yes sorry I am not familiar with the official encyclopedia of physics words

I mean, you have to use words in a way that is conducive to discussing experiments. It needs to be convienient for the experimenter. It makes much more sense to say that the universe's wavefunction is still pure, than to say that "sub-systems are not physical states at all"

Ok, let me try to rephrase that

@Davyz2 No, this is just the non-realism of quantum mechanics: We are not allowed to talk about everything having "definite values" for all measurement operators at all times

that too

this is true in every interpretation that is not a hidden variable theory: There simply are situations where the system is neither in the state 0 nor in the state 1, even if classically it would only have those two possible states

Alice and Bob measure a shared entangled QBit pair. When Bob traces out the second QBit, the resulting state is not a QBit, correct?

yes, but that is because you used the trace operation

that does not mean the system does not "have a physical state", it simply means the "physical state" no longer - in contrast to classical mechanics - contains a list of definitive values for all observables

That's what i am trying to say, using the trace only allows you to define information about your "sub-system", this sub-system you obtain is not a state of the theory

if you insist that a "physical state" should, then you are commiting to only accepting hidden variable theories

a physical state should NOT contain a list of definitive values at all, this is why Bob's subsystem is not physical, Bob is using hidden variables if he assumes that

@Davyz2 No, it absolutely is a "state of the theory". QM theory has no issues with mixed states (and one can define them without every talking about classical mixtures or probabilities or whatever)

No, usually people consider the density operator as the definition of the quantum state. Mixed states are perfectly fine as quantum states in this usual lingo

if you think a mixed state is "not a state", you are simply using non-mainstream terminology

Alright, so are we going to purely reason about what people usually say

@Davyz2 No, but you have to explain what you mean by it not being a state

because you say that as if its meaning should be clear, but in the standard meaning it's just wrong

Also, note how practical it is; you know that the oven spits out quantum particles obeying a certain mixed state kind of distribution. There is no reason why we cannot call that a quantum state.

so I'm trying to understand what you really want to say

It's not a state because it's in a probabilistic mixture of definite values for an observable, so it's equivalent to a random classical state

random classical states do not fully capture all the states of the quantum theory, it's true that they can be incorporated

And note that in the Stern-Gerlach apparatus (with bending the two beams back into one), after measurement or not, you know that you have a quantum state ready for the next Stern-Gerlach apparatus to measure. It can be mixed, it can be pure, whatever it is, it is definitely a quantum state

but they don't represent physical things, they are observer's classical knowledge descriptions

While it is true that density matrices/mixed states are often introduced in that way, there's really nothing inherently classical about them

@Davyz2 Ah, this is already an interpretational decision

No it's not, because if a physical state is in a mixture of definite values, you are in a hidden variable theory which is non-local

@Davyz2 oh, a density operator can well not just be a random classical state. It can be a mix of classical ignorance AND having quantum correlations

approaches to QM that do not rely on quantization (and hence have no notion of classicality) still need mixed states (because of the partial trace operation producing them from entangled states)

@Davyz2 no; you would find it difficult to prove these assertions of yours

There is nothing "classical" about the mixed state resulting from entanglement

(The whole point of Bell's theorem is that entanglement isn't classical!)

yes, but the "sub-systems" of Alice and Bob behave like classical states, they have the same outcome distribution of a superposition, but their density is different

in a sense, what ACM and miao miao is trying to tell you, is that we actually know a whole lot more about the nature of quantum theory now than the pioneers knew.

@Davyz2 I don't know what "behave like classical states" means here.

They have definite values before the measurement

But...they don't

before anything is measured, you have a global state like $\lvert 0\rangle_A \otimes \lvert 0\rangle_B + \lvert 1\rangle_A\otimes \lvert 1\rangle_B$

nothing definite about that, it's either 0 or 1 :P

How? A density like |0><0| + |1><1| is either in 0 or 1, but you don't know which is which

that's a hidden variable description, as such this state cannot be a physical thing?

