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12:18 AM
@AccidentalFourierTransform More or less. Is this about physics.stackexchange.com/questions/338790/… ?
If so we are on the matter.
 
 
5 hours later…
5:20 AM
@peterh Yeah, see this for example:
20
Q: Are Germans a different kind of human?

JeromeRMy title is facetious, but my question is serious: I'm being told by a client's business-analyst team that Germans WANT to read every detail on a website and WANT to read every word in a contract—more so than people from other cultures. Research done in North America shows that there people sca...

But that's OK...
I think humanity needs that kind of humans too, to advance physics, math, philosophy and to build better machines...
 
Haha, I'm 3/16 German at last count. It would certainly explain a thing or two.
 
interesting question haha]
 
5:38 AM
@Mostafa Also I think the same. Btw, US people are in fact a large part German, and they are similar also in this aspect, but not so strongly.
 
5:50 AM
@ACuriousMind The idea is not so young, there were a Second Space Odyssey, written in 1982 and the idea may root from here. I think it would be a more nice idea to peel its athmosphere somehow, and then a big, big, Earthlike planet could remain after a little terraforming.
 
6:08 AM
German ancestry in the US is the largest, right?
 
6:37 AM
@Avantgarde As I know, yes. I've just got a half hour ban, for not watching my steps enough carefully.
 
Is there a name for the rule that says that the period of a swinging pendulum doesn't depend on the angle that it's perturbed?
 
@DawoodibnKareem not that I know of. In any case this is an approximation using $\sin\theta \approx \theta$ and isn't true for large $\theta$. The period does depend on the amplitude.
 
True. OK, thanks for that @JohnRennie
 
 
2 hours later…
9:01 AM
@peterh Yeah...for example Trump is German on his father's side ;)
 
no racism pls
 
@Kenshin pls = please?
 
yeah
 
We're talking about *facts*.
Not personal opinions.
 
A "fact" can be racist
 
9:13 AM
But that's still a fact.
 
all I said was no racism
 
9:48 AM
Nobody here?
 
@bolbteppa : Why the long face?
 
I wrote "left as an exercize" in an answer
Hopefully that will be good enough
Couldn't be arsed to write down a full proof
 
 
2 hours later…
11:51 AM
@Mithrandir24601 Hi, if you have a chance please see the following query: In the following uploaded picture, the von Neumann entropy is introduced, can you maybe see why a complete measurement (defined as each measurement operator projects the system onto a pure state) implies (2.17)?
Also, do you know what is meant by the last statement "The value of the lower bound is achieved by a von Neumann measurement in the eigenbasis of the density matrix". As I understand, the lower bound $S(\rho) := -Tr[\rho ln \rho]$ refers to $\rho$ before measurement?
@Mithrandir24601 I think I understand the last statement, is the last statement stating that if we diagonalize the density operator $\rho$, then the measurement operators which correspond to projections onto the pure states (eigenstates of $\rho$) are the von Neumann projections of the eigenstates with probability given by the eigenvalues. Hence the entropy of this is $S(\rho)$?
 
 
2 hours later…
2:03 PM
@Qmechanic good proof, good references, but don't fully get it yet, apparently the original proof is here archive.org/stream/TheTheoryOfSpinors/… according to physicsforums.com/threads/… which seems less metaplectic but still trying to get them :(
 
 
2 hours later…
user228700
3:42 PM
@JohnR: Hi! :-)
 
@Kaumudi.H Afternoon ...
That was good timing, I've only just got back from doing the shopping :-)
 
user228700
@JohnRennie Ah, I see :-) Nothing at home after having lived at ur mum's for two weeks huh?
 
@Kaumudi.H I did the emergency shopping, e.g. milk and bread, when I got back yesterday. This was just routine shopping.
 
user228700
Ah, OK...
 
Though since I am 5kg heavier than when I arrived at my Mum's, food shopping is not a priority right now :-)
 
user228700
3:49 PM
Ah, right, right :-)
 
user228700
Went out with friends today. We went to this place called the Writers' Cafe for lunch and it was so. good.
 
user228700
I can't even pronounce the names of the food we had but they were all great!
 
user228700
(Italian, I think but I can't be sure).
 
Writers' cafe? Is it a hangout for local authors?
 
user228700
@JohnRennie I dunno...
 
user228700
3:52 PM
It looked like this on the inside:
 
user228700
 
user228700
 
It certainly has a literary theme ...
 
user228700
Certainly, yes. And there was a reading room upstairs as well.
 
Were there any scruffy impoverished young people scribbling furious in notebooks :-)
 
user228700
3:54 PM
@JohnRennie Hehe :-) I saw one girl just at it on a napkin but nothing else.
 
user228700
I'll post a few pictures of the food we had once I get hold of them, if u want!
 
@Kaumudi.H yes, why not. Can you remember the names of any of the dishes?
 
user228700
Hang on...
 
user228700
Dang, this might take a while. I'll just show them to u tomorrow, then...
 
No hurry, I was only idly curious.
@Kaumudi.H: Any news on the college? Have you been formally accepted?
 
user228700
4:05 PM
@JohnRennie OK...
 
user228700
@JohnRennie Nope, not yet. The rank list is s'posed to come out in 8 days...
 
