Does a static magnetic field necessarily imply a static electric field, and vice versa?

why is the identity operator included in pauli strings

or i guess i am a little bit confused

lol okay i figured it out you do need the identity to span su(2)

@SillyGoose the identity is not in $\mathfrak{su}(2)$

There's an online advanced GR course starting today if anyone is interested.

Registration for it has closed, but it looks as if the lectures will be freely available on YouTube:

youtube.com/playlist?list=PL04QVxpjcnjhVn7CJbY5W4tZbJtQeNGG0

The course is a tribute to Raychaudhuri to mark the centennial of his birth.

wait what how come

hm, so if we include the identity with the pauli sigma matrices, it seems like we generate all 2x2 hermitian matrices. then if we exponentiate the identity with the sigma pauli matrices, do we still get SU(2)?

@SillyGoose $\mathfrak{su}(2)$ consists of traceless antihermitian matrices

(The hermitian/antihermitian comes from the $i$ in the exponential map, what physicists use is $i\mathfrak{su}(2)$)

It's almost the equinox: 2023-Mar-20 21:25 UTC. Using my script from astronomy.stackexchange.com/a/49605/16685

Here's a Horizons query: ssd.jpl.nasa.gov/api/…

Oh my, I suddenly remembered something that puzzled me a year ago. In order for the coupled Maxwell action to be gauge-invariant, $\partial_\mu j^\mu=0$ has to be true off shell

Hi

Could things like p-adic numbers make qft rigorous

why would they?

They give correct values of divergenr sums without having to deal with calculations involving infinities

I think re-normalisation is perfectly sound, EXCEPT the intermediate steps sort of involve infinities

Not that i find that to be a problm

I think its already pretty rigorous. Mathematicians just need to invent some system which axiomatizes it.

e.g. in the Casimir effect, you need to subtract infinities after introducing a regulator to get -1/12

p-adic numbers r known to give the correct answers of similar sums without using a regulator

Tho the example sum that i saw using p-adics was 1+2+4+8....=0. I dont think this sum is relevant in physics

@RyderRude renormalization in perturbative QFT is understood pretty well rigorously

it's called Epstein-Glaser renormalization or "causal perturbation theory"

and there are no "correct" values of divergent sums - that's what "divergent" means

what value you want to assign them depends on the context

it is not clear to me at all why you'd hope that something you computed in the p-adic numbers is useful in physics

the problem with rigor in QFT is really not the renormalization part :P

Yeah, it's the Haag's theorem stuff

Correct?

I had heard of "causal perturbation theory" in the context of Algebraic QFT. I don't know much about it tho

I think of lattice QFT whenever I want to make renormalisation sound rigorous

But that comes with technical problms like Fermion doubling

e.g. Haag's theorem doesn't matter if you try to do everything via path integrals

Yeah, then the measure becomes the problem

there it's more establishing the existence and the desired qualities of the integral in the cases that interest us

@ACuriousMind this is what i had read in the "causal perturbation theory" approach. They were trying to sideline the path integral itself and just writing down the essential properties that they want from the path integral.

I was thinking more about the approach of actually properly defining the integral

Glimm and Jaffe wrote a book about how to do that in the 2d and some 3d cases but I'm not aware of anyone succeeding in extending this to 4d or even arbitrary dimensions

Whats wrong with the naive method where u do the path integral on the lattice and take the continuum limit

I don't really know any details of Epstein and Glaser's approach, I just know that there is consensus that it is rigorous :P

@RyderRude that limit doesn't mean anything mathematically

Lol

I was thinking it can b simulated on a computer

People study the behavior of QFTs using lattice simulations

see this answer of mine for how the mathematically rigorous path integral actually works

@RyderRude so? That doesn't mean "take this vaguely defined continuum limit" defines a measure on the space of field configurations of the continuum theory in any mathematically rigorous sense

Yeah

also: lattice methods aren't inherently related to path integrals

Making it rigorous is a different issue from extracting predictions frm the theory

@ACuriousMind yeah, u can just do the hamiltonian lattice formulation too. Even Haag's theorem may stop applying

I had read that answer of yours back when u posted it

Ultimately, i think this rigor problm is mostly a puzzle for mathematicians to b entertained

Cuz physics can b done with simulations, at worst

do you know how hard it is to write good numeric code?

