The h Bar - 2018-09-05

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12:09 AM

lol

I might try...but it's pretty hard to install stuff around here

danielunderwood

I think it would just be a conda/pip install like other python libs

And it looks like dask supports dataframes that you can do parallel operations on...hmmm

I think I may have cooked up a weekend project

hmmm

pip install dask?

danielunderwood

Yeah that should work. You may just need dask-core, but I'm not sure what all tpot needs dask.pydata.org/en/latest/install.html

(and a pip install --upgrade tpot)

1:17 AM

0

Q: Glitch in reputation points?

Today, my question at Can a substance have zero vapour pressure above absolute zero? had 3 upvotes and my reputation was 597 points. Later it had 4 upvotes and my reputation was 602 points. Later it had 5 upvotes and my reputation was 606 points and when I opened the reputation menu, I saw that i...

1:42 AM

sciencedaily.com/releases/2018/09/180904114733.htm

Make that macroscopic, and we will talk

4 hours later…

5:34 AM

@SirCumference aka nested logic :-)

or the very definition of complex, as in a complex fraction

5:53 AM

\o @Loong

hi

how goes the radiation business?

6:12 AM

It's in decay.

:-D

1 hour later…

Captain Bohemian

7:41 AM

usually when I say I like school, I mean I like research university, not any school prior to university, but people on the street may misunerstand me into liking schools prior to university. There is not any school prior to university which gives me good impression; some of them was even like a hell for me.

5 hours later…

Anonymous

12:27 PM

50

Q: Ignore Users Script

IMPORTANT - Read before using... This script is not meant to be a solution for bad behavior on Stack Overflow. If you feel somebody is violating the rules of this site, please contact the proper authorities (team@stackoverflow.com). Remove any instance of a particular user or group of uses fr...

discussion browser-enhancement ignore

Anonymous

Wow, no idea that this existed

1:47 PM

man

i'd forgotten how much i hate clebsch-gordan coefficients

1:58 PM

^ ::chuckles::

danielunderwood

2:11 PM

If it makes you feel any better, I hate anything to do with angular momentum. Someone asked a question about torques on here and I was thinking of it in terms of force on infinitesimal bits of mass

I guess that just means I need to learn it more though. Or keep doing problems like that and decide that it's not worth the extra work

What I'm trying to remember right now is how stuff like $\mathbf{J}_2=\mathbf{J}-\mathbf{J}_1$ works

for J=J1+J2, I know how to write down a state |jm> in terms of |j1m1>|j2m2> tensor products using C-G coefficients

but I'm blanking on how it should work for J1=J-J2

1 hour later…

Captain Bohemian

3:20 PM

I like angular momentum in GR for its particular relevance to gravitational wave and quantization of gravity.

3:57 PM

@Semiclassical They're still a major annoyance, but I find the representation-theroetical approach the clearest: Clebsch-Gordan coefficients are just a way of decomposing a tensor product of representations into direct sums of irreducible representations, which can be done diagrammatically via Young diagrams. E.g. the explanation of why a spin-0 occurs only if the two particles have the same spin becomes relatively straightforward, cf. physics.stackexchange.com/q/202707/50583

@Semiclassical What do you mean by that? You can express states with values for $J_1,J_2$ in terms of states with values for $J$ and vice versa, but you can't express states with values for $J,J_2$ purely in terms of states of $J_1$ - the latter are only states of the subsystem on which $J_1$ acts

A state with only values for $J_1$ is underspecified in the context of the larger system

I think notationally what I have in mind is this

If I combine two particles with spin j1,j2 to get spin j, I can write the total spin state using C-G coefficients $(j_1 m_1,j_1 m_1,JM)$ as $$|(j_1j_2)JM\rangle = \sum_{m_1 m_2}|j_1 m_1,j_2 m_2\rangle (j_1 m_1,j_2 m_2|JM)$$

with $m=m_1+m_2$ being enforced by the Clebsch-Gordan coefficient vanishing when this isn't fulfilled

blah. "C-G coefficients $(j_1 m_1,j_2 m_2|JM)$"

What I had in mind was something like $|(j_2 J)j_1 m_1\rangle$. But I'm not sure that actually makes sense.

