okay that makes sense. so can i say that SvN tells us that the hilbert space is fully determined by the CCRs of the theory? @ACuriousMind

the part of it where position and momentum act, yes

okay and besides the rotation group is there any other part of the space that needs to be accounted for? @ACuriousMind

usually not

okay and in your answer, you write a term $L^2(D'(\mathbb{R}^{n-1},d\nu))$ what is $D$ here?

i see that you have said a suitable space of distributions but im not sure what is meant by this

what about the phrase is unclear? I deliberately didn't want to be all too specific about it because the whole point of my answer is that it doesn't really matter

well i was trying to get more at what about qft gives rise to the inability to describe the entire theory on a hilbert space. but now im thinking maybe that paragraph doesnt matter and maybe i can appeal to the fact that the various "theories" in qft are governed by different CCR, so they do not necessarily share a hilbert space. but then this doesnt really prove that they dont share a hilbert space so id still need to find a way to show that

As I also say in the answer, the crucial point is Haag's theorem and its related statements, which is somewhat the opposite of the SvN theorem for finitely many CCRs

yes i am looking into Haag's theorem but i got even more confused. iirc, @Mr.Feynman said qft uses interaction picture but Haag's theorem says we cannot use interaction picture? so i was trying to figure out my misunderstanding there as well

where the SvN theorem guarantees that all representations of $[x_i,p_j] = \mathrm{i}\delta_{ij}$ are isomorphic, Haag's theorem essentially says that two representations of $[\phi(x),\pi(y)] = \mathrm{i}\delta(x-y)$ which have different Wightman functions (i.e. vacuum expectation values $\langle \phi(x_1)\dots\phi(x_n)\rangle)$ cannot be isomorphic as representations of those CCR

@ACuriousMind how can two representations have different expectation values?

I mean the Wightman functions are what (via LSZ or whatever else) yields the scattering amplitudes, if they were fixed by the representations of the CCR then how would it matter whether there's an interaction term in your Lagrangian/Hamiltonian or not?

it's sort-of obvious that they have to differ between free and interacting theories, the non-obvious part of Haag's theorem is that this means the representations of the CCR in the two cases are inequivalent - in contrast to the finite-dimensional cases, where all the interaction Hamiltonians and scattering amplitudes can live on the same unique representation of the CCR

the "intuitive" explanation of Haag's theorem is that in the infinite-dimensional case the interacting vacuum is always "polarized" - it always has "infinite energy" with respect to the free vacuum before renormalization - so those two objects, the free and the interacting vacuum, cannot exist inside the same Hilbert space

well i didnt expect them to be fixed by the CCRs but i thought representations should yield the same result upon acting on something because i thought representations are just a nonunique way of expressing something (in this case the fields) in a way that they can act on elements that we want them to act on?

I'm afraid I don't really know what you mean by that

if you didn't expect the expectation values to be fixed by the CCRs, then why did you ask "how can two representations have different expectation values?"

if all representations of the CCR had the same Wightman functions, then that would be exactly what we would mean by "they are fixed by the CCR"

okay so i think i was confused because i know that if we have an operator, there are lots of ways to represent that operator. but what we measure in physics is something very specific, so it should be that we can recover that result with our formalism. so i think this means that physics restricts the representations we can use?

what i said above was that i expect there to be a unique expectation value since that is what we can measure, so i was thinking that all representations should recover that expectation since there isnt a unique representation, but i think that is wrong @ACuriousMind

don't forget that an expectation value is relative to a state

the vacuum in QFT is very special because it is the natural state from which all the other states are created by acting on it with the field

so it's natural to look at the vacuum expectation values in every representation of the CCR, and in fact it turns out that knowing the Wightman functions suffices to know everything there is about the QFT

in the finite-dimensional case we don't have this "special state" - sure, you can always look at the $\lvert 0\rangle$ ground state of the QHO, and the expectation values of $x$ and $p$ are always the same, but if you aren't looking at a QHO system, why would that matter?

