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user223626865

06:00

\o @SirCumference how goes the grad school blues, pal.

🎓

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user223626865

06:21

user 85795

07:16

in English Language & Usage: Multi-Layered Discourse Room, 27 mins ago, by CowperKettle

Philip Goff is a philosopher who thinks consciousness pervades the universe. Keith Frankish is a philosopher who thinks consciousness* doesn't even exist. From their very different perspectives, Keith and Philip interview leading scientists and philosophers of consciousness, engaging and debating in a friendly way in pursuit of truth

Cc: @RyderRude^ et al

Mind Chat aims to be highly accessible, allowing those with no background in science and/or philosophy to get a grip on the cutting edge of the field.

(*To be more precise, Keith thinks *phenomenal* consciousness doesn't exist; listen to find out what this is.)

Slereah

My consciousness is phenomenal

amazing consciousness

SnoopyKid

@JohnRennie is there an instance where electrons can indeed completely turn and change into photons and vice versa?

John Rennie

An electron and positron can annihilate to create two photons.

Slereah

"a cat must be distinguished from the corresponding amount of feline tissue, for the former can survive the annihilation of certain parts (the tail, for instance) whereas the latter cannot by definition (Wiggins 1968; see also Doepke 1982, Lowe 1989, Johnston 1992, Baker 1997, Meirav 2003, Sanford 2003, and Crane 2012, inter alia, for similar or related arguments)."

John Rennie

@SnoopyKid A single electron cannot change into a photon because that would violate the conservation of charge i.e. an electron has charge -e and a photon has charge zero so the charges before and after wouldn't be the same.

An electron and positron can change into two photons because a positron has charge +e so the total charge before is -e + +e = 0 i.e. the charge is zero both before and after.

Silly Goose

07:29

physics.stackexchange.com/questions/759217/… Is there some other definition of point-like particle? I feel like point-like is by definition an idealization of reality...

John Rennie

"Point like" doesn't really mean "like a point" except in the popular consciousness.

It really means "has no substructure".

Silly Goose

oh XD gosh well here I go to delete my answer

John Rennie

Fundamental particles are Fock states or superpositions of Fock states for localised particles. There isn't any sense in which they are a geometric point.

ACuriousMind

see physics.stackexchange.com/a/119802/50583 for details on what we mean by the "shape" of a quantum particle

"point-like" in this context has a well-defined meaning

Slereah

The second and more general response on behalf of EM is that the appeal to Leibniz's law in this context is illegitimate. Let ‘Tibbles’ name our cat and ‘Tail’ its tail, and grant the truth of
(30) Tibbles can survive the annihilation of Tail.

There is, indeed, an intuitive sense in which the following is also true:
(31) The lump of feline tissue constituting Tail and the rest of Tibbles's body cannot survive the annihilation of Tail.

ACuriousMind

07:36

it just may not mean what you want it to mean, but given that quantum states aren't really solid objects with a shape in the classical sense anyway it's unclear what else it should mean

Slereah

Why do they hate cats so much

Sir Cumference

@user223626865 I'm hanging in there

how's life for you

SnoopyKid

@JohnRennie my English skill is limited so if I understand you correctly, both of the electron and positron completely change into photons?

John Rennie

Yes i.e. the reaction is e⁻ + e⁺ ⟶ 2γ

Electron–positron annihilation

Electron–positron annihilation occurs when an electron (e−) and a positron (e+, the electron's antiparticle) collide. At low energies, the result of the collision is the annihilation of the electron and positron, and the creation of energetic photons: e− + e+ → γ + γAt high energies, other particles, such as B mesons or the W and Z bosons, can be created. All processes must satisfy a number of conservation laws, including: Conservation of electric charge. The net charge before and after is zero. Conservation of linear momentum and total energy. This forbids the creation of a single photon. However...

This

SnoopyKid

@JohnRennie thanks John. You mentioned about "point like" is basically saying "it has no substructure". Is it logical to say within the elementary particles, there is no difference between the void and what is inside those elementary particles?

Silly Goose

07:51

hm now why should we map the quantum analogue of the distribution of charge to the "shape of a particle" where by particle I just mean say an electron whatever an electron physically may actually be

even in a classical case the distribution of charge of an object may not align with the "shape" of the object right? :0. I guess a simple example also would be a chargeless fundamental particle?