310

Q: Should we stop commenting altogether?

Yes, that's a pretentious title. I love commenting. I'm on my way to having posted 15,000 comments. I want to help people improve their question so it won't be closed, or help them find a duplicate that answers it. But I notice useful, non-harmful, non-unwelcoming comments disappearing left a...

discussion comments welcoming

Still wondering why SE hasn't addressed this

(with the quantum numbers in parentheses labelling those of the subsystems)

@ACuriousMind hmmmm. I guess part of the upshot there is that the only rotational invariant possible on a symmetric 2-tensor is the trace.

I feel like I'm missing something easy here.

4:15 PM

@SirCumference What's there to address from SE's side? "What have you tried?" comments have long been delete-worthy unless they specifically and constructively point out what is missing from the question, and "This is a simple X operation" is either a partial answer, which doesn't belong in comments, or just noise. So the comment was potentially subject to deletion long before any of the recent events.

@ACuriousMind "What have you tried so far" is also problematic in that it encourages the OP to respond in the comments rather than edit the question.

From my viewpoint, this comment deletion discussion has the same resolution as almost all the other discussions about deleted comments in SE history: Users need to stop assuming that anything they type into the comment box will be preserved for eternity.

yeah

by definition, a comment without lasting value isn't worth keeping

@Semiclassical Indeed, which is why I try to remember to remind OP to edit their question if they're new when I leave a comment asking for more information.

yeah

it's different in chat, of course, where "What have you tried so far" is useful in clarifying HW questions

4:20 PM

Chat plays by different rules, since it's not intended as a permanent record of useful information like the Q&A is

yeah

different context, different rules

4:31 PM

@ACuriousMind Following up on this, suppose I take the case of adding two spin-1 particles together i.e. $D_{1}\otimes D_1 = D_0\oplus D_1\oplus D_2$

I guess one would have $\Lambda^2 D_1=D_1$ and $S^2 D_1 = D_0\oplus D_2$?

Though, in the case of tensoring $D_{1/2}$ with itself, the $j=0$ is the antisymmetric one

@Semiclassical No, it isn't

@Semiclassical Yes, exactly ($\Lambda^2 D_1 = D_1$ is the statement that 3d bivectors are equivalent to vectors)

Isn’t it? The singlet state is antisymmetric under exchange of the two spins

@Semiclassical Ah, but that's not the kind of (anti)symmetry I'm talking about

Hmmm

Wait

I completely misinterpreted what you said :D

4:40 PM

Lol

You're correct

My explanation there doesn't work for half-integer angular momentum

Ahhh

That's because the half-integers are not proper linear representations of the rotation group

Yeah, it seemed like something was off in that case

Still, the general representation-theoretic decomposition works

4:41 PM

Sure.

That certainly explains why you need j1=j2 for the singlet state

But the connection to the trace now seems mysterious

(In the half-integer case)

@ACuriousMind The problem is how subjective "not relevant" and "unfriendly" are.

@SirCumference It's exactly as subjective as "rude" or "no longer needed" or "not constructive" have been before as far as I can see.

@Semiclassical Yeah, maybe I need to think about my answer there again.

Kk

@Semiclassical Fun fact: the meaner cousins of CG coefficients, the Wigner symbols, are what led me to join physics.SE to ask my first question

Lol

4:50 PM

Holy cow, I've been here for more than 4 years now

In one way, seems like it was yesterday, in others, like it was a lifetime ago

@ACuriousMind Perhaps it's my perspective, but "unfriendly or unkind" seems much more lenient than "rude or abusive". If you're not directly insulting the OP, you shouldn't need to sugarcoat legitimate criticism so that it is more "kind".

