2:03 PM
@ACuriousMind the answer may be incorrect! and I just meant to give him a hint.

@samjoe Whether the answer is correct or not does not matter at all for the purposes of the HW policy

fine.

And whether you meant to give a hint or not is likewise irrelevant - what you did is write down something that could be the actual, full answer to the question.
In any case, we would generally prefer that you don't answer such homework-type question at all. The policy is in place to discourage such questions. Try to find better questions to answer.

Get rekt

In unrelated news, I see 53 items in the close review queue. >3kers, please review!
4

2:10 PM
Is it a legitimate use of the board to open up a chat room for a personal theory?

@ACuriousMind does that include me?

@MikeDoonsebury If you have enough reputation to create a room, you can create it for any purpose as long as you abide by the Be Nice policy.
@0ßelö7 Yes, I can, one moment
@0ßelö7 Why would it not?
@SirCumference Unless you're well-rested, I command you to turn back and go to bed now!

@ACuriousMind Don't worry, I got 8 hours of sleep. I'm sick XD

why are you so concerned with his health

@0ßelö7 First Law.

2:16 PM
@ACuriousMind don't talk about fight club?

@ACuriousMind A body at rest must stay at rest? So I should sleep?

lol
Neither of you has the correct reference

@SirCumference lmao he wants you to be at rest for infinity, i.e. die

> i.e. die

yes?

2:18 PM
Wow, I didn't know what this was, ACM

Anonymous
If the potential function is finite in a region why should the wave function and it's first derivative have to be continuous in that region? (I'm talking in reference to the Kronig-Penney Model)

@SirCumference What what was?
As for the "First Law": A hint is that @JohnRennie will probably know what I'm referring to.
4

Here we want to give an easy mathematical bootstrap argument why solutions to the time independent 1D Schrödinger equation (TISE) tend to be rather nice. First formally rewrite the differential form $$-\frac{\hbar^2}{2m} \psi^{\prime\prime}(x) + V(x) \psi(x) ~=~ E \psi(x) \tag{1}$$ into the int...

what is $\mathcal L^1_\mathrm{loc}$

@0ßelö7 Everything relevant is hyperlinked, no?

@ACuriousMind I think he means $L^1_\mathrm{loc}$

2:24 PM
I think he doesn't care about calligraphy that much :P

Anonymous
@ACuriousMind Thanks!

@ACuriousMind it's an important distinction!
mathcal L could be something else, like a Campanato space
@ACuriousMind do you think I should bother citing the original papers here or just the books I get the material from

Anonymous
As usual I didn't understand many parts of the answer due to the math. But I at least got a basic idea now. :P Mostly looks like Linear Algebra and Analysis though.

@Blue it is entirely analysis

[Some time travel comments] Since in the previous paragraph, we have explained how travelling to the future will not necessary result in you to arrive in the future that is resulted as if you have never time travelled (via twin paradox), what is the reason that the past you travelled back, has to be the past you learnt from historical records :?

2:27 PM
@0ßelö7 I would probably cite whatever you think is best for people to read if they actually want to follow your references.

it is kind of important to note that being in 1D is crucial to the argument, @Blue

@Secret What previous paragraph? Reminder that this is not your blog.

Anonymous
@0ßelö7 I see. As of now I'm dealing with only 1D SE

@ACuriousMind Right, good point. The books it is. When I mentioned giving this talk people said they knew the books but never read them

2:28 PM
@Blue it's why PDE is very much harder than ODE

@0ßelö7 Ah, yes. I kinda assumed we're in ordinary intro QM territory here where all the potentials are 1d wells

Anonymous
@0ßelö7 I've heard that

In 1D, if you know $u''$ then you know $u'$ and $u$. But in $\ge$2D, if you know $\Delta u$ then it's HAAAAAAAAAAAAAARD to get info about $u_{ij}$, $u_i$ and $u$.

Anonymous
I'll start learning ODE and PDE during my winter vacation this year. :)

Anonymous
@0ßelö7 Yup, I get that

2:30 PM
I seriously doubt you'll learn elliptic regularity theory

Sounds like we're doing cool stuff

Schroedinger operators are even worse though, because you want $V$ to live in a really bad space usually
And the domain is typically all of $\Bbb R^n$
@Blue If you want an idea, read the book "Introduction to Spectral Theory" by Hislop and Sigal

Anonymous
@0ßelö7 What are the prerequisites for reading that?

Basically none, the appendices look very through
I doubt they teach you measure theory though
you'll spend a few weeks reading the appendices if you know none of this material though

Anonymous
@0ßelö7 I see. Thanks for the suggestion. I'll surely look into it during my vacations.

