The h Bar

Semiclassical

3:00 PM

Those are based on the charges which generate the electric field

user929304

@Semiclassical aha I see

Balarka Sen

If an electric field is monopolar or dipolar, that means the vector field has one or two (resp.) singularities, that's all.

lılostafa

@ACuriousMind That's not a good habit especially when others are around you, calculating.

Balarka Sen

Those singularities correspond to position of the test charge that generates the field

ACuriousMind

@lılostafa Obviously I tone it down when others are nearby

Semiclassical

3:01 PM

Suppose I put a collection of positive charges near the origin. If I zoom out, I’ll just see field lines emanating from the origin

user929304

@Semiclassical very good. So as a sanity check for myself: when we say magnetic fields are dipolar, the field itself is still a vector-valued function of position, but the source of the field is dipolar

ACuriousMind

Although my former roommate once knocked on my door, asking who I was talking to. I was just tackling a particularly stubborn exercise.

Semiclassical

So I can characterize the behavior far away in terms of how far away I am from the origin. All other sense of directionality gets washed out

John Rennie

@user929304 Yes

Slereah

You can have energy without a Noether charge

Semiclassical

3:03 PM

By contrast, suppose I put a positive and a negative charge near the origin and zoom out.

ACuriousMind

@Slereah So then it's the generator of time translation still.

Slereah

true

well, depends on the definition I think

Semiclassical

No matter how far out I go, I’ll still have a preferred direction: the line connecting the two charges

user929304

@JohnRennie oh so the different orders of poles can be thought of as a elementary decomposition of the field in a basis (I mean similar to vector decomposition)

0celo7

@ACuriousMind autocorrect went insane

I said you're a mathematician

user929304

3:05 PM

@Semiclassical sigh... now I m losing you again.... sry

ACuriousMind

@0celo7 I guessed, but mttaaician is an odd thing to autocorrect to ;)

John Rennie

@user929304 Exactly. It's called a multipole expansion and is widely used in physics.

0celo7

@ACuriousMind I thought so too.

Semiclassical

it helps to draw pictures here

Namely the field lines for a positive-negative charge pair

user929304

@JohnRennie Great, thanks. God... self learning is not going well, so much ground to cover ( I never had formal training in physics). But I love it so will never lose my ambition of wanting to learn more about it :)

@Semiclassical it s true. For a closure, to avoid extended discussions here, let us take an example: we know magnetic fields have dipolar sources, and GW's are quadrupolar, from a physics point of view, what is the main contrasting feature between the sources of the two latter fields?

ACuriousMind

3:09 PM

@user929304 This is chat, extended discussions are not really a problem here (unless the participants in it tire of it)

Semiclassical

well, let's simplify this a bit

let's compare the pictures for a monopole charge config, a dipole charge config, and a quadrupole charge config

that's the field of a monopole: field lines emanating from a single positive charge.

that's the field lines in the plane, anyways, but you should imagine them coming out of the page and into the page as well

user929304

@ACuriousMind alrighty great :) by the way I ve been meaning to ask you something about QM and information: specially in the new Q-information field, one hears a lot people saying "information is conserved". I understand they refer to the unitarity of the evolution of the whole system,

but do I understand correctly that there's no "law of conservation of information" in physics right? I mean for one thing, we have the second law of thermodynamics telling us all systems evolve in producing more information/entropy.

Semiclassical

a key point here is that if i were to rotate this picture a bit, it wouldn't change. so all I need to know is how far I am from the charge, and what the strength of the charge is.

by contrast, here's the field of a pair of positive and negative charges:

user929304

@Semiclassical aha, makes perfect sense.

Semiclassical

that's a bit different!

the positive/negative charges are at (0,1) and (0,-1) respectively

user929304

3:16 PM

@Semiclassical indeed, now we see a circular pattern

ACuriousMind

@user929304 Whenever I hear "information is conserved", I just substitute "time evolution is unitary".

