The h Bar

0celo7

6:03 PM

To get matrix elements

Say I have an operator like x_1x_2

AccidentalFourierTransform

you have to integrate over spin too

Slereah

"integrate"

AccidentalFourierTransform

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

0celo7

@Slereah sums are integrals

The old physicist saying is backwards

Saying integrals are sums is stupid really

@AccidentalFourierTransform I'm neglecting spin

Slereah

Sums are more fundamental to define in the definition of $\Bbb R$

0celo7

6:09 PM

Saying sums are integrals is more correct than the reverse.

AccidentalFourierTransform

$$\sum_{x\in\mathbb R}$$

deal with it

$$\int_{x\in\mathbb N}$$

2

he, this is fun

Kiarash

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

0celo7

@AccidentalFourierTransform nothing wrong with that

It's just an integral wrt. counting measure on N

AccidentalFourierTransform

you dont know that

$$\int_{x\in\mathbb N}\frac{1}{x}$$

0celo7

@AccidentalFourierTransform this is pretty egregious

Slereah

6:12 PM

$$\int_{n \in \Bbb N} \frac{1}{n} dn$$

0celo7

What are you people doing

Just stop

You're embarrassing yourselves

AccidentalFourierTransform

$$f(x+\mathrm dx)=f(x)+f'(x)\mathrm dx$$

0celo7

...

Slereah

$$\aleph_0 + \omega^\omega = \infty^2$$

AccidentalFourierTransform

$$\int_{x\in\mathbb R}f(x)\mathrm dx=2\pi\delta(i\partial_\epsilon)f(\epsilon)$$

^ that one is pretty cool actually

Slereah

6:16 PM

$$\Gamma = ☎^{-1} \Psi \Psi^-$$

AccidentalFourierTransform

everybody knows that $\Psi \Psi^-$ is not in the domain of $☎^{-1}$

you have to consider the self-adjoint extension

0celo7

@AccidentalFourierTransform what?

Slereah

$$\int \delta^2(x) f(x) = f^2(0)$$

AccidentalFourierTransform

@0celo7 have fun

0celo7

That's so physicist I don't even know what's happening

Lol

Do physicists even give a shit?

Slereah

6:19 PM

We do not

ACuriousMind

@AccidentalFourierTransform What the...

0celo7

What is a delta function of an operator lol

ACuriousMind

$\delta(\partial_\epsilon)$

AccidentalFourierTransform

amazing innit

ACuriousMind

I have now seen everything

0celo7

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)

Amazing

Slereah

Part of my master thesis hinged on the equality $dt^2 = dx$

0celo7

The authors say these equations hold in a distributional sense

Is the delta there supposed to be a distribution on the space of operators??

AccidentalFourierTransform

sure why not

0celo7

Whatever that means

John Rennie

@ACuriousMind question to a moderator: I reopened this question because it had been wrongly closed as a duplicate.

4

Q: Can heavy elements be fused?

Yes, 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:

8

Q: Why only light nuclei are able to undergo nuclear fusion not heavy nuclei?

Is 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.

nuclear-physics fusion binding-energy

0celo7

6:23 PM

@ACuriousMind does that paper make sense to you?

ACuriousMind

$\sum_n \mathrm{e}^{-\lambda_n \partial_\lambda}\delta(\lambda) = \sum_n \delta(\lambda - \lambda_n)$??? Is this supposed to be obvious?

@0celo7 not at all

John Rennie

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

Slereah

Obviously it means $$\int \delta (\partial_\epsilon) d\partial_\epsilon = 1$$

AccidentalFourierTransform

@ACuriousMind it is a translation duh

@Slereah that looks like mathematica code

ACuriousMind

@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

0celo7

6:26 PM

@ACuriousMind is that from your paper?

ACuriousMind

@0celo7 What? It's from the one AFT linked and which we were, uh, "discussing"

John Rennie

@ACuriousMind shrug :-)

0celo7

I'm on mobile! Hard to read fucking papers sorry

Lol

This paper is the pinnacle of physicist math

ACuriousMind

@AccidentalFourierTransform How did you find this?

0celo7

@ACuriousMind ah!

It's the translation operator

ACuriousMind

6:29 PM

4 mins ago, by AccidentalFourierTransform

@ACuriousMind it is a translation duh

0celo7

He's Taylor expanding a Dirac delta holy shit

4

Slereah

what is the taylor expansion of the dirac

AccidentalFourierTransform

@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

ACuriousMind

@AccidentalFourierTransform Suuuure, you don't remember. I'm on to you, Mr Morales!

0celo7

What?

AFT's name is Morales?

@Slereah I guess you can do a distributional Taylor series

Polynomial in what though...