John Rennie

@SnoopyKid That's quite a complicated question. We describe a particle by its wave function, so "inside an electron" would mean at some point inside the region of space covered by the wave function. The problem is that no-one knows what the physical significance of the wave function is. That is, the wave function could just be a mathematical description and not actually a physical object.

So I don't think your question can be answered.

SnoopyKid

@JohnRennie ok it's fine, thank you for your responses John.

John Rennie

You're welcome :-)

Leaky Nun

08:08

@ACuriousMind and therein lies the problem right, because the collapse itself doesn't follow from the Schrodinger equation

@ACuriousMind "we're not treating the apparatus as a quantum object correctly" how would we treat it correctly?

like, the problem is that even if we treat the detector as a quantum object, then we just get decoherence, which just gives you discrete probabilities

it doesn't explain why we ever only see one outcome

Silly Goose

to my understanding the resulting (possibly) mixed state you get as a result of decoherence will only have one observed outcome? as in your "decohered state" which is a mixed state which is a statistical mixture of pointer states each of which are "classical states" and we will only observe one such state at the probability given by the mixed state

John Rennie

@LeakyNun Any one version of you only sees one outcome. It's a bold step to conclude there is only one outcome.

Silly Goose

but idk i would like to learn more about decoherence :0

Leaky Nun

@JohnRennie and then the question is why do we only see one version of ourselves?

John Rennie

It's far from clear to me how a sentient organism could experience being two versions of itself.

Leaky Nun

08:18

well because the time of observation is different from the time of being two versions

John Rennie

I have enough trouble just being me. The thought of being a tensor product of me with myself is too horrifying to contemplate :-)

Leaky Nun

if a double slit can "see" a photon being at two places at once, why can't we?

John Rennie

A slit is not conscious so it doesn't experience anything - it just exists.

But then no-one knows what consciousness is anyway.

Leaky Nun

@JohnRennie have you heard of the Afshar experiment?

John Rennie

No, I hadn't heard of it before.

Silly Goose

08:25

is this idea of complementarity widely accepted? I thought I just read in Schlosshauer's book on Decoherence that complementarity is not true (at least in the theory of Decoherence, which I thought was widely accepted)

Leaky Nun

Silly Goose

Leaky Nun

@JohnRennie he set it up so that the photon detector 1 only detects photons that pass through pinhole 1, and photon detector 2 ... pinhole 2

so he claims that the which-way information is preserved

but then he also claims to demonstrate that the two "photons" still interfere before the lens

by placing a grid before the lens

Silly Goose

oh hm i must be misreadeing

nickbros123

09:50

@ACuriousMind, My main problem with the dirac delta function coming into the pic when discussing point charges' divergence: I understand that pictorially, the divergence doesnt appear to be 0, and moreover, the flux integral comes out to be 4pi no matter what sphere we use, so it mustve pushed physicists to say that the integral of divergence of E to be 4pi also? and i suppose all the stuff with the delta function came

but my issue is, why should the gauss divergence theorem hold in the first place? isnt div E a discontinuous function?

Slereah

Some version of Gauss theorem still holds for distributions

nickbros123

it hinges on the fact that div E is volume integrable right?

Slereah

12:09

Why are there so many axiomatizations of mereology

"More generally, it appears that (P.6) would force one to accept the existence of a wealth of “scattered” entities, such as the aggregate consisting of your nose and your thumbs, or the aggregate of all mountains higher than Mont Blanc."

I just want to do geometry dang it

I'm not trying to solve the ship of Theseus here

Ryder Rude

13:11

@Slereah Y do u want to build geometry frm this stuff?

Geometry is just like a formal manipulation system

Slereah

@RyderRude Points are of the devil

Ryder Rude

Devil's anus

Slereah

'The pointless line we wish to characterize weíll label G, for "gunky"'

gross

DIRAC1930

13:29

Do we know exactly what goes on when a position and electron anihilate to form 2 photons or is all we know that it's a discrete event involving the immediate exchange of quantum numbers. Experimentally, do we have no information about exactly what goes on in an interaction?

Slereah

It is generally considered in some asymptotic manner so what happens in between tends to be a bit vague

DIRAC1930

Have we observed what happens at the point of interaction?