Semiclassical

sure. moreover, though, there's a preferred axis

no matter how far out I go, i'll always have field lines running up the y-axis

ACuriousMind

As for how the second law of thermodynamics is compatible with a reversible (unitary) time evolution, that's Loschmidt's paradox

Semiclassical

and keep in mind that this is just a cross-section of the 3d picture

user929304

@ACuriousMind yes :) I do the same. How do interpret the unitary here without resorting to notions of "information"?

Semiclassical

3:18 PM

if i imagine rotating my system along the y-axis, i'd find that the system wouldn't change

the positive and negative charges stay in the same place, so nothing actually changes

by contrast, if i rotate around any other axis, i definitely change where the particles are and therefore change the picture

ACuriousMind

@user929304 It's meant literally, in the mathematical sense of "unitary": The time evolution operator is a unitary operator on the space of states

user929304

@Semiclassical yes I see. so in a sense you re pinpointing the different symmetries of the generated filds

Semiclassical

something like that

what that means is that, if I want to describe the field far away from the origin, I now need to know not only how far away I am and the strength of the charges involved

i also need to know what that axis of symmetry is

and that axis is characterized by a vector pointing in that direction

user929304

@ACuriousMind aha, so for example in the sense that: if I know the state of my system at anytime t, I know its full history and future?

ACuriousMind

@user929304 Yes, precisely, and from any point in the past or future, you can also know all of that

Semiclassical

3:21 PM

@ACuriousMind so long as no measurements are made :)

ACuriousMind

@Semiclassical Sure, the measurement problem rears its ugly head :P

John Rennie

@Semiclassical that's a decoherent argument!

Semiclassical

now for the quadrupole case, and this one is a doozy

to get a dipole field instead of a monopole field, I picked my charges so that they had opposite signs

hence the net charge was zero.

to create a quadrupole, I'll take two dipole configurations of equal strength but opposite direction

(there are other ways to do it, but this is a simple one)

user929304

@ACuriousMind and naively, why do we say that time evolution should preserve the normalisation that we have chosen for our states?

@Semiclassical nice picture, very helpful!

Semiclassical

user929304

3:26 PM

@Semiclassical wait a minute, are you generating these in mathematica as we speak? :) wow

Semiclassical

in this pictures, there's positive charges at (1,1) & (-1,-1) and negative charges at (1,-1) & (-1,1)

yup

the joys of StreamPlot :)

user929304

@Semiclassical how do you generate them so quickly?

Semiclassical

experience

i know what the functions are for the vector fields as functions of position in the plane

and to get the effect of all four charges together you just add those functions together

so it's actually easier than you might expect

user929304

@ACuriousMind I mean, it seems that unitary "evolution" of a system, is not really an evolution, as nothing seems to be changing, as in everything's stationary

Semiclassical

anyways

key point here is that we've now got this weird four-lobed creature

ACuriousMind

3:29 PM

@user929304 Frankly, I don't like that argument for why time evolution is unitary. The better argument, imo, is that the Schrödinger equation tells us that $U(t) = \exp(\mathrm{i}Ht)$, and if $H$ is self-adjoint (as it has to be, since the Hamiltonian is often the energy, which is an observable) then $U$ is unitary.

user929304

@Semiclassical ye I see, I ll try to play with these on my own in mathematica. Yes

Semiclassical

kk

ACuriousMind

@user929304 What do you mean "everything is stationary"? A superposition of stationary states with different eigenvalues is not stationary.

Semiclassical

StreamPlot[q{x - h, y -k}/((x - h)^2 + (y - k)^2)^(3/2),{x,-3,3},{y,-3,3}] plots the electric field lines of a charge q at (h,k)

you'll need to pick values for q,h,k for that to run, of course

then you just take different such functions and add them together before doing streamplot

user929304

@ACuriousMind that s true, I meant in the sense that we say: the probability to find the state $\psi$ in $\phi$ at $t_0$ should be the same as finding the evolved state $\psi$ in the evolved state $\phi$ at $t_1$. Which then prompts one to ask: then what has changed in the system from $t_0$ to $t_1?$

Semiclassical

3:31 PM

I should emphasize again that this is just a 2D cut, though

there's not a stream-plot 3D that I know of

ACuriousMind

@user929304 What has changed is the probability to find it in state $\phi$ (not evolved).

user929304

@Semiclassical sure, thanks a lot, very nice explanations.

