The h Bar

1:43 AM

@grand_unifier Are you using "lecturer" in the sense of junior, full-time, college faculty (what an American would call an assistant professor) or in the sense of someone whose main job is to teach at the college level (a teaching professor)?

I don't really think there is such a thing as a "theoretical particle physics teaching professor", you're just a teaching professor with a background in theoretical particle physics (but you generally teach everything).

For the former, the critical thing is to get positive attention from well-know people (other than your advisor and mentor) in the subfield while you are a PhD student and post-doc.

You have to be a leader. You have to be seen and heard. You have to do work which is built on sound foundations and shows both skill and creativity (even if it is wrong, though it is better to be right).

@grand_unifier The APS does surveys. For physics in general about 1 in 7 PhDs end up as full time faculty at "mid career". I think it is a little lower for particle physics.

Mikhail

5:51 AM

Naveen Balaji

6:24 AM

Does anyone know any book(s) which contains detailed info regarding black hole embeddings?

bolbteppa

8:33 AM

When you do an integral in momentum space in QED, how in the world can you actually do integrals to $\infty$ when the momentum can never be higher than the speed of light, or does that make sense

Mithrandir24601

9:10 AM

@bolbteppa velocity can never be $>c$. Momentum includes $\gamma$, so automatically has this restriction included (although now you mention it, I don't actually recall seeing $\gamma$ anywhere in QFT...)

bolbteppa

Oh yeah, clearly need a break if I'm forgetting gamma's :p

Mithrandir24601

@bolbteppa Or convince yourself that absentmindedness is the perfect property to be a physicist :)

Kiro

9:45 AM

Stupid question, but why exactly are optical parametric processes called "parametric"?

Slereah

2:09 PM

@Mithrandir24601 QFT uses the proper Lorentz transform :p

Although you could technically write it using $\gamma$

Since $\gamma = \cosh(\chi)$

with $\chi$ the rapidity

Mithrandir24601

@Slereah fair - it's been over 2 years since I did QFT, so I've long forgotten :/

Slereah

QFT generally just uses the exponential map of the Lorentz group

So you can do the conversion from $\exp(\omega_a \tau^a)$ to $\cosh(\xi)$ to $\gamma$ if you're feeling like it

The rapidity is best for theory but the Lorentz factor can be more practical for experimental things

Semiclassical

2:38 PM

Rapidity is to the Lorentz factor as angles are to slopes, basically

Slereah

pretty much

vzn

3:24 PM

@grand_unifier this has colorful bios of many top physicists of the 20th century, inspiring but daunting at same time amazon.com/Age-Entanglement-Quantum-Physics-Reborn/dp/…

Secret

We master the field of the imcomprehensible in order to turn all that is comprehensible into the incomprehendable

Those who resists will be dissolved into the formless unknown

For the absolucidity of Slereah Spacetime opens the new bpath to the new world, and the old reality that is left behind to be dissolved into The Formless

curare.tumblr.com/post/89580615077/…

Semiclassical

3:48 PM

An idle observation: curare is a toxin which produces weakness of the skeletal muscles and, at high enough dosage, causes death by asphyxiation due to paralysis

So curare=“death by paralysis”

Secret

it is also one of the first analgesics introduced to america in history if I recall

o btw, this is cool:

Two Vortex Rings Colliding in SLOW MOTION - Smarter Every Day 195

►

Semiclassical

yeah

that's pretty cool

the spontaneous generation of the little red-blue clumps is my favorite part of that I think

Secret

Yeah same, it's just so sublime looking at them

AccidentalFourierTransform

chat sessions are back?

or did anything important happen?

enumaris

weez

John Rennie

3:59 PM

The chat sessions have been comatose for a while. Just like us really :-)

Semiclassical

snore

Anonymous

Let's chat something...lol :P

Semiclassical

I could ramble on about bipartite correlations, but I suspect the only person I'd be satisfying would be myself :P

Anonymous

Oh, see, Terry is here! :P

Semiclassical

@Blue oh, did you see my comments re: electron spin precession

Anonymous

4:02 PM

@Semiclassical Which portion?