@SirCumference Why not? What harm is there in being kind? If there is no way to express your criticism kindly...maybe it is not legitimate criticism. Let's be clear here: "X is wrong, actually it is Y" is not unkind.

"it would be better if this question included X" is also not unkind. What kind of statements exactly are you concerned about being no longer able to make?

5:07 PM

@ACuriousMind The harm is, again, that "kind" is very subjective. Do you consider this comment with 24 upvotes unkind?

No gravity on the moon?! Seriously?! Derps are gonna derp I suppose... — knucklebumpler Jul 8 '11 at 19:40

Yes.

Worse, it's also useless.

Is this comment "not relevant" since it doesn't answer the question?

have you tried looking up elementary particles on wikipedia and looking at which has the largest number under "mass"? — Jim Apr 11 '14 at 17:20

So it could have long been removed for being noise regardless of its kindness.

@SirCumference Yes.

@ACuriousMind I think this comment sums up the problem perfectly

Why does it seem like more effort is required to post a comment than is required of people wanting expert help? — Plutonix Aug 1 at 12:22

Telling someone to do basic research before asking here is not "irrelevant". It is the minimal effort they should put in.

Hell, hovering over the downvote button says "this question does not show any research effort"

@SirCumference Because comments are not what you should be here for. The main site is not a social network, not a chat, not a subreddit. It's a site for questions and answers, in which comment fulfill a very specific role: Improving the post being commented on constructively. If you think a question lacks basic research effort, downvote it and move on, there's literally no need to post a snarky comment about it.

5:12 PM

Am I really supposed to downvote when a question does no research, but not explain why I downvoted? How does that help the OP improve?

How do you know they did no research

I am using Jim's comment as an example, which ACM said was "not relevant"

@SirCumference How does a "And did you think to check the Wikipedia article?"-type comment help OP? Do you seriously believe there are users who are not aware of Wikipedia? Do you think they will jump up and down in delight at being told there's such a wonderful site and thank you?

@ACuriousMind "Downvote it and move on" is literally less useful than explaining why you downvoted.

Disclosure: I posted such comments in the past. Never did I receive any kind of positive feedback on them - just a lot of upvotes from other disgruntled users who liked the snark. These comments are a venting tool and not helpful to the asker.

5:16 PM

@ACuriousMind If they are aware of Wikipedia, they should be doing at least minimal research before asking for expert help. Telling them to do that at least offers more constructive feedback than a downvote.

I gotta go to class

@SirCumference Given that the downvote tooltip you just quoted says that the question lacks research effort, I don't see what the comment adds over that.

Let me also point out that this is also not new at all - lmgtfy links have been banned for years because they add nothing but condescension

I really recommend reading the original Yang-Mills paper, it's tiny and amazing

@ACuriousMind And is a new user expected to read the tooltop and understand why they were downvoted?

The jist of it is given here applet-magic.com/yangmills.htm

Not to mention down votes are obviously used for other purposes, e.g. if the question does not belong on the site

5:24 PM

@SirCumference If you absolutely insist, a comment like "Your question would be better received if you indicated what prior research, if any, you have done, e.g. checking relevant Wikipedia articles" is fine. Polish that, stick it in the AutoComments tool, and comment away. But "have you tried X?" doesn't spell that out, and it is almost always phrased in a condescending manner.

5:37 PM

@ACuriousMind one possibility I had in mind was the argument in Bohm’s (orthodox) QM textbook for the case of two spin-1/2 particles: books.google.com/…

Maybe I’ll play with that and see if there’s any mileage

@Semiclassical As a possibility for what?

Deriving arbitrary CG coefficients?

6:00 PM

Deriving that all values of $m_1=-m_2$ are equally likely if I entangle two particles in a singlet state

The argument Bohm gives is intended for two spin-1/2 particles . But I think I wasn’t thinking properly anyways

6:15 PM

Making isotopic spin invariance local in a nucleon wave function amounts to defining the difference between a proton and a neutron at each point, when you make phase invariance local to get the EM field, does the choice of sign in the $e^{\pm i\alpha(x)}$ in $\Psi' = e^{\pm i \alpha(x)} \Psi$ at each point amount to defining the charge at each point to be positive or negative?