Slereah

it's too smol

the notion of a particle probably doesn't make sense in this context

DIRAC1930

Ah okay thanks

Slereah

13:54

"One mereological notion is that of a fusion or sum: the whole composed of some given parts. The fusion of all cats is that large, scattered chunk of cat-stuff which is composed of all the cats there are, and nothing else. It has all cats as parts. There are other things that have all cats as parts. But the cat-fusion is the least such thing: it is included as a part in any other one."

There's a lot of cat examples

"It does have other parts too: all cat-parts are parts of it, for instance cat-whiskers, cat-quarks."

Ryder Rude

@Slereah Is this cat stuff like a third foundation of mathematics othr than set theory and category theory

I sounds like u r doing philosophy

Slereah

It is at the boundary of the two

Ryder Rude

Oh

@Slereah ok but how does this define the boundary of a cat

DIRAC1930

On second thoughts, these discrete jumps (i.e. the word quantized) are kind of what QM and hence QFT is about

Ryder Rude

Isnt this a huge philosophical problem of precisely defining categories

Slereah

14:03

@RyderRude just look at it

Mr. Feynman

Slereah, what kind of pure Math rabbit hole have you stepped in this time?

Slereah

@Mr.Feynman Pointless math

Whitehead's point-free geometry

In mathematics, point-free geometry is a geometry whose primitive ontological notion is region rather than point. Two axiomatic systems are set out below, one grounded in mereology, the other in mereotopology and known as connection theory. Point-free geometry was first formulated in Whitehead (1919, 1920), not as a theory of geometry or of spacetime, but of "events" and of an "extension relation" between events. Whitehead's purposes were as much philosophical as scientific and mathematical. == Formalizations == Whitehead did not set out his theories in a manner that would satisfy present-day...

Ryder Rude

@Slereah I can see furry quantum fields

Wherr to draw the boundary

This cat fuses into space. It is a part of a large hilbert space

Becuz its interacting with the background field, u cannot isolate this cat

Mr. Feynman

@Slereah Why have people created so many things? I wish I were born in 1600 and know everything they had discovered back then

Or just to be immortal... :P

Slereah

@Mr.Feynman But then you'll have to deal with talking about wrong things mostly

Also the literature and pedagogy of the era was pretty bad

Ryder Rude

14:10

Even Newton was doing alchemy

Slereah

Just gotta read Aristotle's Organon until you get it

DIRAC1930

One day people will think that about our era

Slereah

I already think it

DIRAC1930

lol

Ryder Rude

I think it is 400 years before quantum gravity

I mean our era

Slereah

14:12

I don't even know how people taught classes back then

In the 1600's blackboards didn't even exist yet

and paper was pretty expensive

Ryder Rude

I think they just used books

ok but how wud they practice math

Slereah

Books were also very expensive

Ryder Rude

Oooh quills

@Slereah i mean the teacher had books

Slereah

And then what, everyone huddles around the one copy?

Ryder Rude

No, the teacher just explains stuff and everyone tries 2 remember

And wut about quills

U write on non paper things with quills

Ooh U can use cheap cloth to write on! @Slereah

I mean cheap clothes produced 4 this purpose

Idk how cheap clothes cud b back then

Slereah

14:19

I guess everyone had a book

and a hat

and one guy is having a nap

Things haven't changed much

Ryder Rude

And everyone's 70 years old here

DIRAC1930

QFT courses should be 70 years long

With no mention of scattering amplitudes from asymptotic states

Except as a footnote

Mr. Feynman

@Slereah I wouldn't know :P

Slereah

1600's are an exciting time I guess

that's when you get analytic geometry properly

Mr. Feynman

And without a theory of everything I don't really feel any better :P

Slereah

14:28

They already had a theory of everything back then, it was the bible

Mr. Feynman

Lol

Galileo's biggest mistake was to deny the Bible as a scientific authority

We'd still have a ToE

Slereah

that wasn't the big issue of the objectors of Galileo overall

I can almost read Descartes' original geometry book

It's a bit old timey french

Mr. Feynman

I can barely read books of the XX century, how can you read Descartes? :P

Slereah

It's mostly basic algebra and geometry :p

Ryder Rude

I tried to read his meditations. First two steps were logical and then he brings God out of nowhere

I stopped right there :P

Mr. Feynman

14:37

@Slereah did Descartes use LaTeX though?