Semiclassical

I think that's the simplest (though presumably not the only) analogue of the transition matrix in the photon case.

Anonymous

I didn't manage to read the whole conversation yesterday

Terry Bollinger

@Blue Hi Blue, all!

Anonymous

Checking

Semiclassical

kk

Terry Bollinger

4:04 PM

@Semiclassical so hey, what are they at least? The name is intriguing...

enumaris

I typed a partial answer to Blue's GR question yesterday...then someone else answered it and I deleted my answer XD

Semiclassical

I think I'm dubious that the M-Z interferometer is a useful example of non-classical behavior---or, at least, I'm dubious that one can take that as 'obvious'

Anonymous

19 hours ago, by Semiclassical

So that precession apparently eliminates all your information about up and down.

Anonymous

This part? ^ @Semiclassical

Anonymous

19 hours ago, by Semiclassical

I want to say that the point goes like this: If you measure an electron as spin up on the z-axis and then apply a magnetic field in the x-direction for a quarter period, then you're as likely to have Sz=up as Sz=down

Terry Bollinger

4:05 PM

@JohnRennie zzzzz.... snort... hmm?... Oh, Hi John!

Semiclassical

Right. Though for context I was specifically referring to a quarter-precession of the spin

oh, you linked that

Yeah. I think that's the simplest electronic instance of $U=I+i \sigma_1$

Anonymous

19 hours ago, by Semiclassical

But if you were to have done two such precessions to begin with, then that amounts to flipping the spin

heather

::yawns:: morning everyone =)

Anonymous

What do you mean by "flipping the spin" tho?

heather

or afternoon or evening or whatever greeting is appropriate for your time zone =)

enumaris

4:07 PM

mornin

Anonymous

evening

Semiclassical

$U^2 |\uparrow\rangle=|\downarrow\rangle$ and $U^2|\downarrow\rangle = |\uparrow\rangle$ up to phase factors

Anonymous

Ooh, okay

Anonymous

Reading on

Anonymous

19 hours ago, by Semiclassical

But, eh, that doesn't really seem so different from a coin flip. If you could track it in mid air, and always measure it when it's in the same orientation it started with, you'd be guaranteed to get the same measurement you started with.

Anonymous

4:09 PM

19 hours ago, by Semiclassical

But if done twice (prior to measurement) it's guaranteed to reverse the original measurement

Semiclassical

To clarify that just a bit: Suppose you didn't have a small coin, but a big heavy disk

with a bit of practice, you could consistently toss it in the air in such a way that it rotates at a constant rate

If you then time it so that it consistently lands face down (assuming it starts face up in your hand) then your transitions are up -> down and down-> up

Anonymous

@Semiclassical I'm not sure I understand this statement. What is done "twice"?

Anonymous

@Semiclassical That makes sense, yes

Semiclassical

if you instead timed it so that it landed on an edge, then it'd be basically a fifty-fifty chance

I guess I also need to insist in the first scenario that it makes exactly a half-rotation

e.g. not two-and-a-half flips

anyways. since I said it was turning at a constant rate, the time I should wait in order for it to land on its edge is half the time if I want it to land face-down

enumaris

woooo finally getting my alienware

Terry Bollinger

4:17 PM

Ah well... given the lack of any response by anyone to my attempt at a friendly reply to @Blue, I'll take "Oh, see, Terry is here! :P" as (somewhat perplexing) genuine animosity and take my leave now from h-bar... probably permanently, since I only come here for fun. You had no idea you had such power, 'ay, Blue?