the thing that is more wobbly in QFT is that it doesn't work like ordinary QM where for each individual system we look at the structure of the eigenstates of the Hamiltonian of that system etc

instead - at least in relativistic, hep-th QFT - we just look at what happens to the vacuum expectation values, more or less

@ACuriousMind oh no wait sorry im still stuck on this. i still dont see why the reps of the CCRs "have to differ between free and interacting theories" :,P

@Relativisticcucumber note that I said that this is the non-obvious part of Haag's theorem!

in fact, it is so non-obvious that most standard treatments of QFT entirely ignore this fact :P

the "they" in the "sort-of obvious" part of my statement refers to the Wightman functions

@ACuriousMind oh bleh i misread

so i guess its good that i didnt get that bc it would have been wrong lol

@ACuriousMind oh no why should a vacuum state ever have anything different about it in any theory?

meow @naturallyInconsistent

er, what i meant is why arent all vacuum states the same?

@Relativisticcucumber because the interacting vacuum is filled with vacuum bubbles (divergent Feynman diagrams with no external legs)

@Relativisticcucumber meowwww jigglypuff

in the naive interaction picture approach, you need to divide by an expression that represents those several times, in particular in equations that relate the free to the interacting vacuum, so since this expression is divergent, the two cannot actually be related

are fields thought to be physical or just mathematical tools? im inclined to ask how it is specified in a vacuum physically that it should give rise to specific particle types and not others

you should not try to understand any of what I'm saying right now before you have seen the standard derivation of QFT scattering amplitudes via LSZ

@Relativisticcucumber I don't think this is an answerable question :P

what does it mean for something to be "physical" as opposed to "just a mathematical tool"?

are classical Lagrangians physical or "just" mathematical tools?

why does the distinction matter?

okay too philosophical. so iiuc, SvN theorem tells us that the representations of the CCRs in QM are isomorphic to each other and then Haag's theorem tells us that in QFT this is not the case and actually they must be distinct and thus we have independent hilbert spaces for our theories in QFT.

is this correct? @ACuriousMind

@ACuriousMind i guess it tells me whether the question is worth pondering or not :PP

@Relativisticcucumber the first part yes, I have no idea what you're trying to say in the part about the rotation operators

the space $S$ is determined by the intrinsic total spin of the particle (e.g. 1/2 for elementary fermions like the electron), and being spin-z up or spin-y up or spin-up w.r.t. any other axis is just a specific state in that spin-1/2 space - don't forget that $\mathbb{C}^2$ is just the Bloch sphere

okay so you mentioned there is a representation space $\mathbb{C}^2$ and there are specific operators in this space that are given by a representation map, e.g. pauli matrices. i feel like somewhere between having this representation space and making a measurement, there is some assumption being made, but im not exactly sure where

because it seems that, on paper, i have a representation space, and i can choose any representations given by a rep map, but when i make my measurement in the lab, it encodes some specific information that limits this freedom, right? @ACuriousMind

have you not dealt with intrinsic spin in QM?

like, how the state of spin-1/2 particle lives on the Bloch sphere?

that depends on what you mean by "dealt with"

because this is an issue completely disjoint from anything in QFT

yes i agree with that

there's nothing different about measuring spin from measuring e.g. position conceptually

I don't really understand what the confusion is

i think i found out how to explain my confusion. so i need a representation to express my operator. my expectation values require an operator. if the representation is not unique, then theoretically i should be able to get many different expectation values. however, if i go into a lab, i do not encounter this issue, right? i just get an expectation value with respect to my states?

so it would seem somewhere along the way an assumption was made, but where?

@Relativisticcucumber did you notice how I always specifically said stuff like "for elementary fermions with spin-1/2, that representation is the Bloch sphere?"

the information about which representation of the rotation group to use is the intrinsic spin of the particle/system under consideration

@Relativisticcucumber You get to choose a representation of an operator, but that choice also changes the representation of the quantum state. The (transition/probability) amplitudes are independent of such choices.

(and thus the expectation values are also representation-independent.)

