The h Bar

3:18 AM

@vzn just take that tree and call it a weird pendulum system. Easy!

@danielunderwood lol? maybe understand your joke but not sure. anyway there is a lot of crosspollination between Computer Science + Physics in topics in here sometimes. the interdisciplinary/ crosscutting elements are some of my favorites, its showing up in my longterm research, and shows up eg in cutting edge stuff like quantum-computing etc :)

John Rennie

6:14 AM

@pZombie if a single coil generates a field $B$ then $n$ coils each generate a field $B$ and they sum to give a field of $nB$.

ayc

6:30 AM

@JohnRennie could you join the pss-room!

John Rennie

@ayc I'm in the PSS room ...

ayc

@JohnRennie 15 min...i have to leave...!.

ill come back

John Rennie

OK see you later

Rumplestillskin

8:19 AM

@CaptainBohemian thanks mate. I was thinking something more along the lines of Feemi-transport and local tetrad but I’ve realky no idea!

Slereah

8:47 AM

morning

Jake Rose

10:33 AM

Can anybody help with this?

Part b is the bit I’m struggling eith

When I state the first law should by $\Delta Q =0$?

undefined

how is this hat on top of letters called?

$\mathbf{\hat{z}}.\mathbf{\hat{r}}$

Jake Rose

Like that

it doesn’t work particuarly well for some reaons

undefined

11:07 AM

never mind, it's called Circumflex

and has different meanings in different mathemathical topics

Jake Rose

11:26 AM

Oh I thought you meant call it as in get it for latex

yeah, most common use in physics is operator algebra

undefined

thanks :)

GodotMisogi

1:17 PM

@JakeRose I'd think unit vectors would be the most common use of hats/circumflexes

@undefined Notational advice: Use \cdot: $\hat z \cdot \hat r$

undefined

1:40 PM

@GodotMisogi thanks for that advice

Emilio Pisanty

@Blue Do you know when the HNQ update will be posted on Meta? Has it been posted already?

undefined

1:59 PM

greetings @EmilioPisanty I would like to thank you for your nice answer to the question "why-do-atoms-emit-a-certain-colour-of-light". Really impressive

you inspired me to read much more about that

Avnish Kabaj

Hey I think I remember you

Did you get about learning the basics again?

undefined

2:14 PM

@AvnishKabaj hello :) yes, indeed, I have to admit I skipped some steps again :/

Avnish Kabaj

Heh heh :P

undefined

I cant help, those topics are just too fascinating for me

PranshuKhandal

@rob will you keep the answer at physics.stackexchange.com/a/464303/221932 closed?

Emilio Pisanty

@undefined happy to hear it

@PranshuKhandal The question itself is likely to be deleted automatically, unless it gets fixed and reopened.

398

Q: How does deleting work? What can cause a post to be deleted, and what does that actually mean? What are the criteria for deletion?

What circumstances can cause a question or answer to be deleted, and what does that actually mean? How can a post be deleted? When can't I delete my own post? Can I still see my post even after it's deleted? Can I see a list of my deleted posts? How can I undelete one of my posts? What does de...

I would suggest letting that answer go, and focusing on answering questions which are not off-topic

undefined

@EmilioPisanty regarding to your sentence "(Basically, it requires you to discretize a certain differential operator into a big matrix, and then to diagonalize that matrix.)" - Do you know a link or paper in which one can see such a calculation?
I already googled, of course but without success

Avnish Kabaj

2:46 PM

Big Win For Open Access, As University Of California Cancels All Elsevier Subscriptions, Worth $11 Million A Year

Emilio Pisanty

@undefined If you go directly to a paper, then it will either be modern and intensely technical, or pretty ancient and very hard to understand.

You should look for textbooks on atomic physics or quantum chemistry

@AvnishKabaj already on the star board ;-)

PranshuKhandal

@EmilioPisanty thanx for help :) but is the problem in question or answer?

Avnish Kabaj

@EmilioPisanty whoops

Emilio Pisanty

@PranshuKhandal The problem is the question. But more specifically, it's that the question is so bad that we do not want it to get answered. It encourages the asker, and others who see it, to keep posting bad homework questions.

