The h Bar

02:10

@user16217248-OnStrike under what condition?

03:20

@naturallyInconsistent At, say, 20 C as a starting temperature, but I would not be surprised if compressing water increased the temperature.

naturallyInconsistent

@user16217248-OnStrike are you intending for this to be fast? Or incredibly slow? On constant temperature? Or constant entropy? Or constant what?

Do you have an infinite set of different constant pressure atmospheres?

(bye for now)

Silly Goose

10:13

what is with free pdf books from springer having strange latex rendering 0.0

Relativisticcucumber

11:32

what does this statement mean "Quantum spins are notoriously difficult objects to deal with in many-body physics, because they do not behave as canonical fermions or bosons." ? how do they behave then? the book doesnt seem to say, as it just goes on to discuss how spin one half behave as fermions, which i thought was the definition of fermions (having half spin)?

@SillyGoose what is it with YOU 0.0

naturallyInconsistent

@Relativisticcucumber might need a LOT more context

Relativisticcucumber

@naturallyInconsistent well i am currently reading a chapter to understand the jordan-wigner transformation, and it was stated at the beginning of that chapter and not returned to :(

ACuriousMind

@Relativisticcucumber The usual terminology is that a boson has commuting c/a operators and a fermion has anti-commuting c/a operators. In relativistic QFT in more than 2 dimensions, the spin-statistic theorem says that being a fermion means having half-integer spin and being a boson means full-integer spin

naturallyInconsistent

But it kinda makes sense. If there are just one orbital, then skewsymmetry is pure skewsymmetry, bosons are bosons, very simple. But when you have 3 electrons, they cannot all stay in the ground state, and then you have automatic skewsymmetry for the pair that are in the same orbital, and the 3rd electron will still be skewsymmetric w.r.t. each of the pair, but it is quite horrible how to get the exactly correct wavefunction.

ACuriousMind

this fails in non-relativistic QFT and in 2 dimensions - spin and statistics are not intrinsically related there, and in 2 dimensions you get anyons that are neither half-integer nor full-integer spin

naturallyInconsistent

11:39

In particular, one single Slater determinant can get you the correct skewsymmetry, but it will necessarily have no correlation by definition, and that cannot be correct.

@ACuriousMind And that is part of why I was complaining that NRQM is inconsistent, yesterday

ACuriousMind

depends on what you mean by that

Relativisticcucumber

@ACuriousMind and which regime does quantum many body systems usually refer to?

ACuriousMind

@Relativisticcucumber depends on the system!

if you're looking at some weird material where the excitations are confined to a plane you might actually have to think about anyons, if you're looking at a regular 3d crysttal you're probably doing non-rel QFT in 3d, etc.

fqq

@Relativisticcucumber of you're doing jordan-wigner the spins are just spins on a lattice, not particles moving around

Relativisticcucumber

@fqq so what does this imply about their statistics?

fqq

11:44

Usually you are modelling something like the nuclear spins in a solid, which are decoupled enough from the other DoF to just study the Ising/Heisenberg Hamiltonian

There's no statistics

ACuriousMind

you just have one spin per lattice point, they don't get created or destroyed so talking about statistics doesn't quite make sense

Relativisticcucumber

okay and when we say one spin per lattice point this means there is one particle at each point or there is one state at this point? sorry this lattice model is slightly unclear to me

ACuriousMind

I mean we don't really need to pick a particular physical representation of this to talk about the general theory but yes you can imagine some spinful particle at each point

naturallyInconsistent

11:59

I think the context will be helpful here. When you have a solid lattice, the valence electron orbitals overlap very significantly, i.e. the electron orbitals are very delocalised, and so fermion skewsymmetry must be obeyed. But the nuclei are incredibly separated from each other, and effectively behave as if they are not at all overlapping, i.e. Boltzmann statistics.

Slereah

One thing I thought of about that whole decay of particles in NRQM is that it may fuck up with the mass superselection rule

Silly Goose

i did not know that local hamiltonians is part of the assumptions of QFT 0.o.

or maybe it is apart of all textbook physical theories...

Relativisticcucumber

@SillyGoose coffee video my butt

Silly Goose

this is break time too @Relativisticcucumber

Relativisticcucumber

lmfao

this is like those couples who cheat on eachother on dating apps but they accidentally match with eachother XD

"im going to work" "im going to break" sees eachother in hbar

Relativisticcucumber

12:30

@ACuriousMind ah i think this goes back to a confusion about how i should interpret the field operators. i know that they are defined as $\psi(\xi) = \sum_j \psi_j(\xi)b_j$ for some operator $b_j$ and that this can be interpreted as acting as a creation operator? but i cannot see how this should be interpreted and why the components of these operators are what we are putting on a lattice

