The h Bar

Alfred Centauri

01:32

@ACuriousMind I won't disagree that that is true in a reasonably broad sense. Still, I'm not satisfied with that bromide. There's more there than meets the eye. What is one's job?

DIRAC1930

02:14

The annoying thing however is when people from different fields arrogantly assume they know everything about your field

Classic one is hearing 'how can you do research without doing experiments?'...

I usually give the example of Einstein which quietens them up but you can tell that inside they still believe you can't

antimony

02:59

of course you can do research without doing experiments

antimony

03:10

ofc experimentalists are a bit more touchy these days, as lockdowns have hurt them alot ;)

^ i have to joke about it otherwise i'd cry

DIRAC1930

03:55

Okay so in terms of risks with declining this PhD here is where I stand

I got rejected from all Hep-th PhD's which is what I expected but just so I had no regrets I applied anyway

For CMT (which I believe is probably what I'll fit in with better), I applied to 2 places, got accepted to both however only one had funding.

I think if I were to wait and apply to 10 more schools in CMT, there's a good chance I'll get at least 2.

The issue is, I run the risk of not finding that project interesting

antimony

how different is the general topic you want to do from your Masters?

DIRAC1930

My masters was in hep-th. I asked my thesis supervisor to suggest a topic that was relevant to CMT so I think that looked favourably

DIRAC1930

04:21

But honestly, I'm happy to work on anything quantum to do with more exotic systems

But it seems like most of the projects that have definite funding are more in the material science/real world application realm

Which really makes me lose all interest

antimony

04:36

may i ask why it makes you lose all interest?

DIRAC1930

04:49

I don't know. I'm just interested in general structures and foundations which in the fields that disinterest me, there's not really any aesthetic consideration which just makes me lose interest. But it's more than losing interest. It makes me anxious and makes me feel like I'm wasting my life. Of course those subjects are of great importance however.

Hep-th is too abstract for me however there are certain areas of CMT that I feel that I can get what I'm looking for out of them

antimony

05:31

ah thats fair

ACuriousMind

07:53

@AlfredCentauri Depends - sometimes it's just the thing you do to get money. Sometimes it's something you enjoy doing, sometimes it's something you're reasonably good at but secretly hate, sometimes it's your "calling", something that fills you with purpose.

But even in the latter case I would argue that there is more to the human experience than doing that (non-exhaustive list: Friendship, love, art, music, rollercoasters, a good night's sleep, tasty food, a bright summer's day) that should not be sacrificed to the "job".

John Rennie

08:16

Did someone mention food?

2

Slereah

Calm down Scooby Doo

DIRAC1930

10:54

Is non-relatavistic spin a natural consequence of a 2-dimensional Hilbert space?

Since the identity matrix and Pauli matricies form a basis for any Hermitian operator

But then why are those observables singled out instead of a linear combination e.g. $\sigma_1 + 3\sigma_2$

ACuriousMind

@DIRAC1930 the 2-dimensional Hilbert space is just the representation space for spin 1/2

a spin 3/2 particle has no 2-dimensional Hilbert space, yet it has spin

there's nothing special about the spin 1/2 case non-relativistically, except that it's the "simplest" non-trivial case

DIRAC1930

Sorry I meant spin 1/2

If experimentally I find that I have 2 states

ACuriousMind

in that case I don't understand the question :P

DIRAC1930

I have a 2 dimensional Hilbert space

ACuriousMind

@DIRAC1930 how do you experimentally find this?

DIRAC1930

11:09

I have no idea lol

ACuriousMind

it's not that there are "just 2 states" - there is still an infinity of different states made out of their superpositions

we don't do experiments to find the Hilbert space and then somehow figure out the rest

what you see in experiments are some hints at how many independent states there are, and some values for observables

bolbteppa

In non-rel QM, even if you try to stick to normal wave functions, when you study problems involving rotations e.g. like the hydrogen atom via the Schrodinger equation you're really working with representations of the rotation group, the angular parts of the wave functions are eigenfunctions of representations of the rotation group, if you try to find them from the Schrodinger equation directly there's a step where you force $m$ to be an integer,

ACuriousMind

but e.g. if you want to figure out spin, you'd likely do Stern-Gerlach as your experiment, where indeed you see curious pairs of spins - but you also know you're applying a magnetic field, so it is natural to think that they come from something "like" angular momentum

bolbteppa

But the rotation group representation theory admits half-integers for irreducible representations, so one needs to consider the more general representation theory and ask how it reduces down to the usual Schrodinger equation stuff as best it can

ACuriousMind

i.e. you don't figure out that it's about $\mathfrak{su}(2)$ from the "2 states", you figure it out because spin behaves like angular momentum and $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ is the rotation algebra!