This one isn't quite as bad as others

but it's still pretty bad, I should think

undefined

@EmilioPisanty yes, I guess you are right. I'm just curious how such a calculation would look like and to see which parts are all involved in it.
I started to read a paper called "A new multiplet table for Fe I" which was one of the only ones I found about this topic. But it doesn't seem to include the way described by you

Emilio Pisanty

2:56 PM

@undefined as I said, you're unlikely to find suitable introductory material in a paper

undefined

@EmilioPisanty I know and it doesn't need to be introductory. Even if I wont understand all of it, it's still fascinating for me and I would just like to see/read/learn. But it's alright, I dont want to bother you further. Thank you for your answers :)

Emilio Pisanty

Try e.g. chapters 6 and 7 of archive.org/details/…

undefined

aw thank you!

PranshuKhandal

thank you from me also!

Emilio Pisanty

no worries

If you're going to do a deep dive, I should point out that the rough way I pointed out in my answer is typically only a part of the picture

if you've got a multi-electron atom and you want a reasonable approximation to the eigenenergies that's as simple as reasonably possible, then you're probably using Hartree-Fock theory

this does involve diagonalization

but it's also a self-consistent theory

this means, basically:

1. you produce some guesses for the orbitals (which will typically be as a superposition of gaussians of different sizes)

2. you calculate the hamiltonian in that basis, including single-electron effects via single-electron matrix elements, as well as multi-electron effects included at the mean-field level, i.e. you replace the inter-electron-interaction two-electron operator with a single-electron operator based on the charge density of the existing orbitals

3. you diagonalize that hamiltonian

4. you update the charge densities that make up that hamiltonian based on the eigenfunctions you just got

5. you diagonalize again

and then you keep going until it converges (or fails)

Hartree–Fock method

In computational physics and chemistry, the Hartree–Fock (HF) method is a method of approximation for the determination of the wave function and the energy of a quantum many-body system in a stationary state. The Hartree–Fock method often assumes that the exact, N-body wave function of the system can be approximated by a single Slater determinant (in the case where the particles are fermions) or by a single permanent (in the case of bosons) of N spin-orbitals. By invoking the variational method, one can derive a set of N-coupled equations for the N spin orbitals. A solution of these equations...

undefined

3:17 PM

thank you very much, I really appreciate that you take your time to write that.
Maybe that sounds a bit silly but you almost made my eyes wet by talking to me about stuff like that. If someone takes his valuable time to write me something like this, I feel valued and taken seriously. And this means a lot for me, thank you

pZombie

4:08 PM

@JohnRennie There is no doubt that more of the same coils will generate a stronger field but here i am arguing a single much bigger coil, taking up the same space as n smaller coils. In fact, since the smaller coils are usually insulated, the bigger coil taking up the same space should contain more copper. More copper means more electrons. So again, why would i use many smaller rather than one big coil if all i wanted to was to generate a large magnetic field with DC current?

John Rennie

@pZombie the field at the centre of your loop is $\mu_0 I/(2R)$, so if you have $n$ loops the field is $\mu_0 I/(2R)$. So to get the same field you could use $n$ loops with a current of $I$ or one loop with a current of $nI$.

You want to keep the current as small as possible because large currents mean large resistive losses in the rest of your circuit, so it's better to use many loops and a small current than one loop and a big current.

Bearing in mind that the number of loops will typically be in the hundreds, maybe even thousands, the different in the currents can be very large.

pZombie

@JohnRennie Current is the number of electrons going through a surface area per second if i am not mistaken. So of course, since my large coil has even more than n times the surface area the smaller coils would have, the current would also have to be more than n times that

@JohnRennie there has to be another reason why many coils rather than 1 bigger coil is used when trying to create a strong magnetic field using DC current. It cannot just be that you would have to use n times the current going through the larger coil. That is just logical since the surface area of the larger coil is also more than n times that

in the end, both with the n smaller coils or the one big coil you should end up pushing the same amount of electrons through a total surface area per second, requiring the same amount of work

John Rennie

@pZombie no it doesn't.

