« first day (2599 days earlier)      last day (2347 days later) » 

7:00 PM
@EmilioPisanty The color segment changes color depending on whether it's a git repo or not
and the status of the repo
 
hmmmm.
so
 
(Getting a screenshot)
 
there's a ~/.vim_runtime/.git/ folder?
and in that screenshot the terminal is at ~/.vim_runtime/ ?
 
@EmilioPisanty Exactly
Here's an e.g.
 
vzn
lol popsci articles on physics breakthru (shrilly) equated with "vandalism + spam"... o_O :P
 
7:03 PM
Ah, I also use tmux
So it's alacritty running tmux running zsh
But I'm moving to fish
 
@vzn That doesn't make popsci. That's a regurgitation of hype straight from a university PR office. If you want to post links here, post links to pieces that actually talk about science.
 
@vzn If I were you I'd just stop posting links here, the governance clearly isn't capable of admiring your art and hard work and willingness to share
 
A one-time screw-up doesn't rate. But this is consistent, repetitive behaviour that has been pointed out to you as problematic and unwanted on this chatroom by multiple people, with detailed explanations of why they're problematic that you've systematically refused to discuss.
So yes, that makes it spam and vandalism. Please stop it.
 
@EmilioPisanty I have to say, the deal breaker maker for me were the completions
 
If you're not willing to spend five minutes of your time out of consideration for a communal space, then leave that communal space as is.
 
7:06 PM
Let me show you god
 
@BernardoMeurer deal breaker?
completions?
you sure you don't mean deal maker?
 
For leaving bash
 
@BernardoMeurer does it do hg as well as git?
 
7:07 PM
@EmilioPisanty Let's try, do you have an hg repo I can clone at hand?
 
@BernardoMeurer hmmmm lemme see
have you got a bitbucket account?
 
@EmilioPisanty I just created one
And yes it works beautifully!
Check it out
 
hmmmm, nice
it looks like it's a little bit too liberal with adding whitespace
but yeah, I am interested in them config files
 
@EmilioPisanty Alrighty, I'll post them on gist for ya
 
@BernardoMeurer long time no see
 
7:13 PM
@0celo7 My body has been claimed by alcoholism
 
:(
 
@BernardoMeurer cheers
I'll see whether they take on my linux box
how much sudoing do they take?
 
@EmilioPisanty Nothing, just a couple packages. I can craft an install cmd for you. Are you on Ubuntu?
 
@BernardoMeurer it's a managed machine and I don't have sudo
but yeah, it's ubuntu
which packages are they?
 
Ah, then check that you have zsh and tmux
That's all you should need
 
7:15 PM
cool
I'll ping IT on Monday if I have time
 
You will need powerline fonts
That can be installed locally
 
if I wanted to install them system-wide, how does that work?
 
And if you want alacritty you can also install it locally
I'm not certain, but it must be documented in that GH page
This is also a good reference:
 
do your scripts run on the standard ubuntu terminal?
 
What scripts?
 
7:19 PM
configs
 
Ah, yeah, they do :)
 
I just like alacritty because it's faster
All they need is a terminal that supports 256 colors, which most of them do
 
whaddaya mean, faster?
how is the standard terminal slow?
 
It renders faster, literally
It's real slow
alacritty uses the GPU for the text rendering
 
7:20 PM
I guess I'll ask IT to install the thing and see if I notice a difference
I suspect I won't
 
So if you're not printing asynchronously
the speed of your code is actually dependent on the speed that your terminal can push out lines
 
yeah, well, then again I don't code on the terminal
do I look crazy?
=P
 
Well, neither do I most of the time
But most of my code runs on the terminal
e.g. hemoglobin
 
oook
I suspect your requirements on that kind of front are just generally higher than mine
 
It's not really that, I just like to complicate things
2
 
7:30 PM
hello
 
@BernardoMeurer don't we all
but you do do that more than the rest of us
 
@EmilioPisanty I think it's mostly because I get bored a lot
Like now, I'm kind of bored when I'm not at the bar
So I'm re-writing all my scripts from bash to fish
For no good reason
I actually have something I'd like to build
A short program to setup my dev environment
I did it once but it's painfully out of date
I don't even use atom anymore
Nor fish
I have to change kernels, this ck patch-set is not working as I expected
 
8:05 PM
Indian food all day today at the beach yeeeeehaaaaa
 
Anonymous
@Cows What type of Indian food?
 