AccidentalFourierTransform

6:32 PM

@ACuriousMind lol busted

0celo7

@ACuriousMind can you access [7]?

AccidentalFourierTransform

Hi everybody, Im Alejandro Morales and I live in LA

0celo7

I don't have access at home

@Slereah this is actually a really interesting question.

ACuriousMind

@0celo7 It's on arXiv

0celo7

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.

AccidentalFourierTransform

6:38 PM

eq 13 is something I definitely see myself using in the future

0celo7

in which paper?

AccidentalFourierTransform

5 mins ago, by ACuriousMind

@0celo7 It's on arXiv

0celo7

lol

$\delta[J]$??

AccidentalFourierTransform

also, eq 22

0celo7

@ACuriousMind I'm not familiar with tempered distributions, but do they always admit some sort of derivative?

AccidentalFourierTransform

6:42 PM

yes

$T'[f]\equiv -T[f']$

0celo7

Is that actually a tempered distribution?

AccidentalFourierTransform

yes because of the minus sign

0celo7

what?

AccidentalFourierTransform

$T'[f]\equiv \color{red}-T[f']$

it cannot diverge weakly if you multiply by $i^2=-1$

0celo7

I don't think you're making sense

ACuriousMind

6:44 PM

@0celo7 All distributions that are defined on differentiable test functions are differentiable simply by defining $\delta'(f) = -\delta(f')$.

AccidentalFourierTransform

have you heard of the Euclid-Weierstrass theorem for temepered distributions?

0celo7

@ACuriousMind I know about on test functions, I'm not asking about that.

ACuriousMind

Then I don't know what you're asking about

0celo7

I'm asking if anyone in this chat has actually verified that $T'\in\mathscr S'$ if $T$ is.

ACuriousMind

Oh, you're asking whether the derivative of a tempered distribution is tempered?

0celo7

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.

ACuriousMind

Oh, I see what you're worried about and I say that I happily leave that to people who actually enjoy analysis :P

0celo7

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??

Sir Cumference

@JohnRennie I could write up something... :)

ACuriousMind

@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

0celo7

@ACuriousMind I'm blaming the whole field for this atrocity.

Sir Cumference

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

0celo7

@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

ACuriousMind

Dear lord, how did this question go unprotected for so long?

Emilio Pisanty

@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

0celo7

> Assume now that g is a suitably well-behaved function.

Emilio Pisanty

there were something like five obvious occasions

0celo7

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.

Emilio Pisanty

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

0celo7

I don't know about you, but I definitely glow

maybe you should use some moisturizer

Emilio Pisanty

@0celo7 a proper radiant glow?

have you checked whether you're pregnant?

0celo7

Just the usual American food baby.

@AccidentalFourierTransform After some thought, it should be $$\int_\Bbb{N}a_n\,\mathrm d(\mathrm{count})(n).$$

ACuriousMind

@0celo7 Locally, it holds for all analytic functions

0celo7

7:05 PM

@ACuriousMind I meant reciprocal, not inverse, apologies.

confusing notation above

ACuriousMind

Then that's also true, cf. math.stackexchange.com/q/42591/143136

0celo7

@ACuriousMind Ideally it should work everywhere

ACuriousMind

Well, it can't because $1/g$ is undefined where $g$ is zero, no?

0celo7

Obviously, so take $g$ nonzero

So like what about $x^2+x+1$

Is $1/(x^2+x+1)$ analytic?

ACuriousMind

@0celo7 then the second answer in the thread I linked says it works everywhere

0celo7

7:08 PM

Ah, I did not see the second one

AccidentalFourierTransform

$$\int f(x)\mathrm d\lceil x\rceil$$

0celo7

@ACuriousMind nice google fu

AccidentalFourierTransform

rather standard notation tbh

0celo7

What is that supposed to be?

AccidentalFourierTransform

$\sum _n f(n)$

0celo7

7:10 PM

Mine is actually correct though

$$\delta(-\mathrm i\partial_x)\delta(x)=\frac{1}{2\pi}$$

Slereah

What is this madness

0celo7

This is the new $\mathrm e^{\mathrm i\pi}=-1$.

AccidentalFourierTransform

that is disgusting indeed

0celo7

Is this a troll paper?

AccidentalFourierTransform

$\mathrm i$ instead of $i$

ewwww

2

Slereah

7:13 PM

have you ever seen the Bogdanoff papers, @0celo7

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

Slereah

The Bogdanoff brothers are two dudes who managed to get a physics PhD by publishing complete nonsense

0celo7

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?

Jaime Gallego

AccidentalFourierTransform

in mathOverflow

0celo7

7:16 PM

@Slereah what university?

