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12:53 AM
@EmilioPisanty i do that all time. most commonly: "no, you moron" with an arrow pointing to the offending item in my notes.
 
good to know, y'all
=)
 
1:35 AM
sigh. when you spend several hours on a proof that ends up being 2 lines
no words can describe that feeling
@EmilioPisanty oh actually this fits
 
 
1 hour later…
2:43 AM
@EmilioPisanty I think I'm more prepared for the discussion of compound observables now. My impression comes from this http://www.statintquant.net/siq/pdf/park_simultaneous_measurability.pdf
(Jut see pg 281 of book or 71 of pdf) for conclusion). Do you know any better source to read about these "compound observable"?
I'm under the impression we don't know when some operator is a compound observable or not
@EmilioPisanty Also the reason I asked for one instrument was because I didn't even know one.
 
 
1 hour later…
4:14 AM
2 much math in physics nowadays
 
 
4 hours later…
8:18 AM
Having a lot of math is much more preferable to the alternative
 
8:56 AM
@EmilioPisanty as appointed I am pinging you (after the two day expired), since the question didn't get any more attention then from the start. An answer would be immensely useful for me.
 
9:19 AM
@RyanUnger That guy who started the great Twitter/Blog fiasco of April 2018 – now he is presented on the SO Blog again.
 
For the non-relativistic QM 1-particle problem (without EM field) we say the state function $\psi$ is only determined up to a constant complex factor $e^{i\varphi}$ since $\psi$ is no observable but only $|\psi|^2$ is. Why is the state function $e^{i\varphi(r)}\psi(r)$ with a non constant complex phase factor then not equivalent (describing the same state) to $\psi$ if nevertehless $|e^{i\varphi(r)}\psi(r)|^2=|\psi|^2$ ?
 
@Rudi_Birnbaum $\hat{p} (e^{i\phi(r)} \psi)$ is not equivalent to $\hat{p} \psi$
Only if the phase is a constant is it true for all measurements
 
Yes I know that. But that means that the argument is not sufficient
, right?
 
Sure
 
What would be the proper argument then?
 
9:32 AM
You use the projective Hilbert space because $$\forall O \in \mathfrak{B}(\mathcal{H}), \langle \psi, O \psi \rangle = \langle e^{i\alpha}\psi, O e^{i\alpha} \psi \rangle$$
that is true because those operators are linear
 
I see!
cool!
So you really need to invoke the statement about all possible operators
 
Well
It's QM
 
the density operator is just ONE of them
 
So you only need it for hermitian operators
But that is also true for all operators
 
@Slereah = possible
A pitty that this is usually thought like this and not like you write ...
 
9:53 AM
@Rudi_Birnbaum needs one more day to be bountyable, I think
 
@EmilioPisanty Ah now I see! Thank you!
 
@EmilioPisanty 3 hours and 7 minutes, to be precise
 
i.e. the part that is not connected with oermanent magntic moments like spin and angular momentum
Anothe QM question: again electrons in the magnetic field the hamiltonian conatins $(p + A)^2$ Then on usually say the terms linear in A govern the interaction with spin and angular momentum = paramagnetic and the term in square governs the diamagnetic interaction. Now in standard PT theory for electrons in the magnetic field there only apear the linear terms as perturbation operator, but the theroy anyways decribes (exclusively) what is called diamagntism,
this confuses me
Not sure if the question is clear ..
 
10:30 AM
Ugh, working with MCNP is such a pain.
 
 
2 hours later…
12:44 PM
> bad trouble in subroutine source of mcrun

you need a source subroutine.
:-(
 
 
2 hours later…
2:54 PM
Has anyone got a "are you a bot" message when posting a question?
 
3:06 PM
@MoreAnonymous : Only when posting an answer :)
 
3:22 PM
@Qmechanic To be fair, there is reasonable suspicion that you actually are a bot
 
I mean he didn't say what his answer to the question was
 
Rob specifically said "This bot" when referring to him, and sarcasm is impossible through text... so I think the evidence is quite stacked against him
 
QMechanic gets into every nook and cranny.
Rob is the best Mod, though.
And Kyle Kanos is possibly the strictest user.
Wait. ACuriousMind is very good, also. But he's really orthodox.
 