Semiclassical

(though there is VectorPlot3D)

ACuriousMind

If $\phi$ is the eigenstate of some observable you want to measure, that's a very significant change.

Semiclassical

anyways. key point here is that there's no longer a single preferred axis

you do have the lines y=x, y=-x

user929304

3:32 PM

@ACuriousMind ohhhhh I see now where I was going wrong!!

ACuriousMind

This is the difference between the Schrödinger and Heisenberg pictures: In the Schrödinger picture, the state of the system evolves but the states we want to measure do not. In the Heisenberg picture, the state of the system doesn't change but the states we want to measure do.

Semiclassical

so now if you want to characterize this system, we're not going to be successful just by knowing the charges, the distance, and a single vector

we're going to need a bit more info than that.

the information one needs is characterized by the rank-2 quadrupole tensor

user929304

@ACuriousMind so as to preserve overlaps (inner products), true true. Physically, what gave us the hint to expect unitary time evolution?

ACuriousMind

@user929304 I don't think anything gave us the hint. It just follows from the Schrödinger equation.

Semiclassical

there's still a definite notion of geometry here---the orientations of the four lobes---but that geometric info isn't as simple to characterize as just "what direction does it point"

user929304

3:35 PM

@ACuriousMind I mean people often give arguments that, the probability of finding the electron cannot be 0, as it MUST be somewhere, right? and then we impose that \sum p_i = 1 as a normalisation (or more correctly the squared modulus being summed)

Balarka Sen

Are we asking why the Hamiltonian is self-adjoint again

Semiclassical

why time evolution of the wavefunction should be unitary

Now I'm wondering what VectorPlot3D would say

user929304

:)

Balarka Sen

@Semiclassical That follows form self-adjointness of the Hamiltonian though

I don't know the physical meaning of that

ACuriousMind

@user929304 Yes. But that's a silly argument because there's no obvious reason for why a non-unitary operator would necessarily mean some probability becomes 0, and the normalization argument is also silly because there's a version of the Born rule that computes the probabilities also for non-normalized states (you just divide by the norms of the states involved).

user929304

3:38 PM

@BalarkaSen that eigenvalues should be real? (as in measurable experimentally)

ACuriousMind

@user929304 Yes, exactly.

Balarka Sen

@user929304 There are lots of non-symmetric matrices with real eigenvalues

0celo7

@BalarkaSen it's not always

ACuriousMind

@BalarkaSen The crucial condition is "measureable experimentally" - physical observables are by definition self-adjoint.

0celo7

There are those God Awful classical things that quantize to God Awful operators

@JohnRennie Galloway wrote back...

ACuriousMind

3:40 PM

QM gives you no recipe for "measuring" non-Hermitian operators.

@BalarkaSen A different take on the physical meaning is "the system is closed". Open systems can have non-Hermitian Hamiltonians and therefore also non-unitary time evolution (parts of the system escape into the environment or come into it)

Balarka Sen

@ACuriousMind Hm. So somehow being self-adjoint relates to measurability? Should the connection be intuitively clear?

0celo7

@JohnRennie @Slereah He doesn't know the answer :/

user929304

Good remark about closed systems, I tend to forget that.

Semiclassical

$V(x)=2e^{ix}+e^{-2i x}$ :>

John Rennie

@0celo7 oh well, at least you know it's a genuinely hard problem.

ACuriousMind

3:42 PM

@BalarkaSen I don't think there's an intuition other than "It's not clear how we're supposed to measure non-self-adjoint operators".

Balarka Sen

@ACuriousMind I see. Interesting.

user929304

@ACuriousMind that said, then please please please, can I ask what we mean by closed here? (in QM) because it seems to be going back to that argument you didnt like about the electron

Balarka Sen

@ACuriousMind Got it.

Slereah

@0celo7 For the strongly causal thing?

Semiclassical

then $V(x)^*\neq V(x)$ for real $x$, but $V(-x)^*=V(x)$ for real $x$

Slereah

3:42 PM

Try asking Ellis maybe?

