i was watching MIT's Quantum I lecture 2 in which the professor claims that Bell's inequality (at least the form he wrote down) suggests that classical probability is incorrect. I have not heard such a strong consequence of Bell's inequality before watching this video. Is this claim true? Or perhaps what I am asking is not a well-written question.

otherwise it was a nifty video to see some experimental results to call upon for quantum mechanics I

@SillyGoose Bell's theorem says that no local hidden variable theory can reproduce the predictions of quantum mechanics. In particular, "classical" usually means "local hidden variable", so classical theories cannot describe the real world. What's the question?

I understood what he was claiming to be that classical probability theory, the axiomatic mathematical theory, is incorrect due to Bell’s inequality and its empirical violation

So i am asking if the claim is true

But this sort of claim didn’t really make sense to me so i probably have misunderstood it

it's just Bell's theorem

So bell’s theorem says that probability theory as constructed from the kolmogorov axioms is wrong?

there's a large body of literature on which of the assumptions you'd need to relax to evade its conclusions

and yes, quantum probabilities do not obey Kolmogorov's axioms

I don’t get how a theory of mathematics can be wrong so long as it is (beyond really foundational and perhaps borderline philosophical issues) self-consistent

well i am not saying if quantum mechanics uses classical probability theory it is clear that it is different

but the prof seems to claim that classical probability theory itself is wrong. But i don’t get what it would mean to say that “abstract algebra is incorrect”

@SillyGoose what do you mean? The claim is just that quantum probabilities are not classical probabilities,

If that is the claim i understand

but i don’t think that’s all he is claiming

what else could he possibly be claiming?

well if he is to be making a correct statement yes that narrows it down to what you said

"Logic and integers and adding probabilities is misguided..." is part of what he says verbatim

it is misguided when you want to describe the quantum world

what's your problem?

i think the absence of the qualification "if you want to describe quantum mechanics"

you're worried that a lecture on quantum mechanics is not qualifying its statements that it applies to quantum mechanics???

if it applies to, I have no care if the qualification is omitted or not. it applies only to, though, in which case I think it's nice to qualify such a statement

what exactly is the statement you want here?

classical probability theory fails in quantum mechanics because the axioms of probability theory (or whatever does breakdown) are not satisfied

i don't know about logic at all so I'm not sure what the statement would be there

you mustn't forget that people frequently learn about "quantum mechanics" without any education in probability theory

just like you probably did

"I want my lectures to pre-emptively discuss every possible conflict with topics I probably don't even know that this point" is not a consistent pedagogical standpoint

What's the deal with $\vec{\nabla}\cdot \vec{B} = 0$

my prof said that while it's not been found to exist, magnetic monopoles aren't ruled out or something

how they would imply charge is quantized and we know it is but we don't know if they exist. also something about how the magnetic component of the lorentz force wouldn't exist without SR or something

would existence of magnetic monopoles change anything? i suppose there'd be another term on top of $q(\vec{E}+(\vec{v}\times\vec{B}))$

maybe $+b(\vec{B})$

my first sentence sounds like a terrible opener to a comedy routine held at a physics themed bar

probably called the h-bar

@SillyGoose Classical probability theory, the rigorous mathematical theory formulated in terms of Kolmogorov's axioms, has a wide range of successful applications, so much so that people would have intuitively suspected that it would work in all scenarios. This is false, because experiments have shown otherwise: the experiments that show quantum mechanical behaviour, in particular, require that probabilities can subtract in the phenomenon called destructive interference, thus violating

Kolmogorov's axioms. Thus, the logical deduction is that classical probability theory is a wrong theory if used to describe our universe.

You should be completely happy with the above description. However, it is so long-winded, that you can also understand why it is that the professor would elect to take a shortcut in exposition. It really says nothing much more than "Shockingly, our intuitive expectation that classical probability theory works always, is actually not correct for our universe"

@Obliv Correct; to date, no experiment has yet uncovered a magnetic monopole. We do, however, have condensed matter games that produced some magnetic monopole -like stuff in the lab, and so it really shows that the mathematics should work out.

