Godel's theorem is surprising @DIRAC1930

Tobias Fünke

but phonon-phonon interactions are very important in general (because they lead to anharmonicity, which is necessary for e.g. the expansion of a solid under heating); also, electron-phonon interactions are important (think of superconductivity)

Allie

ah, okay so does the phonon-phonon interaction come from deviations from the hooke's law model?

Tobias Fünke

16:44

mmhhm

Allie

yay

im trying to think of how to phrase thisn ext question without sounding dumb

Tobias Fünke

actually, I think I got it the other way around: you need to take the anharmonic terms of the lattice potential into account to get phonon-phonon interaction

Allie

yeah thats what i figured

anharmonicity leads to phonon-phonon interactions

Tobias Fünke

yes, exactly (because taking only the harmonic terms of course leads to the non-interacting phonons)

Allie

beautiful

but yeah, when a photon inelastically scatters of a crystal to create a phonon, of course the entire crystal is not going to start vibrating immediately

(unless it is and im mistaken. it just seems that that wave would have to propagate)

Tobias Fünke

16:48

Yes, physically I'd say so: Any interaction or distortion is "mediated" by a finite speed

ACuriousMind

well, "a photon" isn't an infinite plane wave nor something localizable to a point, either, even though almost everyone idealizes it one way or the other :P

Tobias Fünke

true

Allie

okay good point

ACuriousMind

so don't put restraints on the picture of "phonon" you're not prepared to also put on "photon" (and due to their usually non-relativistic nature, I'd argue it's probably easier to make sense of phonons :P)

Allie

hmm

im just thinking like

maybe i should just follow the old adage of shut up and calculate and ill understand it deeper later

because i'm shown the equation $\Delta \mathbf{k} = \mathbf{K} + \mathbf{G}$ where delta K is the change in photon's wave vector, K is the wave vector of the phonon created, and G is any reciprocal lattice vector

i dont know how to phrase my question but i do but im afraid it will show how misguided i am lul

ACuriousMind

16:57

If you're still stuck on the "local" aspect - it's just like in ordinary QM: If you deal in momentum eigenstates (as talking about wave vectors implies), then you are fully delocalized in position space. "Real" states are superpositions/wavepackets of momentum eigenstates that have some localization in both position and momentum space.

Allie

okay, that sounds like what im thinking

ACuriousMind

And because they are superpositions, you can understand their behaviour if you understand the behaviour of all the eigenstates - that's the power of the linear character of QM, you really only ever need to solve the behaviour of your system in one basis to have implicitly solved it for all possible states

so objections like "a real state will never be a momentum eigenstate!" are both fully correct and fully miss the point

Allie

like okay to illustrate this, im imagining a photon that is somewhat localized interacting with the crystal

only the atoms on the surface feel that interaction initially, right?

ACuriousMind

that's...not how I would try to think about anything here

Allie

okay

@ACuriousMind the phonons are analogous to eigenstates you mean?

ACuriousMind

17:02

@Allie There's no analogy: You talked about "the wave vector of the phonon", meaning you assume they have definite momentum/wave vector, so they are momentum eigenstates.

Allie

ahhhhhhh

okay that is a good point

and the same point applies to the photon

Tobias Fünke

From a purely operational point of view, I'd say (now a bit simplistic) that the scattering of light which can experimentally observed (in some regime), is, to some extent, well described by the phonon model

ACuriousMind

@Allie Yes; note that as I stressed with the superpositions before, it is not necessary to assume that every scattering has a photon of sharp momentum incoming and a photon and a phonon of sharp momentum outgoing. The general case is that some horrible mess of wavepackets comes in and some horrible mess of wavepackets goes out, but if you understand the scattering in terms of the bases of the incoming and outgoing states, you can in principle compute everything

Tobias Fünke

ACM: I agree to that, but as far as I know, you really associate (in the scope here) to the excitation of the solid some phonon; i.e., there is some energy and momentum transfer, which is the energy and momentum of the phonon

similarly for electronic band structures

and in this way, you can actually measure the phonon (or--mutatis mutandis-- electron) band structure

what you measure is some (double differential) cross section, related to the dynamical structure factor

e.g. in neutron scattering experiments

ACuriousMind

@TobiasFünke isn't it more that you measure the momentum of the outgoing photon and thereby force it (and the phonon) to be in definite momentum states at that point?