The n coils fill a toroidal region of basically the same cross sectional area as a single wire. A bit large, due to the voids between the wires, but not much larger.

probably_someone

@pZombie I'm confused as to what you mean by "a large coil" and "multiple smaller coils." It seems like what you're actually trying to compare are two finite solenoids of equal length and radius, where one is made of wire with a larger cross-section and has fewer turns, while the other is made of wire with a smaller cross-section and has more turns. Is this correct?

pZombie

4:23 PM

@probably_someone by large coil, i mean a single turn with a thick copper wire which takes up the same space as n coils with thinner wire

probably_someone

@pZombie "n coils" = "n turns"?

pZombie

@probably_someone since all i need in order to create a magnetic field is for electrons to run around in a circle, a single big wire should be enough rather than many smaller ones.

yes, "n turns". In the case of the larger wire, just 1 turn

that came out wrong. In order to create a magnetic field associated with n turns

we could forget about n turns and just compare a thick copper wire vs a copper wire with many insulated smaller copper wires inside bundled together

John Rennie

@pZombie if you ask any electrical engineer they'll tell you that working with large currents is more hazardous than working with small currents. Keeping the current small is the main reason for using multiple coils not one big coil.

pZombie

i would think that the voltage is more important. I can touch an ultracapacitor capable of sending many thousand amperes, but if it is only 12 volts then it won't send any dangerous current through my body. Same reason why i can touch both poles of a car battery even though a car battery can send many hundred amperes with the right setup

similarly, if my one big coil was driven by 12 volts, there should not be any danger for me to get toasted. Not any more danger than running 12 volts through n turns

John Rennie

@pZombie the power dissipation in the rest of the circuit, i.e. the bits that supply your coil, is $W=I^2R$ where $R$ is the external resistance. Unless you use huge conductors for the rest of your circuit this means unacceptably high losses and resistive heating.

Jake Rose

4:35 PM

You defeated your own argument by saying if it’s only 12v. Your body resistance is on the order of 100kOhm thus you won’t get a large current through your body anyway

its fundamentally not about what current goes through the circuit when you’re not touching it,

It’s about what’s the current going through when you do touch it

which depends on the resistance of your body

pZombie

Jake Rose it seems you are replying to someone invisible. I was making exactly the same argument that because of only 12v, no matter how many amps a battery or other source can deliver potentially, there won't be any deadly current going through our bodies if we touch the poles because of our bodies' resistance not allowing for any high current to flow when only 12v are applied

Jake Rose

Having large currents is dangerous for numerous other reasons, such as fire hazards.

pZombie

yes, that could be why many turns might be safer

one smaller wire in the chain failing just cuts off the whole thing usually, before anything bad happens

Jake Rose

I mean yes, but it’s better to up the turns and reduce the voltage

pZombie

while with one big coil, if for some reason you end up with more current flowing through it than specified, it could lead to some serious fire

i am not sure you would have to up the voltage. The resistance of one large wire is much smaller, allowing for a higher current to flow when the same voltage is applied. While not sure, i would guess that the same magnetic field is generated when using the same voltage in both cases

Jake Rose

4:48 PM

B field depends only on the current

so unless the currents are the same you’ll have different fields.

better to reduce voltage for safety concerns and up the turns to increase the field

pZombie

no, when talking many turns, then the B field does not only depend on the current but also on the number of turns.

Akash. B

Hi

pZombie

longer thinner wire means higher resistance, so the same voltage would results in a lower current vs 1 thick wire. A higher current would flow in the thick wire at the same voltage but with the coiled up n times thinner wire you would generate n times the magnetic field

i would actually have to do the full math to figure out the exact field in both cases

Jake Rose

The maths is quite simple

I’m not talking about thickness of wires

in general, a smaller voltage will mean a lower current and thus less of a safety hazard. But to compensate to keep the field the same you need more coils

Hi @Akash.B

pZombie

5:15 PM

the current depends on the resistance of your wire and on if your source can send the required current at all. So you cannot really generalize it.