2 hours later…

8:26 PM

So, the way we determine these colors is by looking at the primary button color (in this case a dark red/pink) and lightening it. We can get rid of the pink but we'll also need to change the button colors for the Ask Question button and the other buttons around the site. — Catija ♦ 17 mins ago

damn

designing themes is hard

At least if you're doing it programmatically. I wish they had the capacity to actually design each theme individually, but all this redesign is because they don't :/

well, depends what you mean by capacity

technical capacity, in principle yes

designer man-hours to back it up properly, apparently not

I sure wish they responded to the ugly logo SVG answer though

Novalium Company

Hmm, I'm learning about impulse and I was thinking, when something has a momentum (e.g has mass and is moving with a constant velocity) in space for example, in order to bring it to stop for example, we must apply some force in the opposite direction of the object's motion in order to change it's momentum (p = mv) and bring it to stop. That force multiplied by the time it was acting on the object is called the impulse and is equal to the change in momentum? Is this true?

@NovaliumCompany yes

Remember

@EmilioPisanty Yeah, that.

8:35 PM

$$ F = \frac{dp}{dt} $$

force is the rate of change in momentum

therefore $\Delta p = \int_{t-\Delta t}^t F \, \mathrm dt$.

Fun fact: In German, momentum is Impuls and there's no proper term for impulse, although some books call it Kraftstoß

so for a constant force $\Delta p = F \, \Delta t$.

Novalium Company

@EmilioPisanty I'm still learning calculus but I get the idea.

@ACuriousMind True fact: nobody should ever ever learn or use scientific German.

'impulse transfer', perhaps?

8:38 PM

@Semiclassical Yes, Impulsübertrag probably does the job

@ACuriousMind and it isn't even wordy by German standards!

What does Kraftstoß translate to?

@Semiclassical "push of force"

:/

i guess that doesn't read as strangely in german as in english tho

@ACuriousMind from 'stoßen'?

8:39 PM

@EmilioPisanty Yup

@Semiclassical not as much, no

@ACuriousMind hmmmmm

@Semiclassical You could also say "force push", but that just sounds like a Jedi power

is that what it says on doors? I swear it wasn't.

Ziehen vs Stoßen?

No, on doors it's drücken

@ACuriousMind Now I find myself wanting to make an 'action at a distance' joke

8:40 PM

so what's the difference between drücken and stoßen?

History nytimes.com/2018/09/05/opinion/…

@EmilioPisanty drücken is more like "slowly pushing something while being in continuous contact", while stoßen is "pushing something with a forceful motion so that it continues moving after you lose contact"

oh f*s

that's just ridiculous

Novalium Company

$F = \frac{dp}{dt}$ I suppose this is calculus and it says that the force acting on something is equal to the change in momentum. The $\frac{dp}{dt}$ parts basically says that if we have a momentum-time graph, using differentiation we can find the slope (rate of change) of the momentum at any point on the graph. Pff I don't know if this is right, I'm still learning calculus...

push vs. shove, maybe

8:43 PM

@Semiclassical yeah, I guess

though that's a matter of connotation rather than denotation

Yeah, "shove" is close to drücken than "push", indeed!

@NovaliumCompany if you're not too hot on calculus, you can just replace all the $dp/dt$'s with $\Delta p/\Delta t$'s without a problem

it only works for constant accelerations (or only approximately for non-constant accelerations), but that's fine for a first brush with newtonian kinematics and dynamics

Novalium Company

Yes but that will give us the average force over some period of time? Using the dp/dt would tell us the force at an exact "moment"?

yep, that is correct.