Slereah

Université de Bourgogne

ACuriousMind

@0celo7 That's a pure math question :)

0celo7

@ACuriousMind Pure physics math

Emilio Pisanty

@0celo7 that's horrifying

0celo7

@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.)

Emilio Pisanty

7:18 PM

"If you assume <insert Cthulu-level horror>" is not a good way to start

0celo7

Hey, this is physics

the stuff they do in QM is not really much worse

Slereah

Dialetheism

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...

0celo7

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

Emilio Pisanty

@0celo7 what on Earth is $\delta(i\partial_x-y)$?

0celo7

@EmilioPisanty Ah, well. That's a good question.

Emilio Pisanty

7:21 PM

also, formula is wrong

0celo7

You write $\delta(x)$ using the usual Fourier integral

then you substitute in $\partial_x$ for $x$

@EmilioPisanty why?

Emilio Pisanty

@0celo7 automatic consequence of the use of $\mathrm i$ instead of $i$.

0celo7

@ACuriousMind can I call him names?

ACuriousMind

Why all the hate for upright $\mathrm{i}$ here? :/

Slereah

$\imath$

is the proper latex symbol

Emilio Pisanty

7:23 PM

@Slereah is not

ACuriousMind

@0celo7 If you don't mean actual names like "John" or "Rupert", no

0celo7

Who the hell is called Rupert

Emilio Pisanty

@0celo7 uh... this guy?

Jaime Gallego

Murdoch

Danu

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.

Slereah

7:24 PM

0celo7

$$Z[J]\propto \mathrm e^{\mathrm i\int J\phi}\delta\left(\mathrm i\frac{\delta}{\delta\phi}-J\right)\mathrm e^{\mathrm iS[\phi]}.$$

ACuriousMind

Rupert Giles

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,...

0celo7

^new way to compute path integrals

ACuriousMind

Enough Ruperts for you? :D

Emilio Pisanty

seen here on his ride:

0celo7

7:25 PM

@ACuriousMind fine

Emilio Pisanty

apparently also this guy

DI Rupert Lestrade

0celo7

This is pretty amazing, I can't even lie

Emilio Pisanty

@0celo7 what, the existence of four different people named Rupert?

0celo7

$$\int_\Bbb Rg\,\mathrm dx=2\pi \,\delta\left(\mathrm i \frac{\mathrm d}{\mathrm dt}\right) g(t).$$

Emilio Pisanty

The trick is that they're actually all the same person

@0celo7 maybe some \mathopen{}ness there?

0celo7

7:29 PM

what's that

Emilio Pisanty

$$\int_\Bbb Rg\,\mathrm dx=2\pi \delta\mathopen{}\left(\mathrm i \frac{\mathrm d}{\mathrm dt}\right) \mathclose{}g(t).$$

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

0celo7

This paper claims it does. I think the whole thing is nonsense.

Danu

That's some ridiculous identity

0celo7

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.

Emilio Pisanty

where presumably $e^{-k\partial_x} f(x) = f(x-k)$?

0celo7

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

Emilio Pisanty

@0celo7 well, that can presumably be extended into something that vaguely makes sense?

0celo7

@EmilioPisanty Perhaps. But they try to apply it to path integrals

So in the end nothing is defined :D

Emilio Pisanty

@0celo7 ah, well, then that's a situation I usually model as "the faster you jump a sinking ship, the better"

0celo7

It was a good distraction from multipole expansions

Emilio Pisanty

@0celo7 why would you veer off from physics into nonsense?

0celo7

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

Emilio Pisanty

yah, well, mathematicians do tend to get sore when physicists' manhandling of the math turns out to work just fine

AccidentalFourierTransform

you mathematicians should be proud

0celo7

@EmilioPisanty Not sore as much as bewildered

AccidentalFourierTransform

you made something very robust

Emilio Pisanty

@0celo7 "we're not sore about it" is not the oldest lie ever, but it's close to

0celo7

7:41 PM

Ok, I'm not sore.

Emilio Pisanty

dammit, dupehammer still takes me by surprise

someone double-check this

0

Q: Shor's Algorithm vs. No-Communication Theorem

I 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 -->...

quantum-mechanics quantum-computer

0celo7

@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

Emilio Pisanty

@0celo7 what people really mean by that is that the equation is only true in the distributional sense

0celo7

Maybe I'm missing something

@EmilioPisanty Even in the distributional sense, what does it mean?

ACuriousMind

@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.

Emilio Pisanty

7:48 PM

@0celo7 essentially, that you're only meant to do the $k$ integrals last

0celo7

@ACuriousMind The $x$ signifies the com of the measure, I have no problem with that.

Emilio Pisanty

$\delta(x)$ 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 $$\delta_0(f) = \int \delta(x) f(x) \mathrm dx = \frac{1}{2\pi} \iint e^{ikx}f(x) \mathrm dk \mathrm dx $$

0celo7

I think you want to interchange the measures.

