The h Bar

DIRAC1930

14:00

If I have an electron in front of me, why isn't it a dressed particle?

ACuriousMind

@DIRAC1930 what does that mean

it is "dressed" in the sense that it has its renormalized and not some "bare" mass, to the extent that that is a meaningful statement

DIRAC1930

A one particle eigenstate of the full interacting Hamiltonian

Oh, that's what dressed means

Ryder Rude

@Amit But what u cudnt say is that the metric field is emergent out of the other fields. This is becuz there r metric-less theories

I agree that processes subsume a manifold

But they dont sumsume a metric

ACuriousMind

@DIRAC1930 why would it be an eigenstate of the interacting Hamiltonian? There's no law that says stuff must be in energy eigenstates

DIRAC1930

I measured it to have energy $E$

ACuriousMind

14:05

what does that mean? what apparatus did you use that you expect to be measuring the energy according to the interacting QFT Hamiltonian and not the free Hamiltonian of which the asymptotic states approximate eigenstates since they're plane waves in momentum space

DIRAC1930

I dunno

A state is a superposition of energy eigenstates of the full interacting Hamiltonian

with maybe some other quantum numbers too

Ryder Rude

I think these apparatuses do measure the full Hamiltonin energy. It's just that the free energy plus asymptotic dressing energy is a good approximation to the full energy asymptotically? @ACuriousMind

Becuz u can never expand an interacting theory wavefunction in an eigenbasis of free energy operator. So these apparatuses cannot measure the free energy

ACuriousMind

@DIRAC1930 sure, and the asymptotic space is a subset of the full interacting space consisting of the scattering states that actually has the structure of a Fock space of free particles

@RyderRude what apparatuses?

Ryder Rude

Classically, if u measured the energy of an asymptotic particle, u wud measure the particle's energy plus the dressing energy

@ACuriousMind i mean the energy measurement apparatus in scattering experiments

ACuriousMind

@RyderRude calorimeters are not "I have an electron sitting in front of me and measured it to have some energy", they are very much rather destructive with respect to the incoming state

the outcome isn't so much some eigenstate of energy and more that your high-energy particle hit the calorimeter and disintegrated into a lot of products from which you re-engineering the original energy of the incoming particle

DIRAC1930

14:18

Is the issue that current calculations in QFT can't really do anything other then $t=-\infty$ to $t=+\infty$ scattering

ACuriousMind

what issue

DIRAC1930

Well everything

ACuriousMind

I feel that you keep gesturing at problems but I'm not really getting a clear picture of what concrete problem there is other than that QFT doesn't work like you expected it to :P

DIRAC1930

Well QM seems to answer questions about how a particle behaves better than QFT (ignoring particle production/destruction processes)

ACuriousMind

Well, the QM description is certainly usually more tractable

in that you actually know what your Hilbert space looks like (praise be to Stone-von Neumann) and can just do perturbation theory without being plagued by Haag's theorem or infinities or whatever

Ryder Rude

14:33

@ACuriousMind If a QFT lattice is infinite, does Haag's theorem still apply?

ACuriousMind

again, lattice theory doesn't work like you think it works

Ryder Rude

Yes, it has othr issues like Fermion doubling and derivative discretization. But is Haag's theorem still an issue there?

Assuming lattice has no boundary

DIRAC1930

What about this en.wikipedia.org/wiki/Haag%27s_theorem#Work-arounds

ACuriousMind

but no, lattice theories are not subject to Haag's theorem, just like they do not produce "infinities" during renormalization

Ryder Rude

I think since it's infinite, Haag's theorem is still an issue

Yeah, there r no operator valued distributions anymore

I was thinking that Stone Neumann theorem may not apply to infinite lattice

@ACuriousMind so we can say that these r tied : making products of distributions well-defined and Haag's theorem

ACuriousMind

14:36

@RyderRude what would the operators be to which you want to apply Stone-von Neumann?

you don't have continuous position or momentum on a lattice!

Ryder Rude

Like there wud still b an infinite number of canonical commutation relations whose representation has 2 b unique

It's just that the RHS wud b Kronecker delta

ACuriousMind

@RyderRude canonical commutation relations between what

you're on a lattice, how are even constructing a Hamiltonian theory that gives you some canonical conjugate to your field

Ryder Rude

The field amplitude and momentum at each lattice site

Amit

@RyderRude Clearly there is something right about this metric business, which also implies time really is not independent from space... and yet we perceive rocks and not metrics lol.. also we perceive objects and not the space they occupy. We always perceive in 2d as well :)

ACuriousMind

I'm not saying there aren't approaches that actually do this, I'm just pointing out - again - that putting a field on a lattice is much subtler than you keep pretending

@DIRAC1930 what about it?

