The h Bar

Slereah

07:38

"Imagine you have some experiment that is not an S-matrix experiment. Imagine a robot doing the experiment. Assemble such a robot and some Hydrogen bombs from cold asymptotic particles, let the robot do the experiment, beam the results out in a photon beam to infinity, the detonate the H-bombs, so that the whole thing is blown into cold asymptotic particles. The S-matrix for that process includes the result of the experiment."

Silly Goose

08:12

do yall see value in learning physics in the order it was discovered

Slereah

It is generally assumed to be more intuitive, but I'm not sure that's true

you hammer in principles into students and tell them that they are true and then tell them psych, actually it's not true

Silly Goose

i was thinking that what one loses might be being exposed to the path of discovery

and perhaps how messy it was to get to a simple result or etc

which cxould give students more appreciation o f the ideas but idk

ACuriousMind

I think in order to appreciate the evolution of the ideas it is much more valuable to first learn them in a straightfoward modern fashion

i.e. I think there is value in learning the historical sequence of discoveries and arguments, but I don't think it is usually the best way to learn a subject the first time you're learning it

early quantum theory in particular was a mess

Silly Goose

ah hm i think i agree

im trying to learn more about the mess xD

also do yall have a resource recommendation on matrix mechanics specifically

Slereah

From what I can see there's a ton of students still thinking about mass, energy, frames, time, space, etc in a Newtonian manner and they struggle to understand modern physics because of those assumptions

Silly Goose

08:18

that seems to be happening in my modern phys class xD

i feel it would do some people good if my school had a separate SR course

to get people used to new ideas

ACuriousMind

@SillyGoose now see, "matrix mechanics" is just the idea that observables are linear operators and that we can do QM by essentially linear-algebraic methods

in early quantum theory, it made sense to contrast this with "wave mechanics" (which is essentially just doing everything in the position representation) because people didn't know the connection between the two yet

but the two concepts as distinct "mechanics" don't really make sense from a modern standpoint

it's just abstract linear algebra in infinite-dimensional spaces vs. doing it in a specific basis

Slereah

Some paper claims that a natural choice of the Weyl scale is the Planck mass

Ouhlala

Color me interested

ACuriousMind

but in order to understand that you need tools like the Stone-von Neumann theorem, which came several years after people started doing QM

Silly Goose

i feel every time i learn about some mathematical object or structure in QM a million others behind it show their faces xD

ACuriousMind

that feeling is largely true :P

Slereah

08:26

It never ends

Silly Goose

unfortunately not enough time to learn it all

ACuriousMind

most of the more complicated math is more useful to know about than really understand in this context, I feel

it's important to realize that QM is not as simple as finite-dimensional linear algebra

it's often less important to understand exactly how some issues are resolved by actual mathematicians - but it's valuable to be aware of them

Silly Goose

hm that's interesting since i was thinking about this earlier today--how much of the intro. group theory that im learning in this math course will actually carry over into understanding physics better

ACuriousMind

oh, imo, group theory and representation theory you absolutely need to learn from mathematicians because every physics text I've ever seen absolutely sucks at explaining representation theory :P

Silly Goose

ah i might not even take representation theory (it's a separate course for us). i have one math elective undecided

but i think i should xD

ACuriousMind

08:31

In the context of QM and "complicated math" I was more thinking about all the functional analytic issues like what $\lvert x\rangle$ actually are or why $x$ or $p$ sometimes give nonsense when applied to perfectly good states (domain issues), etc.

Silly Goose

so investing in functional analysis won't shed much light on QM?

ACuriousMind

I mean...it will shed light on its mathematical structure

I don't think it does anything for understanding the "core physics" involved

because the physical essence of QM - non-commutativity of observables - you can already see on finite-dimensional vector spaces (like state spaces of qubits) where no functional analysis is involved

because we found the double slit first intro QM tends to focus on wave mechanics but I really think better pedagogy would be to start with finite-dimensional systems

Silly Goose

is the non-commutativity of observable equivalent to the inherent uncertainty in QM? and is that what you are saying the physical essence of QM is?