We have also worked out the maths in theoretical physics that the symmetries of Maxwell's equations means that if magnetic monopoles exist, they would have this and that consequences. i.e. we actually know a lot about and around magnetic monopoles. It is just that we have no experimental evidence of its existence.

@ACuriousMind i actually took prob theory before learning quantum heh

Note that it is that "electric (monopole) charge" and "magnetic (monopole) charge" form a 2D space that has a (pseudo-)rotational symmetry in Maxwell's equations. It is that Maxwell's equations will have to be augmented with some extra terms. The Lorentz force law for electric charges will not change, but because, say, the electron has spin and thus feels the magnetic force directly from there, there will be another force law for the magnetic part. It is a pseudo-rotation because IIRC there is

i don't get why this isn't a bigger deal then, if classical probability and logic fail in one of the should be most important applications of said theories. I mean on one hand it doesn't make sense to say that a mathematical theory is false because it just is what it is. On the other hand, I suppose probability and logic were developed for the sole purpose of being applied.

a complex number coming in, and/or rotates the opposite direction compared to the usual, etc.

@SillyGoose It is a gigantic deal! Physicists are not shy about stating that this is wrong. Sci-fi and mystics keep asserting that the logic we use about the world is wrong; it is just that these folks are even wronger than that.

@naturallyInconsistent what do you mean by "this" in "Physicists are not shy..."

Note that the probability theory for quantum theory is built on top of the foundations laid in classical probability theory. The non-negative-ness is very helpful in that it allows us to prove a lot more nice theorems, and it is also the case that classical probability theory is applicable in a wide range of scenarios and so it is interesting in its own right, and simpler, so we should still study it in detail.

@SillyGoose this = "classical probability theory is applicable in every conceivable scenario in our universe", which is what is wrong.

H O N K

oh, the ditto wants meow meow to not just

B A H

but also to

S C H W E G

bleb another facet of QM that I have not ever thought about 0: well this is interesting

so a proper treatment of QM should develop some "quantum probability theory" and then use this language to produce the results of quantum mechanics?

In a sense, yes, but I don't think people have fully fleshed out quantum probability theory enough to produce the results of quantum theory. That might be way too difficult to do. We are still working with some nice results but ultimately hodge podge of stuff in, say, quantum computing and other fundamentals of quantum theory.

hm i see. so the textbook presentation of things is just "what works"

do you know an example of someone doing work in developing quantum probability theory :0

mayn what an interesting topic

@SillyGoose Hans Reichenbach had a small book where he shows that both "X and Y" and "X or Y" can work. Waves and particles, or waves or particles, as long as you are consistent and rigorous.

There is just a lot of small little details hidden here and there. I am not aware of any comprehensive treatise covering quantum probability theory in the style that mathematicians would.

is the book you mention by Reichenbach: Philosophic foundations of quantum mechanics

hm i shall look around. I found one book by Stanley P. Gudder. But all the literature seems quite "old" upon a cursory search (1980s)

Yes

mew mew will go back to purr in bed

Tf probability theory doesn't work!!!!?

@naturallyInconsistent lmfao

(Taken from The 'King)

@naturallyInconsistent SCHWEG

@SirCumference hi. from 7:16 to 12:15, this cosmologist is discussing some ideas about probability i was discussing that day

@bolbteppa classic

some people really have \define{\obsly}{\begin{proof}Obvious.\end{proof}} at the beggining of their latex

i dont know how to escape the latex here

@lucabtz If you use Chrome or Edge there is an addon that renders Latex in the chat.

If you use Firefox it can still be done but it's more work.

no i am using it, but i wanted to type latex and the addon is rendering

Ah, put inside code quotes?