8:45 PM

the net force on an object is the the instantaneous rate of change of its momentum, yeah

@EmilioPisanty In the words of my dean in a welcome session: The international language of science is broken English :P

Novalium Company

Oh god, calculus slowly starts to make sense after months of trying :P

danielunderwood

Just wait until it doesn't make sense again!

the nice thing about F=dp/dt is that it makes Newton's third law into "two-body interactions conserve total momentum"

@Semiclassical the nicest thing about "force is the rate of change of momentum" is that it is a great stepping stone to "stress is the flux of momentum"

but that's maybe a liiiitle bit too much for a newtonian mechanics 101 level

danielunderwood

8:49 PM

Unrelated question: do people often call the electromagnetic field tensor the Faraday tensor? I'm seeing it while skimming MTW, but haven't seen it much elsewhere

just a bit, lol

Novalium Company

$F = m\frac{dv}{dt}$ does that make sense?

@danielunderwood it's relatively common imo

@danielunderwood They do, for some values of "often" :P

@NovaliumCompany that's correct, yes

8:50 PM

@NovaliumCompany for constant $m$ yes

danielunderwood

Ahh good. It's a lot easier to say Faraday!

but I'd encourage you to thing of $F= \frac{dp}{dt}$ as the primary version and $F=m\frac{dv}{dt}$ as a consequence of that.

danielunderwood

Do people also call its dual the Maxwell tensor? I find that one a bit odd

@NovaliumCompany That's only valid if the mass is constant over time, though! (e.g. not for rockets...)

@danielunderwood I've never seen either term in use, I don't think.

but I don't do relativistic EM for my day job.

8:51 PM

@danielunderwood When in doubt, play it safe and say field strength tensor and dual field strength tensor

I mean, honoring scientists is all well and good, but why not use a name that's at least a little bit descriptive?

danielunderwood

Because Faraday is slightly shorter than field strength :D

@ACuriousMind on that note

magnetic B-field vs. magnetic H-field

good luck finding consistent terminology for that

danielunderwood

Although calling it the field strength tensor isn't bad. I've always heard electromagnetic field tensor, which is a bit more of a mouthful

Novalium Company

Well, in the equation $F = \frac{m\Delta v}{\Delta t}$ we can basically view the top part as the change in momentum since $p = mv$. And we get $F = \frac{\Delta p}{\Delta t}$. But that is just the average force over some period of time. The instantageous force would be $F = \frac{dp}{dt}$?

danielunderwood

8:53 PM

You mean the Henry field!

@danielunderwood Replacing common sentences or words by their hash value under some algorithm is probably also shorter, yet we don't do it :P

(Griffiths has B as the magnetic field and the other as the H-field iirc. He does do electric displacement for D tho)

the IAU is holding a public consultation on whether to rename Hubble's Law to the Hubble-Lemaître Law

... in some of the stupidest news I've heard this week

danielunderwood

@ACuriousMind I once tried to do language processing with document hashes...it did not go well

@EmilioPisanty Just call it the cosmic expansion law. Down with idolatry!

8:56 PM

and that includes a peaceful student demonstration getting attacked with stones and knives by guerrilla shock groups at my uni back home

I have seen Faraday in a couple of sources

danielunderwood

Tell me that that demonstration wasn't because of the possible name change...that's honestly what I thought at first

@danielunderwood I've spent all week figuring out efficient ways to process syntax trees. I'm tired of staring at tables full of hash values :P

Maxwell tensor could refer to the energy-momentum EM tensor

@danielunderwood It wasn't. They were protesting the poor security in their campus, including the recent murder of a student.

Novalium Company

8:58 PM

In the case of $F = \frac{dp}{dt}$ the $\frac{dp}{dt}$ part basically says that the momentum is a function of time, $p(t)$? I guess that's just another way to write it?

For more than one particle, both $m$ and $v$ can change with time, $F = \frac{dp}{dt} = \frac{d}{dt}p(t) = \frac{d}{dt}[m(t) v(t)] = \dots$

danielunderwood

@ACuriousMind that sounds odd. Are the hashes just for storage/access and they're reversed for processing or are you somehow directly processing the hashes?