DIRAC1930

11:13

What if I had a two-state system that had nothing to do with spin

ACuriousMind

then you have a 2-state system that has nothing to do with spin :P

DIRAC1930

Lol but the basis of the Hermitian operators are the Pauli matricies and the identity

ACuriousMind

sure, and these Pauli matrices simply have nothing to do with the physical notion of rotation

the physical meaning of mathematically identical operators can differ in different contexts!

whether or not it is useful to describe your system in terms of the Pauli matrices in this case in the end will depend on whether or not the Pauli matrices correspond to easy-to-measure/meaningful observables

DIRAC1930

How do I know what the physical meaning of a random Hermitian operator is?

ACuriousMind

that's an input to the theory

the operator itself can't tell you

most "random" operators won't have a nice meaning - consider that while $x$ and $p$ have obvious physical meanings, all their linear combinations $x+\alpha p$ for real $\alpha$ are Hermitian, too, but have no evident physical meaning

DIRAC1930

11:28

So in general, how do I go from experiment to theory?

Slereah

If you want an example just look at isospin

Isospin is unrelated to rotation, yet it is a little pair of wavefunctions invariant under SU(2)

@DIRAC1930 A very good question

Most experiments aren't even close to measuring actual "nice" operators

Some of them are kind of good approximations for the position operator

ACuriousMind

@DIRAC1930 with an educated guess :P There's essentially two basic ways of doing theory: You dream up a theory, look at what it predicts, and then do the experiment to see if it's right, or you have results of an experiment you can't explain and try to make a theory that does explain them.

bolbteppa

A good question to ask would be how does this all go e.g. in 11D where we don't have an accidental isomorphism like that between $\mathrm{su}(2)$ and $\mathrm{so}(3)$

ACuriousMind

In practice it's a very messy mixture of the two, e.g. the Higgs mechanism was devised to explain a lot of observations about particle masses, but it also predicted the Higgs boson which wasn't yet to be seen in any data

@bolbteppa what exactly is the question? There, too, is some lowest-dimensional representation of the rotation algebra, it just won't be 2-dimensional

Slereah

Example of a pretty cool measurement

You can roughly assume that the tracks are about measurements of the position

Up to some smearing

since this involves like bubbling which is a lot bigger than the particle itself

DIRAC1930

11:41

Are the observables just the generators of a symmetry transformation?

Slereah

They do not need to be, no

ACuriousMind

as Hermitian operators, they all generate some unitary transformation (via their exponentials), but it's only a symmetry if it commutes with the Hamiltonian

Slereah

You probably don't need to overthink observables too much rly

They are just operators that will give you a number

[a distribution of numbers anyway but close enough]

In an experimental setting that will be some number you read from an instrument

That's the basic idea behind observables

ACuriousMind

well, the issue may be that not all "observables", i.e. Hermitian operators, actually have a corresponding instrument

DIRAC1930

Okay so say if I found 3 conserved quantities of a particular measurement experimentally, will the Hilbert space be 3 dimensional?

ACuriousMind

11:46

I mean, they theoretically have, but in practice it might not exist

@DIRAC1930 no, what do the conserved quantities have to do with the dimension?

Slereah

@ACuriousMind Just make a machine to measure the position and momentum and write a program to combine them idk

ACuriousMind

any time you have a proper osition operator your Hilbert space is infinite-dimensional, regardless of what conserved quantities or whatever you have

DIRAC1930

Will the irreducible representation that the states transform under will be 3 dimensional?

