« first day (2324 days earlier)      last day (1551 days later) » 

6:03 PM
To get matrix elements
Say I have an operator like x_1x_2
you have to integrate over spin too
Mar 13 at 14:11, by AccidentalFourierTransform
an integral is just a special case of a sum
also, idgaf about causality
continuous spin is fine
@Slereah sums are integrals
The old physicist saying is backwards
Saying integrals are sums is stupid really
@AccidentalFourierTransform I'm neglecting spin
Sums are more fundamental to define in the definition of $\Bbb R$
6:09 PM
Saying sums are integrals is more correct than the reverse.
$$\sum_{x\in\mathbb R}$$
deal with it
$$\int_{x\in\mathbb N}$$
he, this is fun
hello. Does anyone know a good textbook on statistical mechanics? with a more mathematical point of view (some remarks on dynamical system, information theory and etc.) I have seen Reif and Pathria but I am looking for a more rigorous book
@AccidentalFourierTransform nothing wrong with that
It's just an integral wrt. counting measure on N
you dont know that
$$\int_{x\in\mathbb N}\frac{1}{x}$$
@AccidentalFourierTransform this is pretty egregious
6:12 PM
$$\int_{n \in \Bbb N} \frac{1}{n} dn$$
What are you people doing
Just stop
You're embarrassing yourselves
$$f(x+\mathrm dx)=f(x)+f'(x)\mathrm dx$$
$$\aleph_0 + \omega^\omega = \infty^2$$
$$\int_{x\in\mathbb R}f(x)\mathrm dx=2\pi\delta(i\partial_\epsilon)f(\epsilon)$$
^ that one is pretty cool actually
6:16 PM
$$\Gamma = ☎^{-1} \Psi \Psi^-$$
everybody knows that $\Psi \Psi^-$ is not in the domain of $☎^{-1}$
you have to consider the self-adjoint extension
@AccidentalFourierTransform what?
$$\int \delta^2(x) f(x) = f^2(0)$$
That's so physicist I don't even know what's happening
Do physicists even give a shit?
6:19 PM
We do not
@AccidentalFourierTransform What the...
What is a delta function of an operator lol
amazing innit
I have now seen everything
6:20 PM
@ACuriousMind I have to send this to my advisor, he'd get a kick out of it
Look at equation (30)-(32)
Part of my master thesis hinged on the equality $dt^2 = dx$
The authors say these equations hold in a distributional sense
Is the delta there supposed to be a distribution on the space of operators??
Whatever that means
@ACuriousMind question to a moderator: I reopened this question because it had been wrongly closed as a duplicate.
Q: Can heavy elements be fused?

TobiYes, I know, in stars, fusion occurs up to Iron(-56); but, I want to know if fusion past this nucleus can happen at all. If so, the daughter element would move to the right of the peak and its parents (away from the point of Iron-56), on the following graph (meaning additional energy is required...

But David Hammen points out it is actually a duplicate of:
Q: Why only light nuclei are able to undergo nuclear fusion not heavy nuclei?

Monia RezaIs it because of the binding energy or the binding energy per nuclei . I am having trouble with this whole binding energy and nuclear fusion concept.