2:35 PM
Normed vector spaces, $L^p$ spaces and integration, linear operators on Banach spaces, Fourier transform, Sobolev spaces, convolution
@ACuriousMind I need a lemma that takes two slides to state lol

@0ßelö7 At that point you should ask yourself whether you really need to tell them exactly what the lemma says

@ACuriousMind I was actually hoping to give a full proof of the lemma, because it's folklore
you know me, I love proving folklore

@0ßelö7 I'd still maybe put a shorter statement of the lemma into the talk itself and put the full statement on a slide after the last slide that you can pull out if someone asks.
(Being less precise/in-depth than one usually would, but keeping the details in reserve in case it becomes relevant is generally a good strategy for talks, imo)

I could naturally break it up into 4 sublemmas
but then my talk has like seven lemmas
er, 3 sublemmas

2:44 PM
@0ßelö7 Well, I'd omit the explanation of the notation on the slide itself, and since there seems to be two pairs of formulae, I'd just put one of the two and then say that there's another one with suitable substitutions.

@ACuriousMind Maybe I should break it up, and prove one of them. Then, if I have time, have slides with proofs of the other parts?

So that reduces the lemma to two formulae, which should fit on a single slide :P

Right, I shouldn't have to explain to these guys what a multiderivative is

@0ßelö7 Is this a folklore proof where everyone competent in the field will know how it's supposed to go or is there some sort of trick that's not evident to an expert?
Because if it is not the latter, there will probably be no one interested in seeing the proof

How am I supposed to judge that

2:46 PM

Hm, ok

I mean, "Hey, I bet you've always wondered how to prove X - here it is" is interesting. "Hey, you know that statement everyone knows how to prove but doesn't bother to write down? Here is the proof written down" significantly less so

@ACuriousMind Yeah good point. Let me write out the proof carefully and see which parts are actually interesting
@ACuriousMind Well that was my point with Neumann regularity. Everyone thinks it's a routine calculation but you run into subtle issues

@0ßelö7 Well, if everyone thinks it routine but it isn't, then that's double interesting!

@ACuriousMind hence my obsession with Sobolev spaces on manifolds :P
Is $\Bbb R_+^*$ supposed to be $(0,\infty)$?

3:21 PM
@ACuriousMind no physicist shall discuss interpretations of quantum mechanics or by inaction allow interpretations of quantum mechanics to be discussed.

hmm?

@JohnRennie Heh, excellent

not having a notation index should be a capital offense

4:12 PM
@ACuriousMind how do I get the half harpoon for restriction?
$\upharpoonright$
nice
@ACuriousMind Is $\{(x,x):x\in X\}$ the diagonal of $X$ or the diagonal of $X\times X$?
just terminology wise

I'd say that of $X\times X$

the y = x line is the diagonal of R^2 not R

4:42 PM
Sorry I have a quick question: For questions like this physics.stackexchange.com/questions/356260/… where the accepted answer clearly does not answer the original question what is the best thing to do; downvote, flag or just leave it?

In a latex thingy, like texmaker, when you press new tab you get a blank file - any way to modify that so it opens a page with a pre-amble and \begin/end{document} already there :p

@bolbteppa use TeXstudio and use the wizards :P

oh thank god

@0ßelö7 :O nice

nothing worse than getting an old ass book and having to decipher the notation

4:57 PM
So this question says express $u^0$ in terms of $u^j$ where $u$ is the four-velocity and I get what $u^0$ and $u^j$ are but I'm a bit confused how to go about this one? I thought maybe using the space-time interval and evaluating for $\frac{dt}{d\tau}$ but it's not workin out for me... :/ Anyone give me a quickie starter please? :p

is the 4-velocity normalized?

If by that you mean $u\cdot u=-1$ yes? but I may've just seemed liek an idiot by saying that

Yes, so you have $-(u^0)^2+\mathbf u^2=-1$

Ahhh okay...
so $u^0=\sqrt{1+u^ju_j}$ I guess.. Thanks :D

5:58 PM
@ACuriousMind what's a more compact way to write $(k+l-n/2-1)\cdots(k+1-n/2)$
I guess it is already pretty compact
might not want to overcomplicate
I feel like it's some combinatorial thing

6:19 PM
@0ßelö7 $(k-n/2+\ell-1)!/(k-n/2)!$
It's a permutation coefficient

Anonymous
@BalarkaSen Wouldn't $n$ have to be even for that to be true?