ACuriousMind

the irreducible representation of what?

bolbteppa

The real question is what does the du Val singularity $x^2 + y^2 + z^2 = 0$ have to do with spin half spinor of the hydrogen atom, or what does $x^2 + y^2 + z^3 = 0$ have to do with $\mathrm{su}(3)$ in the standard model :p

DIRAC1930

11:49

I have no idea what's going on lol

Okay maybe I'll start right from the start

Slereah

I mean if you want an example

ACuriousMind

try testing your ideas against example systems: The spin-1/2 object, the free particle, the hydrogen atom

Slereah

The basic Hilbert space for a single free particle has 3 translation symmetries

$p_x$, $p_y$ and $p_z$

But it is infinite dimensional

The generators don't work on some tuple of wavefunctions here

you just have some function like $e^{i(p_x x + p_y y + p_z z)}$

DIRAC1930

Okay, but when we do rotations, why do we write the wavefunction with 3 components if the Hilbert space is infinite dimensional?

Slereah

We don't!

Not necessarily

ACuriousMind

11:52

@DIRAC1930 it's infinite-dimensional because it's a function

Slereah

Scalar particles just transform under the trivial rep

There is only one component

ACuriousMind

the space of (even scalar-valued!) functions is an infinite-dimensional vector space

Slereah

Easy enough to show since even just polynomials are infinite dimensionals

[although polynomials aren't square integrable but whatever]

bolbteppa

With some weight e.g. as with Hermite polynomials they will be right

Slereah

not quite $L^2$ either but for all $\mathbb{R}$ to $\mathbb{R}$ functions you can just treat every point as a basis vector

DIRAC1930

11:56

If i expand in spin basis $|\psi> = a |0> +b |1>$, where is all the information about position?

Slereah

Although that's a super broad space and not a Hilbert space

But that's an example of a function space

DIRAC1930

It seems dodgy to just project it out

using $\int \mathrm{d}X |X><X|$

Slereah

If you're using the Hilbert space of spin 1/2 particles, $a$ and $b$ are functions of $x$ there

DIRAC1930

So if I also had energy $a$ would be a function like $a(E, X,p,\dots)$?

Slereah

Well no, in the position representation you only use the position for the wavefunction

The energy will be a function of this

usually something like $$\hat{H} =\propto \frac{\partial^2}{\partial x^2}$$

DIRAC1930

12:01

If I were to write this all out in vectors $|0>$ is 2 dimensional and $<X|$ is infinite dimensional

How do I multiply them together

ACuriousMind

I don't really understand what you're talking about there

Slereah

The Hilbert space is something like $$L^2(\mathbb{R}^3) \otimes \mathbb{C}^2$$

DIRAC1930

Ah okay thanks

Slereah

The basis are just a direct product of the basis

ACuriousMind

didn't we have this exact conversation just a few days ago?

DIRAC1930

12:03

So $|0>$ is really $|0> \otimes \psi(X,Y,Z)$

Slereah

Not always, but if you're talking about say an electron, that's about what it will look like, yes

An electron wavefunction will look something like $$\psi(x) = \xi(x) \begin{pmatrix}\psi^+\\\psi^-\end{pmatrix}$$

DIRAC1930

In braket notation what would that be?

Slereah

You can expand $\xi(x)$ in a position basis if you want, and the "spin" part would be $\psi^+ |+ \rangle + \psi^- |-\rangle$

fqq

@Slereah that's not a generic state though

DIRAC1930

Is there a good textbook that explains this fully and completely?

fqq

12:12

it should be in most QM books...

(all?)

Slereah

Not all

But most standard ones

[I mean you Landau Lifshitz]

Landau doesn't talk about Hilbert spaces at all

DIRAC1930

Most QM books are very sloppy

Slereah

Maybe they weren't considered cool in the soviet union

fqq

I like Sakurai for reference. Cohen-tannoudji probably goes in a lot of detail

@Slereah Von Neumann himself did not like them apparently sciencedirect.com/science/article/abs/pii/S1355219896000172

and there are people trying to get rid of them in a pretty cool way cambridge.org/core/books/picturing-quantum-processes/…

Slereah

The dodo isn't the best mascot for a cool new idea

bolbteppa

12:18

That's why they are the best

Slereah

My advice would be that if you're still not 100% sure about how the Hilbert space works, maybe don't worry about spin too much for now

Just work with scalar particles for a bit

still plenty to do with just that

bolbteppa

Even after all that stress about Hilbert spaces, a simple free particle plane wave will laugh in your face

DIRAC1930

I just don't understand how anything is done on a fundamental level in QM

Slereah

Well do something simple then

Particle in a box or the classic step potential

or the slit screen experiment

those are very simple systems that have an obvious experimental equivalent

also on a simpler level, did you study EM waves?