ACuriousMind

all the weirdness of QM is already there and you don't have to lie to student about continuous eigenvectors or whatever

@SillyGoose yes - the r.h.s. of the general uncertainty relation is just the expectation value of the commutator of observables, after all

Silly Goose

Townsend begins with 1/2-spin systems :D

ACuriousMind

08:35

the other weird aspect is entanglement which isn't directly related to non-commutativity (and more to linearity), but that's also fine on finite-dimensional systems

Silly Goose

also...is requiring the rotation generator to be hermitian equivalent to saying space is isotropic

ACuriousMind

the only thing you can't discuss in a finite-dimensional context is quantization

@SillyGoose no, it has to be Hermitian because angular momentum is an observable

doesn't have to be conserved (which would be isotropy)

Silly Goose

i need to learn about entanglement xD

im reading through townsend to begin to understand paige-wooters mechanism

is space postulated to be isotropic in quantum?

ACuriousMind

not particularly, no

I mean, there isn't really "space"

you just have a Hilbert space

there is no $\mathbb{R}^3$ you could talk about being "isotropic"

Silly Goose

ah hm

Slereah

08:40

I think I found a good paper

might help understand what the various parts of the tractor represent

Silly Goose

is phase a continuous symmetry in QM? or does this question not make sense xD

ACuriousMind

depends on what you mean by "symmetry" :P

Silly Goose

i mean to ask if gives rise to a conserved quantity

ACuriousMind

those are two different questions!

Silly Goose

oh no xD

ACuriousMind

08:46

the global "phase invariance" of QM is really because the actual states of the theory are rays in Hilbert space and not vectors, making it similar to a gauge symmetry - the symmetry is in our description of the system, not an intrinsic property of the physical system

you can do QM on the projective Hilbert space and then the "phase symmetry" is just gone

Silly Goose

ah

so it isn't something interesting?

do we lose any interesting structure from working only in the projective hilbert space

well wait im confused though

ACuriousMind

working in the projective Hilbert space is annoying, which is why no one does it :P

Silly Goose

doesn't QM only make sense in the projective hilbert space

oh XD

ACuriousMind

in Hilbert space, all our equations are nice linear equations about linear operators

Silly Goose

why do textbooks work in the projective hilbert space at least griffith's and townsend it seems (and it seems like sakurai as well)?

ACuriousMind

08:49

I think we have different ideas about "working in the projective Hilbert space"

your books still write all equations in terms of state vectors and linear operators, don't they?

Silly Goose

yes

ACuriousMind

I mean, yes, the projective Hilbert space is the "actual space of states" but every computation we usually do is in the linear Hilbert space

Silly Goose

but the states are presumed normalized which i thought implies projective hilbert space

or idk maybe the books are not very strict on that presumption

ACuriousMind

@SillyGoose normalization stll leaves you the phase freedom $\lvert \psi\rangle \mapsto \mathrm{e}^{\mathrm{i}\phi}\lvert \psi\rangle$

but those two things are the same point in projective space

Silly Goose

i thought the fact that we are free to pick phases means that we can normalize kets and call it done. and so if we pick a phase then we throw away all other kets that differ only by phase

but it seems i misunderstand what is being done

ACuriousMind

08:53

as for the phase and conserved quantities: people often get confused about the global phase that's due to the "states are rays" things and try to claim things like that it's the U(1) global symmetry of EM, because that also acts as a phase, and so the associated conservation is charge conservation but that's really not true (for one because the global phase also acts on neutral particles while the EM symmetry doesn't do anything to neutral particles)

Slereah

If you want to pick a simple example you can look at the most trivial example of a quantum theory

which is just $\mathbb{C}$

ACuriousMind

@SillyGoose I think what you misunderstood is what the projective space is - but don't worry about that too much, because no one in intro QM cares about it :P

Slereah

If you remove zero, you can normalize all the states by $z / |z|$

And you are left with a circle, $U(1)$

that circle corresponds to the "phase gauge"