ACuriousMind

@user929304 "closed" means it doesn't interact with its environment in a way that would allow something from the system to get "lost".

Semiclassical

It's like water sloshing back and forth in a closed container versus what happens if the container has a hole in the bottom

user929304

@JohnRennie @Semiclassical and @ACuriousMind many thanks again for your time devotion and help. Really appreciate it. It's so valuable to be able to have these discussions, so glad I found out about chat SE. When do we ever get to ask physicists these kinds of questions in real life? ... never :(

Semiclassical

grad student offices :)

Balarka Sen

I suspect this is related, but not the thermodynamical notion of closedness though?

0celo7

3:44 PM

@Slereah ye

@Slereah He said he'll ask other people about it and now he's genuinely curious

user929304

@ACuriousMind right, yeah. Sorry for all these questions... I know they are extremely naive :(

Semiclassical

blah, vector plot 3D is sorta useless for this purpose

ACuriousMind

@user929304 Don't apologize for asking!

Semiclassical

I think if you want to get a 3D representation of mono/di/quadrupole one usually looks at the electric potential

0celo7

Don't ask to ask, don't apologize for asking...too many rules!

Semiclassical

3:48 PM

since then it's a scalar not a vector-valued function

user929304

@ACuriousMind thanks, appreciate it.

@ACuriousMind by the way, did you ever get to check the Schumacher book I told you about? Quantum Processes Systems, and Information by Benjamin Schumacher , Michael Westmoreland

ACuriousMind

@user929304 I'm afraid not

user929304

@ACuriousMind you can see the contents on amazon for example. Definitely worthy of a check out

@ACuriousMind I think safesphere is doing his/her best to provoke you to write an answer :-)

Bernardo Meurer

4:06 PM

@JohnRennie I guess the most cores a merge sort can use is 2?

God, currency is a brainfuck

John Rennie

@user929304 safesphere has, shall we say, some idiosyncratic ideas. He accuses me of posting nonsense about the Big Bang. These things are sent to try us :-)

user929304

@JohnRennie haha indeed :)

John Rennie

@BernardoMeurer I'm not familiar with the merge sort. Is it the one where you split the arrays into lements greater than the median and less then the median, then recurse to sort each half?

Ah, no it isn't.

@BernardoMeurer currency causes all sorts of hassle in programming. Some systems, e.g. SQL Server, have a specific currency datatype. It's usually handled as a sort of integer.

user929304

4:29 PM

@ACuriousMind by the way, I had a side question I forgot to ask, regarding the unitarity discussion: does the (conservation of information) unitary time evolution in any way link up to the degrees of freedom of a system? As in, in unitary time evolution, a system's degrees of freedom remain unchanged. I guess also partly related to saying the system's "closed".

Bernardo Meurer

@JohnRennie Merge sort is different

It's where you split to singletons and then orderly merge

0celo7

4:51 PM

@Semiclassical What is the best way to aggregate large amounts of literature research?

Slereah

@0celo7 Use Jabref?

Bernardo Meurer

@EmilioPisanty Did you get my facebook message?

Emilio Pisanty

@BernardoMeurer I did

0celo7

@Slereah what's that?

Slereah

A bibliography tool for Latex

Emilio Pisanty

4:57 PM

@user929304 as Semiclassical has already explained, those aren't very useful ways to understand multipolarity

there's better ways

The very easiest way to understand them (I think) is to consider charge distributions that are confined to a spherical shell

so you have $\sigma(\theta,\varphi) = \sigma(x,y,z)$

0celo7

@Slereah I'm not writing anything yet. I have 40 papers that I need to keep straight and manage

Slereah

@0celo7 Jabref does that

You can make a database of the papers

with tags, links, files and biblioref

And only at the end of the project, generate a tex bib page

Emilio Pisanty

in short, a charge distribution (on that shell) is $2^\ell$-polar if and only if $\sigma(x,y,z)$ is a homogeneous polynomial of degree $\ell$ in $x$, $y$ and $z$.

0celo7

@Slereah can I attach summaries with TeX equations?