@danielunderwood Storage/access. You don't want to search for strings when you can search for integers instead.

Novalium Company

Yay, derivatives slowly start to make sense! I can go to bed now in peace. Thanks everybody for the help and goodnight :---)

There's also some number of boolean flags that are encoded in the bits of a single variable, but that's not a hash.

But in the end, my optimizations resulted in a 100-fold speedup over my initial attempt, so I'm quite pleased

danielunderwood

9:04 PM

Ahh I suppose I should learn more about algorithms and data structures. I don't really know what the difference would be searching for an integer from a hash and just casting a string to an integer array

Hashes are shorter. E.g. you can use them as array indices.

danielunderwood

In that case, you just have a dictionary right?

@danielunderwood The problem with strings is that they can share their beginnings and have arbitrary length. The hash value has a fixed length and can basically be compared in a single operation and can also be used as a direct index into an array

i was seeing something earlier which I was trying to find a definitive reference on

@danielunderwood Yeah, a python dictionary is accessed through hashes

danielunderwood

9:06 PM

Ahh right I meant 0-padding the integer array

Which is why you have to implement __hash__ on something you want to use as key to a dictionary

Suppose I take a total of $2j$ spin-1/2 particles, with $j+m$ of them being spin-up with respect to the z-axis and the other $j-m$ being spin-down

danielunderwood

But is a dictionary just an array with a hash index?

I think that's an implementation detail (the standard just says you have to have O(1) access on a dictionary)

If I symmetrize that, do I just get $|jm\rangle? (ignoring normalization for now)

danielunderwood

9:08 PM

I learned about __hash__ the hard way when I had a failing set comparison for weeks

But in practice I'll bet you it's an array with hash indices

danielunderwood

You guys are just a bucket of knowledge!

you know how this room was all like "yeah, all of these physicist chatroom regulars are switching to a programming dayjob, but that's OK 'cause it's not going to derail our regular physics schedule"?

lol

Programming is just applied physics

4

9:10 PM

@Semiclassical Hm, I'm not sure, would you get the proper coefficients out of symmetrization though?

not at all sure.

@Lozansky Or maybe physics is just applied programming...

3

danielunderwood

I've seen a lot of physicists that would have a much better time if they took a couple months to properly learn bits of programming!

I was seeing a version of that claim that didn't include symmetrization, which was nonsense

@Semiclassical I seriously doubt it

there's not enough 'handles'

9:12 PM

I was thinking that to express $\lvert j m \rangle$ in terms of the individual particle states it would involve the Clebsch-Gordan coefficients, which I don't think you'd get from symmetrization. But I'm not sure.

i.e. it doesn't account for combinations that make up for a non-maximal $J$

it should be relatively* straightforward to test numerically on Mathematica, in any case

I mean, it works at the level of $J_z$

* ok, not that easy, but doable

@Semiclassical but not at the level of Beyoncé?

5

badum-ts

wow

sorry, couldn't resist

9:14 PM

that was breathtaking

@EmilioPisanty You know, the more I think about it, the more I think that the fact that this room is full of people who started as bright-eyed physicists who have now turned programmers is trying to tell us something about the state of academia :P

i.e. $J_z|jm\rangle = m |jm\rangle$ since $m=(1/2)(j+m)+(-1/2)(j-m)$

That pun though

danielunderwood

I bet somewhere there's a bright-eyed programmer turned physicist

maybe one or so

Or a mathematician turned billionaire

9:16 PM

the real question is whether $J^2|jm\rangle = j(j+1)|jm\rangle$ tho

@danielunderwood I do know a mathematician turned stock broker who got so rich and so bored that he now funds his own chair in a physics department

one has $J^2=\sum_{k=1}^{2j} J_k^2+2\sum_{k<l=1}^{2j} \vec{J}_k\cdot \vec{J}_l$

@Avantgarde I give you James Harris Simons

@ACuriousMind Yes, I know haha. That's why I said it ;)

from the first set of terms we get $\sum_{k=1}^{2j}J_k^2|jm\rangle = (2j)(3/4)|jm\rangle $

9:20 PM

His company seems extremely specialized with high-profile (and very few) employees.