Sure, if not you can replace the factorials by the gamma function

Anonymous
@BalarkaSen Okay, yes!

6:53 PM
@BalarkaSen yeah, but that's not any more compact
ah, but that does help for the convergence proof, thanks

kewl

7:26 PM
@BalarkaSen this is almost done, do you want to look over it

2 hours later…
9:03 PM
Although a physics question, this is still important to chemistry. The delocalized electric field is related to the force (and therefore the repulsive potential) between two electrons. This in turn is what we need to solve the Schrödinger Equation to describe molecules. Short answer: You can calculate the expectation value of the corresponding operator, which comes close to the mentioned superposition. — Feodoran 13 hours ago
:((((
still got closed anyway
0

If we take an electron that's delocalised w.r.t position, how can one evaluate the electric field over some space? Is it some superposition or a sort of field with all the charge at the expectation value of the position?

@ACuriousMind ok it's done
do you want me to resend it?

9:48 PM
Why do you have to go to the eigenbasis of $M^i$? Why cant you just simply observe that the kronecker which it is equal to is either going to be zero, +1 or -1

@Phase How is that observation supposed to relate to the eigenvalues?

hang on
what if you let i = j, so then you end up with $M^2 + M^2 = 2M^2 = 2\delta ^{ij}I$ so then you get $M = sqrt I$ and get +I or -I with eigenvalues +1 or -1
or am I butchering all of maths by doing this
Because if i != j then you get zero, which doesn't really count as a solution afaik

@Phase Well, indeed you get $2M^2 = 2I$, so $M^2 = I$. How does that prove something about the eigenvalues of $M$?

Because if M = sqrt(I), that means that the eigenvalues of sqrt(I) would be the eigenvalues of M right?

What is $\sqrt{I}$?
These are matrices, not numbers

9:56 PM
Idk, I guess +- I?

@Phase Well, did you just define that? That's not a proof!

i
uhhhhhhhhhhhhhhhhhhhhhhhhhhhh
confuse

@ACuriousMind So, I take it you don't want the full version?

@Phase (I suspect your intuition is correct here, but pretty much everything you'll do to make it into a coherent argument will involve going to the eigenbasis)

If $M^2$ is in the question, doesn't that mean that $M^a$ is a valid operation as long as $a \in R$?
Idk how to do the nice font

9:58 PM
@Phase Jesus

@0ßelö7 i know i know i make your skin crawl with my lack of rigour

@0ßelö7 I just looked back at chat and noticed Phase's question, I wasn't purposefully ignoring you - do you want me to look over it? Because I don't think I'll gain much personally from reading the slides.

@ACuriousMind do you know this already?

@Phase Uhhh. You can raise a matrix to an integer power - just multiply it $n$ times with itself, but it is not clear at all what using a non-integer power is supposed to mean
There is such a thing as a square root of a matrix, but I'd expect you to define how to obtain it before using it

Oh. I guess I just felt that since I was quite an easy thing to imagine a square root of, that I could use that
r i p

10:00 PM
@0ßelö7 No, but I won't know it after reading the slides either unless I go read the references, will I?

Im not entirely sure I understand the question then. If it's quick could you please just write a few lines about how you'd start it?

@Phase Well, use the hint: Go to the eigenbasis of $M$, then write out what $M^2 = I$ means in components.

@ACuriousMind I think you were right, anyone who is good at doing PDE type analysis could prove these lemmas. So in that sense it's quite complete.

..without imagining the square root of I? just break it into its column space and do it like that?

The only problem is I did all of the proofs in my notes by hand, but didn't use one of the lemmas
Well, I did, but the lemma is ridiculously general

10:02 PM
@Phase Just write down how it looks when you write out the matrices.

@ACuriousMind I sent it to you and would appreciate general feedback
I fear it's too long, going to read the whole thing out loud now and see

@0ßelö7 Okay, will read it soon

@ACuriousMind Depending on who shows up I might have to review some basic geometry
this is a new seminar so no one knows who will show up

Hurricane hunters aboard a WP-3D aircraft, about to hit the eyewall and penetrate the eye of Hurricane Katrina ^

@Phase You could start by writing out what "$\lambda$ is an eigenvalue of $M$" means

10:10 PM
This is fascinating

@BalarkaSen that it's the image of an normalised-eigenvector in a diagonalised basis?
$Mv - \lambda v = 0$

For some $v$, yeah.
Feed that to $M$ again. What do you get?

$\lambda ^2$?

Well, you don't get a number. What's $M(Mv - \lambda v)$?

0?
the null vector?