Emilio Pisanty

@0celo7 use zotero

Slereah

5:00 PM

@0celo7 i don't recall

0celo7

@EmilioPisanty what is that?

Emilio Pisanty

@0celo7 lmgtfy.com/?q=zotero =P

a reference manager

0celo7

@EmilioPisanty dead link

Emilio Pisanty

@0celo7 zotero.org

Bernardo Meurer

@EmilioPisanty Did it tickle you and make you giggle?

0celo7

5:03 PM

@EmilioPisanty 🙇🏻

Emilio Pisanty

similar to Mendeley but without the evil

0celo7

That's a worshipping man

@EmilioPisanty evil?

Emilio Pisanty

@0celo7 theguardian.com/higher-education-network/blog/2013/apr/10/…

techcrunch.com/2013/04/08/…

Bernardo Meurer

@0celo7 Halp

Emilio Pisanty

Mendeley is part of Elsevier, has been for about half its lifetime

0celo7

5:05 PM

@EmilioPisanty I own 3 Elsevier books. Am I part of the problem?

Slereah

but I hate elsevier!

ACuriousMind

@EmilioPisanty It's a shame it's not pointed out more often that spherical harmonics are just these polynomials.

Emilio Pisanty

@0celo7 did you explicitly choose those books over functional equivalents because they were Elsevier?

@ACuriousMind 100% agreed

Bernardo Meurer

Show that for any $n\in\Bbb N_1$ $$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5} + \cdots + \frac{1}{(2n-1)(2n+1)}=\frac{n}{(2n+1)}$$

0celo7

@EmilioPisanty No. Reed-Simon and Adams have no equivalents.

Bernardo Meurer

5:07 PM

I forgot how to do induction proofs

0celo7

They are standard references and will continue to be for the rest of our lives, probably.

Emilio Pisanty

@ACuriousMind the way spherical harmonics are normally taught seems explicitly designed to make people hate them

Anonymous

@BernardoMeurer $1=\frac{3-1}{2}$, $1=\frac{5-3}{2}$,...

Emilio Pisanty

@0celo7 that was kind of a rhetorical question

unless you did, then no, you're not part of the problem

Elsevier is

ACuriousMind

@user929304 Not really - the degrees of freedom of a system are fixed by what you define to be the system, but in an open system, one might say that you chose to ignore some relevant degrees of freedom (for whatever reason). Nothing to do with unitarity, though.

Emilio Pisanty

5:09 PM

@BernardoMeurer well, you just need to show that if you add the next term the sum simplifies to what it needs to

Anonymous

@0celo7 I've started reading the first chapter of Reed-Simon. It's nice so far. :) I got stuck a few times though in that Zorn's lemma stuff

0celo7

Zorn's lemma is false

2

Semiclassical

@EmilioPisanty it doesn't help that one of the main places you're forced to get good at using them is in graduate level electrodynamics

Emilio Pisanty

so $$\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)} = \frac{n+1}{2n+3}$$

Semiclassical

which means you're probably going to be using Jackson :P

0celo7

5:10 PM

It depends on evil AC

Bernardo Meurer

@EmilioPisanty Aaaaah, right!

I have to show for an $n$

and then show for $n+1$

0celo7

@BernardoMeurer actually has a proof that ZFC is inconsistent

Emilio Pisanty

@BernardoMeurer no, you assume your formula for $n$, and then you use that to prove for $n+1$

Bernardo Meurer

@0celo7 That is true, ZFC is inconsistent

ACuriousMind

@0celo7 The traditional view-point is that the well-ordering principle is obviously false, the axiom of choice is obviously true, and no one can tell for Zorn's lemma :P

3

Semiclassical

5:12 PM

that's a fun joke

though my favorite mathematical 'joke' is the following test for whether you're a mathematician or a physicist

0celo7

@ACuriousMind they're equivalent so by basic logic Zorn is false

Semiclassical

"If $f(x,y)=x^2+y^2$, then $f(r,\theta)=$?"

user929304

@ACuriousMind aha I see. But isn't for instance the number d.o.f's connected to the dimension of Hilbert space in which our system's state is unitarily evolving?