Anonymous

@Semiclassical Agreed. So the electron spin example doesn't differ with the coin tossing example, at least in this case

Semiclassical

Yeah, it doesn't seem like it

That doesn't mean there isn't a difference, but it's not obvious to me

enumaris

o.o

Anonymous

@TerryBollinger Umm? It was just meant as a joke! (since your arrival is an indication that chat time has begun)

Anonymous

Don't take it seriously, lol :P

Anonymous

4:20 PM

Feel free to initiate a conversation @TerryBollinger. At the moment, I and SemiC are continuing a previous discussion on the difference between a qubit and a coin

Anonymous

@Semiclassical BTW does the loss of determinism due to a measurement of the qubit count as a difference in your opinion? I guess it doesn't for exactly the reason AFT mentioned

heather

@TerryBollinger i don't think anyone was intending any animosity. sorry if the lack of response came across that way - let me try again: hello and welcome! =)

Semiclassical

I'm not sure tbh

heather

please do stick around.

AccidentalFourierTransform

is AFT me?

what did I mention?

Anonymous

4:24 PM

@AccidentalFourierTransform Yes! ;)

AccidentalFourierTransform

btw, this guy is weird: physics.stackexchange.com/…;

Semiclassical

My main thing is that saying that "you can take the square root of a NOT gate" as a touchstone of quantum behavior seems sorta silly since (at the level of outcomes) the scenario that DW gave in his answer seems entirely explicable

Anonymous

121

Q: What makes a theory "Quantum"?

Say you cook up a model about a physical system. Such a model consists of, say, a system of differential equations. What criterion decides whether the model is classical or quantum-mechanical? None of the following criteria are valid: Partial differential equations: Both the Maxwell equations ...

AccidentalFourierTransform

oh

is that question still a thing?

it was a rather silly question

Semiclassical

Eh, I think it's a good one

Anonymous

4:27 PM

@Semiclassical I'm not exactly sure if it's possible to replicate square root of NOT as a transition matrix for a coin though (probably missing something important here)

Semiclassical

I think my correlation rambling might provide a useful example, though, of how the two-qubit case works

@Blue well, what's the goal exactly? If it's to produce the same probabilities/outcomes as quantum does, then I think the 'tracked coin flip' experiment gives the same ones as the square-root of NOT would

hmm, though...hmmmmmm

one way in which an electron spin definitely differs from a coin is the the whole "you need to rotate by 4pi to get back the initial state" thing

AccidentalFourierTransform

what is this meta guy doing here?

shoo

enumaris

what's weird about that guy

Semiclassical

he's meta, therefore beyond us

@Blue I guess my issue is whether or not the transition matrix as such can be directly measured. The transition probabilities can be, certainly

Anonymous

@Semiclassical Ah, yes, as far as outcomes are concerned (for the quantum case we simply consider the square of amplitudes as probability aka 2-norm) while for the classical case we take 1-norm. So as far as probability of different outcomes are concerned, they match up

Semiclassical

4:35 PM

Right.

Anonymous

@Semiclassical I don't think it can be....we can only find the probabilities afaik....

Anonymous

But....

Semiclassical

That's what I'd tend to say as well.

Anonymous

I honestly had gotten a bit excited seeing DW's answer. But he doesn't really touch upon the main point i.e. if we can't measure the transition matrices directly then there's really no way to distinguish it from a coin

Anonymous

Maybe I should leave a comment about this

Semiclassical

4:37 PM

One thing I will note, in relation to @Mithrandir24601's point about HOM-dip

in that context, the significance of the dip is that it's purportedly twice as large as would be expectedly classically

Novalium Company

Evening'.

Semiclassical

so in that context it would seem that the analogy with a coin flip should break down

(lemme source that quick)

"
A classical analog to the HOM effect occurs when two coherent states (e.g. laser beams) interfere at the beamsplitter. If the states have a rapidly varying phase difference (i.e. faster than the integration time of the detectors) then a dip will be observed in the coincidence rate equal to one half the average coincidence count at long delays. (Nevertheless, it can be further reduced with a proper discriminating trigger level applied to the signal.) Consequently, to prove that destructive interference is two-photon quantum interference rather than a classical effect, the HOM dip must be l…

from the wikipedia page

If one could translate that classical bound to the coin flip experiment, then that might be persuasive.