A lot of QM will go down smoother if you worked on EM decently

DIRAC1930

Formally I've done up to BRST quantization

Slereah

12:27

Seems a weird order to do QM in

DIRAC1930

I didn't do well in that at all though

fqq

Maybe have a look at some quantum info/computing books, they go through tensor products etc in finite dimension which might be conceptually clearer

DIRAC1930

The thing is you can even study advanced GR without knowing the where contra and covariant vectors come from

Slereah

the slit screen experiment is the best one for an introduction to QM as a physical theory, I would say

It's pretty easy to understand

I mean do a free particle first maybe

it is also important, if a bit less quantumy

and exactly solvable, too

DIRAC1930

But I did all these things years ago

Slereah

12:32

What has you confused then

DIRAC1930

Specifically the connections between representation theory and QM

Slereah

It's not something specific to QM

There is also a connection between representation theory and classical mechanics

The representations are different, but roughly the same principle applies

DIRAC1930

Maybe that's a good place to start

Slereah

It could be!

You can look up how symmetries work in Lagrangian/Hamiltonian mechanics

Or even just in general in a space without worrying about the dynamics

fqq

@DIRAC1930 to be honest the stuff discussed above (hilbert spaces, tensor products etc) is more basic than representation theory of Lie groups

Slereah

12:36

One thing that's confusing in physics regarding representations is that a lot of the groups have a canonical representation, so it is easy to confuse groups and their representations

ie $SO(3)$ has one for $\mathbb{R}^3$ in a trivial way

But that's not the case for every group

and it's not the only rep

something to keep in mind

it can be a bit risky to think of Lie groups as their matrix representations because of this

bolbteppa

The idea is, e.g. in Newtonian mechanics we're just using e.g. position vectors, so we can just apply the rotation matrix to it to rotate it. In QM we have a wave function representing a system which is also a function of a position vector, so we now need to allow for the wave function to change when we rotate the position vector, thus the wave functions have to live in a representation of the rotation group

DIRAC1930

@fqq Knowing what a Hilbert space is and actually constructing one from scratch are two seperate questions

Slereah

you can also use that principle in classical mechanics by using fields, too

It will help out a bit

there's a lot to learn for QM by doing classical field theory

Fields also have less trivial behaviour than points under symmetries

Balarka Sen

>classical field theory

>https://en.wikipedia.org/wiki/Class_field_theory

hi all

Slereah

Wrong fields :p

Balarka Sen

12:44

Field's metal

Sir Cumference

13:26

@BalarkaSen eyy how's life

Physics Meta

14:43

0

Q: Should cosmetic edits be made to a closed question?

Sometimes when I am going through the reopen queue I come across questions that have been edited by another user who is not the author of the question. These edits are mostly or entirely cosmetic in nature (fixing grammar, typos, formatting, etc.), but because the question has been edited after c...

Bohemian relativist

15:08

[We realize that quantum phases of matter (i.e. the zero-temperature phases of matter) can be divided into two classes: long range entangled states and short range entangled states.](https://en.wikipedia.org/wiki/Topological_order)
Why is the zero-temperature phases of matter called quantum phases of matter? quantum phenomenon of matter only appears when it is at the absolute temperature 0? I don't think so.

Slereah

15:31

The quantumness usually peaks at 0K?

DIRAC1930

So how would I calculate something like this $<\xi_1, \xi_2, \dots \xi_n|A, \xi_2, \dots \xi_n>$

Slereah

What do you mean

DIRAC1930

Say for $\xi_1 \in \{0,1\}$ and $A=0$

Slereah

I mean what is everything

DIRAC1930

The other $\xi_i$'s are other commuting observables

Slereah

15:37

Are those operators or wavefunctions?

ACuriousMind

@Bohemianrelativist I think the idea is that that 0K, all the phases are purely due to quantum effects - classically we would not expect any interesting things to happen at 0K.

DIRAC1930

They correspond to the eigenvalues of the operators $\hat{\xi}_i$

Slereah

Alright, also what Hilbert space are you doing here

Is it just a direct product of the Hilbert spaces

DIRAC1930

Just a physical Hilbert space of one particle

ACuriousMind

the notation $\lvert a,b,c\rangle$ usually means that this is a state with joint eigenvalues $a,b,c$ of some operators. Since eigenvectors are orthogonal, the inner products $\langle a',b',c'\vert a,b,c\rangle$ are just $\delta_{aa'}\delta_{bb'}\delta_{cc'}$, assuming there is no degeneracy and you've chosen a CSCO as the operators to characterize your states with.