Silly Goose

oh interesting

Slereah

since the states $z$ and $e^{i\alpha}z$ are equivalent

So your actual projective Hilbert space is just a single point

This is the trivial Hilbert space of just a single vacuum state

So you have $\mathbb{C} \to U(1) \to \{ 0 \}$

ACuriousMind

08:56

0 is a particularly bad choice there :P

since the thing on the left should be $\mathbb{C} - \{0\}$

Slereah

Well 1 if you prefer

or $\{ * \}$

To be all categorical about it

The Hilbert space of a single thing

Silly Goose

so in this case everything maps to a single element because everything will be = 1 in $\mathbb{C}$ post-normalization?

Slereah

Silly Goose

err wait

Slereah

Yes

ACuriousMind

08:58

normalization is not fixing the phase

Slereah

That is the smallest Hilbert space you can write that isn't empty

ACuriousMind

normalization is just saying you want your state vector to have norm 1

Silly Goose

so normalization is just choosing to look at the normalized ket that already exists in your space?

ACuriousMind

$\mathrm{e}^{\mathrm{i}\phi}$ is a normalized complex number for any $\phi$

Silly Goose

wait what is $\{1\}$ here?

what is it the identity of?

if it is an identity

Slereah

09:04

It's the set containing a single point

And we can pick any value of that circle as a representative

usually 1

Silly Goose

ah ah ah okay

or i guess i should just say 1 is really an equivalence class of things that we can represent with just one element

Slereah

Yes

Silly Goose

i was confused about calling it a point but it makes sense now!

Slereah

Although in a sense, that's what your actual (projective) Hilbert space is

The rest is just redundant information

But the zero dimensional case is easy enough, though

this process doesn't give things as pleasant in many dimensions or infinite dimensions

You can do it for $\mathbb{C}^2$ though : en.wikipedia.org/wiki/Bloch_sphere

Which describes a quantum system with two states, like spin

Silly Goose

wait now im confused again are there two equivalence classes being made

1) for $ze^{i\alpha} \equiv e^{i\alpha}$
2) for the resulting normed vectors?

Slereah

09:10

Yes

Silly Goose

ah i see i see

is the 2nd equiv class created by equivalizing states with same norm

Slereah

I mean all states have the same norm

It is 1

Silly Goose

err then what is the condition of the second equiv class

Slereah

The phase

Phase equivalence

Silly Goose

oh so like $e^{i\phi} = e^{i\alpha}$?

err \equiv instead of =

Slereah

09:13

If you have two points in your Hilbert space, they are related by a phase

is the intent

you could technically work directly in those spaces, but the benefit of working in the actual Hilbert space is that they are vector spaces

You can add and multiply in them without issue

Silly Goose

ahh i see

Slereah

This happens a lot in physics where the "real" spaces are very nasty looking so instead we just use a vector space and then have redundancies

Silly Goose

it seems potentially problematic to expand your space and only introduce redundancies :o

as in hard to control what you introduce? but maybe not

Slereah

It certainly is

Silly Goose

that process sounds very interesting

Slereah

09:32

that's already sort of what you do with analytical geometry in a sense :p

Instead of having lines and circles that you have to translate around and compare and whatnot, you just have numbers that you add and multiply

Slereah

09:48

@ACuriousMind If I have a map $c : \mathbb{R} \to \mathbb{R}^n$, and I have a measure $\mu$ on $\mathbb{R}$ and a map $\mu^n$ on $\mathbb{R}^n$, is there a notion of how those measures relate?