$a^b$

Yes, that works

ah right

code in the above message was supposed to be \define{\obsly}{\begin{proof}Obvious.\end{proof}}

@SillyGoose There's a bunch of old work on "quantum logic" etc. but it wasn't really all that useful. The replacement for "classical probability theory" is just to do quantum mechanics, i.e. stubbornly apply the Born rule to compute probabilities for measurement results. There's no direct need for anything else.

I am told that the cool new version of quantum logic is linear logic

who tells you these things?

It is the Urs

Cool physicists, presumably, though that seems awfully close to an oxymoron.

The ones that deal with negative Kelvin temperatures :^)

@Slereah This?

Maybe not the best introduction for it but yeah

From what I've read it's a more dynamical version of quantum logic pretty much

in that truth values of propositions can "update"

which makes it a bit more interesting than quantum logic

Since you apparently can't use quantum logic for much

@ACuriousMind Does $SU(3)$ have a double cover?

@DIRAC1930 No, all SU(N)s are simply connected

Sorry I meant is $SU(3)$ the double cover of any group?

no, because it does not have a central subgroup of order 2, since the center of SU(N) is $\mathbb{Z}/n\mathbb{Z}$

Okay thanks

it is therefore a triple cover of some group, though

What group is that?

I don't think it has a special name

Ok thanks

Isn't there that thing where the standard model's gauge group is possibly $SU(3) \times SU(2) \times U(1)$ quotiented by one of its center since they have the same algebra

It could be a potential gauge group of QCD I guess

@Slereah no, the center is pretty relevant in certain contexts; it has physical effects

see e.g. this answer of mine on center symmetry

Does it have experimental effects that we know of in this case?

How does the equation (2.1.5) follow: what exactly is this unfolding trick?

hi. u can write $\int _{-\infty} ^{\infty} dx g(x) $ as a sum over areas over sub-intervals of size 1. @Sanjana

@Sanjana What exactly is unclear to you? Note that $\int_{-n}^n g = \sum_{i = -n}^n \int_0^1 g(x+i)$, this is just the limit for $n\to\infty$

@RyderRude Hmm?

@ACuriousMind How to prove this identity? Is this something elementary?

the integral over the infinite interval is the whole area over the line

and the sum divides that area into areas over sub-intervals

for e.g. $\int _0^1 g(x+2)dx =\int _2^3 g(x)dx$ @Sanjana

@Sanjana it's very easy to prove, it's just basic properties of integrals: $\int_a^b = \int_a^c + \int_c^a$ and $\int_a^b f(x) = \int_{a+c}^{b+c} f(x-c)$

we hav $\sum _n \int _0^1 g(x+n)dx= \sum _n \int _n ^{n+1} g(x)dx =\int _{-\infty} ^{\infty} g(x) dx$

@ACuriousMind Got it...hadn't used that 2nd property in years!

@RyderRude Yes...I proved it this way too...thanks

Nice team work, for a change :p

Hello All...

Hi @naturallyInconsistent and @RyderRude . Hope you are both doing good. Sorry I was busy yesterday . I just read your conversation. Which really helps me to understand better N3L.

What I concluded from @RyderRude example it is also possible to apply different forces on each other 10N & 15N in punch and kick example. From @naturallyInconsistent confirmation about k&k example I found if both apply force at same point then the only possible force can apply is lower force which is 10N in tug war example.

@123 also note that the net forces applied by two bodies on each other must be equal and opposite, which also holds in the punch-kick example (25N each)

@ACuriousMind What are your thoughts on the top of page 4 of this pdf jetp.ras.ru/files/JETP_7_96.pdf?

regarding the interpretation of resonances

@RyderRude yes person 1 net force 25N , which is action of punch 10N by person 1 and reaction 15N of kick by person 2 .

@RyderRude is there any other different case in N3L?

these are not really cases. N3L holds separately for the punch action-reaction pair and the kick action-reaction pair

Ooh okay.. this is how we treat punch-kick example.