6:23 PM
@ACuriousMind does that paper make sense to you?
$\sum_n \mathrm{e}^{-\lambda_n \partial_\lambda}\delta(\lambda) = \sum_n \delta(\lambda - \lambda_n)$??? Is this supposed to be obvious?
@0celo7 not at all
However I can't now vote to close as a duplicate of that second question. Can anything be done or do I just shrug?
I realise (too late) that I should have used the new feature to edit the duplicate - oh well
Obviously it means $$\int \delta (\partial_\epsilon) d\partial_\epsilon = 1$$
@ACuriousMind it is a translation duh
@Slereah that looks like mathematica code
@AccidentalFourierTransform lol
That's...probably actually what they mean
@JohnRennie I was just about to say that. No, except for letting others close it again if they agree with the new duplicate there's nothing you can or should do
6:26 PM
@ACuriousMind is that from your paper?
@0celo7 What? It's from the one AFT linked and which we were, uh, "discussing"
@ACuriousMind shrug :-)
I'm on mobile! Hard to read fucking papers sorry
This paper is the pinnacle of physicist math
@AccidentalFourierTransform How did you find this?
@ACuriousMind ah!
It's the translation operator
6:29 PM
4 mins ago, by AccidentalFourierTransform
@ACuriousMind it is a translation duh
He's Taylor expanding a Dirac delta holy shit
what is the taylor expansion of the dirac
@ACuriousMind I dont remember, I found it last year
I think of it from time to time
I want to use it in some assignment so that my professors think Im retarded or something
@AccidentalFourierTransform Suuuure, you don't remember. I'm on to you, Mr Morales!
AFT's name is Morales?
@Slereah I guess you can do a distributional Taylor series
Polynomial in what though...
6:32 PM
@ACuriousMind lol busted
@ACuriousMind can you access [7]?
Hi everybody, Im Alejandro Morales and I live in LA
I don't have access at home
@Slereah this is actually a really interesting question.
Huh. I swear I googled
Ah. They're interpreting the delta function as a formal series
So in some sense that might not be complete nonsense.
6:38 PM
eq 13 is something I definitely see myself using in the future
in which paper?
@ACuriousMind I'm not familiar with tempered distributions, but do they always admit some sort of derivative?
6:42 PM
$T'[f]\equiv -T[f']$
Is that actually a tempered distribution?
yes because of the minus sign
$T'[f]\equiv \color{red}-T[f']$
it cannot diverge weakly if you multiply by $i^2=-1$
I don't think you're making sense
6:44 PM
@0celo7 All distributions that are defined on differentiable test functions are differentiable simply by defining $\delta'(f) = -\delta(f')$.
have you heard of the Euclid-Weierstrass theorem for temepered distributions?
@ACuriousMind I know about on test functions, I'm not asking about that.
Then I don't know what you're asking about
I'm asking if anyone in this chat has actually verified that $T'\in\mathscr S'$ if $T$ is.
Oh, you're asking whether the derivative of a tempered distribution is tempered?
6:45 PM
On $\mathscr D'$ it's not so hard.
@ACuriousMind Yes.
It's clearly in the algebraic dual.
Checking continuity would involve horrible things though.
Oh, I see what you're worried about and I say that I happily leave that to people who actually enjoy analysis :P
According to Yosida, it does. You can all rest easy.
Apparently $\phi\mapsto D\phi$ is continuous in the semi-norm topology on $\mathscr S$.
@Slereah Ok, so we can't Taylor expand $\delta$ when regarded as a distribution
But $\delta\in \mathscr S-\mathscr D$ might make sense
Or, rather, $\delta\in(\mathscr S\cap C^\omega)'$
So we have $$\langle \delta,f\rangle=\langle \delta,\sum_{n\ge 0}\frac{f^{(n)}(0)}{n!}x^n\rangle$$
Hmm. Maybe this doesn't work :P
This paper is pretty infuriating. Do physicists not understand that convergence of a series has to refer to some kind of topology? Writing $g(x)=\sum a_nx^n$ does not mean $g(\partial)=\sum a_n\partial^n$ has to make sense.
And what is $1/g(\partial)$ supposed to be??
@JohnRennie I could write up something... :)
@0celo7 "physicists"? Only one of the three authors lists an affiliated institute that's a physics institute