No :P

there's a lot of rigorous math involved in making sure discrete approximations and algorithms actually converge to the desired continuum limit

it's pretty easy to put some equation "on a lattice" and then write code that solves the discrete version on the lattice that just produces garbage instead of the actual solution for certain inputs

I once spent about a month writing Monte-Carlo code from scratch and my grand achievement was in the end that all the observables went to zero in the continuum limit

turns out there was a little bug in the description the course had us follow so while the algorithm looked okay, the way in which it scaled when you added points to the lattice just made stuff vanish on larger lattices

in another case I tried to reproduce results from some paper from the 70s and also just got mostly nothing, to this day I don't know if my implementation was buggy or the algorithm they described didn't actually work

(to be fair I didn't really care about that one after the course was over since there were plenty of newer results with other methods available in that area)

Oh. That sounds ....terrifying

Ur course had u simulate QFTs?

Maybe a phd course

It was called "Quantum Fields on the Lattice" or something like that

and there were only like 10% of the visitors of the first lecture left after that lecture started with "So, you all can write C code, right?"

Lol

I wud never consider joining that course

I think the lecturers didn't consider that "programming knowledge" as a prerequisite in most other physics courses meant something like being able to write a few lines of Python scripts

I barely kno any python either :P

Ok so there are exreme problms in writing effective algorithms to get predictions from lattice QFT

I shud've expected this, given the computational complexity

oh, what I've said is not exactly specific to lattice QFT

Yeah

the thing about having to make sure your discretization actually converges to what you want is a general issue in numerics

and then you also have to make sure it converges against that in a reasonable number of steps

plenty of correct algorithms that need to run until the heat death of the universe to actually compute their answer out there :P

In QFT simulations, does the problm come from choice of the grid structure, or the numerical computation techniques? I expect it to be the latter

Or maybe both

I think there's not much choice in the grid structure tho, beyond the lattice size

Im just making sure that, conceptually, i can spend my life thinking of QFT as a theory on a rectangular grid.. even tho it doesnt work for practical computations

@ACuriousMind but this sentence of urs is making me think that there r bad choices of the grid structure

So i cannot think about it in terms of a rectangular grid, even conceptually

@RyderRude it's not about the grid structure, it's about how you discretize your equations to work on that grid

simple example: what do you do with a derivative? it needs to become a finite difference but there essentially are infinitely many of these finite differences that converge to the derivative in the continuum limit

but some of them may be good and others may be bad choices

Thanks. I understand it now. I had read about this in the context of fermion doubling, i think

It was smthin like : "The central derivative gets rid of doubling"

e.g. you can discretize a unitary time evolution equation in a way that isn't unitary anymore for finite steps (this problem is classically solved by symplectic integrators).

While your discrete equation then still converges to the continuum equation in the limit, your solutions might "lose/gain energy" (or normalization or whatever) in each time step, so that they are pretty useless if you care at all about the original conserved quantity of your equation

I was naively thinking that it was a deterministic procedure to discretize the equations

this is all pretty generic numerics stuff and there's known rigorous solutions to most of these problems, but it shows that "just put it on a lattice and let a computer do the math" is still a far more involved and less straightforward procedure than non-computational physicists and their handwavy continuum limits tend to suggest

Will the 4pt function have anything interesting about it's pole structure like the case for the 2pt function?

Doesn't it indicate like the masses of bound states?

IIRC the structure is like a pole at the sum of masses (for free particles) and then poles below that for bound states

Do non-unitary and non-anti-unitary operators really have no place in quantum?

I am thinking okay so you want to use unitary (or anti-unitary) operators to essentially preserve probability. but surely there is some phenomena which includes changes in probability which can then be modeled by acting with a non-unitary operator?

well i guess wave function collapse :P

Many operators aren't unitary

hm what are some examples

for more context: i just started reading nielson and chuang's QI/QC book and they say that a quantum logic gate must be a unitary operator acting on your state, and i am trying to understand why it must be unitary

@SillyGoose how can u ever hav changes in total probability

Then it wont b probability

@SillyGoose the time evolution operators for open systems aren't unitary

i.e. cases where you're just modeling part of a larger system and stuff can "escape" from your description

That makes sense @ACuriousMind

well maybe your space of outcomes changes @RyderRude

oh i see

so is unitarity only important in the sense that it is an artifact of always considering closed systems :P

@SillyGoose but in the context of quantum information and quantum gates: A quantum gate just changes the state of a qubit to another state of a qubit. The total probability for the qubit to be in any state must remain 1

so there is an assumption that you want to remain in a fixed hilbert space?

don't take this as some deep statement that there are never non-unitary operators in this context or whatever, they're just saying that what a quantum gate does is usefully modeled by unitary operators

@SillyGoose we usually do not change Hilbert spaces

the Hilbert space is supposed to be the space of all possible states of your system

if you can leave it you messed up the part about "all possible states" :P

heh but isn't it possible that your hilbert space changes over time. i guess depends how you define your system. since if you define your system as this cubic centimeter of space no matter what fills it at each time t, then your hilbert space will change with time in general? but maybe that is an unusual definition of a system

that's a pretty silly system :P

as in, that's not what we usually mean by "a system"!