I think there is a fundamental confusion about density matrices here, but I'm not quite sure what it is

@Davyz2 I don't understand where the hidden variable is supposed to be, nor what a "physical thing" is

@Davyz2 what you have written down is the identity operator; while it is a valid density operator, it is not portraying what you think it is

this is just a density matrix that you chose to write in a specific basis (funnily enough, in 2d it looks like that in every basis, it's the identity)

1/2(|0><0| + |1><1|) represents a scenario in which I send you a |0> particle or a |1> particle and you don't know which one I sent you, the whole state is a description of your knowledge with respect to this particular experiment

it's not a state of anything physical, you are describing your probabilities with respect to an experiment in which receive randomly 0 or 1

@Davyz2 no, that is the wrong way around: That scenario is one way to provoke me into producing that density matrix as its description, not that every density matrix that looks like that has to describe that scenario

that was my point about producing mixed states from entanglement: Just because you can produce density matrices from classical lack of knowledge does not mean that all density matrices have to always represent a lack of knowledge in the classical sense!

Yes, whatwever you prefer, the point is that it's an informational state with respect to an experiment, that's very different from writing |0> + |1> which is an actual particle state

For example, if you were instead originally given $\frac12\begin{pmatrix}1&1\\1&1\end {pmatrix}$ then you will also seem to measure $\left|\uparrow\right>$ or $\left|\downarrow\right>$ with equal probability, but actually this density operator is a pure state always giving you $\left|+\right>$

The density matrix description depends on the observer knowledge about the experiment, if I tell you that I am going to send 3/4 of the time the state |0>, you have to adjust these probabilities

but I am still sending you the same particles

@Davyz2 No, the distinction matters: You are trying to claim that just because sometimes a density matrix arises just because I lack classical knowledge, it is always just a lack of knowledge and not a genuine state. I say this is wrong: Either you have to accept that sometimes (when produced from entanglement) a density matrix is a genuine description of a physical state or you have to accept that subsystems of entangled systems don't have states

@Davyz2 What you are saying is an "actual particle state" is really just a pure state. But mixed states are perfectly fine as quantum states.

@ACuriousMind Subsystems of entangled systems don't have states, they can't

@Davyz2 then what do experiments producing entangled stuff do?

@Davyz2 Okay, you should use something other than "state" for that because the vast majority of physicists at least uses the word differently even if they may agree with you, but yes, that's a consistent viewpoint

Before Bob receives his part of the entangled states, he is not writing the proper quantum state of his subsystem, because this state doesn't exist, he is writing a density which describes his experimental knowledge

there is no state, as you said before, Bob is part of the entangled global state, it's just that his experimental knowledge is described by the tracing operation

you are creating a strange hybrid view of quantum states here

people usually either fully commit to the ontic ("all states are real") or the epistemic ("all states represent knowledge") viewpoint (where I use "state" for both pure and mixed)

which explains the miscommunications so far :P

yeah.

Yes, both are wrong in my opinion, if you take QM to hearth, one state is real and highly entangled, observers are "sub-systems" of this state describing other sub-systems via tracing the global state

I mean I guess there may be MWIers that have no problem with only calling the global state "real" and the rest representations of knowledge

Like, it is perfectly fine that from Bob's PoV his best attempt at describing the quantum subsystem that he received is that mixed state, but he himself would know that, if he widens his perspective, there would be a pure state that describes the Alice+Bob system

but it's not usually the point of contention

Yes, that was my point

I think it's the only satisfactory explanation to be honest

for me at least

It also explains why everything around us is classical, being a sub-system of the highly entangled state, I can only share classical knowledge via mixed states between other sub-systems

what

now you've lost me again :P

also clearly not everything around us is classical, we can clearly do QM experiments within the global state, how else did we figure out QM...

Yeah no sorry just a rambling, i can't really talk

i meant something different no worries

i'm glad we agree on what we talked before tough

@ACuriousMind yes this is correct

Now, imagine if in the standard curriculum we have everybody understand these basics.