It means if point of application of force is same then smaller force is the action-reaction pair.

a simpler example might be two big bodies exerting gravity on each other. each particle of the first body exerts a force on each particle of the second body, and all these forces separately obey N3L

and from the separate N3Ls, u get the N3L for the net force between the two bodies

@123 did they specify the condition that the rope must not accelerate?

Ooookay.. that's nice explanation what I have read this topic before in books.

if the rope doesnt accelerate, then the total force on it must cancel. hence the pull force of person 1 and 2 must be equal there

@RyderRude No k&k not said anything about rope acceleration, what I remembered. Otherwise I need to open my laptop to see the example again

@RyderRude let me open my laptop to double check and send you a picture.

yeah

Pls confirm , is it true if point of application of force is same then smaller force will be the action-reaction pair?

i cant understand this question without context

If person 1 and person 2 both punch each other at same fist but person 1 apply 10N and person 2 apply 15N. Then what will be the action-reaction pair?

@RyderRude Pls read the above picture example of tug war from knk.

@123 they have written about the condition of zero acceleration

@RyderRude I am again reading the example after your discussion. But diagram shows acceleration of rope $a_r$ in the direction of astronaut "$B$".

read the next page. theyve shown the acceleration must be 0 because the mass of the rope is 0

@RyderRude Okay i am reading further.

theyve shown the derivation why the lower force must be the force on both ends. this derivation is for this very specific experiment.

dont generalise it too much

@RyderRude Pls explain it to me more clearly.

:65262734 pls forget the rope at the moment which is making more confusion. Put example this way. both persons punch each other on fist with different forces.

Basically from this i am trying to understand law of conservation of momentum from N3L results.

@123 a simpler example here can be two connected springs of different spring constants, both slightly elongated

@123 did u read the full derivation for the rope experiment

@RyderRude In law of conservation momentum(LCM), In isolated system of two particles (A & B) having different momentum and both collide each other in opposite direction. So total momentum before collision and after collision remains the same. It means both momentum is contributing during collision. From this information it seems to me from N3L person 1 with 10N force and person 2 with 15N both must contribute to valid LCM

the action reaction forces at the punch-point must be equal and opposite

@RyderRude What is the result. Person 1 punch with 10N force and person 2 punch with 15N force at same punch-point. What will be the action-reaction pair?

the formulation is slightly inconsistent. if person 1 applies a 10N force, how can person 2 apply a 15 N force in the first place

i believe the action reaction pair becomes 25N in ur case @123

@RyderRude To make N3L consistent with law of momentum they both occur at same time.

yes..but the applied forces cant be different in the first place

@RyderRude person 1 is weaker than person 2 then both can have different amount of force.

In my example if the action-reaction will be 25N. Then N3L will be consistent with law of momentum.

yes. the two people just cant apply different forces on each other at the same point, even if one is stronger

@RyderRude Pls clear this point with force amount.

p1 punches with 10N force, and also applies the reaction to the 15N that he got punched with. so the total force applied on p2 is 25N

@123 does the book later talk about any generalisation of that result

@RyderRude Ookay.. It is confirmed the net force in this situation will be 25N for P1. which is the $P1 action 10N + P2 reation 25N = -(P2 action 15N + P1 reation 10N)

My confusion is that if both persons force contributed then N3L consistent with Law of momentum.

$F_{12} = -F_{21}$

$\frac{\Delta{P_{12}}}{t} = -\frac{\Delta{P_{21}}}{t}$

$\Delta{P_{12}} = -\Delta{P_{21}}$ cancelling "$t$" from both side

yes, the conservation of momentum always holds

$m_1V_1 - m_1U_1 = -(m_2V_2 - m_2U_2)$ where $U_1$, $U_2$ initial velocities of both objects and $V_1$, $V_2$ final velocities

@ACuriousMind Did you look at that paper?