John Rennie

@JakeRose the resistance of n thinner coils will be n times greater than the resistance of a single coil containing the same mass of copper. So if we keep the voltage constant the current $I_n$ will be $I_0/n$.

The field if proportional to $nI$, so for $n$ coils the field is some constant times $nI_0/n = I_0$ or exactly the same as the field from one single coil.

So if we keep the voltage constant we get the same field from n thinner coils as we do from one fat coil.

pZombie

when considering AC voltage i read that in such a case it makes sense to use many smaller coils as in an attempt to increase the surface area. AC voltage is supposed to have the electrons traveling on the surface rather than within the wire. I don't understand why so i have to consider it as a given, which is why i specified DC

John Rennie

@pZombie that's the skin effect. At mains frequencies it isn't a big effect.

pZombie

5:32 PM

@JohnRennie yes, that was it, i remember now.

However, i don't understand the formula P = V*I in this regard. Since I seemingly would be much higher in the case of one fat coil, given the same voltage applied to both, the single fat coil and the multiple n coils with the thinner wire, it would appear that the fat coil would draw much more power, yet generate only the same field?

Maybe i am missing something here

IT makes no sense because in both cases the same amount of electrons are pushed through a total surface area per second

or maybe not, not certain about this statement

enumaris

weez

Sir Cumference

Ugh, I don't know how anyone is expected to learn QM at 9am

Captain Bohemian

5:58 PM

but there is a high possibility that a QM course is arranged at 9 am.

Sir Cumference

@CaptainBohemian It's the worst

I'll never understand why attendance is mandatory for these classes. As if I'm going to learn anything that early in the morning.

Would've been better if I could get some sleep and hit the textbooks later, but whatever.

pZombie

öl..................................

Captain Bohemian

@SirCumference is the attendance to this kind of course compulsory? It's almost always selective besides the exam days in my schools.

Sir Cumference

@CaptainBohemian Yeah, for some reason it's worth like 10%

I can't wrap my head around it

My guess is that very very few people would show up otherwise

Captain Bohemian

@SirCumference really?just attendance no matter whether you understand the course would earn you 10% score of the course?

rob

6:08 PM

@SirCumference US schools have attendance-reporting requirements attached to financial aid. If a student stops coming to their classes, the school can clawback money that has already been disbursed. This has led to a resurgence of attendance-taking in college classes.

Captain Bohemian

we have never had this kind of rule. For this kind of course, only actual understanding can earn sccore.

Sir Cumference

@CaptainBohemian Yeah, but the other 90% is purely exam questions.

Which is probably one of the most stressful grading schemes I've ever seen

@rob I know. And I think it is the dumbest thing I've ever heard.

As if I need to be told how to learn as an adult, like seriously

I'd be far more productive just using a textbook than listening to someone talk at 500 words per minute for 1.5 hours

rob

@SirCumference It's a global solution to a local problem (that is, a tiny fraction of students who might squander a term's worth of financial aid money).

Captain Bohemian

@SirCumference our professors for this kind of course have never taken students’ attendance; they don't want to spend time to do that kind of thing. They start the lecture as soon as they enter the classroom, so they never know whether you attend the course. So you can only attend the exam day of the course if you don't like to go to the course no matter for what reason.

Sir Cumference

@CaptainBohemian Lol dude, many physics courses here test your 8th grade algebra skills more than physics knowledge. More often I'm concerned about making a mistake in the ungodly amount of algebra than figuring out how to solve the problem

Welp at this point I'm just venting

rob

6:15 PM

There's also enough evidence that lecture attendance is strongly correlated with exam performance that encouraging students to attend is part of "encouraging students to succeed." Which is the kind of marketroid-pap that administrators say, but then turn into school-wide policies with consequences.

Captain Bohemian

@SirCumference don't know what is 8-grade algebra. We don't that kind of course called algebra in high school; we only have the courses called math and scientific math in high school.