I mean collider experiments are also always phrased as if the inputs and outputs are momentum eigenstates, but the reason this is a valid assumption on the outputs is because we're surrounding the interaction site with momentum measurement devices

Tobias Fünke

17:18

yes, in general such scattering experiments measure the outgoing particle (whatever it is). I guess there are some semi-classical explanations going on here

I agree to what you say... I have to think about this a bit more, though

E.g. Chaikin and Lubensky in section 7.7.6 describe the operational (inelastic) neutron scattering quite well --this is used to measure the phonon spectrum.

or perhaps the other way around: We can explain what we measure with our concept of "phonons" ... and similarly with the electronic band structure, so in general with the concept of quasi-particles

User198

17:47

In what kind of relationship are sympletic manifolds and Lie groups?

ACuriousMind

@TobiasFünke I mean all I was saying is that I don't think anyone would insist on "real-world" phonon states necessarily being momentum eigenstates, even if 99% of the time you discuss them you talk about the momentum eigenstates like with any other particle in QFT

@User198 Why do you think "symplectic" is particularly relevant? Again, the geometrical "prototype" for Lie algebras and groups are vector fields and their flows, you don't need a symplectic structure for that at all

Ryder Rude

@User198 given a group, u can define a group action which is the action of a group on a set. Now, Lie groups can be defined to act on symplectic manifolds. the connection is not deeper than that, i think

but also, they act in a way that is compatible with the symplectic structure

that is why they are called the group of symplectomorphisms

so it is somewhat deeper than just group action

in general, we have the notion of flow of a manifold generated by vector fields. this is called the diffeomorphism group

User198

@ACuriousMind Ok thanks.

@RyderRude Ok thank you.

ACuriousMind

18:06

@User198 Careful: The diffeomorphism group is not the group of flows. What is true is that the identity component of the diffeomorphism group is finitely generated by flows, cf. this MO answer. It's another one of these cases where there the relationship between a group and an algebra is more complicated in the infinite-dimensional case than we'd like.

User198

18:26

When you say this: "the geometrical "prototype" for Lie algebras and groups are vector fields and their flows".
Do you mean like (group = circle), (algebra = tangent lines)? Kind of like vorticity?

ACuriousMind

no, not at all

User198

:/

ACuriousMind

I mean that while the modern definitions of "Lie algebra" and "Lie group" are abstract, the motivation for these concepts are algebras of vector fields (with the original Lie bracket) and their corresponding groups of flows.

User198

Ok thanks

Ryder Rude

i think the best way to learn lie groups is to exponentiate rotation generator 3x3 matrices

i think Lie got the idea from there

ACuriousMind

18:28

if you don't know enough differential geometry to understand what I mean, you should learn at least enough differential geometry to understand the notion of a flow

Ryder Rude

@ACuriousMind one can also start with matrix lie groups

manifolds are not necessary

i think Lie got the original idea from matrix lie groups

ACuriousMind

@RyderRude Do you have any evidence at all for this ahistorical claim?