Bernardo Meurer

AoC is BS

ACuriousMind

@user929304 Yes, the number of d.o.f. is the dimension of the space of states.

@0celo7 That's the joke :/

user929304

5:13 PM

@ACuriousMind ok, and isn't the that dimension also preserved by unitary evolution?

0celo7

@ACuriousMind You don't think I know that?

I know the joke, said it many times myself

Semiclassical

You're the one who insisted on explaining it :P

0celo7

I wasn't trying to explain

I was a bit miffed that ACM tried to joke about a serious issue. @Blue needs to know that many people disregard Zorn

All reasonable people do

Semiclassical

riiight

ACuriousMind

@user929304 I'm not sure what you mean by "preserving" the dimension.

user929304

5:17 PM

@ACuriousMind ah, just that the space of states cannot shrink or expand

ACuriousMind

If you mean that it's bijective, i.e. the image is still the entire space, then yes, that's implied by unitarity.

0celo7

"Sov. Math. Dokl."

Crap.

That's not going to be in a language I understand.

Emilio Pisanty

@0celo7 so, you're OK with the statement "for every vector space $V$ over a field $F$, $V^*=\{f:V\to F : f\text{ is linear}\}$ has at least one nonzero element" being false?

user929304

@ACuriousMind exactly

ACuriousMind

@user929304 The space cannot shrink or expand simply because there's no way for it to do so in the formalism. The space of states is fixed, always, before we even talk about time evolution

0celo7

5:18 PM

@EmilioPisanty We've discussed AC to death in this room

Emilio Pisanty

@0celo7 but are you OK with that?

Semiclassical

@0celo7 actually, Doklady usually has English translations

0celo7

I don't LIKE any of the weird shit you get without AC

But I refuse to believe well ordering

Semiclassical

not always easy to find online, but your math library likely has them in their collected periodicals

Bernardo Meurer

@ACuriousMind Did you like my DRM essay? :P

@dmckee Are you around?

Emilio Pisanty

5:19 PM

@0celo7 you might be surprised. I was hunting a Danish paper the other day and it turned out to be in German

0celo7

@EmilioPisanty I also don't claim to understand infinity and if you do I think there's an issue

Cantor went insane because of this shit

ACuriousMind

I understand infinity.

Emilio Pisanty

@0celo7 I'll cop to not understanding infinity but I'll keep that fact under the rug if you don't mind

Bernardo Meurer

@0celo7 Cantor was born insane

@ACuriousMind You're not human

user929304

@ACuriousMind thanks, understood.

0celo7

5:20 PM

@ACuriousMind Are you parodying a certain Duffield?

ACuriousMind

@0celo7 No, just playing the contrarian to you :P

Anonymous

@0celo7 Proof? :P

Semiclassical

mary mary quite contrary

ACuriousMind

@Semiclassical The physicist replies "$r^2$" while the mathematician complains about undefined and overloaded symbols?

Emilio Pisanty

@ACuriousMind and we thought 0celo7 was in a joke-explaining mood

Semiclassical

5:22 PM

well, or the mathematician insists that $f(r,\theta)=r^2+\theta^2$

0celo7

I'm not. I'm srs

@Semiclassical Hey! Don't I get any credit here?

ACuriousMind

@Semiclassical Ah, heh

0celo7

I agreed with you!

Semiclassical

heh

yeah, and I do appreciate that

i mean, I can understand the objection: $f(r,\theta)=r^2$ is not sensible if we insist on $f$ as a mapping from $\mathbb{R}\times \mathbb{R}\to \mathbb{R}$

but if we're talking about an experiment, what matters (for instance) is what electric potential would find at each point in space.

the particular coordinate system we use to represent that relation really does not.

and insisting that one nevertheless use different symbols to denote the same output is in that respect just silly

blah blah blah

0celo7

@Semiclassical just come up with a name for the function that takes polars to rectangular and then precompose

That's mathematically correct

Semiclassical

5:34 PM

yeah.

this is one place where I really do like bra-ket notation, namely being able to write $\langle x|\psi\rangle =\psi(x)$ with $|\psi\rangle$ being the fundamental object

Emilio Pisanty

anyone ever seen a zebra centaur?