a system is something whose state I can describe with a fixed number of variables

the space of all possible values for these variables is the space of states

it doesn't change

I guess it is also possible that such a system will be inconsistent with the uncertainty principle

I mean the cubic centimeter one... as a reason for why it doesn't make sense to think of such a system in QM...

hm but if your system is so large like all space in europe and you would like to characterize such space, then it would help to take representative samples as representing your system

@SillyGoose we're not sociologists

or geologists or whatever

"representative sample" is not exactly a technique physics uses :P

and again, "all space in europe" is a pretty weird system

what are the state variables here?

a system is something like "N particles in a box". Nice and tidy

U wud b doing some very weird theory if the sample space changes with time. One example i can think of is a particle in an expanding box. But then, the correct way to think of it is to consider the entire real line as the unchanging hilbert space, and then modeling the expanding box using an infinite potential

@ACuriousMind Is it sufficient to just say that we need to make sure that whatever wavefunction represents the system we need to ensure that the system is sufficiently "large" so that it goes to zero at the boundaries of the system?

I don't like arguing about specifics of the wavefunction in position space

effectively the value of the wavefunction at the boundaries doesn't mean anything since values at single points don't change integrals

and the standard lore of "the wavefunction goes to zero at infinity" is a lie anyway, there are $L^2$ functions on $\mathbb{R}$ that don't go to zero

Ah I see that's a good point

I'm stuck with some kindergarten notions of QM :D

But wait -- you mean they don't go to zero in the sense they don't attain zero? Or that they don't approach zero?

Oh oh I see I think you mean ones like $x^2\exp(−x^8 \sin^2 x)$ , keeps oscillating and yet $L^2$ integrable apparently

i guess there is some research on non-unitary quantum circuits: nature.com/articles/s41598-021-83521-5

The second part of L&L 4 on radiative corrections has so many insights that you can't seem to find anwhere else

formally, when we want elements of the group SU(2) to act on vectors in a hilbert space, we still have to define a group action, right? it is just that matrix multiplication gives rise to a natural group action?

can two distinct observables be isospectral?

well more so i am asking: is an observable roughly characterized by its spectrum? or maybe a weaker condition is true: its dimension?

@SillyGoose think about ordinary position and momentum

as in classical momentum and position?

@Amit that function has limit zero at $\pm\infty$

@fqq Are you sure? I tried it with WolframAlpha, it didn't give me a straight answer lol

L&L 4 eqn 109.22 is the key to the whole of QFT I think

@Amit no, I was wrong, it's a good example

The above is the Källén–Lehmann spectral representation but without the fourier transform from $\mathbf{r}$ to $\mathbf{k}$ I think i.e. just with $t \rightarrow E$

Shows that rel QFT and non-rel QFT are essentially the same

but taught differently for some reason

@SillyGoose no, as in quantum position and momentum in 1d

both position and momentum are observables whose spectrum is non-degenerate and the entire real line

hm i see

what if we specialize to the hamiltonian? is there any way to distinguish it from other operators?

in particular, let's say we are given a hilbert space and a hamiltonian over the hilbert space. why should operators isospectral to the hamiltonian also be identified as being the hamiltonian?

i mean i guess from one perspective it is the same operator and all you are doing is changing basis by unitarily conjugating the hamiltonian.

Just need to figure out if this branch cut exists in non rel QFT

Just out of interest, do we ever use the Pauli–Lubanski pseudovector in QFT as an operator?

do any of you all use tensor flow :0

tensorflow is evil

@SillyGoose they should not.

They use it to create thinking machines and other dangerous stuff

@SillyGoose who claims this?

again, position and momentum are isospectral, so $p^2$ (the free particle Hamiltonian) and $x^2$ are isospectral, too

but no one would claim that $p^2$ and $x^2$ should be "identified as being the Hamiltonian" of the free particle

but if you just look at this as the theory of abstract operators on a Hilbert space, then the Fourier transform provides an isomorphism between the $p^2$ Hamiltonian and the $x^2$ Hamiltonian, i.e. if you know how one of these systems behaves then you know how the other behaves

so the point about isospectrality is, I think, less that you should think of all systems with isospectral Hamiltonians as "the same" and more as "alike" in the sense that if you understand how to solve one of them then you know how to solve all of them

I guess from this observation. But i think perhaps the interpretation I provided is not really the right way to think about it physically

Does anyone know about how to deal with bound states?

I don't see why these shouldn't also pop up as poles

in the 2pt function