> Google originally intended to use a 72 qubit chip (Bristlecone) where qubits were essentially directly connected to each other. They then switched to an architecture where qubits were connected indirectly via a coupler. The coupler requires a control line, so this increased the number of wires per qubit. The fridge was only setup for so many wires, so the total number of qubits was reduced to 54 to fit. Then one of the qubits didn't work, so it was excluded from the experiment leaving 53.

lol

$m_1U_1 + m_2U_2 = m_1V_1 + m_2V_2$

Total momentum before collision = Total momentum after collision

@RyderRude It means i am concluding what ever momentum object 1 and object 2 has before collision is equal to the after collision. It means both objects momentum is contributing during collision, no mater both has different momentum Where i am doing mistake in understanding this topic.

@123 i dont understand. u just concluded that the total momentum is conserved, while the individual momenta changed. that is correct

@RyderRude I am comparing this result (total momentum of both objects before collision) with force of both objects (10N & 15N). Is this right way to think or wrong.

@123 i think it is best not to associate each punch with individual forces of different magnitudes. only the net force applied by punch 1 on punch 2 and vice versa is well defined

punches can be associated with different momenta

@RyderRude Pls explain your statement with force magnitudes.

i imagine the situation as two punches heading toward each other and when the distance between them becomes small, two repulsion forces F12 and F21 kick in

such that F12 = -F21

@RyderRude Okay... What if p1 try to punch with 10N force and p2 with 15N. What will be the result?

10N the smaller force or net force 25N

i just dont know how to make that question well defined. punches come with momenta, not forces.

@RyderRude yes the forces have to be equal and opposite. But confusion is what force need to take smaller 10N or net force 25N.

Because in law of momentum , both objects momentum contributing . Can i compare this result with this statement that both forces contributing?

in some cases, the smaller force is the deciding factor. e.g. take a bridge which can hold a max weight of 10N. if u put a 5N weight on it, the action reaction force between the bridge and the body becomes 5N @123

I am considering ideal case both objects are isolated in free space and there is no other interaction between them beside applied force.

i cant help much with the punches example, as i dont know what the two different forces mean there. maybe try asking on the site @123

@RyderRude Thanks for your time and contribution. I have just posted this question on quora.

great :)

@RyderRude Pls read the chatGPT answer. Do you agree with that?

no, never ask chatgpt about mechanics

@RyderRude :D Oooookay. It means this is incorrect answer.

@DIRAC1930 seems to be a completely standard fact about resonances; e.g. Weinberg chapter 3 discusses this in passing, too

ACM if your time permit , can you pls spare small time to answer my question.

@ACuriousMind Not the part about it's relation to complex energy solutions in standard QM

See Landau QM section 132

L&L 3 QM

@DIRAC1930 I'm not sure what you mean by that

These states exist in QM

nothing around eq. (21') or eq. (22) in the paper seems unusual to me

I'm not sure what states you're talking about

"This result can be interpreted in the following way: the state $\Psi( 0)$ contains, with amplitude $c_p$, a packet describing a quasiparticle with energy $E_p$ and damping $\Gamma$."

that's just a reformulation of the form of the time evolution

it's just eq. (22) in words

Yes but how do I interpret a complex energy solution in QM

as they say: as damping

if this were a harmonic oscillator, it would be damped with damping factor $\Gamma$

more abstractly, this is an "open system" where your states can decay and just vanish

Hmm I need to think about this more rigourously

again, you didn't need to dig up old papers from the 50s to see this, Weinberg makes much the same point (unsurprisingly) in the section about resonances

really, anyone who discusses resonances should mention this in some way or another

I think first it's better to understand this in the simple QM identical particle manner (which is already pretty complicated) which is why I brought up the paper from the 50s. Understanding this just from the poles of the propagator/S-Matrix in a QFT setting only gives you an understanding through analogies.

Weinberg's chapter 3 is precisely on scattering theory in QM before he introduces the machinery of fields

he does exactly what you want!

I think I'm going to understand L&L section 132 first

Classic weinberg

@ACuriousMind hm well i guess i am just surprised that there is no formalization (in the math sense) of this new probability theory.