Ryder Rude

Some people say only the first two steps r worth anything

Mr. Feynman

I think they only had TeX back then

Slereah

@Mr.Feynman You be the judge

ia803200.us.archive.org/BookReader/BookReaderImages.php?zip=/5/…

Ryder Rude

1600s had terrible hygiene

U wud hav maggots in ur armpits

Mr. Feynman

@Slereah someone needs to get acquainted with the equation envinroment

Slereah

14:39

Paper's expensive back then

probably not economically viable to have pages full of space for readability

Mr. Feynman

Imma go back to 1600 and teach em how to format books until my laptop battery runs out

Ryder Rude

They wud witch hunt u

For knowing future stuff

Slereah

knowing future stuff was a whole industry back then

Astrology and such

Ryder Rude

If u drown, u r spared

Mr. Feynman

Last week I overheard someone in my department mistaking astronomy and astrology

1600 is still here

Ryder Rude

14:45

Tbf to astrology, planets do exert force on us

Slereah

also some people change their behaviour due to astrology, making their behaviour influenced by the stars

including Ronald Reagan

For all we know the Iran-Contra affair may be due to the stars

Ryder Rude

LOL

I now wonder if Astrology can actually help u due to placebo effect

Ignorance is bliss

Slereah

Official white house astrologer

Love how the paper is done with old timey printers

doing her charts on some MS DOS astrology program

Ryder Rude

Wow. I didnt know the White House had this. Explains a lot of decisions

Slereah

well only Reagan as far as I know

and even then because of his wife

"Descartes wrote La Géométrie in French rather than the language used for most scholarly publication at the time, Latin. His exposition style was far from clear, the material was not arranged in a systematic manner and he generally only gave indications of proofs, leaving many of the details to the reader.
His attitude toward writing is indicated by statements such as "I did not undertake to say everything," or "It already wearies me to write so much about it," that occur frequently. Descartes justifies his omissions and obscurities with the remark that much was deliberately omitted "in ord…

How little things change

Ryder Rude

14:59

"This is trivial"

Y books gotta insult u

At least say "exercise for the reader"

ACuriousMind

15:21

@LeakyNun sure, that's the measurement problem - I'm not saying it's clear that there is a unique "correct" way, I'm just saying that this is the way some interpretations attempt to resolve the problem. I was just pointing out that I don't think any interpretations assert collapse somehow acts on the wavefunction at all times

@nickbros123 there is a rigorous theory of distributions in which the claim that the divergence of the Coulomb potential is the delta function is true, but physics texts don't usually develop that theory so you get handwaving justifications. I wouldn't worry about it too much if you're just learning EM

Sir Cumference

16:13

@Slereah Why does it not surprise me that this is what our tax money is being spent on

the money spent on astrology could've gone to astronomy instead lol

Ryder Rude

16:35

@ACuriousMind pls give some thoughts on this physics.stackexchange.com/questions/746454/…

I wrote an answer but idk if it's correct

ACuriousMind

Lots of things to say but mostly I think the question suffers from an oversimplification of the idea of classical limit

since $\langle V(X)\rangle \neq V(\langle x\rangle)$ in general, Ehrenfest's theorem is not enough to establish the classical limit, the idea of the correspondence principle is that you need to find specific "limiting states" for which this equality holds approximately

Ryder Rude

Yeah, the coherent states. The Guassians?

I mean we want both the field value and the field momentum to be approximately well defined

ACuriousMind

the question also suffers from an oversimplification of renormalization

I don't understand the alleged limit $\lim_{\Lambda \to 0} m(\Lambda)$ - you have to specify a renormalization scheme here for this to make sense, and also in the usual notation I would expect $\Lambda$ to be a momentum cutoff and hence be *raised to $\infty$" instead of lowered to 0.

the question is "not even wrong" - it's so vague that I can't really tell what kind of procedure it actually imagines for the classical limit and where the specific problem lies

Ryder Rude

@ACuriousMind I mean that the correspondence principlr for non rel. QM works out becuz the coupling parameters of both the classical and quantum theories r the same

ACuriousMind

the answer likewise makes assertions so vague that they are impossible to judge, such as that we get large negative masses: Have you done any actual computation for a limit that yielded this result? Or did you just guess that this has to happen because it conformed to whatever "intuition" is going on in your head?