Ryder Rude

is it possible to have lorentz invariance but non locality in a field theory

@Slereah one has to derive why the relatvistic QFT applies even at non relativistic speeds

Silly Goose

it always seems like locality (in the sense of nearest neighbor interactions are the only interactions allowed) is an assumption motivated by experience

i learned today that supposedly weinberg sets up some axioms to derive that hamiltonians in QFT must be local in this sense :0. but the axioms/assumptions going into the argument are indeed a lot: lorentz invariance and the cluster decomposition principle

Ryder Rude

weinberg wants to derive why we r using fields in the first place

ACuriousMind

@Relativisticcucumber The field operators will be useful because you can write the Hamiltonian in a neat form in terms of them

Ryder Rude

he wants to set up fields as merely tools for convenience because of locality and lorentz invariance

naturallyInconsistent

12:37

@SillyGoose It is one of those things that are rather important. Many texts will not emphasise the cluster decomposition principle, but if you do not have that, one cannot do physics in the usual sense. isolation becomes impossible.

Ryder Rude

Schwartz mentions that not every theory with locality and lorentz invariance needs to be a field theory. so weinberg is unsuccessful in reducing fields to convenience tools

naturallyInconsistent

Also, the book, PCT, Spin, Statistics and All That, is a very nice read that also covered the importance of locality. If you just want to have widely separated stuff not interact with each other, that is sufficient to be the microscopic spacelike separation must also not interact. There is a lot of known stuff in QFT that is simply not taught to the beginner.

Relativisticcucumber

@ACuriousMind yes i see that but im just not sure how to interpret them

ACuriousMind

@Relativisticcucumber why do you need to interpret them?

at this stage of second quantization they're literally just a repackaging of the other operators, there is no obvious significance to them except their computational convenience

naturallyInconsistent

The QFT field operators $\hat\psi(\vec x)$ (or of $\vec p$) and its conjugate have a simple interpretation only in the free field. Once interaction comes in, life is confusing

Relativisticcucumber

12:42

@ACuriousMind well because i should know what it is i feel like XD because i do have a hamiltonian written in terms of them that i need to perform a jordan-wigner transformation on but i cant understand this transformation right now so im trying to learn it

Ryder Rude

@ACuriousMind this is only true in one of the schools about second quantisation. in the other school, fields are not a re packaging of more primitive operators

the other school starts with fields

Relativisticcucumber

@naturallyInconsistent bah

ACuriousMind

@RyderRude r.c. literally wrote down $\psi = \sum_i \psi_i b_i$ as the definition, so what I'm talking about is the approach their text is taking. What's your point?

Ryder Rude

@ACuriousMind since they want to interpret the field operators, they are not satisfied with the text. so we should mention the other approach

ACuriousMind

@Relativisticcucumber I don't think there is a general "interpretation" for field operators. If you have a concrete physical system described by the theory, you can often give the field some meaning, but in the abstract there isn't really a lot one can say

Ryder Rude

12:45

in the other approach, we apply CCR to classical field theory @Relativisticcucumber

naturallyInconsistent

@Relativisticcucumber Look, you can familiarise a lot with the mathematical toolset of QFT whilst working in the free-field, non-interacting limit. If Im not completely off-base, your Jordan-Wigner transformation is being done in the free field context, isnt it?

Relativisticcucumber

@ACuriousMind bah so then i have another confusion. when i solved the dirac equation awhile back i thought that the solutions were wave functions and so i treated their modulus squared as a PDF and the results looked like atomic orbitals when i did the single electron atom case so what is up with that XD

ACuriousMind

@SillyGoose "QFT" is not a single axiomatic framework, there are so many different approaches to its formalization that this statement doesn't even make sense in some of them (e.g. the Wightman axioms don't talk about Hamiltonians at all)

Relativisticcucumber

@naturallyInconsistent well i dont think so. i think there are electrodynamics

naturallyInconsistent

@Relativisticcucumber That is RQM, and a tolerable interpretation inside the weird shit that is RQM. You do not have to beat yourself up over it.

@Relativisticcucumber Thou hath mine condolences.

Ryder Rude

12:48

Wightman axioms try to reduce QFT into merely a study of scattering

ACuriousMind

@Relativisticcucumber The Dirac equation pops up both as the equation for a single wavefunction and as the equation for a field operator

in fact, it's a general principle in QFT that the one-particle wavefunctions will obey "the same" equation as the field

so you did nothing wrong

Relativisticcucumber

ah how fortuitous for old me since that was my modern physics final project xD

@ACuriousMind ok so in the J-W case can we have an interpretation or still no?

ACuriousMind

@Relativisticcucumber I mean J-W is just a mathematical transformations it doesn't contain any specific physics

Relativisticcucumber

god

ACuriousMind

you have something that behaves like spin operators - you map it onto fermionic c'/a operators, J-W tells you how

J-W doesn't care what your operators "mean"

Silly Goose

12:54

@ACuriousMind well cripes ! are all the approaches not equivalent?