@EmilioPisanty are you around, by chance?
 
@heather yes
what's up?
 
so, i've been working on that problem of calculating the matrix for spontaneous parametric down conversion
 
@heather Don't trust him it's an imposter
 
8:16 PM
and i think i'm doing it way wrong, because no matter what i do, it's inconsistent.
so, there are three cases that i know of.
1: the pump photon starts as horizontally polarized, and produces a signal and idler photon which are each vertically polarized.
 
1 hour ago, by Emilio Pisanty
@BernardoMeurer does it do hg as well as git?
that should be a plenty-strong credential, shouldn't it?
 
Lol, proof it's actually him indeed
 
2: the pump photon starts out vertically polarized, and produces a signal and idler photon which are horizontally polarized.
 
@Blue chicken tikka massala. Lol i think that's what its called hehe
 
3: the pump photon starts out in the state 1/sqrt(2)|H> + 1/sqrt(2)|V> and produces signal and idler photons that are each in the state 1/sqrt(2)|H> + 1/sqrt(2) e^(i phi)|V>
 
Anonymous
8:19 PM
@Cows I see. That sounds really tasty
 
@JohnRennie making Jambalaya round 2 tomorrow
 
@heather ooooof
 
@EmilioPisanty ooooof indeed
 
no
as in oooof, you're doing this wrong
 
oh.
how so?
 
8:20 PM
so, first thing
 
i mean, i know it's really that there's $0|P>_H + 0|P>_V +...$
 
if process A can happen and process B can happen, then the response to inputs A+B is the sum of the responses
it's a bit more complicated in nonlinear optics
but with the formalism you're on, it's still linear
so there's no point in describing 45° polarizations if you already did H and V pumps
 
i'm a bit confused by what you mean.
 
@heather what I mean is that quantum mechanics is linear
how much does that do to clarify things?
 
well, nonlinear optics isn't linear
so that seems like there might be a bit of a problem.
 
8:24 PM
@heather indeed it isn't
 
is that what I'm doing wrong?
treating nonlinear optics as linear?
 
what that means is that the hamiltonian has changed
 
ehm...i don't suppose that can be represented by a matrix?
 
i.e. linear optics has a hamiltonian of the form $H=\hbar \omega a^\dagger a$
the NLO you're doing has an additional term in the hamiltonian
of the form $a_1^{(\dagger)}a_2^{(\dagger)}a_3^{(\dagger)}$
 
Anonymous
@Sid You there?
 
8:26 PM
where the bracketed daggers may or may not be present depending on what process you're describing
 
I hate physics, by gods
 
Sid
@Blue about to sleep. Sorry, I couldn't be here today at the scheduled time. Something came up
 
and two of the bosonic operators might coincide if it's a degenerate process (like SHG seen as degenerate SFG)
@heather it can
 
Anonymous
@Sid Oh, did you watch lectures 3 and 4?
 
Anonymous
They're pretty short
 
8:27 PM
it's the matrix elements of that hamiltonian with three bosonic operators
 
Anonymous
Just 30 mins in total
 
Sid
@Blue Haven't watched 4 yet. Will do that tomorrow
 
@EmilioPisanty sorry, what is a bosonic operator? the $a^\dagger$ things?
 
Anonymous
@Sid Okay. That's just 8 mins. See you tomorrow then
 
Anonymous
Goodnight
 
8:28 PM
@heather yes.
 
Sid
Gnight.
 
the creation and annihilation operators, $\hat a$ and $\hat a^\dagger$
 
learned something new.
so is $H = \hbar \omega a^\dagger a$
 
(like all good quantum physicists, I oscillate between using hats on operators and not. sorry about that.)
 
saying that a particle is being created and another annihilated?
 