I'll note, though, that the HOM dip is still a two-photon effect

So that doesn't contradict the idea that you need at least two qubits to see non-classical probabilities

Mithrandir24601

@Semiclassical Exactly

Semiclassical

if I were to guess, the issue is that a single qubit space is just too small

and that you need to have a bit larger Hilbert space in order to see something interesting.

I'm not entirely confident on this, though.

yeah, definitely not confident enough :P

enumaris

hmmmm

looks like Spacy is not built for GPU optimization at inference time...means it will still take quite a while to do analysis on my 3M+ comments...

Semiclassical

4:51 PM

That's a lot of comments.

enumaris

yep...

but at least the new Alienware will be a hugely upgraded processor...from a i5 2.5Ghz quad core I'm working on now to a i7 3.6Ghz 16-core :D

I just hope the security team hasn't bloated it to death by installing every security software known to man...

Semiclassical

Fingers crossed

Secret

5:07 PM

0

Q: Is it possible to make a reasonably strong electromagnet at home?

I wish to add a rope/thread in my shorts. The traditional way to do this is to put the rope (along with a pin because a pin is easier to grab) inside the waist of the shorts and pull it (the pin) all the way around. However, that method is slow and boring. The way I did it is using strong ne...

electromagnetism electricity magnetic-fields electric-circuits everyday-life

This is a very "strange" question

Semiclassical

Well

the question itself isn't so strange

the context, however...yeah, that's pretty goofy

Secret

yeah

enumaris

any ferrous material works

just put a chunk of iron

wrap it with wires

plug in wires to outlet

watch as the short blacks out your neighborhood

Semiclassical

lol

enumaris

get electrocuted and die - worth it for science

Semiclassical

5:09 PM

Sounds about right.

hmm, brain fart

Anonymous

@Semiclassical Why did you insist it shouldn't be like "two and half spins" though ?

Semiclassical

Suppose I have a qubit with density matrix $\rho$ and two observables $A,B$. I can compute the expectation values as $\langle A\rangle = \text{tr }(\rho A)$ and $\langle B \rangle =\text{tr }(\rho B)$

@Blue because then 'half that' time isn't going to be a quarter of a rotation

the better example is one-and-a-half, though, since then half of that is 3/4 of a rotation

which still accomplishes the same thing, I guess.

so I suppose it really doesn't matter.

enumaris

wewt

computer here

Semiclassical

(cntd from above) To obtain $\langle A\rangle $, I'd repeatedly prepare the state $\rho$ and measure $A$; the mean value of the samples of $A$ would converge to $\langle A\rangle $, and similarly for $B$.

Here's what I'm forgetting: If for each trial I instead measure $A$ and then measure $B$, what does the mean value of these sample products converge to?

It won't be $\langle A B\rangle$ in general, since $AB$ isn't Hermitian unless $A,B$ commute

Anonymous

5:25 PM

I'm a bit confused how that matters. I thought we're trying to constrain a coin to have the transition matrix:
$$\left(\begin{array}{cc}
1/2 & 1/2 \\ 1/2 & 1/2
\end{array}\right)$$

Semiclassical

@Blue yeah, I don't think it actually matters.

Anonymous

We just need it to land on an edge :)

Semiclassical

right

it won't matter which edge

Anonymous

Yup. But lol, DW might say "coins can't land on edges" XD

Anonymous

I'll say sufficiently thick disc instead

Semiclassical

5:27 PM

heh, yeah. for a more 'realistic' setup, one probably would take the initial orientation of the coin to have a small uncertainty

So something like: Pick a number in $(-\epsilon,\epsilon)$, then add 1/2 and round to the nearest integer.

You're just as likely to get an even result (0) as an odd result (1).

If you add 1/2 twice, though, you're definitely going to get an odd result.

(I guess one doesn't need to do even/odd here, but I have in mind taking the results mod 2 so as to capture the whole "two flips = no change" idea)

A simpler version of this would be to take a spinner and allow it to rotate through some angle

Which I'm pretty sure I've seen used in discussions of Bell's inequality, come to think of it...