DIRAC1930

15:42

So can I can write $|A, B>$ as $|A>_i |B>_j$ and $<A, B|$ as $<A|_i <B|_j$ where only the same indices hit eachother?

$<A ,B| 0 , B>= <A|0>$

is that right?

ACuriousMind

I would personally prefer to write it as proper tensor products instead of the weird physics notation, but yes, that works :P

Slereah

The Hilbert inner product for direct products of Hilbert space is just a sum IIRC?

DIRAC1930

Okay sorry, I meant how to calculate $<\Psi| 0, \xi_2 ,\dots>$

Where $|\Psi>$ is a general state

Slereah

$$\langle A \otimes B | C \otimes D \rangle = \langle A, C \rangle + \langle B, D \rangle$$

Maybe up to some normalization factor

$1/n!$ or some nonsense

DIRAC1930

This would just be $<A,B,C ,\dots| 0 , \xi_2 , \xi_3,\dots>$

ACuriousMind

15:49

@DIRAC1930 well, in general, there is nothing to compute

if you just know it's a "general state", how do you expect to compute anything?

DIRAC1930

for $|\Psi> = |A,B,C,\dots>$

Bohemian relativist

@ACuriousMind really?

ACuriousMind

you can of course expand it as a direct sum of basis vectors

DIRAC1930

Yeah

Why does the tensor product appear though?

Nevermind

So all I need is a CSCO then I know everything

RewCie

@DIRAC1930 That's the physicist swag

doesn't needs to be written everytime

can be omitted

DIRAC1930

15:54

What can be ommitted?

RewCie

tensor product

Slereah

the symbol

RewCie

in einsteinian notation

Slereah

Although beware because sometimes it can mean different things!

ACuriousMind

@DIRAC1930 the operation between your $\lvert A\rangle_i$ and $\lvert B\rangle_j$ when you write them next to each other is the tensor product

Slereah

15:55

Like the symmetrized direct product

RewCie

yep

DIRAC1930

My question is how do you go from the wavefunction $\psi_{nlm}$ or whatever to a tensor product of kets

RewCie

wait

that's different!

Tensor product in Hilbert space is used to manipulate dimensions

uh!

I forgot how to write kets in MathJAX

one second

ACuriousMind

@DIRAC1930 essentially by writing it as a product of functions - the wavefunction decomposes into a radial part $\psi_n(r)$ and a spherical harmonic $Y_{lm}$

DIRAC1930

$\psi_{nlm}$ is just $<\Psi | n,l,m>$ right?

RewCie

15:59

@DIRAC1930 I don't use that notation,,, (I don't know what you are trying to ask here...)

ACuriousMind

well if that's what you mean then that's what it is, but that's not a function of position

Slereah

\langle and \rangle

RewCie

@Slereah very much thanks :)

ACuriousMind

usually by $\psi_{nlm}(x)$ people mean the wavefunction associated to the state $\lvert n,l,m\rangle$, i.e. the function of position $\psi_{nlm}(x) = \langle x\vert n,l,m\rangle$

DIRAC1930

Does the position operator commute with everything?

RewCie

16:00

Remember the Theorem below (In using direct product in that context)

ACuriousMind

@DIRAC1930 that's a very strange question :P why would it?

Slereah

it famously doesn't

RewCie

No

DIRAC1930

Oh yeah I forgot about that special commutation relation

RewCie

@DIRAC1930 what do you mean by comma here?

Slereah

16:02

It does in the $\hbar \to 0$ limit, but then again everything commutes in the classical limit

DIRAC1930

But then why aren't the basis states $|n,l,m,x>$

ACuriousMind

@DIRAC1930 why would they be

RewCie

basis states aren't denoted using comma

Slereah

@DIRAC1930 The radial and angular eigenstates are functions of $x$

The dependance on $x$ is implied

DIRAC1930

What if I wanted to make it explicit?

Slereah

16:03

Well just write it

DIRAC1930

How?

ACuriousMind

then you do $\langle x\vert n,l,m\rangle$ to get the wavefunction, as I wrote above!

Slereah

$$\psi(x) = R_n(r) Y^m_l(\theta, \varphi)$$

DIRAC1930

but how is $|n,l,m>$ dependent on $x$ explicitely?