@naturallyInconsistent what? No, there aren't even the same kinds of quantum states in representations of the rotation group with different spins

I mean, these are the kind of stuff that if you would simply go out and compute some examples, it will become exceeding clear.

like how for spin-1/2 you only have +1/2 and -1/2 but for spin-1 you have -1,0,+1 as possible $m_z$s

@ACuriousMind within the same spin subspaces... not different spins... why ACM attacc meoww...

sadddd

I'm talking about, like, Dirac gamma matrices, you can choose to use mass basis, or chirality basis, or what have you, and when you choose that, you simultaneously choose the Dirac spinors. In the end, the physically meaningful contracted stuff will not depend upon the representation chosen.

I think we might be using "representation" differently, but I'm not sure which meaning the cucumber is stuck on

when I say "representation" anywhere in the preceding conversation, I mean "representation up to unitary equivalence", i.e. I don't really care about the specific form of the operators chosen

I'm thinking that the Lorentzian vegetable is stuck on something trivial, like when you rotate an experiment. Then the spin half Pauli matrices mix up between themselves

Or when you choose between using position rep or momentum rep.

Things of that form

i feel like you could talk about all this independent of a choice of vector space basis which seems to be a needless detail,no ?

@SillyGoose that's literally what I meant when I said I'm talking about "representation up to unitary equivalence"

it's also the sense in which the SvN theorem says that there is only one representation of the CCR

which is why I thought it implied we're not worrying about expressing operators in different bases, since this conversation started with the SvN theorem

@SillyGoose That is what I was talking about too, lol

@ACuriousMind but nooo, the Lorentzian vegetable was specifically saying that it is time to slightly shift away from SvN and discuss a bit about choice of representations of operators.

i think i need to read up on rep theory in quantum. i will start with spin i suppose

sorry i was thinking for awhile

i think the issue is i have only seen spin 1/2 so i think im overgeneralizing some things and i need to clarify more about spin

@Relativisticcucumber The nice thing is that you only need to know spin 0, half and 1. Spin 0 is trivial, so hey, you just need to learn about spin 1, and it is just vectors! Familiar stuff! Just maybe presented differently because spherical harmonics are wacky

@ACuriousMind you recently closed a question about virtual particles deriving Coulomb force law as being a duplicate. I think it should be closed for being philosophy rather than about the specific route to the derivation. He isn't asking how the maths works. He is asking how fictional stuff can give rise to actual effects.

in the following, i will represent transpose connugate by * and adjoint by $\dagger$. if any operator on a hilbert space has real eigenvalues, and it transforms $\psi $ as $e^{iOa}\psi$, then it transforms $\psi ^*$ as $e^{-iOa}\psi $. this implies $\psi (a) \psi ^ (a)= \psi^* e^{-iOa} e^{iOa}\psi = \psi ^* \psi$

this means such operators leave $\psi ^* \psi$ invariant. but boost doesn't leave this invariant. instead it leaves $\psi ^{\dagger}\psi$ invariant

so boost is not one such operator, which means it's not write-able as $e^{iOa}$ where $O$ is a real eigenvalued matrix

is this correct? im talking about the infinite dimensional boost.

but if we switch to a non rel normalised basis, the boost leaves $\psi ^* \psi$ invariant, as now it's the same as $\psi ^{\dagger}\psi$. so this means boost is now write-able as $e^{iOa}$

so write-ability as $e^{iOa}$ depends on choice of basis?

this doesnt make sense as $A=e^{iOa}$ is a basis independent equation. it shud hold in any basis

correction to first message :then it transforms $\psi ^*$ as $e^{-iOa}\psi^* $. this implies $\psi ^* (a) \psi (a)= \psi^* e^{-iOa} e^{iOa}\psi = \psi ^* \psi$

pls help

Gotta prove that 4-manifolds can't be classified so I need to bring up bloody Turing machines

if $A$ is a matrix of real eigenvalues, then $e^{iA}$ leaves $\psi ^* \psi $ invariant. is this incorrect?