Ryder Rude

16:45

@ACuriousMind It's becuz the parameter gets carried over from the quantum theory. In QED, it's known that u get large negative masses for the lattice theory

And infinite negative mass for the continuous theory

ACuriousMind

@RyderRude I don't know any claim so general as "the correspondence principle for non-rel. QM works". I know several specific examples of a correspondence principle for specific systems, but I know of no general theorem that establishes a correspondence principle exists for general quantum systems

@RyderRude If that is "known" I would expect a reference for this claim and also some sort of argument that this is a "physical effect" and not an artifact of the lattice

limits of lattice theories are notoriously subtle

Ryder Rude

Yeah. This is y Connor Behan suggested to make this argument with Scaler fields instead of Spinors

For scalars, I hav an argument. Lemme explain

ACuriousMind

If that argument contains no actual computations of expectation values or amplitudes and actually computing limits I'm not interested

Ryder Rude

@ACuriousMind When u take coherent states with low uncertainty in both position and momentum, u can start with Ehrenfest theorem and do stuff like $\langle V(x) \rangle =V(\langle x \rangle) $

ACuriousMind

@RyderRude what is a "coherent state" in this context?

Ryder Rude

16:49

@ACuriousMind I mean the low uncertainty Gaussian syate for both field value and field momentum

Becuz theyre like harmonic oscillators

ACuriousMind

your question is about a generic quantum field theory, even figuring out what the asymptotic states even are is subtle e.g. for non-Abelian gauge theories

Ryder Rude

Ok so it's very technical

ACuriousMind

e.g. if you're being naive and don't know about confinement you'd try to construct some sort of coherent free quark state which makes no sense

Ryder Rude

Yeah

ACuriousMind

you can't just conjure the states necessary for the correspondence principle into being by assumption, you have to construct the states

Ryder Rude

16:51

Do these states even exist even for the Scalar field theory?

ACuriousMind

for what scalar field theory?

Ryder Rude

I mean Kleim Gordon coupled to Maxwell

And without any phi^3 or phi^4 stuff

ACuriousMind

I don't know since I don't exactly know what states you mean

Ryder Rude

Just $A^{\mu} J^{\mu}$

ACuriousMind

that's what I mean when I say this idea of a classical limit is too vague to judge - since we generally don't really have access to a useful description of the interacting Hilbert space of the theory, I'm not even sure what states you want to try to use Ehrenfest's theorem on here

Ryder Rude

16:54

@ACuriousMind i wud need to think about this

ACuriousMind

@RyderRude so how do you expect anyone to answer your question if you don't even know what you mean :P

Ryder Rude

@ACuriousMind Let's say we take the Scalar field theory on a latticr to avoid the Haag's theorem stuff

Now we can construct Gaussians like we do in finite dimensional QM

ACuriousMind

we can?

have you ever actually done any lattice theory?

as in, computed anything with a lattice model?

Ryder Rude

Well... It's just decoupled Harmonic oscillators :P

ACuriousMind

I suspect lattice theory in practice does not actually work like you imagine it does

for instance, do you know how you have to describe a gauge field like the 4-potential on a lattice?

do you know what a comparator, a link or a plaquette are?

Ryder Rude

16:56

No:P

ACuriousMind

do you know what a Wilson line is?

Ryder Rude

Idk any of the specifics

I just thought i wud descretize the function

ACuriousMind

then frankly you have no idea how lattice QFT actually works

Ryder Rude

So is this question too vague to handle?

ACuriousMind

To me, absolutely

Ryder Rude

16:58

And my idea using Ehrenfest theorem is not even the gist of the solution?

Like a roadmap

ACuriousMind

but I know I'm on the stricter side when it comes to demanding more technical detail :P

@RyderRude I couldn't say

to me the classical limit of QFT in general just works by first taking the non-rel. QM limit of QFT and then doing whatever classical limit you want on that non-rel. QM system

Ryder Rude

Yeah. U told me about this. We get the Coulomb potential. This way

ACuriousMind

or you have something specific like free EM ("quantum optics") where we then can actually construct things like coherent photon states and show those result in the usual sharp values for electric and magnetic field operators in the case of high occupation number average

Ryder Rude

But idk. The argument to go from Coulomb potential to Maxwell's equations wud need additional assumptions?