Sir Cumference

@rob Personally I have never been able to learn something out of lecture. It's not a matter of the professors, it's simply that I cannot process new concepts meaningfully at that speed. I've gotten through almost all my classes by hitting the books and going at my own pace

Which indeed means I'm spending ~9 hours a day studying, but it's the only thing that happens to work for me

Classes just get in the way at that point

@CaptainBohemian Rearranging terms and such

JMac

I would generally just look up stuff I wanted to understand during lectures, usually trying to look at something related to the lecture topic

rob

9 mins ago, by Sir Cumference

I'd be far more productive just using a textbook than listening to someone talk at 500 words per minute for 1.5 hours

@SirCumference I assume that is an exaggeration. But I'm curious: are your fast lecturers Powerpoint users, or chalk-and-talkers?

Sir Cumference

@rob The latter

JMac

6:21 PM

holy crap, that reminds me of my heat transfer professor who had powerpoints made for every lecture; but they literally overlapped completely with the lecture before and after, you could attend every second lecture and only miss a couple anecdotes about each concept, it was great

Sir Cumference

Honestly, if I'm given a well structured textbook and all the hours in the day, I'll learn the material very strongly. I'm used to working for very long periods of time

But lectures just detract from that time

rob

@SirCumference Interesting. I find that the chalk-and-talk approach forces me to go a little slower and think on my feet with the students, while having Powerpoints makes me skim over things because they are printed on a slide already.

Sir Cumference

@rob Maybe it's just my ADD that makes it harder for me to gain anything out of lectures. But it does feel like some lecturerers here teach almost frantically

Either way, as long as I learn the material I can't fathom mandatory attendance

ACuriousMind

Man, just spent 40 minutes trying to get my computer to boot into a GUI again. I'm never trusting the "automatically install update before shutdown" option again :P

enumaris

6:37 PM

just work from command line

ACuriousMind

I didn't intend to work

(It's 7pm here)

Mithrandir24601

@rob black/white-board lectures were better than powerpoint lectures, definitely

pZombie

the c64 never needed an update. Wonderful silent computer without making any noise or catching dust inside and giving you an operating system within seconds of hitting the power button

since then, personal computers have only devolved

ACuriousMind

Updates are good. Updates that manage to break a dependency chain and install only half of the needed components are bad.

John Rennie

@ACuriousMind which distribution?

pZombie

6:43 PM

that's destined to happen at some point. You cannot prevent mistakes from happen at some point

Mithrandir24601

@pZombie Umm... I so thoroughly disagree with that statement that I don't even know where to start :P

ACuriousMind

@JohnRennie Debian stretch

But I think the issue was with the proprietary nvidia drivers. Reinstalling them (after breaking everything else :P) fixed it. Anyway, it's working again.

John Rennie

Ah, I've had that happen to me, but it was a lot of versions ago. Five years or so.

In the end I gave up, wiped the disk and reinstalled from scratch :-)

ACuriousMind

It was not reassuring when at some point in between trying to boot into gnome just made the console flicker ominously

pZombie

I try to make a complete image backup of my OS after cleanly installing with all drivers and software i use and have a separate partition for all my data, such that i can just apply the image whenever i manage to shoot down my OS

ACuriousMind

6:49 PM

@JohnRennie Yeah, I was about 10 more minutes away from that

probably_someone

@ACuriousMind Same issue happened to me on Ubuntu. Switching from X11 to Wayland seemed to fix things enough that I could reach the desktop

pZombie

with fast nvme SSDs those days, applying a backup image back onto your OS partition is a matter of seconds

John Rennie

huh?