Ryder Rude

14:40

@Amit remember that if u go this route, u wud need to deny th existence of elementary particles like the electron in the hydrogen atom. U only ever measure its properties in a measurement apapratus, similar to how u can measure the energy of a graviton

U never see any elementary particle

A graviton will carry energy-momentum that can b measured just like any othr particle

Amit

I don't deny but I do like to distinguish direct perception from the models we invent

DIRAC1930

How is the universe translationally invariant if it has a boundary?

ACuriousMind

@DIRAC1930 who says the universe has a boundary?

DIRAC1930

Well if it started from the Big bang, at some point it would have had a boundary

Amit

The atom is an interesting case, on the one hand it's a model, on the other things like that "a boy and his atom" movie are quite direct :)

Slereah

14:43

It's not time translation invariant, if you mean the big bang

It has no boundary in the directions where it is invariant in

Ryder Rude

@Amit i mean electrons

Slereah

Universes that are translation invariant in all directions don't have such boundaries

ie Minkowski space, De Sitter and Anti De Sitter

ACuriousMind

@DIRAC1930 no

Ryder Rude

If u go ur route, u will only accept classical entities that r directly correlated to our mental states. U wud need to deny the existence of elementary particles that make up those classical entities @Amit

Amit

It's not a matter of denying. Distinguishing models from measured quantities is important 'cause models are questionable.. measured stuff aren't to the extent your appartus functions correctly

DIRAC1930

14:46

Also, isn't translational invariance spoiled by mass curving the space in one region

Ryder Rude

In fact, one may go ahead and just believe that only their brain exists. This is the thing that is directly correlated to mental states. Everything else is a model

Amit

I'm not a solipsist

Newton's model of gravity turned out to be wrong

Ryder Rude

I'm just saying that this discussion is being unfair to the metric field. The same discussion applies to all aspects of our models

Amit

Same with bohr's atom

Ryder Rude

@Amit yes, if u dont single out the metric field, then ur PoV is correct

Slereah

14:47

@DIRAC1930 Yes, that's because people have very broad notions of "translation invariant"

Amit

Oh yeah, all the same

Ryder Rude

Then i agree

Slereah

The universe is homogeneous on the scale of super clusters

Amit

Maybe I just entered this discussion with insufficient context, sry

Ryder Rude

It's all good. It made me refine some concepts about ontology :)

Amit

14:49

Cheers

Ryder Rude

@ACuriousMind lattice QFT isnt about canonically quantising a lattice classical field?

ACuriousMind

@RyderRude not really?

like, a lot of lattice field theory is just Monte-Carlo evaluations of discretized path integrals

doesn't even mean really anything to ask whether that's "canonically quantized" or not

Ryder Rude

Ooh that. But that's also the same as quantising a lattice classical theory. It's just that we're doing path integral quantisation

But... path integrals assume a Hilbert space wavefunction. That's what the path integral evolves in time. To define the Hilbert space, we need representation of the CCR? @ACuriousMind

Obliv

15:08

Is the jacobian of an integral the total differential ?

ACuriousMind

@RyderRude I don't think about lattice theory as its own completely well-defined physical theory

you don't define the path integral in some physical sense via states or whatever

you just take the continuum path integral, write down a more-or-less discrete version of it that you can reasonably believe to converge against the continuum version, and off you go doing Monte-Carlo

there is of course more physical approaches to lattice theory where you take it seriously as a physical theory and not just as a computational tool, it's not all like that

but you can't just assume what people mean when they say "lattice theory" :P

for instance, if you want to have a Hamiltonian formulation, which you'd like in a proper physical theory, you more or less need to not discretize time

but of course computers (the main reason we do lattice theory in practice!) can't do continuous time steps, so you have some sort of "double discretization" where you first put the theory on a spatial lattice and then have the additional numerical approximation for the time steps

which is confusing to interpret because the code sure looks like you put the theory on a 4d lattice, but what you're really doing is numerically approximating the exact theory on a 3d spatial lattice

the naive belief that this should always be the same thing is the same as the naive belief that all lattice theories should have the "obvious" continuous versions as their continuum limits, i.e. it's wrong :P

Ryder Rude

I like lattice theory mainly to get rid of Haag's theorem. So i prefer the physical version as a Quantum theory with a countaby infinite number of CCR

But lattice theory has other problems like derivative discretization and fermion doubling

I still think it's worth it to get rid of Haag's theorem.

ACuriousMind

15:23

honestly you probably only like lattice theory because you've never actually done it

what you like is the simple idea of it, not its actual practice

Ryder Rude

Yeah i just like to think about it, as a way to think about QFT concretely. Otherwise, Haag's theorem makes everything mysterious and unknown

I just need smthing concrete

ACuriousMind

yes but stuff like fermion doubling means your naive idea about lattice QFT is equally wrong!

that you think Haag's theorem is a showstopper but the various issues with the continuum limit aren't is not an attitude I can follow

Ryder Rude

People hav made peogress on Fermion doubling. It can go away with derivative discretization. Still, Haag's theorem is an issue for EVERY interacting QFT, not just fermions!