Anonymous

The main problem in this answer is that, although the transition matrix in the case you mentioned i.e.

$$U=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}
1 & i \\ i & 1
\end{array}\right)$$

doesn't occur in the classical case, it doesn't mean the *effects* of the transition cannot be replicated for a classical coin. **After all, the transition matrix cannot be measured directly**. All we can measure is the outcome probabilities!

Take the transition matrix $U$. It basically says:

If you begin in the state $0$ the probability of ending up in the state $|1\rangle$ is $\frac{1}{2}$ (and $|0\ran…

Anonymous

Does this summary look okay?

Semiclassical

sounds right, though it's a bit long for a comment

Anonymous

I'll break it up into two or more :P

Semiclassical

heh, okay

To be clear, my mental splitting of the events goes like this

T: Wait for the coin to make a quarter rotation

M: measure whether the direction vector of the coin has z>0 or z<0

And the point is that doing T,M gives 50-50 odds while T,T,M guarantees that you get the opposite result you started with

maybe Q for quarter would've been better, lol

in which case Q^4 = 1

oh well

Anonymous

5:51 PM

On a side note: One of AFT's points struck me as interesting: "Conversely, you don't need complex numbers to formulate standard QM"

Anonymous

63

Q: QM without complex numbers

I am trying to understand how complex numbers made their way into QM. Can we have a theory of the same physics without complex numbers? If so, is the theory using complex numbers easier?

quantum-mechanics history complex-numbers

Semiclassical

well, you don't need complex numbers to have $z^2=-\text{id}$

the matrix {{0,-1},{1,0}} works just fine for that

Anonymous

True, true

Semiclassical

of course, if you want some analogue of {{1,i},{-i,-1}} you'll presumably need 4-by-4 matrices instead

in which case you're doing gamma matrices

Anonymous

BTW

Anonymous

5:56 PM

What do you mean by "fundamental"? Mathematically, it's because we have to describe quantum mechanics using a richer mathematical structure than classical (as proven by this square root of not, not that this gate is particularly special: you can replace the X by any stochastic matrix with a negative eigenvalue). In terms of physics, well physics is just the working theory that describes experimental outcomes (such as square root of not). In terms of some underlying explanation of why the world is the way that it is, who knows? — DaftWullie yesterday

Anonymous

"we have to describe quantum mechanics using a richer mathematical structure than classical "

Anonymous

Is it really richer than CM, I wonder

Semiclassical

I can sorta give evidence for that, in the context of Bell's inequality

Anonymous

Okay? Like violation of CHSH?

Semiclassical

But my problem with such reasoning is that what Bell's inequality shows to be forbidden is not 'hidden variables' as such, but local hidden variables

I think so, but that one is a bit tedious.

For a relatively simple one, take the case of two spatial observers with access to identical measurement devices

specifically, let me take them to be Stern-Gerlach devices oriented in the xy-plane at 120 degree angles

Mithrandir24601

6:02 PM

@Blue I've seen it happen! To make things more ridiculous, the guy tossing it predicted it would do just that. It was so funny, I still remember it about a decade later

Semiclassical

I'll denote the corresponding observables as $S_1,S_2,S_3$

Mithrandir24601

If only one of us recorded it :/

Semiclassical

One can then show that, if the observers each receive half of a singlet state, then $\langle S_i^1 S_j^2\rangle = 1$ if $i=j$ and $-1/2$ otherwise

Anonymous

The Impossible Coin Toss

►

Semiclassical

(note that $\cos(120^\circ)=-1/2$, so this is just a claim about the dot products of the measurement axes)

Mithrandir24601

6:05 PM

@Blue This was the first attempt though :P

Semiclassical

where the 1,2 superscripts are for the observer

I guess maybe it's easier to just have the observables of Alice be A1,A2,A3 and the observables of Bob be B1,B2,B3

Anonymous

@Mithrandir24601 Yeah, some people can make coins lands on the edge at will :P I've seen one or two such instances too (but they were mostly Rs. 5 coins - a bit thicker than normal)