Don't let your prejudice blind you :P
@ACuriousMind I'm blaming the whole field for this atrocity.
6:55 PM
Actually, second thought, I don't think it'd be possible with JS alone
If you want to see other peoples' colors
You'd need some kind of server
@ACuriousMind Is $1/g(\partial)$ supposed to be the Green's function lol
I wonder if there are existence/uniqueness theorems for infinite order differential operators
Dear lord, how did this question go unprotected for so long?
@JohnDoe I think this didn't get answered. It's the same notation as here. $\pi$ transitions are linearly polarized and $\sigma$ transitions are circularly polarized. It's a bit of an abuse of notation to talk about $\pi$ or $\sigma$ polarizations, but it happens.
@ACuriousMind who knows
> Assume now that g is a suitably well-behaved function.
there were something like five obvious occasions
7:00 PM
I think for any of this to make sense, $g$ has to be a polynomial.
Aha, they want $g$ and $g^{-1}$ to be analytic.
@ACuriousMind What kinds of functions satisfy that?
Constant ones? That's about it.
I guess you can blame both me and you since we're on record deleting one of those answers when we had the protection privilege
> Well we're really made from stars, so since stars glow at all wavelengths, shouldn't we too glow at all wavelengths?
Brilliant zinger from @KyleKanos
I don't know about you, but I definitely glow
maybe you should use some moisturizer
@0celo7 a proper radiant glow?
have you checked whether you're pregnant?
Just the usual American food baby.
@AccidentalFourierTransform After some thought, it should be $$\int_\Bbb{N}a_n\,\mathrm d(\mathrm{count})(n).$$
7:05 PM
@ACuriousMind I meant reciprocal, not inverse, apologies.
confusing notation above
@ACuriousMind Ideally it should work everywhere
Well, it can't because $1/g$ is undefined where $g$ is zero, no?
Obviously, so take $g$ nonzero
So like what about $x^2+x+1$
Is $1/(x^2+x+1)$ analytic?
@0celo7 then the second answer in the thread I linked says it works everywhere
7:08 PM
Ah, I did not see the second one
$$\int f(x)\mathrm d\lceil x\rceil$$
@ACuriousMind nice google fu
rather standard notation tbh
What is that supposed to be?
$\sum _n f(n)$
7:10 PM
Mine is actually correct though
$$\delta(-\mathrm i\partial_x)\delta(x)=\frac{1}{2\pi}$$
What is this madness
This is the new $\mathrm e^{\mathrm i\pi}=-1$.
that is disgusting indeed
Is this a troll paper?
$\mathrm i$ instead of $i$
7:13 PM
have you ever seen the Bogdanoff papers, @0celo7
I like how one of their hypotheses was that the function had to be analytic, then they substitute in $\delta$
truly amazing
@Slereah no, what are they?
@AccidentalFourierTransform that's wha she said
The Bogdanoff brothers are two dudes who managed to get a physics PhD by publishing complete nonsense
Actually, they need $1/\delta(x)$ to be analytic!
@ACuriousMind Can I ask on the main site what the power series of $1/\delta(x)$ is?
in mathOverflow
7:16 PM
@Slereah what university?
Université de Bourgogne
@0celo7 That's a pure math question :)
@ACuriousMind Pure physics math
@0celo7 that's horrifying
@EmilioPisanty If you assume $\delta$ is an analytic function and so is $1/\delta$, you can derive amazing things. (Actually they both have to be polynomials, but that's not relevant.)
7:18 PM
"If you assume <insert Cthulu-level horror>" is not a good way to start
Hey, this is physics
the stuff they do in QM is not really much worse
Dialetheism is the view that some statements can be both true and false simultaneously. More precisely, it is the belief that there can be a true statement whose negation is also true. Such statements are called "true contradictions", dialetheia, or nondualisms. Dialetheism is not a system of formal logic; instead, it is a thesis about truth that influences the construction of a formal logic, often based on pre-existing systems. Introducing dialetheism has various consequences, depending on the theory into which it is introduced. A common mistake resulting from this is to reject dialetheism on...