So lattice theory is relatively much better than having Haag's theorem around

The new problms r somewhat smaller

ACuriousMind

@RyderRude but Haag's theorem is also not a problem in practice!

it's just a problem if you try to take the interaction picture seriously

Ryder Rude

well... Idk

The interacting continuum theory also has infinite parameters. It's just very hard to concretely think about

ACuriousMind

15:30

like fermion doubling, it's not that the problem isn't fixable, it's that the solution isn't what you'd naively expect

@RyderRude causal perturbation theory/Epstein-Glaser renormalization knows no infinities

you only get infinities if you insist that you can multiply distributions at a point like numbers

Ryder Rude

@ACuriousMind I once read in an answer of urs that non-uniqueness of the representation of interactinf theory shows up as a problm with the path integral measure

Am i remembering correctly

ACuriousMind

it's possible

I don't know about it off the top of my head

Ryder Rude

So all this stuff is connected : Haag's theorem, Distribution multiplication, Path integral measure. There's also a fourth thing: the Hilbert space inner product, which also amounts to choosing a representation

The fourth thing i read in a Valter Moretti answer

It's just that, with interacting QFT, u dont hav uniqueness. So we need experiments to choose the right theory

This is also what renormalisation does

ACuriousMind

uniqueness of what?

or, rather, what does the "interacting" have to do with uniqueness of anything?

Haag's theorem also tells you that the representations of two free fields with different masses are inequivalent

Ryder Rude

Uniqueness of let's say : the representation of CCR, or the definition of operator distribution product, or the inner product on the hilbert space

Or the path integral measure, i guess

Choosing any of this stuff amounts to choosing the right QFT

@ACuriousMind Like how renormalising the parameter using experiments is the same as choosing a well defined definition of the operator distribution product

ACuriousMind

15:39

that's...not exactly how it works

I think I said that once and you seem to have taken it as gospel that there is a direct map between renormalization parameters and choosing a product

like everything, it is not that simple

when I mention these things, I generally do that so that people start looking into them if they're interested in the details, but you just seem to ignore all the details I can't fit into a few chat messages and then start wildly speculating

I find that frustrating

there are bodies of literature about every single one of these issues

Ryder Rude

Yeah i will need to study this

I found the Valter Moretti post : physics.stackexchange.com/questions/92549/…

In the last line, he says choosing a $\mu$ is the same as choosing one of the unitarily inequivalent representations

This $\mu$ here is related to the Hilbert space inner product

I shall try to find ur post about the path integral measure

ACuriousMind

@RyderRude And have you understood what role the $\mu$ is here and what representations he is talking about?

Ryder Rude

I think it was one of ur answers on wavefunctionals

PM 2Ring

@DIRAC1930 No, because the Big Bang happened everywhere. See physics.stackexchange.com/q/136860/123208

ACuriousMind

hint: he's not talking about any interacting representations!

Ryder Rude

15:47

iirc, the $\mu$ here is an positive definite inner product between classical field trajectories

ACuriousMind

i.e. this post isn't saying that there is some sort of general equivalence between the choice of an inner product and the CCR representations

Ryder Rude

It is inherited by the Hiilbert space inner product after we quantise

ACuriousMind

but that's what you're trying to cite it as

@RyderRude What I'm after is this: The $\mu$ is a generalization of the classical mass (hence the symbol!) and the representations constructed here are the "free" representations of mass $\mu$

there is no claim here that in general representations of the CCR would be equivalent to choices of arbitrary products on this space

Ryder Rude

Oh. So he is only dealing with the free theory representations with different masses

ACuriousMind

so this is not, in fact, a reference for what you claimed it is a reference for

this is why I insist that the details matter

Ryder Rude

15:50

So we rule this out this inner product thing :P

I read this long ago. I think i forgot it was about free theories

I thought this was deeply connected to Haag's theoren

ACuriousMind

it is in the sense that he shows a generalization of the part of Haag's theorem that says that representations of free fields with different masses are inequivalent even in a setting where we aren't in Minkowski space with global Poincaré invariance

Ryder Rude

Ok i found ur answer about the measure : physics.stackexchange.com/a/721726/156987 @ACuriousMind

Ryder Rude

16:45

@ACuriousMind I was wondering that non. rel. QFT has continuum operators. So it must b subject to Haag's theorem. But then again, non. rel QFT is equivalent to non rel QM, which is not subject to Haag's theoren

non. Rel. QM has Stone Von Neumann theorem applicable to it

i dont mean that non. Rel. QFT is always equivalent to non. Rel. QM. It's not in cond. matter

I mean that non. Rel QM can b translated into the language of non. Rel. QFT with a Schrodinger field. There, Haag's theorem would say that there r inequivalent representations

imbAF

19:59

@Allure It's the formula for the solid angle when considering the Rutherford scattering of alpha particles in a gold sheet

imbAF

20:12

For some arbitrary propagating particles, I have noticed that people use the concept of current density and probability current density with no distinction. Isn't that wrong?

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