ACuriousMind

ignoring several subtleties, $\lvert x\rangle$ is one possible choice of basis, $\lvert n,l,m\rangle$ is another. What exactly is the issue you have with that?

Slereah

16:05

Do you know how a basis works in a function space

like in a Fourier sum for instance

RewCie

DIRAC1930

Yes

Slereah

the Fourier modes are just a list of numbers, but they are interpreted as a function

RewCie

of course direct product omitted

sorry

DIRAC1930

But where are those numbers in $|n,l,m>$?

RewCie

16:06

reverse I meant

Slereah

n, l and m are the numbers

DIRAC1930

oh nevermind

Slereah

each of them point to a specific function

RewCie

@DIRAC1930 what are nlm?

Slereah

For instance $Y^0_0$ is the constant function

RewCie

16:07

if they are components of hilbert space, then you need to take inner product wrt each coordinate to find the components

ACuriousMind

@DIRAC1930 the numbers are not in $\lvert n,l,m\rangle$ - a general state in the Hilbert space can be written as $\sum_n\sum_l\sum_m a_{nlm}\lvert n,l,m\rangle$. The coefficients $a_{nlm}$ are the analogue to the Fourier mode numbers Slereah is talking about

Slereah

Spherical harmonics

In mathematics and physical science, spherical harmonics are special functions defined on the surface of a sphere. They are often employed in solving partial differential equations in many scientific fields. Since the spherical harmonics form a complete set of orthogonal functions and thus an orthonormal basis, each function defined on the surface of a sphere can be written as a sum of these spherical harmonics. This is similar to periodic functions defined on a circle that can be expressed as a sum of circular functions (sines and cosines) via Fourier series. Like the sines and cosines in Fourier...

You can see some examples here

ACuriousMind

when you want to go to position space/talk about functions explicitly, you look at $\langle x \vert \psi\rangle = \sum_n\sum_l\sum_m a_{nlm} \langle x\vert n,l,m\rangle$

RewCie

If you can clear the context

ACuriousMind

The $\langle x\vert n,l,m\rangle$ are functions $R_n(r)Y_{lm}(\phi,\theta)$, i.e. products of the radial Hermite (I think?) polynomials and spherical harmonics

RewCie

16:09

are you asking then in terms of dimensions of hilbert space or bloch sphere or what?

special harmonics?

Has anyone used LibrePCB for designing chipset? How does it compare to OrCAD?

I see it does already has some library of popular ICs, TI-x, Mosfer, Arduino and all.

BioPhysicist

@ACuriousMind Been able to play any good board games lately?

ACuriousMind

@BioPhysicist not yet, but now that most of my friends are fully vaccinated I may be soon!

BioPhysicist

That is pretty exciting

RewCie

I have 64 days more for my second doze

I had some fever on first two days and a lot headache and weakness

but okay now

BioPhysicist

How long between your doses? Mine was just 3 weeks.

PM 2Ring

16:20

@BioPhysicist Some editors of closed questions can be very persistent, even after you try to explain that their edits have a bad effect. Eg,

May 5 '20 at 5:09, by PM 2Ring

Dear @HarishChandraRajpoot. Once again, you've edited a closed question. The first edit of a closed question sends it to the Reopen review queue, as I explained here (you should read the other answers there, too). Why do you think this question now qualifies to be reopened? — PM 2Ring 4 hours ago

satan 29

@BioPhysicist 84 days

RewCie

@BioPhysicist 12 weeks for maximum efficiency

BioPhysicist

@PM2Ring Yeah, I think I tried to tell the same user the same thing in the past.

PM 2Ring

He must've edited dozens of such questions. :( Fortunately, he slowed down once he got full edit privileges at 2k.

ACuriousMind

The moderators may also have sent them a message telling them to knock it off - remember that if you see persistent behaviour from users that you think is detrimental, you can raise a custom flag

DIRAC1930

16:25

Does $<n | m >$ make any sense when talking about $n,l,m$?

RewCie

I don't understand why colleges force every student to memorize all important ICs and with their stupid random serial numbers, one can always refer chips when they are working datasheet or maybe, memorizing them doesn't sounds cool in any context

ACuriousMind

there's no guarantee we'll do something, but it almost never hurts

@DIRAC1930 yes - you can just talk about the space of radial functions and then you just have $n$

PM 2Ring

@ACuriousMind Yeah, ok. I might have raised a flag, as well as mentioning it in this room.