(source)

Semiclassical

that looks like something out a SCP story

0celo7

@Semiclassical Now that I vehemently object to.

Semiclassical

lol

0celo7

Bra-ket notation is NOT fundamental, it's a mnemonic system.

Semiclassical

5:36 PM

meh

Keep these mind

@0celo7 depolar

Semiclassical

my point is that notationally it treats $|\psi\rangle$ as more basic than $\langle x | \psi\rangle$

Emilio Pisanty

@Semiclassical so does $f$ and $f(x)$

(just saying)

Semiclassical

hmm

i suppose the comparison is that no physicist would expect that $\langle x|\psi\rangle = \sin x\implies \langle p|\psi\rangle =\sin p$

and moreover that the rule for how the latter is obtained from the former is 'suggested' by the notation itself: $\langle p |\psi\rangle = \int \langle p | x\rangle\langle x|\psi\rangle \,dx$

but I think that reflects how comfortable I am with $\int |x\rangle\langle x|\,dx=\mathbf{1}$

Captain Bohemian

mnemonic means what?

Semiclassical

5:42 PM

memory

0celo7

@Semiclassical and you have to do a lot of analysis to make sense of that. If you do QM like in Reed and Simon, you never need it though

Semiclassical

I sorta object to it being called just a mnemonic system, but I'm not sure what else to call it

0celo7

Rigged Hilbert spaces are for people with strong emotional attachment to bra-ket notation

You can (and should) learn how everything works without it

Semiclassical

lol

tbh I largely agree with that

0celo7

The best solution is as usual to like both notations and be able to switch immediately.

Semiclassical

5:44 PM

i feel like, for instance, one could do QM just as well by a form of index notation

0celo7

That's true, but I think it's immoral

Semiclassical

pff

and actually I take that back a little. when one is doing finite dimensional Hilbert space stuff, then all that's happening is ultimately just matrices/row vectors/column vectors

in that context, index notation would make sense.

but when you're working with position and momentum representations, that doesn't make a lot of sense.

dmckee

@ACuriousMind Love it. These kinds of "oh, well all know the notational conventions so I don't have to say what I mean" things are (a) really bloody convenient and (b) a potential source of vast confusion.

Semiclassical

ugh. Maxwell keeps spamming MSE chat with the same Schrodinger equation problem

@dmckee my own reaction the first time I saw it was that it was ambiguous

0celo7

And L^2 of a finite set is just R^n. Everything is better with functional analysis.

Semiclassical

5:48 PM

which isn't quite right. the point is that it depends on context

being able to recognize that notation is ultimately a matter of conventions---and, therefore, context---is to me a pretty important research skill

dmckee

@Semiclassical It is. And it is one that becomes so ingrained that when you return to the classroom as a teacher it is easy to forget that beginning students don't have it yet.

That lack of understanding caused some trouble with my first few upper-division tests.

Semiclassical

yeah.

i can understand an insistence avoiding such conventions in that context.

i just think that getting upset at them in practical use is silly.

0celo7

6:10 PM

@Semiclassical what triggered you again?

Semiclassical

someone else indicating that such notation was garbage and to use it was to spew crap

0celo7

6:57 PM

@BalarkaSen My advisor just sent me a Gromov paper. I'm scared

Balarka Sen

7:11 PM

@0celo7 Is this the new paper

0celo7

@BalarkaSen there's a new new paper

Balarka Sen

oh?

0celo7

@BalarkaSen on psc conjectures

@BalarkaSen I'm already doing one presentation on psc in the spring, I bet he thinks I want to do another

problem is it's all ggt/mg or kt

Balarka Sen

7:27 PM

psc?

Semiclassical

lol, the title of this paper:

Galoisian Approach to integrability of Schrödinger Equation

it includes the word 'isogaloisian' which is fun

0celo7

@EmilioPisanty I'm gonna waste hours sorting papers in this damn app

@EmilioPisanty this isn't exactly user-friendly

@EmilioPisanty So I found the place for summaries, can I not put TeX there?