Renaissance Technologies

where the rubber hits the road is those $\vec{J}_k\cdot \vec{J}_l$ products tho

those are pretty miserable iirc

James Simons (full length interview) - Numberphile

Nice interview. I've only seen parts of it myself, though.

@EmilioPisanty just checked $J^2$ on the symmetrization of $|\uparrow\uparrow\downarrow\rangle$ in Mathematica, and the symmetrization does seem to be an eigenstate

So...huh.

How we come up with a mathematical theory for an observation and as per our need, in physics? More specifically, how we came up with the new theory for describing behaviors of a qubit ?

(there's some factors of twos which I'm not confident on, hence why I'm only saying this in terms of eigenstates not eigenvalues)

scratch that. I found the factor of 2 I was doing wrong, and I think I can claim the following example with confidence:

If $$|\psi\rangle = |\uparrow\uparrow\downarrow\rangle+|\uparrow\downarrow\uparrow\rangle+| {\downarrow} \uparrow\uparrow\rangle$$ and $\vec{J}=\vec{J}_1+\vec{J}_2+\vec{J}_3$, then $J_z |\psi\rangle = +\frac{1}{2}|\psi\rangle$ and $J^2|\psi\rangle = \frac{15}{4}|\psi\rangle$

ffs mathjax

there we go

9:37 PM

@Semiclassical In representation-theoretic language, that's the proof that the symmetric and anti-symmetric parts of a tensor product form subrepresentations of the tensor product.

neat

All the (anti-)symmetrizations are eigenstates

in the present instance, it proves that $|\psi\rangle$ has $J=3/2$ and $M=1/2$ i.e. $|\psi\rangle \sim |3/2\,{+1/2}\rangle$

@Semiclassical Chat inserts a breaking space after 50 continuous characters or so. Just use some spaces in your TeX and you're fine ;)

yeah, it's just annoying

@ACuriousMind well, it might be if I had proved it in generality lol

all i've actually done is compute a specific case in mathematica and see that it worked

makes me more confident that what I'm saying makes sense, though

9:45 PM

@danielunderwood I've seen a lot of programmers that would have a much better time if they took a couple of years to properly learn bits of physics.

@Semiclassical what about |↑↑↑⟩?

argh...

well, $|\uparrow\uparrow\rangle$ would just be the $m=1$ triplet state

and then you add another spin-1/2 with spin up to get $|3/2,+3/2\rangle$

in any case, though, mathematica gives back $J^2|\uparrow\uparrow\uparrow\rangle = 15/4 |\uparrow\uparrow\uparrow\rangle$ and $J_z|\uparrow\uparrow\uparrow\rangle=+3/2|\uparrow\uparrow\uparrow\rangle$

so it again checks out

i feel like I remember seeing this in penrose a long time ago

@Avantgarde That's a really interesting interview!

10:25 PM

I'm struck by the odd dichotomy between luck and ability he seems to make. He concedes that a lot of his own success is due to luck, but then constantly makes a record of success the condition for hiring people for his business or supporting them through his foundation

10:58 PM

'In February 1954 Robert Oppenheimer invited Yang to present the work at a seminar at the Institute for Advanced Study (IAS) in Princeton, New Jersey. Wolfgang Pauli was present. He had been pursuing a similar thought, but had quit after encountering what seemed to be a show-stopping issue: in such theories, the mass of such a field has to be zero. Pauli knew that in “Abelian” theories, such as quantum electrodynamics (QED), it’s alright if the force-carrying particle is zero; this is indeed what lies behind our understanding of the zero mass of the photon. But extending field theory to had…

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Sep '185

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