@ACuriousMind so my solution works for the free theory sometimes

ACuriousMind

i.e. the classical limit of free EM matches more or less your vague idea, but the problem I see is that in general it is not clear that such coherent states exist or are useful descriptions of the classical states

for one, the naive idea of coherent state makes no sense for fermions

Ryder Rude

17:02

Yeah. For fermions, we shud better start with rel. QM to get the classical limit

ACuriousMind

then there's the confinement problem (no "coherent gluon beams")

Ryder Rude

In rel. QM, u can hav a wavefunction coupled to the classical EM field

Ok so QED can yield us rel. QM I guess?

And then we take Ehrenfest's theorem?

But it still must be very hand-vawy

Sp instead of doing the Ehrenfest theorem on operator field dirac equation, we instead first obtain the rel. QM Dirac eqn.

And then we take Ehrenfest theorem. This approch works for spinors?

But idk how to go from operator field equations to rel. QM @ACuriousMind Is it the same argument where u dot the wavefunction with the approximate position eigenstates?

ACuriousMind

@RyderRude yes, this is the general argument for one-particle wavefunctions fulfilling the equation of motion of the field, cf. physics.stackexchange.com/a/238761/50583

Ryder Rude

That argument doesnt directly use the operator field equations

@ACuriousMind so wud u say this new roadmap from QED to Maxwell is not vague?

I mean we first get rel. QM and then we do Ehrenfest theorem

To get rel. QM, we dot with the $|x,t \rangle$

ACuriousMind

why do you want to go from QED to Maxwell?

if you want the Maxwell equations with sources, you still have the problem of finding a "coherent state" for the electrons that produces some classical current $j^\mu$

Ryder Rude

17:13

Becuz QED is the deeper theory. Im just interested in how Classical electrodynamics arises

ACuriousMind

nothing you've said so far explains where the classical current comes from

Ryder Rude

@ACuriousMind oh yeh. I have a thing in mind

A roadmap

1. We start with coupled operator field eqns. 2. We do the coherent state thing for the EM field alone and make it a classical field 3. Now we have an operator field coupled to a classical field 4. We dot with $|x, t\rangle$ to get rel. QM of Dirac eqn coupled to classical EM field 5. We take Ehrenfest theorem @ACuriousMind

I know some steps arent well defined. But is this a good roadmap?

ACuriousMind

I mean, it sure is a road map, but I think you've omitted exactly all the things that are hard :P

Ryder Rude

Lol

I do know that there exists a rel. QM theory of Dirac coupled to classical EM. But i just dont know if i can derive this theory starting from QED

I think step 2 in my roadmap is hand wavy

Ooh but step 4 is even more Handwavy

@ACuriousMind I like the other approach where we just go non relativistic. But can Maxwell b determined from Coulomb's law?

I mean if latter add relativity back

I mean after we've already derived the classical theory of particles interacting using Coulomb's law

If we add relativity to this, is Maxwell's theory with charges uniquely determined

120

Q: Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?

As an exercise I sat down and derived the magnetic field produced by moving charges for a few contrived situations. I started out with Coulomb's Law and Special Relativity. For example, I derived the magnetic field produced by a current $I$ in an infinite wire. It's a relativistic effect; in the ...

This means that the non. rel route is the least handwavy one

We just get Coulomb's law using two point function, instead of taking handwavy expected Values on the coupled operator fields

And then we can later add relativity to the Newtonian theory to get back to Maxwell with sources

It's crazy how the roadmaps r completely different for the Free maxwell and the Coupled maxwell theories. We have to take a very indirect route becuz of spinors

ACuriousMind

18:09

I'm reading Moretti's 2019 book on mathematical foundations of quantum mechanics because its table of contents looked relevant to my interests (in particular results about the mere structure of quantum probabilities forcing the use of complex Hilbert space to represent quantum observables and states) and it's an interesting hybrid book that oscillates between pure mathematics and rather insightful physical discussion.

I would recommend it to anyone interested in algebraic formulations of quantum mechanics.

(I initially started looking at it because I wanted a more detailed version of the ideas alluded to in this answer)

DIRAC1930

18:32

@ACuriousMind I think you and @bolbteppa had an interesting point yesterday

What did you mean by experimental uncertainty?

There is a short summary of the Landau paper on the second page 2nd paragraph here tau.ac.il/~yakir/yahp/yh154

Why can't I measure the potential interaction $V$?