ACuriousMind

Most of my data isn't on the same drive as the OS so wiping and reinstalling isn't that terrible

Sir Cumference

@Mithrandir24601 What I find is that for chalk lectures, I can't process what they say because I'm too busy getting it down in my notebook. Though in the end it's rather pointless since a textbook is more structured than my notes anyway

At least presentations are given afterward, so I can listen to what they're saying

Mithrandir24601

6:52 PM

@SirCumference Ah, we always had typed notes (to some extent) as well, which helps mitigate that

Sir Cumference

@Mithrandir24601 Rarely lecturers do that, although it is very appreciated :P

10% of the time they are very nicely typed LaTeX notes. 30% of the time they are mostly illegible hand written notes. 60% of the time you get nothing

HDE 226868

7:23 PM

One of my professors (the only one who lectures in class) is in the habit of spending the first 5-10 minutes writing on the chalkboards without saying anything and letting us copy some stuff down, then going across them and just talking, without much additional writing. It gives us the chance to jot the key bits down before he lectures. But then again, he's said that he's fine with us skipping class occasionally and just reading the textbook, sooooo. . .

Sir Cumference

@HDE226868 I envy you ;-;

HDE 226868

@SirCumference Well, that class is the exception, not the rule. . . I can't exactly skip seminars. :P

Student404Mus

7:51 PM

Why it is not enough to measure entanglement through von-Neumann entropy?

ACuriousMind

@Student404Mus enough for what?

Student404Mus

enough to quantify/measure enetanglement

because other means of measurement are entanglement cost and distillable entanglement

@ACuriousMind I mean what have entanglement cost and distillable entanglement to do with measuring entanglement?

ACuriousMind

8:07 PM

I'm afraid I have no idea what entanglement cost or distillable entanglement are, so I can't help you

Student404Mus

@ACuriousMind Ok, no problem. thanks for your giving attention

Jake Rose

@SirCumference yeah we have nicely typed lecture notes in latex, or maybe a PowerPoint type thing 100% of the time. Even in fourth year I believe.. I think I’m one of the lucky few

@JohnRennie ahhh I see

wait why is the resistance of n thinner coils greater than a single coil with the same amount of mass?

nevermind

Sir Cumference

@JakeRose I have had notes provided for like 8 classes ever. 2 of which were LaTeX

Sigh

Jake Rose

8:23 PM

@SirCumference I pray for you my good man

@JohnRennie I get that if you have a single wire, the. Split it up into N constituent wires that the resistance goes to $R/N^2$

where R is the resistance of the original wire

Emilio Pisanty

@Student404Mus well, each tells you something different about the state, right?

Jake Rose

therefore I should go up

I think I’ve misinterpreted what you meant

Emilio Pisanty

The definitions of those measures are pretty reasonable and transparent

And each has its uses depending on what you're trying to do

You'd hope for those measures to coincide

But they don't

So you're kinda stuck with a zoo of different measures

Ultimately, the core reason is that I the set of all bipartite states, the relationship "A can be obtained from B through LOCC operations" is not a total order

Student404Mus

but it says that they coincide with von-Neumann by a certain condition which I can't recognize now

Emilio Pisanty

That is, you can find pairs of states where neither can be reduced to the other (given a single copy)

@Student404Mus .... and that condition need not be true in all the cases of interest.

What is that condition?

What source are you reading?

Student404Mus

8:31 PM

An introduction to entanglement measures.
Martin B. Plenio and Shashank Virmani

Emilio Pisanty

@Student404Mus I can take a look but it'll take me a bit

Student404Mus

I don't want to waste your time, as long as I still reading. I try to learn more.

Emilio Pisanty

@Student404Mus I mean, presumably the answer to "why are there different entanglement measures" is covered in depth in the introductory chapters of that book

(which I can't find. if you want more detailed help you'll need to ask more specific questions.)

Student404Mus

Ah, good! you mean in other articles of that subject?

@EmilioPisanty I found something here quantiki.org/wiki/entanglement-measure

Emilio Pisanty

8:47 PM

@Student404Mus what?

oh, never mind

I assumed from the title that it'd be a book

@Student404Mus it says where, exactly?

Student404Mus

arxiv.org/pdf/quant-ph/0504163.pdf

Emilio Pisanty

¬¬

where within that paper?

Student404Mus

...

i think page 7

Emilio Pisanty

presumably this?