Semiclassical

blah, I have that wrong: it's $\langle A_i B_j\rangle = -1$ if $i=j$ since A,B must measure opposite results if they both measure the same component in the singlet state

and similarly $\langle A_i B_j\rangle =1/2$ if $i\neq j$

also, the results of either observable is normalized to be $\pm 1$

With that in mind, suppose we wanted to assume the following: When Ai,Bj are measured, the other 2*2=4 observables still have meaningful $\pm 1$ outcomes

so each measurement is really some result like $(A_1,A_2,A_3,B_1,B_2,B_3)=(+1,+1,-1,+1,-1,+1)$ but we only learn about two of the six values.

First off, we note that whenever $A_1$ and $B_1$ are measured at the same time, we get opposite results. same for $A_2,B_2$ and $A_3,B_3$

on that basis, I'll postulate that $B_i=-A_i$

Novalium Company

Guys, for an electric field to form, 2 charges must be present right? Can you have an electric field of only one charge?

Semiclassical

Second, one can check that $\langle A_i\rangle =\langle B_i\rangle =0$

So if $(A_1,A_2,A_3)=(+1,-1,+1)$ is a possible outcome, then $(A_1,A_2,A_3)=(-1,+1,-1)$ must occur with the same frequency.

Secret

6:18 PM

1 charge is sufficient, it's just form a radial field

Semiclassical

(I'm not labeling the $(B_1,B_2,B_3)$ outcome now since I'm assuming it to be superfluous)

Secret

And if you include dynamic electric fields, even 0 charges can do that via faraday

Semiclassical

@NovaliumCompany A single charge produces an electric field. But if there's no other charge in the system, then this field won't have anything to act on.

Novalium Company

@Secret $F = k\frac{q1 * q2}{r^2}$. How can you have a force on only one charge? I mean, the equation $E = \frac{F}{q}$ depends on the force that requires 2 charges?

Semiclassical

The field is still there, it's just not doing anything interesting.

$F_{12}=k q_1 q_2/r^2\implies E_2= kq_2/r^2$

i.e. the field produced by $q_2$ only depends on $q_2$

the point is that, regardless of what $q_1$ is, the electric field it interacts with will be the electric field produced by $q_2$

Novalium Company

6:22 PM

Ahh, got it. Actually, I'm going to read a book on electrostatics and magnetism and then return if I have any questions. Because I've watched all Khan's videos but don't remember most of it, so I'll read.

Semiclassical

and since this field doesn't depend on what $q_1$ is, it's taken to be present even if $q_1$ isn't

Novalium Company

Yep, makes sense.

Semiclassical

that said, you can only ever detect an electric field by putting a test charge into the system and seeing what force it is subjected to

Novalium Company

Ok, so I guess I'll see you later :) My summer holiday started today, I finished 9th grade with straight A's so I'm finally free to read some stuff that I'm not supposed to know yet xD

Anonymous

Congrats. Good time to learn new stuff

Semiclassical

6:29 PM

getting back to what I was saying: given what I've said, it suffices to consider the outcomes $(A_1,A_2,A_3)=(-B_1,-B_2,-B_3)=(+1,+1,+1),(+1,-1,-1),(-1,+1,-1),(-1,-1,+1)$. the other four outcomes for $(A_1,A_2,A_3)$ with opposite signs occur with equal probability.

Since I'm getting rid of B's, I might as well extend this to the expectation values: $\langle A_i A_j\rangle =\langle A_i (-B_j)\rangle 1$ if $i=j$ and $-1/2$ otherwise

So now I've got three random variables $A_1,A_2,A_3$: they have zero mean, unit variance, and pairwise covariance $-1/2$.