ACuriousMind

@SillyGoose no one knows

we do not have a rigorous formulation of QFT in practice

Relativisticcucumber

@ACuriousMind ok i will accept that for now

naturallyInconsistent

I think we do not even know if the approaches that are currently not fruitful are even failures or just temporarily embarrassing road blocks, or totally incompatible with the fruitful approaches. We just have entire libraries of stuff that we do not fully understand.

And of the scheme that is successful enough to not need counterterms, that every comptuation is finite, we know its perturbation series is asymptotic too, i.e. diverges, and it is not at all clear that it would be summable to a sensible theory.

Ryder Rude

everything is finite in effective field theory

but there is also some other approach which makes things finite without using a cut off

ACuriousMind

@naturallyInconsistent I mean no one expects the perturbation theory to not be asymptotic - the folklore is that in that case negative couplings would also yield sensible theories, which doesn't work

Ryder Rude

13:06

by "asymptotic", do u mean things like Landau pole?

ACuriousMind

no

Amit

Dipole dancers earn twice the wage

Ryder Rude

because they can do simple harmonic dance

@ACuriousMind what did u mean by "asymptotic"?

Amit

As long as it's not forced oscillations i'm not against

nickbros123

Is there a cute way to show that $|E|^2$ where E is the electric field, assumes no maximum in a region of no charge (electrostatic situation). Griffiths has an already cool way of doing it using averaging the function, but I was wondering if theres another way of showing it

Ryder Rude

13:10

@Amit lolol

ACuriousMind

@RyderRude see physics.stackexchange.com/a/130235/50583

Ryder Rude

@ACuriousMind if the series cant be convergent, how do we make finite predictions?

ACuriousMind

by not summing the whole series

if you read up on what an asymptotic series is, the first "few" terms are still useful

Ryder Rude

oh

wtf so the theory predicts infinities.

even after renormalisation

and we just take the first few terms in an ad-hoc way?

maybe there is some mathematical justification for it

ACuriousMind

I just said "if you read up on what an asymptotic series is", so yes, of course there's a lot of theory on the behavior of such series

see e.g. physics.stackexchange.com/a/62916/50583

Ryder Rude

13:23

ok so i think it is not ad-hoc. because the theory didnt give us the series as the prediction. the series is only our computational technique, and it is allowed to be an asymptotic series

the theory is fundamentally non perturbative and it does not predict infinities

Ryder Rude

13:54

what would be ur decision in the trolley dilemma?

it is here

walking away is also a decision with moral consequences

Ryder Rude

14:49

in ancient rome, they used to believe that drilling a hole in ur head got rid of headache

several human skulls have been found with those holes

naturallyInconsistent

@ACuriousMind Yes, I know that. It is, however, interesting to note that sometimes the asymptotic series is actually equivalent to another series that is strictly convergent, but in another parameter, maybe $\lambda^{2/3}$ of the old perturbation parameter $\lambda$. If only we can easily get what the nice series are. Alas, life is annoying.

geocalc33

15:06

partial differential equations are used in physics. anyone know if linear third order partial differential equations are used?

Slereah

Korteweg–De Vries equation

In mathematics, the Korteweg–De Vries (KdV) equation is a partial differential equation (PDE) which serves as a mathematical model of waves on shallow water surfaces. It is particularly notable as the prototypical example of an integrable PDE and exhibits many of the expected behaviours for an integrable PDE, such as a large number of explicit solutions, in particular soliton solutions, and an infinite number of conserved quantities, despite the nonlinearity which typically renders PDEs intractable. The KdV can be solved by the inverse scattering method (ISM). In fact, Gardner, Greene, Kruskal...

Oh wait I guess it's not linear

naturallyInconsistent

@geocalc33 Abraham-Lorentz force law, the one with the radiation reaction, is also a 3rd order PDE.

Slereah

eqworld.ipmnet.ru/en/solutions/lpde/lpdetoc5.htm

^bunch there

geocalc33

@naturallyInconsistent gotcha thanks, it's 3rd order linear pde?

@Slereah thanks :)

naturallyInconsistent

@geocalc33 it is an ODE, sorry.

linear in the NR limit

Slereah

15:22

also a bunch of conformal geometry equations are third order

geocalc33

$$r^2 \frac{\partial ^3\Psi(r,s)}{\partial r^3}=s^2 \frac{\partial \Psi(r,s)}{\partial s}$$

for example that

Slereah

the classic conformal ODE equation is the Schwarzian derivative

Schwarzian derivative

In mathematics, the Schwarzian derivative is an operator similar to the derivative which is invariant under Möbius transformations. Thus, it occurs in the theory of the complex projective line, and in particular, in the theory of modular forms and hypergeometric functions. It plays an important role in the theory of univalent functions, conformal mapping and Teichmüller spaces. It is named after the German mathematician Hermann Schwarz. == Definition == The Schwarzian derivative of a holomorphic function f of one complex variable z is defined by ( S f...