That would be a wonderful world.

Hello Everyone...

@Davyz2 u might find this post interesting wrt MWI and Born rule physics.stackexchange.com/questions/762350/…

Why conserved quantities are important in physics? What is the benefit of knowing conserved quantities.

@123 e.g. the work u get from a system is proportional to the fuel. this is important

@123 If you want to make sense of the world, conserved quantities would be some stuff that you definitely want to keep track of.

Thanks can you pls give example

@123 This is a strange question. Surely you have solved exercises where you predict what happens e.g. to two things colliding by applying energy and momentum conservation?

Conserved quantities depend on the mathematical description of some scenario, it just so happens that some of your quantities will remain constant if you assume some evolution law

These can also be non-obvious, in the two body Newtonian problem, there is the Lenz vector for example

(I know you have because you've been coming to this chat for years and you have read standard mechanics texts)

@ACuriousMind yes I solved the problems of conservation of momentum and energy.

@123 so then what is the question? You've obviously seen how one can use those conservation laws to solve problems that are otherwise much more difficult or intractable.

I asked here this question because I can find some great explainations from you.

or could you have done the collision stuff without applying the conservation laws?

I have seen few books derived law of momentum from newton's second law and some books from N3L. Which one is correct?

I am confused

Easy. Any time you see forces as the starting point, it is already wrong and you need to start over.

What is really happening, is that there is momentum (which is conserved) and energy (which is also conserved), and exchanging them is seens as forces in old mechanics.

Once you understand the momentum is conserved, N3L is a free. It is obvious.

How momentum is transferred in time, appears as N2L.

Yes I have seen law of momentum derived from N3L. But then I saw in books they derived law of momentum from N2L by assuming no external force. Which made me confused.

It depends on what exactly people think Newton's laws say in modern language (no, this is not always unambiguous, see physics.stackexchange.com/a/70188/50583 for one discussion) which of the laws derives what; why does it matter - everyone agrees that Newton's laws imply momentum conservation, no?

Look, BOTH are wrong. It is that momentum is conserved, and that is why we have N3L and N2L.

What is meant by "how momentum is transferred in time, appears in N2L". What is the difference between momentum transferred and conserved.

@naturallyInconsistent well, that's the post-Newtonian viewpoint, but not how we teach Newtonian mechanics :P

@naturallyInconsistent Ooh I see

@123 You must first have something that is conserved, before you can think about why it makes sense to track how it is transferred around.

@ACuriousMind And I'm unapologetically interested in teaching modern physics in a proper and modern way, not Newtonian mechanics the Newtonian way.

Not like you dont know that, and that I dont know I'm not teaching it the old way

:P

yes, but I'm not sure 123 realizes that if I don't explicitly state it :P

does anyone have a resource on ramsey interferometry (friendly to a theorist)

@ACuriousMind why ? I have read law of momentum imply newton's law. Isn't it correct?

see what I mean :P

@ACuriousMind Pls explain in easy . My English is weak

@123 physics does not have a unique starting point. some people take some things as starting definitions and postulates while other people take other things

@123 It is correct; Conservation of momentum implies N2L and N3L, but at the same time, N2L and N3L imply conservation of momentum. The problem is that later on in physics, forces is no longer acceptable, and then you have to start over if you started with N2L and N3L. If you started from momentum conservation, then you can continue without problems.

@naturallyInconsistent Ooh I see okay.. thanks

it is (seemingly) natural to want the starting point of physics to consist of as little things as possible. this is the train of thought that naturally's emphasis on conservation laws (over newton's laws) rides on

@SillyGoose thanks understand better.

because if you take conservation laws as the starting point, your framework of physics will be more consistent with more modern theories of physics (theories beyond Newtonian mechanics)

@SillyGoose It is not just "as little things as possible". I'm trying to smoothen the learning journey. If you start with things that will lose their meanings, then you have to keep restarting. I'm trying to start with the best starting point so that we don't need to keep restarting.