But if that's true, then consider the expectation value of $\langle (A_1+A_2+A_3)^2\rangle$.

if you expand by linearity, you get $$\langle (A_1+A_2+A_3)^2\rangle = \langle A_1\rangle^2+\langle A_2\rangle^2+\langle A_3\rangle^2+2\langle A_1 A_2\rangle^2+2\langle A_2 A_3\rangle^2+2\langle A_1 A_3\rangle^2=1+1+1-1-1-1=0$$

and since $(A_1+A_2+A_3)^2\geq 0$, the only way for the expectation value to vanish is if $A_1+A_2+A_3=0$ almost always. (i.e. with probability 1)

but the outcomes I gave only have $A_1+A_2+A_3=3,1,-1,-3$. As such, none of the outcomes with determinate values for $A_1,A_2,A_3$ can ever be consistent with this.

Hence we conclude that the outcomes cannot be of this form.

ugh, typos. second equation in the long line should have $\langle A_i A_j\rangle$ and $\langle A_i^2\rangle$

(I can also do a version of this where, instead of the two observers each having the same three measurement devices, one observer has one measurement device and the other has the other two.)

(blah blah blah blah blah)

Sid

8:11 PM

Argentina qualify in world cup!!

Also, Maradona is a joke these days. The drugs have really taken a toll on him

Avnish Kabaj

8:31 PM

@Sid phew

ACuriousMind

8:44 PM

@Blue "richer", much like "fundamental", is one of these words that don't really mean anything :P

enumaris

man...

ACuriousMind

@enumaris ?

enumaris

Was gonna complain about something

but then I forgot

lawl

ACuriousMind

It can't have been very bothersome, then :P

enumaris

I'm just setting up my new computer atm

so probably just something about the difficulty of set up

heh

ACuriousMind

8:52 PM

If it is already put together and boots, the hard part is done ;P

enumaris

well IT bought me an alienware

so certainly it's put together and boots...but installing CUDA was a huge pain lol

They also bloated it up the wazoo with security software so there's a bunch of issues there too

now I'm running into issues opening SQL server 2014

vzn

9:22 PM

in theory salon, 16 secs ago, by vzn

Real-Life Schrödinger’s Cats Probe the Boundary of the Quantum World / quanta mag

enumaris

anyone here familiar with visual studio?

specifically it's use w/ python...

ACuriousMind

I didn't even know visual studio works with Python :P

enumaris

indeed it does

but I'm unable to figure out how to install a model...within a module...

uh...

ACuriousMind

I don't know what "install a model" means

Lots of things I don't know :P

enumaris

9:42 PM

figured it out :D

i have to open the powershell

and install it from there

makes sense I guess...

I'm a genius

now I gotta get my SQL fixed...

ACuriousMind

@enumaris SQL is demon spawn, and no self-respecting software developer should ever use it.

enumaris

lol

unfortunately it is not my choice

I am just a lowly data scientist

I don't manage the database

it is simply a fact of life

ACuriousMind

You don't have to tell me, I'm being paid to work in a language that basically has SQL embedded into it :P

enumaris

lol

I'm now having them reinstall SQL...and then...I should be set

whoop, should be set

success!

danielunderwood

10:39 PM

Anyone happen to have any tips for studying from books? I made it through undergrad alright using books as a companion for courses, but now having troubles trying to learn from book only. Do people take notes while reading? Problems only and go back when needed? I figure my next attempt will be to write down a list of topics and see where I get without any book, but didn't know if there was a better way

@enumaris if your company is willing, pycharm will probably give you a much better time than vs for python. At least that's how it's been in my experience

enumaris

11:18 PM

I'm used to running stuff on command line...

My company is all about Microsoft

bolbteppa

@danielunderwood youtube lectures

heather

^

my method for learning a topic in an ideal case is to watch a video/lecture on a topic to get an idea for the intuition, look at the book for the formalization, and then practice on exercises from the book/whatever i find online.

generally, though, it's more roundabout.

taking notes always helps too, for video, text, and exercises.

enumaris

ughhhhh wtf, why is python running so slow on VS

what...the crap is this....running the script on command line takes 10 seconds, running the script in VS doesn't end after 5 minutes...nothing changed in the script...wuuuut

forget visual studio

baaaam the GPU deep learning training has begun!

what took my laptop 3 hours takes this desktop 7 minutes loool

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