I can't say I really have much judgement on how relatively "good" mods are, they're all active and the community operates pretty smoothly, so they're all doing good.
7
 
3:54 PM
Yeah, I've never had a bad interaction with a mod here. Gotta give them credit for putting up with some of us during their free time
 
4:29 PM
just going to third that all the mods on Physics SE are in my experience really good and willing to listen to the community - kudos to them for putting up with so much honestly =)
 
 
1 hour later…
5:39 PM
@SirCumference Agreed. When I first learned about Beatty sequences it took me several pages to prove Beatty's theorem. Back then, the Wikipedia page just stated the theorem without proof (or that it's also called Rayleigh's theorem). A day or so later, it finally clicked that the proof could be reduced to a few lines. :)
 
rob
6:10 PM
@WhitePrime Aw, shucks.
We're lucky to have a good team.
 
I have a problem to understand the units in this equation:
Assuming usual conventions the units of J are [C/(s m^2)]
$[\mu_0 = $ H/m and $[B]=$ T
sucht that I get for the right hand side $\nabla\times$ [C/(s m^2)].
But that in turn is $C/(m^3 s)$ in my hands which in in contradiction to the expected $C/(m^2 s)$
(This is about bound volume and surface currents $\bf J$, that shall describe a certain magnetisation $\frac{1}{\mu_0}\bf B$)
What am I missing here?
 
 
2 hours later…
7:58 PM
Am I just being thick, or is this OP's concept of a universal "now" inconsistent with relativity?
Absolute simultaneity is a red herring. The surface that corresponds to the physical 'now' isn't a plane of simultaneity for any particular observer. Imagine your plane of simultaneity is tilted to mine so that at a certain distance I think it is noon on Monday but you think it is noon on Tuesday. Two observers at that point, one moving with your frame the other with mine, will disagree on what day it is, but they will both agree it is 'now'. — Marco Ocram 1 hour ago
 
solved $\mu_0$ has actually N/A^2 then it works ....
 
8:20 PM
@PM2Ring There is indeed a universal 'now', despite the claims of GR.
 
8:34 PM
@WhitePrime what
simultaneity is relative
 
Reality is reality. It doesn't matter that events sometimes take a while to reach the attention of humans. They happened when they happened.
 
I would learn some physics
 
Thank you, mate.
 
Relativity has been tested thoroughly
Your philosophies don't have any experimental verification
Special relativity is a physical theory that plays a fundamental role in the description of all physical phenomena, as long as gravitation is not significant. Many experiments played (and still play) an important role in its development and justification. The strength of the theory lies in its unique ability to correctly predict to high precision the outcome of an extremely diverse range of experiments. Repeats of many of those experiments are still being conducted with steadily increased precision, with modern experiments focusing on effects such as at the Planck scale and in the neutrino sector...
 
8:53 PM
Relativity only describes the universe as humans experience it, not as it actually is.
But look: You guys are far advanced in physics compared to me. I do think about these things a bloody lot, though. A bloody lot.
 
@WhitePrime That's nice. But what sounds good in your head does not necessarily represent reality
 
Lel. Quite condescending, bro.
 
It's the unfortunate truth. It's the reason why it took thousands of years for classical mechanics to show up
And then another several hundred for QM and GR
The universe isn't intuitive, and experiments are necessary to determine how the universe works. That's the point of the scientific method
 
9:09 PM
I fully believe in the scientific method.
I think the universe is intuitive.
Some of you guys even admit that, to learn Relativity, you have to totally forget about intuition and common sense.
 
Perhaps I should rephrase: the universe is not obvious. There are a million possibilities we could think of in our heads for how it could work. The only way to verify which is correct is through experiments
 
Conveniently, it's not yet possible to carry out the experiments required to fully prove Relativity.
To be honest, I'm probably saying stupid stuff now because it's late etc. I bid you goodnight.
 
 
1 hour later…
rob
10:21 PM
"Intuition" and "common sense" are biological phenomena that happen in your brain. There's no reason that this intuition should apply to parts of the universe which are very far from the experiences which shaped that biology. Mathematics is unreasonably effective.
 
@rob I can definitely agree with that.
 
10:36 PM
But (and this will be controversial), we shouldn't throw away intuition, observed reality and common sense for an 100 years old theory. With all due respect.
I'm really not explaining myself well at all, tonight. I shall retire, mates.
 
11:12 PM
@WhitePrime ... Science specifically tries to work exactly within the confines of "observed reality". Common sense and intuition can be tools, but they can also be highly misleading when you're trying to determine what actually happens. This 100 year old theory is still being used (and improved upon) because it fits observed reality, better than previous models.
 
11:55 PM
@danielunderwood added u on google hangouts :D
 
@WhitePrime It is important to note that relativity is not due to the finite time of light propagation. I think this is the misconception you are basing your thoughts on, and it is one I see quite a lot on this site.
 

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