RewCie

7408 quadm 7432 quad, 7400 quad, 747266 Gamma....? what that stupid numbers even for? does it even make sense to byheart something like that?

ACuriousMind

essentially, $L^2(\mathbb{R}^3 - \{0\})\cong L^2(\mathbb{R})\otimes L^2(S^2)$

BioPhysicist

16:27

@PM2Ring Ping the user in a comment, raise a flag, ping moderators in the hbar, make a meta post, good to go :P

PM 2Ring

I was almost tempted to randomly downvote some of his posts, but I managed to restrain myself. ;)

BioPhysicist

As long as you do one per day you should be fine ;P

DIRAC1930

It seems like space is just arbritrarily added

PM 2Ring

@RewCie Well, it does make it more efficient when you're reading schematics, or trying to design a circuit, if you don't have to constantly keep referring to a datasheet. But I suspect that the main reason that schools do it is because it's an easy thing to test in exams. Especially if you're in a country that has a strong "teach to the exam" philosophy.

BTW, transistor & chip serial numbers aren't totally random, they usually have some patterns.

BioPhysicist

It could also help you get a trivia question right at some point in your life. Or make you feel good about yourself when you tell at the TV during Jeopardy :P

Shashaank

16:50

I am having a very silly doubt maybe. Will an accelerometer placed in a frame accelerating horizontally at the rate a, will give a reading of a or 0. I think it should give a and not give 0 like a freely falling accelerometer will give. Anybody, please

ACuriousMind

@Shashaank why would you call it an accelerometer if it wouldn't measure that?

fqq

@Slereah that's wrong, it's a product

Slereah

@fqq Oops

fqq

otherwise the notation without $\otimes$ would make no sense

Slereah

Nothing makes sense

fqq

16:57

also it would mess with the normalisation

Slereah

just look around

Shashaank

@ACuriousMind - for horizontally accelerating (Newtonian Mechaincs) frames an accelerometer will shown non zero reading, a glass filled with water will have water tilted.

But the same effects won’t happen in an inertial frame in Newtonian Mechanics. Whereas in a freely falling frame an accelerometer (in GR) will show zero reading and the one on the earth will show non zero reading. Yet a glass of water kept on Earth (in GR) doesn’t show any tilt in the water level unlike the accelerated frame in Newtonian Mechanics

ACuriousMind

@Shashaank of course the water shows a "tilt" - it's at the bottom of the glass, is it not? :P

in free-fall the water is just a spherical blob not necessarily touching any of the sides of the glass

BioPhysicist

You have to make sure the accelerometer has learned GR first

Slereah

The equivalence principle only works locally, though

Your accelerometer may not be a single point!

Shashaank

17:08

@ACuriousMind Wow. And I was actually going to throw a glass of water down my balcony to see what shape it takes. Phew, surface tension, sphere has minimal area for given perimeter. Grade 10 thing. Thanks for the insight. Amazing and I

*thought I have proved J Synge right by finding a fault in equivalence principle

@Slereah yep. That is why the best resolution to the the paradox of radiation from an accelerated charge in gravitational field that I like the most is the one advocated by Rindler- that Equivalence principle is valid locally and the radiation field extend much outside the local inertial frame.

PM 2Ring

It won't be exactly a spherical blob, because water "likes" to wet glass. That is, there's a slight molecular attraction between the water and the glass. You'd get a cleaner result using mercury. Or replace the glass with a water-repellant polymer, eg Teflon.

Shashaank

@Slereah Actually Synge is right. Equivalence principle is just a very good approximation which will break at the point the the curvature starts producing forces, like the singularity

Slereah

I mean the equivalence principle isn't an approximation in GR

Shashaank

Yeh but in that case of adhesion forces what’re won’t be freely falling. But I am ashamed I didn’t think what A curious mind wrote

Slereah

Bound systems in GR are hard to do

Shashaank

17:20

@Slereah It’s valid locally. Pick a very small inertial frame. Still it will have two ends. Pick the smallest thing you can imagine, let it’s width be of the order of femtometers. Keep it in an very very very strong gravitational field such that the curvature is enough to manifest tidal forces on even two points separated by a femtometer ( like near a black hole)…Boom Equivalence principle will not hold.