10:24 PM
One one hand, yes. On the other hand, if you "distribute" the multiplication by the matrix over the expression...?

$M^2v - M\lambda v$

Given the datas about $M$, which is?

Oh
$Iv - M\lambda v$
$Iv = M\lambda v$

$\lambda$ pops out of the right hand side because of linearity

Sorry, what do you mean "pops out of"?

10:26 PM
So you can simplify it further, given $Mv = \lambda v$.

Oh
$Iv = M^2 v$....?

Where did you get $M^2$?

I replaced the lambda with M

I mean $M \lambda v = \lambda M v =$ ... ?

Oh
$Iv = \lambda ^2 v$?

10:28 PM
Yes. What's $Iv$? :P

There you go.

I see, I got confused because I didn't really see what you meant earlier, thanks though

What up peeps

Feel pretty shameful for needing this spelled up for me
time to commit an honour suicide

10:29 PM
Not before I post my dank meme

One last question I suppose

@lılostafa

Maybe it's just me having not really done much with Eigenbases but I don't recognise where I "put it in terms of M's eigenbasis". I just wrote it down for some vector v, rather than a space that contains all of the vectors v
is that good enough?

@Phase How does the matrix $M$ look in the eigenbasis?

Oh. Like the identity

10:31 PM
@Phase What does that mean?
It's not equal to the identity.

to cleanse the shame of getting everything wrong

Please don't joke like that, confusion is normal when learning things

RIP, I can remove it if it's offensive

I mean, it's not equal to the identity, but it does look, in a certain sense, "like" it. I want you to tell me in what sense exactly.

I guess I don't really understand the casual language because I'm not familiar with it.
I don't really know what a matrix having a "look" means

10:33 PM
Do you know change of basis?

using the matrix of eigenvalues?
*vectors
JEEEEEEZ

@Phase By "looking" in a basis, I mean what its entries in that basis are.

Well
like +-1 in the diagonal

It's not entirely obvious to me what you mean by that but, what I mean is, if you change the basis of your vector space then the matrix changes according to a conjugation formula

@Phase Why +-1?

10:35 PM
I guess you know that

Because... $\lambda^2 = 1$ and if you diagonalise it in terms of the eigenvectors then its diagonal with each entry corresponding to a different eigenvector / value

@Phase Aha!
So, simply by going to the eigenbasis, the equation $M^2 = I$ tells you $\lambda^2 = 1$ and hence $\lambda = \pm 1$.
That's what the exercise intended you to do.
$\lambda^2 = 1$ is a conclusion here, not something you knew beforehand
The sentence "it's diagonal with each entry corresponding to a eigenvalue" was what I wanted to hear as the answer to "how does it look like?"

I'm bringing a new recruit here lol

@JaimeGallego Where is the fresh meat? ;)

What does it mean though

10:40 PM

@Phase What is "it" in that sentence?

If $\lambda = +-1$, does that mean M has the entries sign flipped? Surely the root of I would be +- I though
by sign flipped i mean like $\lambda _1 = 1$ and $\lambda_2 = -1$

was M 2x2

@Phase Why are we still talking about the "root" of $I$? Didn't we settle on that not being a good notion here?

I forgot about the points thing lol
Still, he can do Q&A on the site
So that's good

10:46 PM
@ACuriousMind yeah but I'm just confused
nevermind im dumb

It just means M, in a basis of it's eigenvectors, is a diagonal matrix with plus or minus 1 along the diagonals.

@JorgeCalvar Speaketh

Jesus
How many square roots does $I$ have

There are many many matrices which satisfy $M^2 = I$

@Phase If by "square root" you mean a matrix that squares to $I$, then infinitely many

10:55 PM
If you just take one of those diagonal matrices $M$ with $\pm 1$ along the diagonal entries, take an arbitrary invertible matrix $P$, $P^{-1}MP$ also squares to $I$.
aka "just basechange"

@JaimeGallego Maybe he had his internet destroyed by the mod who shall not be named
the feared one

@JaimeGallego You need 20 rep to be able to talk

@BalarkaSen I see, are there any cool applications of this in Physics?
Any properties that matrices like those have?

@Phase "Bloody Shog9, Bloody Shog9, Bloody Shog9" ghost jumps out of mirror
2
@ACuriousMind Yeah, I didn't know (before I saw his profile page) whether he had the necessary rep.

@JaimeGallego Shows up as 1 rep user to me

11:00 PM
@Phase Applications in physics? Not that I know of
They're a neat class of matrices. That's it :P
Ya boi's favorite isometries of $\Bbb R^n$, reflection along a hyperplane, are involutory matrices

I hear they're basing a lot of the visual effects in the new IT film on the supernatural effects of Shog9 entering a room
They say there's balloons first, then screaming
@BalarkaSen What's a Hyperplane?