8:29 PM
@EmilioPisanty What do hats do?
 
@heather that is the standard hamiltonian for a single mode of light
 
Does QM have Winter Bash too?
 
@heather oooof. that's a tricky statement.
 
see, i now know what all the individual things mean in that, but i don't know what they mean put together.
 
@BernardoMeurer they indicate the fact that the quantity with the hat is an operator
 
8:29 PM
"Make over 300 students fail their midterm to get a hat"
 
@BernardoMeurer!
 
@DanielSank What? :P
 
Get me more hats!
 
Hi, @heather.
 
8:30 PM
@DanielSank how?
 
I dunno...
 
@DanielSank hello
how are you?
 
Pretty good.
 
'cause there's some help I could use to get some hats I want
 
8:31 PM
I will help people get hats
 
Well, technically same here. I need two votes on an old answer to get that awesome universe hat!
 
yooha
 
I owe DS for safeguarding my stereo setup
@DanielSank Link
 
...but I'm pretty sure you guys already voted on it.
 
@DanielSank my thoughts exactly
 
8:31 PM
Some jerks downvoted it though. Without those downvotes, I'd be there!
 
1 more dinny dan
 
though I'm not in the running for the universe hat
 
@DanielSank Edit the answer
 
@DanielSank what answer?
any answer?
 
zomg zomg I only need one more!
 
8:32 PM
@DanielSank unfortunately, I cannot help you any further in pursuing that hat
 
@BernardoMeurer Why?
 
Daneil give me something intelligent to comment I'll bump it into active...
 
@EmilioPisanty I know.
 
@DanielSank bump it
 
@DanielSank It will bump it
 
8:32 PM
@heather This one.
 
boom
 
did you get your hat?
 
Ohhh, so you guys are recommending I get even more shameful than I was already being? ;-)
@CooperCape What boom?
 
You got 100 voties
 
@DanielSank Yep
 
8:33 PM
OH SNAP
 
Hat may take a while.
 
=)
 
@DanielSank I would really like that cosmic brain and galaxy brain hats
they would go mighty well with my ellipses
 
@EmilioPisanty Yeah, me too.
 
8:34 PM
so @EmilioPisanty, I need to find the matrix elements of the hamiltonian with three bosonic operators?
 
Lemme see your profile...
 
I got cosmic from an answer I wrote in 1 minute...
 
@DanielSank 0_o
 
@heather exactly
 
That is an awful answer.
 
8:34 PM
@heather :-D
 
I hate hats
they involve work
 
They invoke $$\int F ds$$
2
 
more specifically, for SPDC, you're looking for the matrix elements of $a_i^\dagger a_s^\dagger a_p$
 
@EmilioPisanty hoo boy, okay. thank you, i will google now =)
 
@CooperCape lol
 
8:35 PM
I'm terrible aren't I...
 
i.e. creation operators on the signal and idler, and an annihilation operator on the pump
 
the annihilation of the idler and signal and the creation of the pump
oh, it's backwards
nvm.
i was a little confused =)
 
@EmilioPisanty SPDC?
 
where $a|n⟩ = \sqrt{n}|n-1⟩$
 
@DanielSank Super Pink Desk Chair
 
8:36 PM
@DanielSank spontaneous parametric down-conversion
 
Learn physics brah
 
@DanielSank spontaneous parametric down conversion
 
Got it.
 
Ahh that old classic
 
and $a^\dagger |n⟩= \sqrt{n+1}|n+1⟩$
 
8:36 PM
@heather are you trying to violate Bell's inequality or something?
 
no, build a quantum computer =)
 
@DanielSank Nika's shipment company did it
 
o_O
 
preferably without needing a dilution refrigerator or a lot of money
 
@BernardoMeurer HAHAHAHAHAHAHAHAHA
 
8:37 PM
@DanielSank :P
 
I am laughing my head off
 
(so, @heather, first exercise: what do those two actions combine to in $H=\hbar \omega \, a^\dagger a$?)
 