Or maybe i should say that such a formalization does not appear to be well known or cared about

@SillyGoose The formalism you're looking for is precisely the theory of the lattice of projectors in Hilbert space

I once recommended Moretti's Fundamental Structures to you when you asked about quaternionic QM, it essentially is about this: Can you start from some weaker assumptions about the structure of physical propositions (i.e. a lattice of observables that's not a priori projectors on a Hilbert space)? And it turns out that under certain assumptions you can't; there is no more general theory of "quantum probabilities" than the theory of density matrices and projectors

@RyderRude I think this wikipedia article saying about mutual actions of two bodies. Can you understand it better?

Understanding QFT means that you are able to reduce everything to non-rel QM

It's basically what they say is going on for the GS superstring here but what the ...

So connecting decaying particles of QFT to L&L section 132

The connection to non-complex energy solutions is done through standard asymptotic states

The last thing to do is this

@123 this references the net forces F12 and F21. again, i dont think it makes sense to talk about punch 1 hitting with 10N and punch 2 hitting with 15N

They discuss two examples of quasi-stationary states in a relativistic context in vol 4, the rest is not just quasi-stationary states in a new language, it's a more fundamental idea which all goes back to what I keep referencing, this quasi-stationary state thing will not illuminate the asymptotic states

@RyderRude Oooh okay... Thanks, but they used the word "mutual actions of two bodies"

i think theyre calling both F12 and F21 as "actions". F12 is the action of 1 on 2. and F21 is the action of 2 on 1

there is no notion of F12 being 10N and F21 being 15N

@RyderRude Ookay...

Which sections in L&L 4 are they?

@RyderRude Pls see the answer from quora.

I think the the total force on each persons will be the 25N. because in N3L pair of forces can not cancel each other. so both persons forces contributing in this. then law of momentum become more sensible this way.

@123 yeah, i think we can trust this person since they have a phd in astrophysics

but maybe they interpreted the forces as being on different points of applications?

thats the only way i could make sense of the question

There simply is NEVER a possibility for 10N one way and 15N the other way.

The very fact that you are asking this, is showing that you do not understand N3L

i think this person is interpreting it in terms of different points of applications @naturallyInconsistent

which is the only consistent way of interpreting it

they use the term "body on fist" and never "fist on fist"

@RyderRude stop confusing people.

it's the only interpretation. this other person too has interpreted it that way

@naturallyInconsistent It means if fist on fist (same point of application) is in collision then smaller force will be the result?

@123 you cannot have bigger or smaller forces. In order to conserve momentum, it can only ever be one force magnitude, seen on both sides as opposite directions.

Either in the fist on fist, it is 10N for both sides, and total is 10N, or it is 15N for both sides and total is 15N. If you get 25N you are just wrong.

If it is 10N, then both sides are fine. If it is 15N, the guy who could only give 10N would have a broken fist. I won't be surprised if it is actually something like 11N

Again, there is NO TWO FORCES. It is only ever one.

That's what K&K is trying to get you to understand.

@123 this may also be relevant physics.stackexchange.com/questions/304317/…

@bolbteppa I'm not sure L&L 9 still references quasi stationary states in their non-rel QFT treatment which was written in 1980

section 62 in L&L 4 specifically references the section on quasi stationary states in L&L 3

@naturallyInconsistent Oooh i see. that's why knk saying extra force will not contribute in N3L.

@bolbteppa @ACuriousMind L&L 4 section 62 talks about this in the context of emission of radiation which is one of the main basic foundational topics that isn't taught anymore

They are basically assuming you are working with solutions of the Dirac equation in an external potential and studying radiation based on the stationary states of those solutions, this is basically modelling a nucleus by a central potential which is ignoring a ton of stuff i.e. it's an approximation, qft involving scattering of asymptotically free particles is ultimately the more fundamental approach

@bolbteppa These poles fundamentally arise in QFT when you have a complex pole in the S-Matrix or propagator. You will observe one of the asymptotically free particles however their will be a spread of momentum free eigenstates you can observe

i.e. the spectral function of the propagator will have a spread of energies

@ACuriousMind hm i see okay thank you

“ In the seventeenth century Descartes developed a theory of vortex motion to explain such things as why light radiated in all directions and the planets moved in circular orbits. He believed that there was no vacuum and any object which moved had to be entering a gap left by another moving object.” this sounds like dirac’s sea :P

The second part i mean

Lol wait Lord Kelvin and J.J. Thomson just coincidentally have the same last name?