Roughly how you get a volume from a set of lengths

ACuriousMind

what do you need the $c$ for?

there's a product of measurable spaces that gives you $(\mathbb{R}^n,\mu^n)$ via the n-fold product of $(\mathbb{R},\mu)$ with itself

Slereah

I am trying to understand what makes the Riemannian measure specific out of all the volume forms

You do have a new measure specifically when you measure the length of a curve which is the pushforward $X^* g$

ACuriousMind

@Slereah it's the volume form where if you take a (generalized, infinitesimal) parallelepiped the volume assigned to the parallelepiped by the volume form is the same as if you compute the Riemannian lengths of the sides and compute the volume from the lengths via "normal" geometry

Slereah

I guess to avoid changes in the length of a curve in certain directions via squeeze mapping

ACuriousMind

there isn't a lot of diff-geo there, this is just linear algebra - when you have a matrix $g$ as an inner product, then the determinant $\sqrt{\lvert g\rvert}$ gives you the associated volume

Slereah

09:56

Is that basically what that kind of measure does, relate the volume form to measures of subspaces?

ACuriousMind

I don't think you need to do anything complicated here, this is really a local condition, i.e. you can express what the volume form is on a single tangent space, no need to talk about curves

Slereah

Yeah but my interest is specifically how the two relate here

how the measurement of a volume relates to the measurement of a length

ACuriousMind

which two?

@Slereah again, it is just linear algebra that an inner product (i.e. a length) $g$ has an associated volume $\sqrt{\mathrm{det}(g)}$

Slereah

Fair enough

Slereah

10:29

Looks like the third component of a tractor is the famed "null direction" that weyl talks about i think

Not quite clear how that works though

You have the "position tractor" (0,0,1) which defines a null ray and then you build your stabilizer group out of that i guess?

I guess I need to figure out what equivalence class of vectors this corresponds to

I guess that's why projective tractors don't have that third component

Although I guess the question is, do conformal spaces that don't have null directions have that?

Tractors look like that because they have the algebra $\mathfrak{o}(n, 2) = \mathbb{R} \oplus \mathfrak{o}(n-1, 1) \oplus \mathbb{R}$, but I guess the conformal group for Euclidian space is gonna be $\mathfrak{o}(n+1, 1)$

Which decomposes as... idk

No apparently that form of the tractor is the same for every signature

It's that weird thing where the Moebius space of conformal geometry already has a light cone even in the Euclidian case

Slereah

13:04

I think I'm gonna have to dip into Schotten again

user250478

13:44

Few weeks ago I did an experiment (I'm mathematician, I'm not a professor) at home with dishwasher soap and a plastic bottle, when the water that fell from the tap formed small soap bubbles that came out of the bottle as a result of the Venturi effect. On the other hand

1)I know (I have been studied it when I was student) the Brachistochrone Problem (as general reference for the audience the encyclopedia Wolfram MathWorld edited an article dedicated to this problem); 2) I know that the soap bubbles are minimal surfaces and that the and that the orifice of the bottle could wish as a border, and 3) the video Self-Assembling Wires from the official

channel of YouTube Stanford Complexity Group (edited seven years ago). I would like to know if it is known if all these seem cases of a same theorem/principle of the mathematical physics: it seem related to a boundary as a constraint, homogeneous particles (length elements, surface elements, particles in three dimentions), the definition of a field and/or provide energy to our system of particles.

Can you (the users of this chat room) clarify if they are examples of the same physical phenomenon? Many thanks.

There is a typo in the phrase "...bottle could wish as a border,...", the right is "it could be considered as a boundary/constraint".

Feynman_00

14:29

There is one thing related to a discussion that happened here in June, I think. I found this related questions, altough it was marked as a duplicate. Regarding the last three comments, why do we "interpret the velocity rather than the momentum (canonical) to transform"? I mean, what is the difference wrt to the classical case for which we need to do so?