Although it's not linear

geocalc33

nice i'll check that out

naturallyInconsistent

@geocalc33 If r and s have units, then this equation is dimensionally inconsistent.

geocalc33

@naturallyInconsistent oh really? does that just mean it wouldn't be useful in physics i suppose?

naturallyInconsistent

15:38

@geocalc33 Dimensionally inconsistent DE will just be suspected when used in physics, though we do sometimes just fix them by throwing in a bunch of constants to absorb the unwanted dimensions. They might, however, be good at approximating reality.

geocalc33

Hm I see

geocalc33

16:07

I'm gonna ask a question about it see if anyone knows of resources or the like

bolbteppa

16:28

@naturallyInconsistent I can't believe you're tripling down on this and trying to use 2D anyons as evidence that non-relativistic quantum field theory (NRQM?) is inconsistent, it's rare to see something this absurd stated in here even with all the woo about alternative interpretations

@RyderRude yes that was a subtle dig at you ;)

@Slereah what's wrong with this:

0

A: $E=mc^2$ in particle physics vs non-relativistic quantum mechanics (NRQM)

It's not quite right that relativity and non-relativistic QM are not compatible. Non-relativistic QM is an approximation of relativistic quantum field theory. For instance if you take the appropriate slow speed limit of the relativistic Dirac Equation or Klein-Gordon Equation, you get the Schro...

naturallyInconsistent

@bolbteppa You may have the last word. Your most beloved Landau does openly point out that a lot of the relations are coming from relativistic quantum theory, even in chapter 9, even its first page. There is nothing more to say to someone like you.

Slereah

@bolbteppa I don't know?

bolbteppa

16:42

@naturallyInconsistent None of the relations come from relativistic quantum theory, the only thing that comes from relativistic theory is the proof of the spin statistics claim that the statistics of a particle is uniquely related to their spin, however you don't need spin statistics to see e.g. a non-relativistic electron has spin half, this can be avoided in practice and it doesn't affect anything in the setup in the chapter, you're desperately cherry-picking at this stage if this is all you've got

Slereah

the spin statistics theorem doesn't apply but that allows too much instead of not enough

You can have anticommuting fermions or commuting fermions

naturallyInconsistent

@Slereah he already knows about anyons but still wishes to claim that NRQM is constructed correctly. You might want to save your breath.

bolbteppa

Yeah from a non-relativistic perspective, different fermions can be taken as commuting or anti-commuting with each other, it's irrelevant non-relativistically, it matters relativistically, if anything it shows you how consistent the material is

It's just compounding things that you apparently think relativistic anyons do not exist for some reason and that the existence of anyons (basically due to the topology of the Lorentz group in 2+1D) somehow says anything about e.g. 3D NRQM/NRQFT or anything really

I'm a big fan of emotional statements that encode wisdom, they just have to be right ;)

We've all been traumitized by QFT, studying the non-relativistic case has lifted that trauma somewhat, so obviously it's not nice to see it being bashed

ACuriousMind

@naturallyInconsistent Anyons exist in 2 spatial dimensions both relativistically and non-relativistically since both $\mathrm{SO}(2,1)$ and $\mathrm{SO}(2)$ have fundamental group $\mathbb{Z}$

naturallyInconsistent

@ACuriousMind I am aware of that.

ACuriousMind

16:59

it sounded like you were saying the existence of anyons was somehow an argument against non-rel. QM

naturallyInconsistent

@ACuriousMind Rather, the point is that without spin-stats, we can have anyons in 3+1D in NR context, but it is just not explored, because, really, the NRQM is only tolerable as a simplification of a SR QFT, in which case we do have spin-stats to reduce the possibilities.

Slereah

I don't believe we can

ACuriousMind

@naturallyInconsistent What do you mean by "anyons" there? I mean "object that has neither full-integer nor half-integer spin", and those definitively don't exist in 3+1d dimensions regardless of relativity

Slereah

The spin statistics theorem isn't directly related to what kind of statistics are allowed at all

ACuriousMind

but you probably mean "object whose c/a operators neither commute nor anti-commute"

naturallyInconsistent

17:03

I'm not sure I should be caring about the difference, since I'm perfectly happy with spin-stats telling us that things will not go crazy in 3+1D.

ACuriousMind

I think the latter are also forbidden in higher dimensions because we're essentially looking at the same topological properties

So I don't think "we can have anyons in 3+1D in NR context" is true in any interpretation of the word "anyon"

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