Emilio Pisanty

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

0celo7

@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 $\mathrm dx\mathrm dk$.

Emilio Pisanty

7:52 PM

@0celo7 that's the thing

the notation formally says first $k$ then $x$

that's what bothers you

0celo7

So $\delta(x)=stuff$ means "multiply by $\frac{1}{2\pi}e^{ikx}$, integrate by $x$, then integrate by $k$."

Emilio Pisanty

@0celo7 no

0celo7

Then what does it mean?

Emilio Pisanty

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"

0celo7

That's exactly what I just said

Emilio Pisanty

7:55 PM

@0celo7 ah, yes

0celo7

I guess you can switch the integrals because $f(x)e^{ikx}$ is indeed $L^1$

Or is it

AccidentalFourierTransform

it is iff $f$ is

Emilio Pisanty

@0celo7 we can switch integrals because we're happy to restrict our hypotheses to whatever is needed for the manipulations to make sense

AccidentalFourierTransform

that was fun to say out loud

0celo7

@EmilioPisanty the manipulations making sense and being correct are different things

Emilio Pisanty

7:58 PM

@0celo7 for the manipulations to be correct

whatever

0celo7

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

ACuriousMind

8:15 PM

Hm, I know of adding bundles, but what is $TY - 3\mathcal{O}$ supposed to mean for a line bundle $\mathcal{O}$?

AccidentalFourierTransform

oh, that is the Gell-Mann-Nishijima formula for bundles, right?

ACuriousMind

@AccidentalFourierTransform :P

0celo7

I can't tell if he's trolling

AccidentalFourierTransform

@0celo7 that is exactly my intention

ACuriousMind

@0celo7 He is

AccidentalFourierTransform

8:20 PM

@0celo7 did you just assume my gender!? I never said I was a he...

0celo7

Yeah I assumed your gender

there are no girls (a) on the internet (b) in science

AccidentalFourierTransform

are there attack helicopters on the internet? and in science?

0celo7

if you sexually identify as an attack helicopter you might have a mental issue

ACuriousMind

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

AccidentalFourierTransform

I... how... you were not there when I said that

ACuriousMind

8:23 PM

On the other hand,

0celo7

he's an AI

ACuriousMind

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

0celo7

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.

ACuriousMind

@AccidentalFourierTransform That's what you think

AccidentalFourierTransform

I dont know what to think anymore

skullpetrol

8:26 PM

::makes popcorn::

0celo7

ACM is a magician

ACuriousMind

::throws rabbit into audience, then vanishes::

AccidentalFourierTransform

:: JR was in the audience, grabs rabbit, mercilessly kills it and eats it raw ::

2

0celo7

I'm sad that my translation operator answer has zero votes

It's incredibly correct :(

skullpetrol

Did you say "sexually identify"? :-O

ACuriousMind

8:29 PM

@0celo7 That's it - it's so correct no one believes it ;)

@AccidentalFourierTransform ...your fantasies disturb me

0celo7

Someone just upvoted it <3

skullpetrol

Of all the identifications why did you choose that one?

AccidentalFourierTransform

@ACuriousMind its a new meme Im trying to introduce here

JR enjoys killing animals

skullpetrol

Where is The Evidence?

0celo7

::gets triggered::

skullpetrol

8:33 PM

::self-bans::

0celo7

Why can't people just give me the hamiltonians?

I don't want to figure out CGS units

$2\pi$ or not $2\pi$

4

ACuriousMind

punny

0celo7

@ACuriousMind It works better in German

well, half-german

ACuriousMind

Hmm, you're right. I pronounced it the German way in my head

AccidentalFourierTransform

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

weird

0celo7

8:39 PM

@ACuriousMind $m_e$ or $m_\mathrm e$ for the electron mass?

AccidentalFourierTransform

1

0celo7

you're a troll, we don't listen to you

ACuriousMind

@0celo7 The latter if you're being fancy, but I normally don'T care enough to do that

0celo7

@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

AccidentalFourierTransform

why dont you set $a_0=1$? Im serious, use the proper (natural) units!

0celo7

8:46 PM

I don't know what a natural unit is

AccidentalFourierTransform

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, $m_e=\hbar=a_0=1$

0celo7

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

Emilio Pisanty

-7

A: On this infinite grid of resistors, what's the equivalent resistance?

This 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?

0celo7

-7 Jesus

AccidentalFourierTransform

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.

ACuriousMind

8:58 PM

@EmilioPisanty Nope, it doesn't answer the question.

Emilio Pisanty

@0celo7 to set the record straight, that's not actually natural units, the proper name is atomic units

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