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@Qmechanic Thank you.

@ACuriousMind about the question here

actually never mind about the factor of two

is $h^{2,1}-h^{1,1}>0$ always?

because I thought the $-3$ case was also considered or something

@NiharKarve not sure off the top of my head, but from the general form of the argument the number of generations would just be $\lvert h^2- h^1\vert$ if that can happen

@ACuriousMind AFAIR the only restrictions on the moduli is that $h^{1,1}\ge1$ and $h^{2,1}\ge0$. Do you think $n_G = \lvert h^{2,1}- h^{1,1}\vert$ then has something to do with mirror symmetry?

@NiharKarve No, it's just that a negative number for $h^2 - h^1$ means that the surviving effective generations come from the "anti-generations", not the "generations"

given any case where the difference is non-zero, you could always swap to the case where it has the opposite sign simply by renaming what you call the generation and what the "anti-"generation, right?

I see

thanks

If $g$ is a symmetry of the Hamiltonian $H$ and $\psi$ is an eigenstate, then $g\psi$ must be an eigenstate as well. I know this to be true, but is it formalized in some famous theorem?

@B.Brekke It's just a straightforward consequence of the symmetry commuting with the Hamiltonian, I don't think it has a special name

@ACuriousMind I just came across Wigner´s theorem, it looks related, but I might be wrong

Wigner's theorem just says that a symmetry will be represented by an (anti-)unitary opreator that commutes with the Hamiltonian

Hello Mars..

Hi Guys..

@Slereah Why "Was"? Are the scandals now over? Did they finally send him to prison?

how do we even show this?

Isnt this just.....trivial? the definition of a derivative?

@satan29 it's definitely not part of the definition of a derivative!

whether it is trivial or not is subjective :P

@ACuriousMind i mean yes, but

a constant vector f--->something that doesnt change with t---->derivative wrt to t is zero

what more rigour do you need lmao

ah, it is certainly intuitive, but that's not a proof

oh okay so

also, you only did one direction there (necessary)

but now that I think about it - what is the definition of "constant" here?

it might not be part of the definition of a derivative, but how do you define "constant" if not as "the time derivative is zero"

@ACuriousMind I mean thats exactly what my problem is

if you define it as the latter (which seems to be the case Always), you are....done

you need to ask whoever gave this problem to you what the definition of "constant" is, then

@ACuriousMind what did you mean by this?

@satan29 your arrows only go one way, but "A is necessary and sufficient for B" means "A if and only if B", not just "A if B"

but since I agree now that the question doesn't make any sense that's not really relevant anymore :P

@ACuriousMind oh

I mean this was a question at an engineering school , and I thought rigour is non existent there

then came this gem

well, I'd say that's one more data point for "rigour is non-existent there" :D

I have a function $x(t)=Ae^{-\gamma t/2}e^{i\omega_1 t}=f(t)e^{i\omega_1 t}$ for $t\geq 0$ and $0$ otherwise. I'm looking for the Fourier transform (FT) of this function, that is $X\left(\omega\right) = \int_{-\infty}^{\infty}{x\left(t\right)e^{-i\omega t}dt}$. I have not studied a lot of FT. Any ideas on how to tackle this? I've been looking at 103 here (en.m.wikipedia.org/wiki/…).

@schn look at 205

your "zero for $t< 0$ and non-zero for $t\geq 0$" is essentially multiplying $x(t)$ with the Heavyside function

I was considering this too, however, the exponent in 205 is not complex (as it is for me), right?

Although probably that gets absorbed by the other $e$.

And so one can maybe use 205.

well, you can of course combine it with other things like 103

just set $f(t) = \mathrm{e}^{-\gamma t / 2} \theta(t)$ (for $\theta$ the Heaviside function)

Right.

So $\nu$ is then $\omega-\omega_!$, or?

The $\nu$ as given in the table.

@ACuriousMind Actually, I'm calculating a power spectrum, so $S\left(\omega\right) = \left|\int_{-\infty}^{\infty}{x\left(t\right)e^{-i\omega t}dt}\right|^2$. Do you know what is meant by the width of $S$ denoted $\Delta \omega$?

@schn typically such widths are fwhm

Hi, what are peoples thoughts on Eric Weinstein?

@DanielAdams lol

Claiming to have some great theory and refusing to even publish it is really stunning stuff

I think he write it up this year apparently.

But agreed I think he takes his gripe with the current system a bit too far

Definitely a bit much claiming his brother and his wife deserve nobel prizes