Turbo

8:03 PM

Hi guys

I see quite a few textbooks and websites saying that "beta decay needs to be explained by the weak nuclear force"

But I don't understand why the weak nuclear force is needed for beta decay

I would be immensely grateful if anybody could explain this to me

Anonymous

6

Q: The role of W bosons in the weak nuclear force and beta decay

I am a beginner Physics student and I am studying the weak nuclear force and how particle interactions work. Now, from my book and the Feynman diagram, I learned that a neutron can change into a proton when it interacts with a neutrino, this happens because W$^-$ bosons are exchanged between $\nu...

Anonymous

16

Q: Weak contribution to nuclear binding

Does the weak nuclear force play a role (positive or negative) in nuclear binding? Normally you only see discussions about weak decay and flavour changing physics, but is there a contribution to nuclear binding when a proton and neutron exchange a $W^\pm$ and thus exchange places? Or do $Z$ exch...

quantum-field-theory nuclear-physics standard-model binding-energy weak-interaction

Turbo

@Blue Ok, thank you for the links

0celo7

8:32 PM

@EmilioPisanty Am I mistaken or is there no option for subfolders? Because that's a deal breaker for me.

0celo7

8:48 PM

wtf it wasn't there a few minutes ago

found it

@BalarkaSen I was made fun of in the topology seminar for being a physicist

Anonymous

@0celo7 wut

Anonymous

Since when are you a physicist?

0celo7

I was reading about conformal structures in spacetimes

that makes me a geometer, not a physicist

Anonymous

@0celo7 Ya. Lol

Anonymous

You should have countered with the GDP argument though

0celo7

9:00 PM

@EmilioPisanty Wow the free storage size is tiny!

bolbteppa

9:12 PM

@Semiclassical pch $r^2 + \theta^2$ be gone...

I've never seen something so unlikeable (in math)

$dr^2 + r^2 d \theta^2$ for $r = 1$ however...

Isogalosian makes up for it

lılostafa

9:27 PM

What is an ellipse with big eccentricity factor called?

Anonymous

degenerate ellipse

lılostafa

Not that big

Anonymous

@lılostafa I've never heard of a term for that. What's the context anyway?

lılostafa

I remember they call ellipses like the rightmost one here with some specific term....:

Those that deviate a lot from a circle (but still not a line)

Anonymous

No idea

lılostafa

9:35 PM

I think I can just say a very eccentric ellipse

@Blue The context is optics. I want to refer to elliptical polarizations that are very close to a linear polarization (but still elliptical)

Anonymous

Invent your own term for that and put it in your thesis. It might just become popular. :P

lılostafa

9:51 PM

Good suggestion :)

Emilio Pisanty

11:08 PM

@0celo7 they're not folders, they're collections

0celo7

Yeah I figured that out

@EmilioPisanty please be a fortran god

Emilio Pisanty

And yeah, you can make sub-collections

@0celo7 yer barkin up the wrong tree

Slereah

Ask @JohnRennie, he's an old man!

He probably knows Fortran

Emilio Pisanty

If it's fortran then you probs want @KyleKanos

0celo7

he hasn't used it since I've been alive

@EmilioPisanty he's not pingable

Slereah

11:10 PM

What FORTRAN do you need, anyway

0celo7

Step size in x, <0.0001,1>, negative to exit:
Step Size: 0.600000 Numerical Result: -3.13600 An. Res.: -1.6667 Err[%]: 88.16000

Step size in x, <0.0001,1>, negative to exit:
Step Size: 0.100000 Numerical Result: -1.96000 An. Res.: -1.6667 Err[%]: 17.60000

Step size in x, <0.0001,1>, negative to exit:
Step Size: 0.010000 Numerical Result: -1.68660 An. Res.: -1.6667 Err[%]: 1.19600

Step size in x, <0.0001,1>, negative to exit:

oh dang

this doesn't display whitespace, damn

I can't figure out how they arranged the whitespace

the whitespace on the 1.19600 is messed up

Transcript for