I would say the reason is that the canonical commutation relations have to remain the same (and this goes for Poisson brackets too in CM) but such commutation relation fix the form of momentum operator since the position is unchanged (I think this is the different part), which is the same for this reason

What I'm saying is that the "status" of infinitesimal generator of translation, which is valid both in the classical and quantum case, constrains the momentum operator to be unchanged in the quantum case, while classically a canonical transformation (e.g. gauge transformation) can be of the form $(q,p)\rightarrow(q,P)$ with $p\neq P$ (in fact the kinematic momentum is gauge invariant classically)

Slereah

14:51

Writing the tangent bundle as $\mathcal{E}^i$

Why oh why

ACuriousMind

@Feynman_00 I don't really understand what the question/statement here is supposed to be

Slereah

15:08

What if

The conformal structure is about the measure on Rn and the projective structure is about the measure on R

Not quite sure what that means yet but I'm writing it down to not forget it :p

Feynman_00

(The primed quantities are in the new gauge, $p$ is the canonical momentum and $mv$ the mechanical momentum here).
After a gauge transformation in QM $\hat{p}=\hat{p}'$ and $\hat{v}\neq\hat{v}'$ while in CM $p\neq p'$ and $v=v'$. Which is what qmechanic meant when they said "interpret the velocity rather than the momentum (canonical) to transform". I wonder *why* there has to be such difference between CM and QM

ACuriousMind

@Feynman_00 I don't think the claim that there is a difference between CM and QM is correct

Feynman_00

Do you mean because the actual measurable quantities are gauge invariant?

Jack Rod

hi everyone I am looking for some good research paper on quantum theory on electricity

Feynman_00

e.g. $\langle\hat{v}\rangle=\langle\hat{v}'\rangle$

ACuriousMind

15:15

In the Hamiltonian formulation the Hamiltonian reads $H = \frac{1}{2m}(p-qA)^2 + q\phi$

Feynman_00

Yes, that is the minimal coupling

ACuriousMind

under $A\mapsto \nabla A$, this is only invariant if you simultaneously map $p\mapsto p-\nabla A$, i.e. the sign difference the question you link claims is just not there

Feynman_00

Isn't this what you do classically? Both $A$ and $p$ change

ACuriousMind

yes, and it's also true in QM - the canonical momentum is not gauge invariant

I don't know where you got $p = p'$ from, but it's just not true

it's $p\mapsto p-\nabla A$ in both cases, classical and quantum

Feynman_00

When I write $p=p'$ I mean that both are $-i\nabla$ in the coordinate representation as coordinates do not change as $p$ is still the generator of translations

ACuriousMind

15:20

that's a silly statement :P

by that logic, $p$ is invariant under all transformations

Feynman_00

According to Cohen, let $U$ be the unitary transformation induced by a gauge transformation $UpU^\dagger$ is not $p'$, which is still $p$

while $UvU^\dagger=v'$

ACuriousMind

I think there's a the conceptual problem here: Yes, if you look at the theory before and after the transformation in the standard representation of its CCR, then sure, momentum is $\nabla$ both times

and then your transformation acts on velocity or whatever

but you're essentially doing a basis change here and trying to hide it:

It is indisputable that $[\partial_x, \mathrm{e}^{\mathrm{i}\Lambda(x)}] \neq 0$ in general, so the operator $\partial_x$ is not invariant under the transformation

Feynman_00

Yes, it is not in the sense $UpU^\dagger\neq p$. Is this what you mean?

ACuriousMind

yes

however, since the result after the transformation is still a canonical momentum for your position, the SvN theorem tells us that there is a unitary map that maps back this new momentum $p'$ to $\partial_x$ in the previous basis and this then "transfers" the gauge transformation onto other things like the velocity

but I think this is a deeply confusing way to think about what's happening since the kinematic velocity must be gauge invariant for very physical reasons

so saying that it's velocity and not the momentum that gauge transforms here is just...????

Feynman_00

That's what kinda drives me nut about what most QM books write. Let me search&post the incriminated part

ACuriousMind

15:32

yeah, that's...possible, but the argument there would have to be expanded by something like what I wrote above. We're not doing this in "the position representation", we're essentially rotating the basis vectors $\lvert x\rangle$ by the phase $\mathrm{e}^{\mathrm{i}\nabla A}$ to make it seem like the momentum doesn't change

i.e. the different gauges are actually associated with different position bases

Feynman_00

This sounds definitely more reasonable and in accordance with the gauge transformation I know from CM

ACuriousMind

It never ceases to amaze me how bad many texts are at explaining how gauge theories work :P

Feynman_00

The thing is basically every text I've read about this does this thing

ACuriousMind

it's folklore

one of them started doing it and then they all just copied it because it is convincing unless you actually think about it :P

Feynman_00

If we didn't do that (books), what would happen to the canonical commutation relations?