If I assume law of momentum as starting point. How this law comes to N2L and N3L. Both newton's laws are different.

I'm prioritising pedagogy, not axiomatisation

@naturallyInconsistent the best starting point would be the axioms of a complete physical theory ;)

@123 Momentum is conserved. The only way for one thing to get more momentum, is for something else to give it the momentum that it had, in which case the giver will have less. There is one act of giving; the one that got more, sees a "force" one way; the other one receiving, sees the opposite force. That is N3L.

well in reality I would think the best starting point is a "mathematically watered-down" recount of an axiomatized physical theory

N2L talk about single object as system. N3L talk about different objects as action reaction force. Both are different. How can law of momentum imply both newton's law?

@SillyGoose I don't think anybody knows how to teach high schoolers quantum field theory, in either of Lagrangian or Hamiltonian form.

@123 I just gave you how N3L is obtained. The transfer of momentum into an object, in time, defines the forces in N2L; i.e. you can use this to define forces, or strictly speaking, forces are not needed in the momentum viewpoint.

@naturallyInconsistent thanks for good explanation. Pls explain law of momentum imply N2L.

@123 $$\frac{\mathrm d\vec p}{\mathrm dt}=\vec F$$

Thanks all of you to clear the confusion.

Do u prefer being busy or free time and why?

Hello, I wanted to ask something regarding the notation used by Rayleigh, when he introduced his Dissipation Function in his original Treatise, Section II. No other book tries to cover it in depth, so I am stuck with the notation

The notation is as follows, he says that, given a coefficient like $\bar{rs}$, is an abbreviation for the "quadratic operator:"

old-timey notation is never fun to decipher

Also think about why

> No other book tries to cover it in depth

btw @AdityaKrishnaPanickar welcome to the chatroom

:-)

Thanks a lot

@ACuriousMind True Sir! But can you give me/ direct me towards the right resources

@user20458579510081670432 Thank you!

@AdityaKrishnaPanickar nope, I haven't read the paper and have no idea what it's doing

@ACuriousMind Bah no i have not seen spinor, but i have started to walk a different route. i do have a BIG cat update.

so i joined my campus' cat caretaking community. there are two female adult cats: cookie and oreo. cookie had 5 kittens. there are two separate kittens living under the building that appear to be close comrades. we have gotten cookie and oreo spayed and are almost done with their vaccinations. we have also adopted out 3 of the kittens. we are now trying to get comrade kittens to come out of the building so we can assess them but they are very shy and only come out at night.

ill send pics of all

@ACuriousMind Ok Sir.

@AdityaKrishnaPanickar really cool

Has anyone here ever heard of “tardyons” and could point me to a source with some sort of basic discussion on the topic?

@RyderRude Hey Ryder! could you help me, if you do understand the notation

they appear related to tachyons but beyond that it seems somewhat obscure.

hello aditya

@AdityaKrishnaPanickar i have no idea...sorry..

oreo loves to bask in the sun where it says "no leaning" lol

@ZeroTheHero isn't that just a jokey name for particles that aren't massless and so are slower than the speed of light? (see en.wiktionary.org/wiki/tardyon)

it's just the inverse of calling a tachyon anything that's faster than light, I think

@Arjun Yo!

do u think "particles can't go faster than light" becomes loosely defined in GR becuz velocities r only local now?

so a “tardyonic representation” would be a rep of Poincaré for a massive particle?

some people have a twisted sense of humour…

@ZeroTheHero wait till you hear from those working on twistor theory

@ZeroTheHero yes, that would be my reading of it

I was about to write a “massive sense of humour” but that was too easy.

wait do u have a cat? @ACuriousMind

lorentzian vegetable, the cats are so cutteeee

@naturallyInconsistent i know. when its my duty to feed them i feel my life has a true purpose.