> This is indeed the case for pure state transformations where $E_D(\rho)$ and $E_C (\rho)$ are identical and equal to the entropy of entanglement

if that is the case, then the cases where the entanglement cost and the distillable entanglement coincide with the entropy of entanglement are indeed too narrow to be really useful

the theory of entanglement is very different for pure states and for mixed states

if all you have on the table are pure states, then yes, the theory is much simpler, and a whole lot of stuff just boils down to von Neumann entropies

however

pure states are boring

they're hardly physical to begin with

the theory cannot be reduced smoothly to a classical theory of correlations

enumaris

man...looking into conversational AI lately...way too much hard programming for my taste

Emilio Pisanty

8:56 PM

there's all sorts of reasons why you want to consider entanglement of mixed states

enumaris

doesn't give me a "AI" or "Conversation" feeling...

Emilio Pisanty

in which case, $E_D(\rho)$, $E_C(\rho)$ and the entropy of entanglement will all be different

when that happens, it means that creating $\rho$ "burns" entanglement

i.e. you can go from $N$ ebits through LOCC to $\rho$, and you can then recover $M<N$ ebits from it, but you'll never be able to get back $N$ ebits once you've made $\rho$

so you need all of those measures.

Student404Mus

which ensures irreversibility of the process

so in conclusion, for pure states, it makes no sense when we talk about entanglement cost or distillable entanglement unless we are concerned with mixed states.right?

as quoted from that article "EC ?
=
ED, i.e. whether entanglement transformations become reversible in the asymptotic limit.
This is indeed the case for pure state transformations
where ED(ρ) and EC(ρ) are identical and equal to the
entropy of entanglement"

Emilio Pisanty

for those two specific measures, yes

if you're working exclusively with pure states, then yes, you shouldn't talk about entanglement cost or distillable entanglement

you just work with the entropy of entanglement and you use the fact that it has the properties of the distillable entanglement and the entanglement cost as and when you need them

if you're working with mixed states, then you need all of those measures

hint: you're probably working with mixed states

Student404Mus

Actually, its a topic of my dissertation which I'm working on

Emilio Pisanty

9:12 PM

@Student404Mus if you want more context on this, I would recommend looking for papers that work explicitly using a Resource Theory framework for entanglement

Anonymous

@EmilioPisanty Nope, I don't think there has been any official update yet (apart from this). And I haven't really been keeping touch with the news.

Student404Mus

Indeed. sometimes I read from Nielsen, sometimes I read from Wilde . Those are studying entanglement as a resource rather a measure

But I don't know how far they touch the subject since their goals deal with quantum information

enumaris

this bot ain't no AI

vzn

9:31 PM

@enumaris why not

enumaris

doesn't feel that way lol

mostly hard coded logic...only the intent detection part is really ML

vzn

@enumaris what pkgs/ systems are you looking at? suspect that many more cutting edge techniques are now being used.

enumaris

this seems to be pretty standard across the "conversational bots"

alexa skills, google actions, etc

I'm not talking about like the core Alexa, but about the skills you can build on top to do other things

vzn

how about this for state of the art in conversational AI? google duplex theverge.com/2018/12/5/18123785/…

enumaris

I don't know if the core Alexa and Google Assistant is built this way

if they are, it doesn't seem too impressive lol

from an AI standpoint anyways

still very impressive engineering wise

vzn

9:40 PM

@enumaris what kind of ML do you want built into the actions side?

enumaris

9:53 PM

Learn from the user's history

trigger actions based on conversational context

responses that are machined-learned rather than pre-programmed

(the actual text of the responses themselves, not the triggering of which response, etc.)

the way these bots are now, there are intents that you build and a set of triggering key-phrases that you provide. The ML is really just to allow a lot of other phrases that would trigger that intent based on the key-phrases that you provided

the rest appears to be hard coded logic

once an action is triggered, you have some pre-set responses you can provide and connect to an external API through a webhook to get other kinds of info or do some other action

the fact that the responses are chosen from a random list is a pretty big weakness imo

I mean, sure, intent detection is an interesting ML problem...but I was hoping a Conversational AI would be more ML than just intent detection

anywayz...

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