I guess for now you just have to chill and try not to think about it ...
 
user228700
Yes, yes...
 
user228700
*Chill*=Eat and sleep, apparently :-P
 
@Kaumudi.H well eating and sleeping are kind of obligatory. I trust there is more to your life than that :-)
 
user228700
4:13 PM
:-) Well, yes, but only almost. I'm meeting friends, reading those books I ordered and playing the harmonica from time to time...
 
user228700
Catching up with friends is taking much more time than I'd anticipated; haven't talked to any of them in over a year, you see...
 
Anonymous
@Kaumudi.H Where are you joining?
 
Presumably your friends are all back from college now?
 
user228700
@blue A govt. engineering college in Kerala. And you?
 
user228700
@JohnRennie Yep yep! :-)
 
Anonymous
4:14 PM
@Kaumudi.H Oh nice :) I've not decided yet
 
user228700
OK...
 
user228700
Travelled the metro today and it was so good!
 
SBM
Hello
 
user228700
Hi :-) How's it going?
 
:(
hey
 
SBM
4:23 PM
There was 5.5 hour long power cut.
 
user228700
Damn :-(
 
SBM
:( power cuts after lunch till dinner time.
 
Hi, everybody.
 
hey Daniel
 
SBM
4:30 PM
Hey Daniel
 
5:12 PM
hey what's up
 
 
1 hour later…
6:34 PM
@JohnDoe Whatever is actually going on here, I'm not too sure at the minute, but to me it looks like the measurement operator is $M = \sum_m \alpha_m |\psi_m\rangle\langle\psi_m |$ as this directly gives $p_m = tr\left(\alpha_m |\psi_m\rangle\langle\psi_m | \rho\right) = \alpha_m \langle\psi_m |\rho|\psi_m\rangle$. But that's what they say is $I$, not $M$... O_o
 
6:56 PM
@JohnDoe After measurement, the density matrix is $\rho ' = \sum_m |\psi_m\rangle\langle\psi_m |\rho |\psi_m\rangle\langle\psi_m | $, which if $\rho = \sum_n\lambda_n |\psi_n\rangle\langle\psi_n | $ (as the measurement is in the eigenbasis) gives $\rho' = \sum_{n, m} \lambda_n |\psi_m\rangle\langle\psi_m |\psi_n\rangle\langle\psi_n |\psi_m\rangle\langle\psi_m | = \sum_{n, m} \lambda_n |\psi_m\rangle \delta_{n, m}\langle\psi_m | = \sum_n \lambda_n |\psi_n\rangle\langle\psi_n | = \rho$
 
7:07 PM
I see Physics passed 100,000 questions, and I see I forgot about it, as I predicted.
This one might be it.
 
@JohnDoe Wait - for the first thing I said - I was being a complete idiot - ignore that bit :P
 
I hope this isn't too OT here, but anyone know of a material (refractory ceramic, anything really) that can handle 2800c, doesn't react with aluminum, is non-conductive, and I can realistically drill a small hole in?
 
7:44 PM
0
Q: Possible classical roots of quantum entanglement

kpvI read this very recent article that is titled - "Classical synchronization indicates persistent entanglement in isolated quantum systems", is published at https://www.nature.com/articles/ncomms14829 If it is within physics SE norms, may I ask QM folks to comment whether this article is credible...

too lazy (sleepy) too read
 
8:09 PM
@JohnDoe This feels a bit weird, but the only thing that I can think of is that $\alpha$ is a normalisation - by definition, making a measurement of an operator $\hat{M} = \sum_m M_m |\psi_m\rangle\langle\psi_m |$ gives the probability of obtaining $M_m$ as $tr\left( |\psi_m\rangle\langle\psi_m |\rho\right) = \langle\psi_m |\rho |\psi_m\rangle$ for orthonormal $|\psi\rangle$, so either it's got something to do with degenerate eigenvalues or non-normalised states as far as I can tell...
this also makes sense when looking at the identity: I = $\sum_m |\psi_m\rangle\langle\psi_m|$ for (orthonormal) $|\psi_m\rangle$ forming a complete basis, unless I'm being an idiot again...
 
 
1 hour later…
9:26 PM
@dmckee A quick comment on your thoughts on the AC stuff. In the case of a non-pure sinusoid signal in the circuit, can we then still use phasor notation? I doubt it, right, since the whole phasor notation is based on the fact that the relative angle doesn't change over time? (All frequencies are the same, so we can always write it as a pure sinusoid)
We can of course write them using Euler
 
@MathematicalRain Each of the Fourier components has constant phase, but to use the phasor notion on the sum you have to keep in mind that the phasor for each component turns at a different speed, so the utility of phasors over doing the complex algebra is lost.
As far as I know everyone reverts to doing hte algebra at that point.
 
I see, thanks
 
> UV complete me: Positivity...
hmmm
 
9:43 PM
Guys, says we are given the following coupled oscillator, where $f$ is the spring constant and $\lambda$ the damping coefficient (so the damping force is $-\lambda dx/dt$. I don’t understand how they get $f(2x_1-x_2)$. I would think that only the difference $(x_1-x_2)$ matters, just as they wrote in my syllabus:
 
=-K(2x_a-x_b)
 
so it was a mistake in my syllabus?
 
hmmm no, why?
$-Kx_a-K(x_a-x_b)=-K(2x_a-x_b)$
 
well they write $-K(x_a-x_b)$
ohhhh
right
 
9:46 PM
okay thanks a lot. I thought I was going crazy
 

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