ACuriousMind

15:37

nothing, since $[x,(\nabla A)(x)] = 0$

Feynman_00

Alright, so $p$ is still the generator of translations as it should be but it is not of the form $-i\nabla$ (?)

ACuriousMind

For any function $f(x)$, you have that $[x,p+f(x)] = [x,p]$

@Feynman_00 well, it generates translations and adds an extra phase $\mathrm{e}^{\mathrm{i}f(x)}$

but since that's a global phase that doesn't matter, it's still "translations"

Feynman_00

OH, that's the thing!

Wait... Isn't that a local phase? Sorry if that is a stupid question

ACuriousMind

global/local has too many meanings

I mean that every state gets the same phase

Feynman_00

Oh ok

I mean that it's position dependent :P

As I'm not accustomed with the SvN theorem, is the meaning of this what you explained below with the phases?

15 mins ago, by ACuriousMind

however, since the result after the transformation is still a canonical momentum for your position, the SvN theorem tells us that there is a unitary map that maps back this new momentum $p'$ to $\partial_x$ in the previous basis and this then "transfers" the gauge transformation onto other things like the velocity

Besides that, this thing didn't convince me for a while and I could never find a source that stressed the problem of the "standard" procedure. I know it wasn't such an insurmountable obstacle, yet I wonder where I should have learned it the way we've just discussed

Relativisticcucumber

17:35

hello - i learned that parity is conserved in situations of even potential and then also that parity is not conserved in the weak force. i am confused about why there are not more cases where parity is conserved. surely we can construct odd potentials as we please.

Slereah

17:47

Hamburger moment problem

In mathematics, the Hamburger moment problem, named after Hans Ludwig Hamburger, is formulated as follows: given a sequence (m0, m1, m2, ...), does there exist a positive Borel measure μ (for instance, the measure determined by the cumulative distribution function of a random variable) on the real line such that m n = ∫ − ∞ ∞ x n...

Burgers' equation

Burgers' equation or Bateman–Burgers equation is a fundamental partial differential equation occurring in various areas of applied mathematics, such as fluid mechanics, nonlinear acoustics, gas dynamics, and traffic flow. The equation was first introduced by Harry Bateman in 1915 and later studied by Johannes Martinus Burgers in 1948.For a given field u ( x , t ) {\displaystyle u(x,t)} and diffusion coefficient (or kinematic viscosity, as in the original fluid mechanical context) ν...

I think there is a delicious connection to be made there

Burgers vector

In materials science, the Burgers vector, named after Dutch physicist Jan Burgers, is a vector, often denoted as b, that represents the magnitude and direction of the lattice distortion resulting from a dislocation in a crystal lattice. The vector's magnitude and direction is best understood when the dislocation-bearing crystal structure is first visualized without the dislocation, that is, the perfect crystal structure. In this perfect crystal structure, a rectangle whose lengths and widths are integer multiples of "a" (the unit cell edge length) is drawn encompassing the site of the original...

Feynman_00

It sounds like someone's hungry

Slereah

🍔➡️

Feynman_00

I'd enjoy a pizza more right now tbh

So look for Pizza's equation

Feynman_00

18:32

@ACuriousMind One last thing about the stuff above. How would you interpret the fact that working out the path integral of the propagator after the gauge transformation yields that extra phase naturally? $(\langle x_2 t_2\lvert x_1 t_1\rangle)'=\exp\left[\frac{-ie}{\hbar c}(\Lambda(x_2,t_2)-\Lambda(x_1,t_1))\right]$, which can be interpreted as $\lvert x t\rangle'=\exp\left[\frac{ie}{\hbar c}(\Lambda(x,t)\right]\lvert x t\rangle$