It seems to be that the longer I leave on a time-independent perturbation, the more it is like it was never there which sounds very counterintuitive. And therefore, $E+\epsilon = E'+\epsilon' + V$ tends to $E+\epsilon = E'+\epsilon'$

This sounds silly but I don't know if it is true

$P(E\rightarrow E')=\sin^2((E-E') t)/(E-E')^2$ goes to $\delta(E-E')$ for $t\rightarrow \infty$ i.e. the state never transitions so it was like there was no $V$ to begin with so the result of the measurement for $E,E',\epsilon,\epsilon'$ becomes exact

here $P$ is the probability density

@ACuriousMind What do you think?

And then the uncertainty in $P$ for low times is due to not knowing if we are measuring $E$ + a bit of $V$ hence we get $\Delta P$

My trouble is that we know what the interaction potential is so we could probably predict it actually

ACuriousMind has given up on me

Mr. Feynman

19:29

@ACuriousMind So you keep reading Physics books :P

Let me check if that's the one I have

The book I own is the (Italian) 2010 version of the (English) 2013 book titled "Spectral Theory and Quantum Mechanics: With an Introduction to the Algebraic Formulation"

The difference is that the last chapter about the algebraic formulation is missing

ACuriousMind

@DIRAC1930 I don't always immediate respond here :P The reason $t\to \infty$ results in the energy not changing is just that if the perturbation was always on, then if you start with an energy eigenstate it won't evolve because it is a stationary state for all t!

@Mr.Feynman it's the first new one I've looked at in quite a while

Mr. Feynman

Anyways, I'm planning to read it next year

ACuriousMind

I was mostly interested in chapters 4 & 5 discussing probability lattices and the no-go theorems like Kochen-Specker; chapter 4 can't in this form be present in a 2013 book you have because it references recent results up to 2019

Mr. Feynman

I should have been more precise: I'm planning to read the one I have regardless of the algebraic formulation :P

ACuriousMind

I understood that but I doubt the claim that the difference between the book is just that the last chapter is missing

Mr. Feynman

19:41

I'm still scared by the idea of becoming a theoretical physicist without knowing rigorous basic QM

@ACuriousMind the difference is between my 2010 edition and the 2013 one

ACuriousMind

ahhhh

I completely misunderstood

Mr. Feynman

Again, I was ambiguous :P

DIRAC1930

Sorry, I meant $E$ is the energy state with no perturbation. Upon adding the perturbation and leaving it for a time $t$, in the limit of $t\rightarrow \infty$, the energy state will have a value of $E$ as it was with no perturbation. Only for short times will the possibility of transitioning to a different energy $E'$ occur.

This appears to be what the equations are saying but it seems kind of dodgy (unless I am misinterpreting the equation)

Hmm it looks like the perturbation is always on

tcm.phy.cam.ac.uk/~bds10/aqp/lec18.pdf

pg 45. here

set $\omega=0$ and you will get the same

It says that the perturbation is suddently switched on

DIRAC1930

20:21

Does this seem wrong to you?

Silly Goose

21:05

@ACuriousMind have you read any papers about what the structure of states looks like when you start with a set of observables that do not form an algebra. I thought this may be of interest to you if not

Like this arxiv.org/pdf/1909.12851.pdf

DIRAC1930

21:27

How does the process of potential energy being converted to something like kinetic energy work in QM?

Also, there is some information here ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/… on pages 90 and the few pages after

When we solve for a finite potential well, we have a set $V$ for the lowest part of the well

In the discussion in the notes and the Landau paper etc. it is suggested that only some of the potential energy goes into kinetic energy etc.

DIRAC1930

21:47

On second thoughts, I suppose $\hat{H} = \hat{H}(X_1) + \hat{H}(X_2)+\hat{V}(X_{1},X_2)$ is a conserved quantity. $\hat{H}_1 + \hat{H}_2$ is not conserved

When we are measuring before, we are actually measuring $\hat{H}_1 + \hat{H}_2$ (the perturbation is off)

The measuring perturbation is switched on subsequently (it is never switched off after) and therefore we are measuring $\hat{H}_1+\hat{H}_2$ after the measurement. As $t\rightarrow \infty$, $\langle{\hat{V}}\rangle \rightarrow 0$

And therefore the eigenvalue of $\hat{H}_1 + \hat{H}_2$ is identical to the eigenvalue of $\hat{H}_1 + \hat{H}_2 + \hat{V}_{12}$ for the state after the measurement