I can't believe someone let them publish this
$$\tilde g(y)=\sqrt{2\pi}\mathrm e^{\mathrm ixy}\delta(\mathrm i\partial_x-y)g(x).$$
@EmilioPisanty ^ new formula for the Fourier transform
@0celo7 what on Earth is $\delta(i\partial_x-y)$?
@EmilioPisanty Ah, well. That's a good question.
7:21 PM
also, formula is wrong
You write $\delta(x)$ using the usual Fourier integral
then you substitute in $\partial_x$ for $x$
@EmilioPisanty why?
@0celo7 automatic consequence of the use of $\mathrm i$ instead of $i$.
@ACuriousMind can I call him names?
Why all the hate for upright $\mathrm{i}$ here? :/
is the proper latex symbol
7:23 PM
@Slereah is not
@0celo7 If you don't mean actual names like "John" or "Rupert", no
Who the hell is called Rupert
@0celo7 uh... this guy?
Pretty weird that under the identification of $\Bbb H$ with $\Bbb C^2$, the standard action of $SU(2)$ corresponds to right-multiplication by a unit quaternion. $\Bbb C$-linearity is weird.
7:24 PM
$$Z[J]\propto \mathrm e^{\mathrm i\int J\phi}\delta\left(\mathrm i\frac{\delta}{\delta\phi}-J\right)\mathrm e^{\mathrm iS[\phi]}.$$
Rupert Giles is a fictional character created by Joss Whedon for the television series Buffy the Vampire Slayer. The character is portrayed by Anthony Stewart Head. He serves as Buffy Summers' mentor and surrogate father figure. The character proved popular with viewers, and Head's performance in the role was well received. Following Buffy's run, Whedon intended to launch a television spin-off focused on the character, but rights issues prevented the project from developing. Outside of the television series, the character has appeared substantially in Expanded Universe material such as novels,...
^new way to compute path integrals
Enough Ruperts for you? :D
seen here on his ride:
7:25 PM
@ACuriousMind fine
apparently also this guy
DI Rupert Lestrade
This is pretty amazing, I can't even lie
@0celo7 what, the existence of four different people named Rupert?
$$\int_\Bbb Rg\,\mathrm dx=2\pi \,\delta\left(\mathrm i \frac{\mathrm d}{\mathrm dt}\right) g(t).$$
The trick is that they're actually all the same person
@0celo7 maybe some \mathopen{}ness there?
7:29 PM
what's that
$$\int_\Bbb Rg\,\mathrm dx=2\pi \delta\mathopen{}\left(\mathrm i \frac{\mathrm d}{\mathrm dt}\right) \mathclose{}g(t).$$
note spacing after the delta
so, wait
you're hoping $\delta(i\partial_x) = \frac{1}{2\pi} \int \mathrm dk \, e^{-k\partial_x}$ makes sense
This paper claims it does. I think the whole thing is nonsense.
That's some ridiculous identity
I don't think they're being 100% serious. What they're trying to say is that these formal equations can be approximated for numerical computations using nascent deltas and cutting the power series for those off.
where presumably $e^{-k\partial_x} f(x) = f(x-k)$?
7:32 PM
@EmilioPisanty Yeah.
@Danu It's not any more ridiculous than the usual physicist math.
It follows completely from relations that everybody knows
@0celo7 well, that can presumably be extended into something that vaguely makes sense?
@EmilioPisanty Perhaps. But they try to apply it to path integrals
So in the end nothing is defined :D
@0celo7 ah, well, then that's a situation I usually model as "the faster you jump a sinking ship, the better"
It was a good distraction from multipole expansions
@0celo7 why would you veer off from physics into nonsense?
7:38 PM
Interestingly enough, $e^{-k\partial_x}$ being the translation operator is nonsense when physicists derive it
The physicist proof requires the function to be analytic, but that operator is not defined on the analytic functions :D
It is defined on $C^\infty_0$, but those are obviously not analytic
yah, well, mathematicians do tend to get sore when physicists' manhandling of the math turns out to work just fine
you mathematicians should be proud
@EmilioPisanty Not sore as much as bewildered
you made something very robust
@0celo7 "we're not sore about it" is not the oldest lie ever, but it's close to
7:41 PM
Ok, I'm not sore.
dammit, dupehammer still takes me by surprise
someone double-check this
Q: Shor's Algorithm vs. No-Communication Theorem