Ryder Rude

i think I read this in Zee's book

but he may have been talking ahistorically

nevertheless, matrix lie groups give u an easy idea of the relation between lie groups and lie algebras (using matrix exponential)

Tobias Fünke

@ACuriousMind yes, I get your point, and I do not disagree on a general note... I want to say that in the usual scattering theoretic language, you do not deal with "phonon states" or so, but with states of the $N$ particle system. ^^

Ryder Rude

i found a paper which says that Lie started by investigating manifolds

Tobias Fünke

18:37

the idea of quasi particles/collective excitations is that you map the excitations of the solid to a system of non-interacting (to a first order) particles, roughly speaking

I have to leave now. See you later

ACuriousMind

@TobiasFünke this might be similar to "resonances" in hep-th. If I had ever really understood what a resonance is, I might have been able to tell :P

Soham Saha

Hi everyone, sorry if I am interrupting any discussion. Has the following already been discussed on PSE? If so, could someone please provide links to the same? External references and sources are also welcome.

TL;DR: Does quantum collapse require a conscious observer, or is the act of measurement itself enough?

I was reading a book which said that the electron double slit experiment did not show interference pattern if there was a measuring device placed at the two slits recording which slit the electron passed through. Now consider this scenario: the measuring device is placed, and it pr…

ACuriousMind

@SohamSaha probably one of the most recurring questions on the site, and the idea you're describing is called the quantum eraser.

Soham Saha

@ACuriousMind Ah, thanks for the keyword. So has any such experiment already been conducted? Is a conscious observer needed, or is measurement enough for the collapse?

ACuriousMind

we have generic questions about observers like physics.stackexchange.com/q/9857/50583, but e.g. the question as you asked it is already ill-defined/on the wrong level, since you treat "collapse" as something that objectively happens, but whether collapse even exists in dependent on your quantum interpretation

@SohamSaha of course, even the "advanced version" with delayed choice has been performed multiple times. But, as I said above, if you expect the outcome of this experiment to be "observers need consciousness" or "observers don't need consciousness", you're not engaging with it on the right level

Ryder Rude

18:53

i think the role of consciousness in collapse is not settled. it can't be determined via experiments

according to current experimental precision, all measurement devices produce something that can effectively be modelled as a collapse

whether or not it is a real collapse is unsettled

so consciousness is not singled out

the only thing that singles out consciousness is that when an outcome becomes subjective experience, you know for sure that it has collapsed

Soham Saha

@ACuriousMind TBH, I am not at all well-versed in quantum mechanics at all, I was just going through John Gribbin’s “In Search of Schrödinger’s Cat”. I was just thinking about what would happen if we actually did the experiment that I mentioned. If the display panel is left unobserved, do we get the interference pattern? Or are there some problems because of which the experiment, as designed, is not possible?

@RyderRude So as far as we can analyse the results from the experiment, whether collapse has happened or not cannot be determined at all?

ACuriousMind

There are many interpretations with wildly varying beliefs about what "observers", "consciousness" or "collapse" really are (among other disagreements). Any that still have any kind of proponents today have their own ways of explaining the outcome of such experiments, i.e. they all agree on the observable facts.

Since the standard intro QM course does not involve lessons on consciousness but does enable you to compute the outcome of the quantum eraser, it follows consciousness can't be intrinsically relevant :P

Ryder Rude

@SohamSaha yes. see decoherence. also, it can be determined in principle but it is not within current experimental precision

one of the predictions is that if there is an objective collapse then quantum computing will fail at some macroscopic level

ACuriousMind

@SohamSaha if you're thinking of a macroscopic display panel, then there's no way to avoid that information "leaking" in subtle ways into the environment, i.e. just closing the door doesn't really make the information in principle inaccessible

Ryder Rude

if there isn't an objective collapse, then we have a consciousness causes collapse theory or something equivalent to it

ACuriousMind

18:58

the delayed-choice eraser is really the setup that does what you want, where the which-way information is fully erased "after" the fact, and it restores the inference pattern

if you read the Wiki articles I linked, you will see that this (and more) is discussed there

Soham Saha

@ACuriousMind and @RyderRude Thanks for helping, at least I now have the keywords to look up :)

Ryder Rude

@SohamSaha great :)