Sorry the last part of the first sentence on line 3 is wrong

bolbteppa

I'm not sure what you're doing with this additional $V(x_1,x_2)$, why wouldn't contributions like that already be incorporated into the final $E$. In addition, modelling the interaction of a measuring apparatus with a closed quantum system through some simple $V$ potential is not practical, the measuring apparatus in reality is composed of billions of particles and one has to account for the classical nature of the apparatus in all this too, is this not all summed up in writing $E + \epsilon$

They explain the measurement process in section 7

There is a fundamental difference between any old perturbation temporarily being turned on, and a measuring apparatus interacting with a system which allows us to draw conclusions about a measurement from it, all tracing back to classical physics, without which we can fundamentally get no conclusions

DIRAC1930

22:02

You are right that $E+\epsilon$ is correct before the measurement. However $E'+\epsilon' +V$ is correct after the measurement but we are not measuring $V$. We are limited to measuring $E'+\epsilon'$ leading to an uncertainty. In the limit of $t\rightarrow \infty$, the $V$ is not important (since all transitions are supressed) meaning that roughly $E'+\epsilon' +V = E'+\epsilon' $

bolbteppa

No, is $E' + \epsilon'$ not the value after the measurement

DIRAC1930

Yes

bolbteppa

$E = T + V$

You're separating out a $V$ for some reason

@RyderRude there are technically infinite 'renormalization' constants floating around in classical electromagnetism, e.g. in the mass terms

DIRAC1930

This whole uncertainty is due to the difference between $E' + \epsilon'$ and $E' + \epsilon'+V$. This uncertainty goes to $0$ in the limit of $V\rightarrow \infty$

Remember that $E+\epsilon - E'- \epsilon'$ can only be verified up to $\hbar/\Delta t$ however $E+\epsilon - E'- \epsilon'-V$ is always equal to $0$

from energy conservation

bolbteppa

@RyderRude Classical electrodynamics exists, instead of deriving it from quantum mechanics, it's actually the reverse, we derive QED based on the existence of the classical theory, even though the quantum theory is more fundamental, it's an unavoidable contradiction (in standard Copenhagen QM). In practice, one can derive the transition explicitly from QED in the case of having the occupation numbers of all states large, the bose operators becoming commutative etc... (it's done in L&L)

I don't think what you're doing with this $V$ is right

DIRAC1930

22:09

I need to think about it some more

bolbteppa

I think the contribution from the $V$ is already included in the $E'$, as well as the $|F_{fi}|^2$ contribution to the amplitude $|a_{fi}|^2$

DIRAC1930

This $\sin^2$ formula suggests that if I leave the perturbation on for an infinite amount of time, it was as if there was no perturbation to begin with since the probability of transition to a different energy goes to $0$

This is why I think $E'+\epsilon'+V$ can be replaced with $E'+\epsilon'$ in the limit of infinite time

bolbteppa

One doesn't need to go through non-relativistic limits to get this, it's all relativistic, the only limiting process is in taking the 'classical limit', which is essential to do things like fix any constants in the Lagrangian (just as we derive Schrodinger, the momentum operator etc... by taking classical limits), at best the non-relativistic limit arises via things like fixing the mass of a particle by taking the non-relativistic limit in the Lagrangian...

The perturbation is periodic (i.e. a Fourier contribution to a general perturbation) so it is on for an infinite amount of time in this derivation technically (depending on what they're discussing, they're only treating the measurement as a perturbation in that section but that's not a big deal I think)

DIRAC1930

$\omega\rightarrow 0$ means that the perturbation is constant, but then in all the sources I've seen with that perturbation, they state that it is off before $t=0$ and then switched on. I think maybe they integrate only from $0$ to $t$ which has something to do with it but I haven't looked into it too much

e.g. page 45 here tcm.phy.cam.ac.uk/~bds10/aqp/lec18.pdf

And page 90 here ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/…

There is some interesting information on page 90 and 91

DIRAC1930

22:40

See also here on page 3 and 4 daarb.narod.ru/mandtamm/mt-eng.pdf

DIRAC1930

23:41

What does he mean by the last statement written in the first column of page 252 starting with 'Hence the probability'?

bolbteppa

23:56

Think he's trying to say one can't misinterpret his (6) because the probability is determined by (42.3) LL

DIRAC1930

The 'nots' in that sentence are confusing

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