Forget $n$, take $\Bbb R^2$. Reflect along a line
That's a linear transformation, and the matrix of it is involutory
Because reflecting twice is the same as doing nothing; $A^2 = I$

@ACuriousMind 50 minutes
perfect

so any matrix that is its own inverse?

@Phase Think about it: By definition, the inverse times the matrix gives $I$. So if $A=A^{-1}$, then $AA^{-1} = A^2 = I$, i.e. $A^2 = I$ is equivalent to being its own inverse.

11:05 PM
that's quite satisfying
What about if it's like a rotation about a third of 2pi? So that $R^3 = I$
does that have it's own class or is it boring

right
yeah these are finite order matrices
not boring

(all of linear algebra is boring)

@0ßelö7 mean
Can anyone answer my electron question? I asked it on hbar, then chem.se and I still haven't had a full answer nor resource recommendations

@ACuriousMind Hmm, the only issue is that I claim that $\int_M 0=\mathrm{vol}\, M$
need to figure out what I did wrong

@0ßelö7 Yeah, that doesn't sound right

11:11 PM
Huh, my main reference has the same typo twice
the recursion relation should start with $u_0(y,y)=1$, not zero

@Phase Which one?

0

If we take an electron that's delocalised w.r.t position, how can one evaluate the electric field over some space? Is it some superposition or a sort of field with all the charge at the expectation value of the position?

@ACuriousMind I think it's pretty good. I think I'll add in the proof of that lemma just to be safe, but otherwise...?
@ACuriousMind Maybe I should add some words about boundaries or noncompact manifolds?

@JaimeGallego Obligatory xkcd

Just reminded me of a time when I was a kid
It was a sleepover and everyone was like 9, and me being an edgelord decided it would be funny to say "candyman" into the mirror in the dark room at night as per the dumb urban legend
a kid cried and i got told off
not the most exciting story but hey ho

11:20 PM
lol
i like it

@Phase I've heard of bloody mary, but not candyman

The idea was that some dude would appear behind you and stick a hook in your back or something
Idk, Cornwall is weird
Gonna play some CSGO because my life is beyond redemption. Peace!

Hi @Avantgarde

Oh hey is CSGO on steam?
:O
@BalarkaSen Heylo

im finally listening to bish bosch

11:26 PM
@Avantgarde it's literally made by valve

noping the hell out

lol let me see
@BalarkaSen what?

last of the walker trilogy

Let me check

dont lol

11:34 PM
Googling Scott Walker gives me an American politician

lol

Google should've known by now what my tastes are, by now
I used 'by now' twice. wow

"I used "by now" twice by now"

You like those vocals?

very much

11:39 PM
spooky

Tilt and (at least one half of) Drift are just awesome
Interesting how a baroque pop artist went down the line of arcane experimental sonics :p

Yeah such changes are interesting

@BalarkaSen Forgive my unworthy attempt at memes

This is great
I absolutely love the promotional tagline

JEE, who shrunk the kids whaa?

11:45 PM
yup that
@Avantgarde it's a reference to the original poster of Invasion of the Body Snatchers

Honey, I Shrunk the Kids is a 1989 American comic science fiction film. The directorial debut of Joe Johnston and produced by Walt Disney Pictures, it tells the story of an inventor who accidentally shrinks his and his neighbor's kids to a quarter of an inch with his electromagnetic shrinking machine and throws them out into the backyard with the trash, where they must venture into their backyard to return home while fending off insects and other obstacles. Rick Moranis stars as Wayne Szalinski, the inventor who accidentally shrinks his children, Amy (Amy O'Neill) and Nick (Robert Oliveri). Marcia...

ah right that's another movie

Hm, seems like I might've heard this phrase somewhere

@Avantgarde One of my favorite tracks from the Tilt (you have heard Farmer In The City, right?): youtube.com/watch?v=wPMsTr3-G3Y

The vocals don't do it for me. Though whichever style that is, he's doing it pretty well. It's very crisp

11:53 PM
True. I can see why you are objecting to the vocals

53 items in the queue, goodness

it's funny because Blackstar uses operatic techniques similar to these albums

heh, i started reviewing at the right time - "thanks for reviewing 20, come back in 3 minutes to continue reviewing"

You got it. Opera ... I can't stand it. It has prevented me from checking out some metal bands that are otherwise pretty good.