Call Martinis, we found how to violate it
Call NASA
 
That was one of the best jokes I've seen on this chat for a while.
 
@EmilioPisanty before that, i'm a bit confused - what is |n>?
 
8:38 PM
Eigenstate of number of excitations in harmonic oscillator.
 
^ that
does that make it clearer?
 
Ironically, that post on which I was just shamelessly asking for votes is relevant to this...
 
here 'number of excitations' $\equiv$ 'number of photons'
 
Context: I had three large, heavy boxes stuck in Lisbon with my ex-gf for a couple months. They had a combined value of >$1000. She arranged for shipment from Lisbon to Santa Barbara, while SB was on fire. The packages went from Portugal to Dan's lab in one day and weren't taxed
 
8:39 PM
huh. $\equiv$ is identical to the smartphone 'menu' button
 
@EmilioPisanty That button is in fact just LaTeX
 
@DanielSank done
 
$$\underbrace{\hat{a}^\dagger \hat{a}}_{\hat{n}} |n \rangle = n |n \rangle$$
 
^ @heather but it's also useful to work it out piece by piece
 
i know what an eigenvalue/eigenvector is; what is an eigenstate?
er, nvm, i'll look it up
 
8:40 PM
@EmilioPisanty You got the hat?
 
@heather same thing as an eigenvector
 
okay
 
in a context where vectors are states
 
Eigenstate means: "A vector representing a quantum state, that is an eigenvector of whatever operator we're talking about"
 
@heather They just like to make up words to trick students
 
8:41 PM
^
 
i can believe that
 
similarly, if you're in a context where your vector space is a space of functions, and each vector is a function, then you use the term eigenfunction
 
Just like "whom", it's a made up word to trick students
 
zomg I'm getting so many votes.
You guys are the best.
Hats!
 
@BernardoMeurer for this specific example, I find that it increases clarity rather than not
 
8:41 PM
I read a book where it replaced the word 'spin' with 'gyre and gimble' for electrons cause it had as much physical meaning.
 
...and the momeraths outgrabe...
 
Bit of a Lewis Caroll fan are we?
 
@DanielSank did you vote on a post of mine?
 
@CooperCape Not really.
 
Faur
 
8:42 PM
@EmilioPisanty Not yet...
 
understandable.
 
@DanielSank ah
 
me neither
sigh what am I upvoting now
 
I did shamelessly edit answers at score 9 and 24
 
okay, yeah, i'm still really confused about |n>
 
8:43 PM
and they got upvoted
 
@EmilioPisanty Hahahaha
 
but I didn't get either the badge or the hat
 
When I got cosmic brain took like 5 mins to come through
Cause badges aren't instant either
 
i need to take an actual physics class so i know what people are talking about =/
 
turns out that I already had the badges on those answers, and presumably someone unupvoted them
 
8:44 PM
Woah, Emilio you got the accepted answer on that question about dimensionless quantities going in exponents and logs.
Nice.
 
ah well. so an eigenstate of the number of photons in a harmonic oscillator
 
@heather exactly
a.k.a. a state with a well-defined number of photons
 
Emilio, just yesterday I learned something about dimensions in exponential functions that blew my brain so hard I had to get up from my desk and go to the office kitchen and think about my entire life.
2
 
@EmilioPisanty as opposed to states with an ill-defined number of photons.
 
8:45 PM
@heather for your purposes right now, though, $n$ is either zero or one
@heather exactly
as weird as that might sound
 
@EmilioPisanty for the record, that was mildly sarcastic =P
 
you can have states like $|0⟩+|1⟩$
that don't have a well-defined number of photons
 
@EmilioPisanty wait what
 
@heather I know. I understand. QM is not exactly the easiest thing in the world to grok.
 
Anonymous
8:46 PM
@DanielSank And that is? Sounds interesting
 
so does |0> mean it's ill defined and |1> that it's defined or something like that?
or am i barking up the completely wrong tree?
 
@heather |0⟩ means that there are precisely zero photons in the beam
|0⟩+|1⟩ is a Schrödinger's-cat state that has both zero photons and one photon at the same time
 
and |1> means there's precisely |1>?
 