@ACuriousMind this conclusion does not follow

the outcome of the quantum eraser is again not sensitive to the exact solution of the measurement problem

so it can't be used to decide what's true

but yes, within current experimental precision, it seems likely that consciousness has no role

effectively, all current experiments can be modelled using the shut up and calculate attitude. consciousness does not factor in the calculations

Tobias Fünke

20:16

@ACuriousMind yes. from my superficial knowledge of hep QFT, I think that e.g. a pole in the Green's function/propagator corresponds to the mass of a particle of the theory, correct (roughly speaking at least)? If so, then things are basically "the same" in the solid state case. For example, we have a solid made of interacting particles, and the poles of the (one-particle) Green's function can be associated to the energy of quasi-particle states.

Signor Feynman

@ACuriousMind On a side note, I'm starting to wonder why I don't care about these aspects of QM. It was probably fascinating to me before I actually learned QM, but after I did, I can't bring myself to be interested in any way to intepretations. I'm "deliberately" ignorant about even MWI

ACuriousMind

@TobiasFünke yes, resonances are poles in the propagator, that's the part I understand

Tobias Fünke

and experimentally the associated spectral function describes how we can excite the system by removing/adding an electron. the peaks in the spectral function give the energy-momentum relation

in the non-interacting limit, we get delta-function peaks at the non-interacting energies

ACuriousMind

@SignorFeynman I hate talking about quantum interpretations, too ;P

(and in contrast to GR and diffeomorphisms, this isn't mostly performative :P)

Tobias Fünke

@SignorFeynman yeah... I anyway have a more "instrumentalist" view on these things. I mean I find QM interpretations interesting, but only to a certain point. ^^

Signor Feynman

20:20

@ACuriousMind Well, you spent some time reading about it if you can reply :P

Tobias Fünke

btw ACM, do you feel better? :) Have you had a movie night? :d

ACuriousMind

@SignorFeynman well, you can only hate what you know :P

Signor Feynman

I thought you had been on reddit!

Tobias Fünke

@ACuriousMind well, that is certainly not true. xenophobia lol

Signor Feynman

@TobiasFünke Probably interesting, but not as a physicist. I have a feeling that QM is way too mystified sometimes

Tobias Fünke

20:22

not the correct term, but I think you get what I mean

ACuriousMind

@TobiasFünke A lot! :) But I spent the last two days mostly sleeping and feeling unnaturally cold, I didn't have the mental fortitude to watch anything requiring any amount of brainpower :P

Signor Feynman

With a "trained eye" I don't consider QM more absurd than CM

Oh, I'm sorry you are/were ill, ACM

Tobias Fünke

@SignorFeynman sure, especially in pop-sci. but I think it is still interesting as a physicist, because I'd bet that any one has a certain mental image when discussing physics, and especially things like QM. depending on ones interpretation, the mental image is different. I don't know what is effect and cause, though

glad to hear that you feel better.

Okay, I g2g again. I have to take care of my sick gf

Signor Feynman

@TobiasFünke why did I imagine this exit in a Superman fashion?

ACuriousMind

@TobiasFünke thanks, I hope she gets better soon as well :)

@SignorFeynman Yeah, one problem is that a lot of the discussion that acts as if interpretations are very important is clearly done by people who don't know anything more than basic QM (if that much), or at least is targeted at such people

Now, that's not to say they are necessarily entirely unimportant, but there's a lot of chaff to get through to find some kernels of valuable discussion

And also the more you understand QM, the more you understand that the interpretation really is fully disjunct from any physical predictions, you don't need to subscribe to any particular interpretation to predict anything (in practice in all cases, and also in theory unless you subscribe to some fringe objective collapse theory), while all the pop-sci presentations act as if understanding that "collapse happens" or whatever is crucial to QM

Signor Feynman

21:36

I guess they need some appeal :P

qwerty

22:31

@ACuriousMind glad you're feeling better, ACM

Silly Goose

23:59

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