the linguist in me feels compelled to point out it's not even a good joke because they didn't translate "tardy" to Greek first :P

the "right" name would be bradyon to contrast with tachyon

@Relativisticcucumber Sadly not, my current flat isn't well-suited for cats; but we had a cat when I was growing up

too much terminology is bad. Words should be self explanatory

@ACuriousMind here comes the brattyon

@ACuriousMind :o what was its name

:/

@Relativisticcucumber uh...Jimmy xD

@ACuriousMind do u have a fish or a dog?

omg

or rabbit/tortoise

Any pet?

he was a stubborn and aloof cat who only came cuddling when either he or I were sick :P

my earliest memory is of a cat. very sad. my mom and i had a cat and in the middle of the night when i was like 4 the cat clawed my face and i screamed. my mom came running in and asked if the cat scratched me. i said no and begged her to believe me. sadly she knew it scratched me and got rid of it the next day :((

in that case it was afloof

@ACuriousMind omg. well at least he came sometimes. when i had snowball nothing in the world could motivate her to give affection

i love cats so much but they despise me no matter what i try

@Relativisticcucumber We got the cat because of me; one of our few vacations were to a hotel in the canaries that was overrun with cats and I liked them so much that I didn't stop talking about cats until we got one :P

@Relativisticcucumber some cats just aren't cuddly cats, but they still love us ;)

I used to have a fish, but he is dead now

And then a sparrow

A sparrow as a pet?

I have hundreds of the little buggers screaming outside my window every morning, can't you adopt some more? :P

He was injured

I took care for two months

I haven't tried big pets yet

...and then he flew away healthy, right? Right?

Yes. And he stopped coming back..

I thought he wud come back forever

I mean...they're migratory birds, probably just followed his migration pattern

maybe he lost the way or flew too far

@ACuriousMind yeah..

now you know how parents feel when their kids don't call ;')

woww.. i hadn't thought of that

I need to call more now

@RyderRude Your sparrow reminds me of an unfaithful girlfriend one my uh uh friends had,that guy took great care of her when she was broken and when she got fixed she flew away ; )

XD

what a weird thing to say

@ACuriousMind what an analogy.. :)

@LuckyChouhan is it even that much of an analogy? Many people consider their pets "their kids" or at least similar to children

@Arjun lol, a guy helped her in fixing in order to get her haha, Intensions Feynman Intensions......

@ACuriousMind yeah, Actually I have never had a pet. Btw do you have pet(s)?

@LuckyChouhan okay people, that's enough weirdness, please stop discussing human relationships in this context

@LuckyChouhan scroll up to find out, I already talked about that :P

@ACuriousMind yeah, how can someone be somewhat excellent in two different fields? I mean there are mathematicians who are physicist as well see Maxim Kontsevich he is field medalist and breakthrough prize laureate in math and fundamental science. And there are many mathematicians who are computer scientist as well so how they manage to study two or more fields at advance level

I'm not sure what that's a response to

@ACuriousMind No no it is not a response. It is just a question that I wanted to ask you :')

why me

@ACuriousMind you seem to be the right person to ask,

I'm not even sure I understand the question - people like Kontsevich do a heavily mathematical kind of physics, it's not as if their physics would be somehow separate from the math they do

you just study what interests you; it if turns out that's the application of advanced mathematics to some other field, you study that

so basically, intersection of two fields.

@ACuriousMind I still have a doubt in mind about what you said about Noether's theorem. You said that we don't transform the coordinates, but only the fields in Noether's theorem. So the integral measure remains same i.e. $d^4 x \to d^4 x$, right? I mean in (d+1) dim. field theory with $d \ne 0$...not point particle mechanics.

it depends on what exactly you mean by "transform"; what is the question?

@ACuriousMind While finding the Noether current, I do the following: promote the global symmetry to a local one and compute the variation of the action. When computing the variation, do I change the coordinates or not in the case of a spacetime symmetry?