ACuriousMind

@Feynman_00 what's there to interpret? I'd say that's just consistent

@Relativisticcucumber did you mean "i am confused about why there are not more cases where parity is not conserved"?

the parity you're talking about with the "even potentials" is a different "kind" of parity than the one we're talking about in the weak force

while both are associated with spatial reflections $x\mapsto -x$ in some sense, in the context of the weak force we're talking in terms of a field theory, not a single particle in some potential

Feynman_00

@ACuriousMind Yes it is consistent, but I mean do you think there is any particular reason why the kets are naturally "rotated" as we discussed before it was our choice to "rotate" them to hide the transformation of $p$

ACuriousMind

I mean...the quantity you just wrote down has to do something to the kets otherwise it's just trivial, no?

I don't think there's something terribly interesting here, this is one of these active/passive transformation issues one can waste a lot of breath on but which I never found particularly relevant

Feynman_00

I don't think it is very relevant either, just that I get confused by this

Relativisticcucumber

could you elaborate on what you mean by "different kind of"

or, also, how field theory versus a particle relates to this difference

ACuriousMind

18:43

@Relativisticcucumber In field theory, the action is a function of fields $\phi(x)$, not of paths $x(t)$. This means that the transformation $x\mapsto -x$ is much more "trivial" - instead of mapping the argument of the action to its negative, it just maps the argument of the argument of the action to its negative, so you should compare it to time reversal in particle mechanics

most "nice"particle systems are time reversal invariant - and likewise one expects most "nice" field theories to be parity invariant

Relativisticcucumber

so parity is not only about the potential then?

ACuriousMind

again, field theory is a bit different from particle mechanics - the "potentials" are likewise functionals of fields and not of $x$ directly

in particle mechanics, the claim about "even potentials" is true, in field theory it is not

Feynman_00

Uh, a message of mine disappeared/was not sent

Relativisticcucumber

bah , im not seeing how time reversal invariance connects to potentials

Feynman_00

I'll rewrite it. I meant that I wondered why path integrals force us to make the rotation, like the formalism takes a stand :P

ACuriousMind

18:52

@Relativisticcucumber it doesn't!

at least not directly

that's why parity in field theory isn't about the evenness/oddness of potentials

Relativisticcucumber

oh i see i see - you mean that this statement is true in particle mechanics but in field theory it is about smth else - ok ok - um but even in particle mechanics

we have potential hills all the time

ACuriousMind

I'm not sure why potential hills are relevant now

Relativisticcucumber

well they are odd right? i think my confusion is id expect odd potential cases to not be rare, meaning no parity in particle mech case, but it seems that parity is always obeyed

so i must be missing smth

ACuriousMind

Who says that parity is "always obeyed"?

if you're doing particle mechanics and don't have an even potential, then there's no parity symmetry

Relativisticcucumber

my professor made it sound like the weak force not obeying parity was a huge thing and im p sure he explicitly said that people thought parity was always obeyed before discovering that

have i been led astray

ACuriousMind

18:57

no

but the weak force is field theory

Relativisticcucumber

well that is just dumb bc he said that after doing the even potential argument

dumb of him not of you or physics but merp i think i understand now the bounds of this statement

ACuriousMind

I suspect he didn't think about the different meaning of "parity" in both situations

Relativisticcucumber

thank you for helping me figure it out !!!

have a nice day

hope you are enjoying grad school @Feynman_00

Feynman_00

Covid denied me last week and probably this one too :P

So I'm enjoying what I've been told about the lectures

ACM, I forgot to thank you for you help today! I hope I didn't bother you

ACuriousMind

19:36

no worries

user4539917

20:13

be happy

Feynman_00

20:25

No

DanielSank

22:41

@EmilioPisanty Do you remember the name of a band/album you recommended to me a while ago, called "Kumbakistan" or something like that?

The album cover is colorful with some kind of mask and I think a lot of green.

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