Isn’t it then just an approximation

Slereah

I mean yeah, but that's what the principle of equivalence is!

It's about local frames

PM 2Ring

The current cutting-edge atomic clocks are so precise that they can detect a difference in GR time dilation at ground level over a height difference of only a few centimetres.

Slereah

The hard part in GR, as with physics in general, is knowing the size of the effect

It's the thing physicists are the worst at

They will write $\mathcal{O}(r)$ and never tell you what size that is

Is it a cm, a km, a light year???

Shashaank

But can’t I put it like - Because of this approximation gravitational field differs from acceleration. It’s not a pseudo force. The 2 situations are really different at the smallest of the scales ( atomic scales)…That at atomic scales and strong gravity regions gravitational field is not really just equivalent to acceleration. It is something in itself, maybe a force field just like other forces are at the quantum level

PM 2Ring

Near a stellar mass BH, the tidal force can be quite strong. There's an answer on the main site that does a calculation of how close a 1m steel rod can get to a BH before getting ripped apart by the tidal tension. But I can't remember the distance offhand.

Shashaank

17:26

Would that be wrong to say

Slereah

I mean talking about GR and the equivalence principle and all that may be a little much really

Maybe you can just look at it practically

Does an accelerometer measure something in free fall

wikipedia says no apparently

Shashaank

Is there research going on as to what effect will Quantum Mechanics have on equivalence principle ( that it will not be really valid at the smallest scales)….is that a part of string theory or loop quantum gravity research

Slereah

I mean sure

IIRC it's one of those differences in the continuous v. discrete theories

Breaking local Lorentz invariance breaks the principle of equivalence I think?

I think that's being tested by the vacuum dispersion of light?

PM 2Ring

@Shashaank No. Gravitation really is completely equivalent to acceleration. If the gravitational field were completely uniform, then there's no experiment or measurement that would distinguish it from uniform acceleration. However, the gravitational field near a planet isn't uniform. It's (roughly) spherically symmetric, and depends on the distance to the centre of the planet.

Shashaank

@Slereah Hang on. Breaking Local Lorentz invariance …..wouldn’t that mean faster than light causation since Lorentz invariance basically comes from there

Slereah

17:38

Who knows

Theories that break Lorentz invariance are pretty weird

But I don't think you can have FTL because of that, no

It's just that they're discrete so you can't have a continuous symmetry on them

Shashaank

@PM2Ring I could equivalently put it like- If the acceleration is varying just like the gravitational field then I wouldn’t be able to distinguish the two. Yet gravity is Generally not uniform whereas acceleration is. What makes gravity Generally not uniform

PM 2Ring

Any local patch is Lorentz invariant. It's just that in general, you can't glue all of those little local patches together into a single frame with Lorentz invariance. That's just like how you can make accurate flat maps of sufficiently small regions of the Earth, but you can't glue all of the maps together into a big flat map.

Shashaank

@Slereah but doesn’t derivation of Lorentz transformations itself lead to a max invariant speed. If we break Lorentz invariance we aren’t we not also just breaking the speed of light postulate

Slereah

@Shashaank Don't get too excited, I think it's just "You can't have a Lorentz transformation in between those two specific values because it's on a lattice"

PM 2Ring

@Shashaank Sources of gravity tend to be spherical blobs, so we use the Schwarzschild metric to describe spacetime in their vicinity. If you could make an infinite disc of finite thickness, it'd have a nice linear uniform gravitational field. But of course that's not physically possible.

Shashaank

17:49

@PM2Ring and @Slereah yeah I your points. Thanks for the nice discussion

PM 2Ring

18:33

I still can't find that post about spaghettification, but en.wikipedia.org/wiki/Spaghettification says that for a 1 m rod, with tensile strength of 10,000 N free-falling vertically towards a 10 solar mass BH, the tidal force disrupts it at 320 km, well outside the Schwarzschild radius of 30 km.

Nike Dattani

My answer to this one was a lot more about the mathematics of Pade approximants, but if anyone has more information on how it's specifically used in density functional theory, the OP might be glad to get more answers!

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A: What is a Padé approximant?

To complement mykd's already excellent answer, I will just add that the approximant we all learn in school (the Taylor approximant) is nice to teach and easy to help students learn the concept of approximations, but in practice it's one of the worst options in terms of it's accuracy-to-complexity...

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