mldI have a question regarding Shor's Algorithm and the No-Communication Theorem (abbreviated NCT). The NCT states that useful information cannot be exchanged faster than light. https://en.wikipedia.org/wiki/No-communication_theorem Now imagine the following: (for technical details, please see -->...

@EmilioPisanty I'm doubtful $\delta(x)=\frac{1}{2\pi}\int e^{ikx}\,\mathrm dk$ makes rigorous sense. The "derivation" requires exchanging some integrals
Since plane waves are not integrable, that's not justified
And they're not real/positive, so Tonelli can't be used either
@0celo7 what people really mean by that is that the equation is only true in the distributional sense
Maybe I'm missing something
@EmilioPisanty Even in the distributional sense, what does it mean?
@0celo7 We-ell, the l.h.s. doesn't really make sense to begin with, you can't feed a $x$ into $\delta$- it's a distribution, not a function.
7:48 PM
@0celo7 essentially, that you're only meant to do the $k$ integrals last
@ACuriousMind The $x$ signifies the com of the measure, I have no problem with that.
$\delta(x)$ is only ever meant to be used as $\delta_0(f) = \int \delta(x) f(x) \mathrm dx$
so really when you say $\delta(x) = \frac{1}{2\pi} \int e^{ikx} \mathrm dk$, you're saying $$\delta_0(f) = \int \delta(x) f(x) \mathrm dx = \frac{1}{2\pi} \iint e^{ikx}f(x) \mathrm dk \mathrm dx $$
I think you want to interchange the measures.
see, that's the thing
that order is what you get by a formalistic reading of the identity that bothers you
and you have the problems you noted
@EmilioPisanty Measures are read from left to right. I know what you're saying.
If you want to do $x$ first, it needs to be $\mathrm dx\mathrm dk$.
7:52 PM
@0celo7 that's the thing
the notation formally says first $k$ then $x$
that's what bothers you
So $\delta(x)=stuff$ means "multiply by $\frac{1}{2\pi}e^{ikx}$, integrate by $x$, then integrate by $k$."
@0celo7 no
Then what does it mean?
when we say "$\delta(x)=\frac{1}{2\pi}\int e^{ikx}\,\mathrm dk$ only holds in the distributional sense", what we're really saying is "you should put in all the integrals, and when you get to the double integral, you should do $x$ and then $k$ for the thing to make sense"
That's exactly what I just said
7:55 PM
@0celo7 ah, yes
I guess you can switch the integrals because $f(x)e^{ikx}$ is indeed $L^1$
Or is it
it is iff $f$ is
@0celo7 we can switch integrals because we're happy to restrict our hypotheses to whatever is needed for the manipulations to make sense
that was fun to say out loud
@EmilioPisanty the manipulations making sense and being correct are different things
7:58 PM
@0celo7 for the manipulations to be correct
That's still not very helpful because it doesn't say what those hypotheses are :P. In any case, it's legitimate for $\delta$ acting on Schwartz functions.
I really have to compute this multipole expansion, cheerio
8:15 PM
Hm, I know of adding bundles, but what is $TY - 3\mathcal{O}$ supposed to mean for a line bundle $\mathcal{O}$?
oh, that is the Gell-Mann-Nishijima formula for bundles, right?
@AccidentalFourierTransform :P
I can't tell if he's trolling
@0celo7 that is exactly my intention
@0celo7 He is
8:20 PM
@0celo7 did you just assume my gender!? I never said I was a he...
Yeah I assumed your gender
there are no girls (a) on the internet (b) in science
are there attack helicopters on the internet? and in science?
if you sexually identify as an attack helicopter you might have a mental issue
Mar 8 at 20:22, by AccidentalFourierTransform
ah, here in this chat it is usually safe to assume that everybody is an 18-year-old boy from India :-)
So, technically, you can't be surprised he addressed you as male :P
I... how... you were not there when I said that
8:23 PM
On the other hand,
he's an AI
Dec 19 '16 at 18:19, by AccidentalFourierTransform
I bet that was not your intention, but people are telling you that assuming someone's gender because of their career choices is very disrespectful, and you don't seem to get it
I don't deal in respect vs. disrespect. I deal in statistics.
If I call people in science "he," I will have to correct myself less than 50% of the time. Works out well for me.
@AccidentalFourierTransform That's what you think
I dont know what to think anymore
8:26 PM
::makes popcorn::
ACM is a magician
::throws rabbit into audience, then vanishes::
:: JR was in the audience, grabs rabbit, mercilessly kills it and eats it raw ::
I'm sad that my translation operator answer has zero votes
It's incredibly correct :(
Did you say "sexually identify"? :-O
8:29 PM
@0celo7 That's it - it's so correct no one believes it ;)
@AccidentalFourierTransform ...your fantasies disturb me
Someone just upvoted it <3
Of all the identifications why did you choose that one?
@ACuriousMind its a new meme Im trying to introduce here
JR enjoys killing animals
Where is The Evidence?
::gets triggered::
8:33 PM
Why can't people just give me the hamiltonians?
I don't want to figure out CGS units
$2\pi$ or not $2\pi$
@ACuriousMind It works better in German
well, half-german
Hmm, you're right. I pronounced it the German way in my head
in Spanish it works better too
actually, I pronounced it in Spanish in my head
didnt notice till you pointed it out
2 in English, $\pi$ in Spanish
8:39 PM
@ACuriousMind $m_e$ or $m_\mathrm e$ for the electron mass?
you're a troll, we don't listen to you
@0celo7 The latter if you're being fancy, but I normally don'T care enough to do that
@ACuriousMind argh
$$\frac{\ee^{-r/a_0}}{\sqrt{\pi a_0^3}}$$
Do I have to care about the angular variables here?
I guess I should check if I need a $4\pi$ to normalize
why dont you set $a_0=1$? Im serious, use the proper (natural) units!
8:46 PM
I don't know what a natural unit is
the units that are natural to the context
in this case, atomic units
where you measure lengths in units of the Bohr radius
and energies in units of the Rydberg energy
so, in essence, $m_e=\hbar=a_0=1$
Hmm. I get $\langle 0|0\rangle =1/2$
Not good.
$\int_{S^2}\mathrm d\Omega=4\pi$, right?
Ah, I did not do the integral correctly.
@AccidentalFourierTransform I don't want to set units to one
A: On this infinite grid of resistors, what's the equivalent resistance?

ShaktyaiThis is a classical problem. The trick is to use symetry whenever it's possible and to connect the symmetric points since they have the same potentiel. First diagonals, etc...

^ is this something we should really keep around?
-7 Jesus
Units are formal variables, and physical expressions are homogeneous functions of them. You lose no generality by dropping some of these formal variables. But ok, do as you wish.
8:58 PM
@EmilioPisanty Nope, it doesn't answer the question.
@0celo7 to set the record straight, that's not actually natural units, the proper name is atomic units

« first day (2324 days earlier)      last day (1551 days later) »