$\psi ^*$ is the transpose conjugate of $\psi$

Man, I hope I'm not being picky but P&S have such a weird way to define covariant derivatives :P

it is extremely weird

complicating a simple thing

@Mr.Feynman it's not weird, but it's underexplained :P

the $U(y,x)$ is sometimes called the "comparator" but what it really is is the parallel transport operator along the straight line from $x$ to $x + \epsilon n$

so it transports the value of $\psi(x)$ at $x$ to a value at $x + \epsilon n$, and then that lives in the same space as $\psi(x + \epsilon n)$ so it's meaningful to take the difference

I've seen this construction various times in physics texts, but it only made sense to me after I learned from the mathematicians what a parallel transport was :P

@ACuriousMind I agree that putting it this way it makes sense. Nonetheless, I think the more appropriate hierarchical structure would be connection->covariant derivative->parallel transport

Why in the world would someone start from parallel transport :P

@Mr.Feynman I don't consider this a hierarchy, it's a circle - you can pick any of these three and then derive the two others from it

particularly in the Ehresmann view of connections there isn't a lot of room between "connection" and "parallel transport", anyway

This day... some interesting has happened some time ago, on a perpendicular time dimension with unknown physical laws.

according to mwi, there r multiple timelines of the universe

if anyone wants to join, i'm reading what is life, and mind and matter

this is the link from google scholar: libarch.nmu.org.ua/bitstream/handle/GenofondUA/7717/…

kinda ackward layout, but fine

i've read it in the past, but could be good to do it again.

this is frm Schrodinger

u may also like wholeness and the implicate order

Bohm talks about an implicate order that is monist and idealist i guess and there's explicate order that is mind-dependent physics

similar to Kant's and Plato's ideas

but idk yet how Bohm actually models the implicate order. it wud b like describing the undescribable

wdym by frm schroedinger @RyderRude?

I'm not able to follow Bohm other than in Youtube talks (like with Krishnamurti and so on.)

it's worth noting there's a few different strands of "pilot wave theory"

in particular there's Bohm's second-order formulation which is trying to make contact with the semiclassical approximation and correspondence principle

(hence why it goes for Hamilton-Jacobi theory as fast as possible)

I found the newer version that I read in the past, the link is indeed to maybe the first revised manuscript (?) or a first version anyways.

whereas Bell's first-order formulation focuses on how to get trajectories from the Schrodinger equation

i like the latter better, and other Bohm people i've spoken with are of that mind as well

u talking to me ? @Semiclassical sorry, i do not know what you mean

i'm talking about Bohmian mechanics, aka the Bohm interpretation of quantum

i haven't looked at the more metaphysical aspects of Bohm in a long while

i'm a chemist, just know some standard quantum physics

but chemists never engage with this topics

most don't yeah

there have been some attempts to do DFT in a Bohm-motivated way

e.g. arxiv.org/abs/0906.0073

@ACuriousMind yes, hierarchy in the sense that I think that is a better order to do things

@ACuriousMind I'm going back to this. I used to find it boring, then I learned about holonomy

@Mr.Feynman But ACM was trying to tell you that there would be other people who would choose a different order because it is actually a circle. You cannot then come back and assert that a "better order" is a hierarchy.

Today I heard a professor saying "global symmetries organize the states in multiplets and gauge symmetries constrain the states to remain singlet under them" or something like that...While I understand the latter part, can anybody give a simple example of the first half of the sentence in the context of field theory?

@naturallyInconsistent I replied to explain that I know those three concepts contain the same information, so I clarified that my perhaps misuse of "hierarchy" should actually be "better order"

Johnson's explanation for my 1st question which I can't understand.

Here's Insert 12.1 which I understand but can't apply this as mentioned above.

Is it wrong to assume that the hermitian Adjoint of this equation: $$d/dt \hat{a}=-i\omega \hat{a} $$ is $$d/dt \hat{a}^{\dagger}=+i\omega \hat{a}^{\dagger} $$

My brain is totally fried due to the Heisenberg picture so please be gentle :P