Thomson's a pretty common name

i see

do you guys think that simple answers are more favorable than non-simple answers? It seems empirically that simple answers are more amenable to change and so more robust to future observations

I have a function (u-energy) dependent on $\theta, \vec r, \phi$ and the weights for this (ie probability density of a particular energy) is $\exp{\frac{u}{kT}}$ , and I want to average this energy over all possible solid angles at a fixed R, let's say, why is it ok to do $$\frac{\iint u \exp{\frac{-u}{kT}}d\Omega}{\frac{\iint \exp{\frac{-u}{kT}}d\Omega}$$, idk if feels right but also wrong for some reason

@SillyGoose i think science is based on simple answers. people try to model the data with the simplest possible assumptions.

@SillyGoose that's called Occam's razor - the problem is people may disagree what "simple" means ;)

i guess i have always heard its statement, but not any justification for why it is an decent principle

also in the context of just textbook special relativity, suppose I have an electron with rest mass $0.511 MeV$. If I give this electron $100 GeV$ of energy in a fixed frame, then is it like effectively massless?

Because if I compute the $4$-momentum of this electron, it will be $(100, 100, 0, 0)$ to great precision, supposing it travels in the $+x$ direction, right?

but this says that $m^2 = 0$

that's sometimes called the "ultrarelativistic limit" when kinetic energy is much larger than rest energy

interesting, i have not encountered this notion before today :P

or maybe i forgot

@SillyGoose this principle is only a rule of thumb. in practice, explanations have to be satisfy a lot more principles. e.g. the simplest explanation of why planets go in circles might be to say that big things in space go in circles. this is what Aristotle did.

@naturallyInconsistent Have you studied emission of radiation in terms of these quasi-stationary states?

scientific principles also have to satisfy reductionism and many more principles i cant enumerate

perhaps philosophers of science have enumerated them

I have a question I can't figure out an answer to, when you gauge fix in electrodynamics using Lagrange multipliers, we call the choice of multiplier $\lambda=1$ "Feynman gauge"; I don't see how we have the freedom to choose the parameter here. The statement in calculus of variations is that if a functional $$F:=\int \mathcal L(x,\dot x,t) dt$$ has an extremum at $\tilde x(t)$ then there exists a constant $\lambda$ such that $$F':=\int \mathcal L(x,\dot x,t)-\lambda G(x,\dot x,t) dt$$

has an extrema for the same $\tilde x$ for some constraint function $G(x,\dot x,t)=0$. I don't see where in this definition we have some arbitrary freedom to choose $\lambda$ if we are also choosing $G$?

Oh wait maybe you do have the freedom, because of the gauge freedom of $A^\mu$, maybe there is some gauge fixed $\tilde A^\mu$ who's Lagrange multiplier for the given Lagrangian with $G(A^\mu, \dot A^\mu, t)=\partial_\mu A^\mu=0$ the constraint is $\lambda=1$.

hmm but then you've made a choice of gauge to fix that, unless it's possible with the residual freedom after applying Lorentz gauge

@Charlie You're correct that you're not really using it as a Lagrange multiplier when you impose specific values such as $\xi = 1$.

The right argument here is not to talk about "Lagrange multipliers", it's to simply state that adding that term to the Lagrangian "does nothing" - quantumly the S-matrix is independent of $\xi$, classically the equations of motion are equivalent to the equations of motion in Lorenz gauge for any value of $\xi$