@RyderRude thank you for those links I'm going to check them today. Will let you know.

you don't need to do that to find the Noether current

the problem here is really that there are so many presentations and tricks when it comes to Noether that I can't be sure what exactly you're doing (and honestly I'm not particularly interested in finding out), see e.g. Wiki for a field theory version of Noether's theorem that is pretty straightforward and does everything correctly

there is an aspect that "transforms the coordinates" but the important thing is that it works a bit differently than it would if we just changed the integration variable by substitution

ok

@Arjun it's ok. Sparrows r meant to fly

@LuckyChouhan great

@ACuriousMind Actually you got to the crux of the problem I am facing. Different sources have different methods of doing this and surprisingly arrive at different forms of currents. I mean there is always a term proportional to the canonical momentum. But in some places there are some extra boundary terms.

@NairitSahoo the boundary term difference is usually due to different definitions of a "symmetry" - they are absent when you demand that the Lagrangian be really invariant, but they appear when you only demand that it changes by at most a total derivative (what some call a "quasi-symmetry")

@NairitSahoo do u do physics as hobby or work?

@LuckyChouhan Witten also has a fields medal because of this

He also has the Dirac medal

@ACuriousMind Yeah. got it.

@ACuriousMind I think I finally get when to transform coordinates and when not to by comparing both the derivations side by side.

If we compute $\Phi'(x')-\Phi(x)$ we are just computing the contributions from the "internal" part. In that case we need to transform the coordinates in addition. But as advised by you, if we do $\Phi'(x')-Phi(x')$ then we automatically get both the contributions from the internal and the external part, so we don't need to transform the coordinates again in which case it will be a trivial change of variables.

As you said everything was hidden in what I mean by "transform". Thanks!

@ACuriousMind XD sorry lol!

Apparently you can derive the cosmological constant and dark matter

arxiv.org/pdf/2402.19459

Also

@RyderRude great, how you got interested in physics and math?

@LuckyChouhan I've always been scientific and philosophical minded, i would say. Math is something i liked after learning about number systems

like infinitesimals and stuff

and also calculus

@RyderRude hey!

@MoreAnonymous hi

If u don't mind me asking are u still in academia?

@RyderRude great,

@LuckyChouhan also, i started learning QM 3 years ago, which i liked very much

@MoreAnonymous he has taken a long break from study like a sabbatical

Ah didn't know ...

@MoreAnonymous i just do physics as a hobby

@MoreAnonymous really interesting

@RyderRude so what kind of job you do nowdays?

@RyderRude I studied with Russo in Nottingham

@RyderRude same

@RyderRude I am a 2nd year UG student. In future I would like to do physics as work.

@MoreAnonymous I feel this is a very misleading teaser to put in front of this article when the authors are very careful to say already in their abstract: "We caution that a greater understanding of this effect is needed before conclusions can be drawn"

@LuckyChouhan i do a variety of stuff... It's hard to explain

@MoreAnonymous great

@NairitSahoo oh

@ACuriousMind guilty but heck saline did that too

oh, just because other people do irresponsible science reporting, you have to, too? :P

@RyderRude oh, so what is your main work? Why didn't you choose academia :)

@MoreAnonymous I remember youtube recommended me this video 2-3 times in past few days.

@ACuriousMind of course

@LuckyChouhan i've always only liked it as hobby. Also, I'm not very good at math

:p

@RyderRude I think many physicists have math envy

@MoreAnonymous yeah

I mean we weren't thought it well enough. For us physicists everything converges cause obviously it exists in the real world no :p

It is enough detail anyway

@RyderRude I'm not good at physics :( Today I was reading this wikipedia article on Proof of Fermat's Last Theorem for some specific exponents I would recommend you because usually wikipedia articles tends to be more advance but this one is really good.

@RyderRude when did you pass out from college?

@LuckyChouhan it looks good. I will check it out

@RyderRude would you like to nominate yourself for MathSE Election?

Given how some transformations act on coordinates, we can find its algebra of generators. And then using method of induced representations we can find the action of the generators on fields. Exponentiating that we find how the fields transform under the transformation. Is there any other option other than the method of induced representations?

@Relativisticcucumber @naturallyInconsistent see miao miao haha

@LuckyChouhan no...