@heather yes
 
and... oh
 
8:47 PM
@heather exactly ;-)
 
what if n > 1?
is that not allowed?
 
@heather yes
it's allowed
 
$n>1$ can happen for bosons, not for fermions, just FYI.
 
but if you're working with single photons, then you want to stick to the $n=1$ layer ;-)
 
Just bumped your helium answer to +25, Emilio. Nice answer.
 
8:49 PM
Did Phase delete his account again...
 
@DanielSank thanks
=)
I agree with your comment
I'll try to look for a reference
 
@Blue Essentially, decay time constants have an implicit dimension of radian attached to them.
 
13 mins ago, by Emilio Pisanty
and $a^\dagger |n⟩= \sqrt{n+1}|n+1⟩$
 
I hate the word 'implicit'
 
14 mins ago, by Emilio Pisanty
where $a|n⟩ = \sqrt{n}|n-1⟩$
 
8:51 PM
@DanielSank No they don't
 
Anonymous
@DanielSank radian is a dimensionless quantity, no?
 
okay, so...the annihilation operator applied to the number of photons makes the coefficient sqrt(n), which in the case of 0 or 1 just stays the same, and then removes one from n in the |>, which makes sense
 
Anonymous
Could you elaborate?
 
Ok @Blue and @Slereah, let's go through this.
 
8:52 PM
and then vice versa in the |> of the creation operator, which makes sense...but why would it become, say, sqrt(2) for n = 1?
 
I have a resonator with a finite value of $Q$, the quality factor. Do you know what that is?
 
Anonymous
@DanielSank Yup
 
Anonymous
Go on
 
Ok actually, maybe what I'm going to show is that $Q$ has dimensions of radian.
 
and then, let's see. $H=\hbar \omega \, a^\dagger a$
 
8:53 PM
@heather exactly
 
so really $H = \hbar \omega \sqrt{n+1}|n+1\rangle \sqrt{n}|n-1\rangle$
so, uh, $\sqrt{n(n+1)} = \sqrt{n^2 + n}|n+1\rangle|n-1\rangle$
well, an annihilation with a creation operator, so is it
 
In culver city now. Going to hit up some restaurant and do comp anal there
 
$H = \hbar\omega\sqrt{n^2+n}|n\rangle$?
@EmilioPisanty
 
@Cows Ayyy I'll be driving past there in a few days
 
Anonymous
@DanielSank I don't know what "dimensions of radian" even means. Afaik $Q$ is $2\pi$ times the ratio of the total energy stored divided by the energy lost in a single cycle or equivalently the ratio of the stored energy to the energy dissipated over one radian of the oscillation. But sure, go ahead
 
8:57 PM
@Blue @Slereah suppose I want a certain frequency detuning $\delta \omega$ to be equal to $$\delta \omega = \frac{\omega_0}{2Q}$$ where $\omega_0$ is a resonance frequency.
@Blue Right! It's [energy stored]/[energy lost per radian] which has dimensions of radian, sort of.
 
@heather no
you can't magically make two kets like that
 
Continuing with my example, it is the case that $Q = \omega_0 \kappa$ where $\kappa$ is the energy decay rate of the oscillator.
So, we get $$\delta \omega = \frac{\kappa}{2} \, .$$
 
you start off with $a|n⟩=\sqrt{n}|n-1⟩$
and then you apply $a^\dagger$ to that
 
This is a very confusing equation.
 
$a^\dagger a |n⟩ = \sqrt{n} a^\dagger |n-1⟩$
 
8:59 PM
Suppose I know that $\kappa = 1/500 \text{ns}$
 
Anonymous
@DanielSank No. It is [energy stored per radian]/[energy lost per radian]
 
@davidphysics nice! It's a pretty chill area. If you want to say hi on your way don't feel shy hehe.
 
but then, the $\sqrt{n+1}$ factor belongs to a different $n$
can you see the connection now?
 

« first day (2599 days earlier)      last day (2347 days later) »