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04:51
@JohnRennie How do you do Dr. Rennie :)
@Obliv Hi :-)
 
4 hours later…
08:44
hi
09:06
is it true (in any reasonable interpretation of the word true) that quantum mechanics reproduces any prediction that classical mechanics makes?
papers like this describe this (traditional) expectation for quantum effects to disappear, seemingly as a function of system size. But this does not a priori seem reasonable to expect. What sort of body of evidence supports this traditional view?
@nickbros123 how do you feel about griffiths electric displacement quantity?
i have a personal distain for the introduction of this entity
@SillyGoose that's not what anyone claims, and is a patently absurd claim on its face: We invented quantum mechanics precisely because we observed things that were not what classical mechanics predicted (cf. ultraviolet catastrophe, double slit, Stern-Gerlach)
@SillyGoose It's not an "a priori expectation" nor a theorem, it's Bohr's correspondence principle, a general description of how quantum systems tend to act, not a physical law
on what grounds was this "general description" founded?
it worked for the atom :P
you must not think of old quantum theory - and the correspondence principle is firmly part of the very first steps toward quantum mechanics - as a body of rigorous reasoning, defined and derived in the clarity in which we find it in modern textbooks; Bohr came up with a bunch of guesses, and they resulted in the best description of the hydrogen atom so far, so people accepted them
but aside from the specific principle: Is it not true that you observe no "quantum" effects at the scales you can perceive with your naked eyes? Is the absurdity of Schrödinger's cat not precisely that we don't expect something as large as a cat to be "quantum"?
09:21
@Relativisticcucumber every student hates when teachers introduce something new for the first time. It is almost always a bad idea; a teacher would only be introducing something if that something is proven to be a good idea. However, Maxwellian fields seem to really be inimical to good understanding.
I'm a bit confused that you seem reluctant to accept the idea that big things tend to be classical, when you've lived every day of your life with overwhelming evidence for this
@ACuriousMind actually, we already have macroscopic versions of quantum behaviour, yet people continue to claim that quantum stuff only happens at micro to mesoscopic scales
@naturallyInconsistent yes, which is why the classical limit is subtle and hard in general, I know
what is an example of macroscopic phenomenon where u cant use classical mech
@SillyGoose classical mechanics is valid within its region of validity, which is quite wide. When establishing a new theory to replace the old theory, it is quite necessary to limit the complexity of generating the new theory, and so it is always nice to have the new theory have some limits such that the successes of the old theory are automatically reproduced. Then you dont have to prove so many things.
09:25
@Relativisticcucumber useful quantity
@ACuriousMind and under no circumstance am miao miao trying to insinuate otherwise. definitely most sure that you already know about it
@nickbros123 curses
why do u think so
@ACuriousMind is the production of light from turning on a light source overwhelming evidence for classical physics :P
@ACuriousMind acm, i had this same convo earlier. the wacky goose also claims to not believe in cause and effect!
join @Slereah in the rabbit hole
@Relativisticcucumber i claim that cause and effect is the best tool we have to probe the world. as opposed to absolute truth.
09:29
@SillyGoose do u believe this for philosophical reasons or for physics or both?
i feel like i tend towards "who cares about absolute truth" i mean at a certain point it seems to become useless to discuss such things
well in a sense the production of "light" from a light source is evidence against classical mechanics and quantum mechanics, no?
@SillyGoose well, see, that's the ultraviolet catastrophe part I already mentioned: I don't mean technology, of course the modern world is suffused with technology inexplicable by classical Newtonian mechanics
and arguably the largest evidence for non-classical physics is the sun in the sky
@Relativisticcucumber im a bit rusty with enm, but i remember it being helpful in circumventing certain calculations with regards to polarisation. D speaks directly to the free charge distribution, thats helpful sometimes. theres a gauss law for $D$, for example. so in case of symmetry, its useful. and for linear dielectrics, theres a cuter relationship between D and E. also The energy for building linear dielectrics has $D$ in it. For systems that extend to infty, theres something neat about D
which I conveniently forgot
09:34
@ACuriousMind well i think this is my point. do people really live "everyday life" with overwhelming evidence for classical physics
@nickbros123 how can YOU be rusty with enm
in the meantime, miao miao just sent 6 papers out to a prof to read, so that he can help us understand what the papers are saying. Which means miao miao would have to dissect them fully first. RIP
@SillyGoose what I meant is that all the mechanics you observe in everyday life are classical: Things are solid and have a definite place and definite momentum
@Relativisticcucumber its been a while since I revised enm
i moved on to other things
@nickbros123 i thought your life passion was enm
gasp
what are you onto now
09:35
goldstein
@RyderRude to remain agnostic on a question that i cannot answer
@Relativisticcucumber you are very wrong. He only had a passion to look up the details of E&M's averaging of the fields problem
and that's the origin of the correspondence principle: The planets orbiting the sun are classical, the electrons orbiting the atom are not; the difference is size (either literally or of relevant quantities to the Planck quantum)
@naturallyInconsistent this is only partially true. I will make a gargantuan comeback to EnM within the next 2 years
09:36
but, again, this is a tendency - bigger things are more often well-approximated by classical physics - not a theorem
@nickbros123 yes, and please do
and not possible to become a theorem, because we already have experimental proof otherwise
i just think all of the polarization calculations are much easier without the stupid D business
but i already have many bones to pick with griffiths
what is the modernized "correspondence principle"? if one exists
@Relativisticcucumber oh no, you are so mistaken. It is way worse
@naturallyInconsistent but i just worked out the entire section without D
it was quite simple
09:38
@Relativisticcucumber when you work out the microscopic fields, where every atom means an infinite spike wiggling of the fields, you will literally cry
@SillyGoose it's still just that "big things" (or "states with large quantum numbers") tend to behave classically
@Relativisticcucumber if i remember 4th chapter correctly, every problem in it had 2-3 methods of solving. Usually the ones appealing to symmetry (and ones that have to do with linear dielectrics), using D leads to quicker solutions
again, this is just a heuristic; the validity of this heuristic has not changed, though we probably understand more about its exceptions - there is no general theory of a classical limit applicable to all quantum systems at once
@nickbros123 hm i guess i only did a problem that involved finding the field using laplace's equation from the section, which doesnt really rely on D at all once you have derived the equation
so maybe i should try another problem?
please see "Mirage". it is about how a storm connects two different points in time, at the same point in space. then the future can influence the past and vice versa
09:41
and deriving the equation without D is quite simply
simple*
this chapter was a bit of a drag ima be honest, but section 2.3 sent me down a rabbit hole i still havent climbed out of
well in any case BLEH to griffiths
@nickbros123 what specifically?
my never ending qualm of enm has been with boundary conditions across a surface
@Relativisticcucumber the whole issue with macroscopic enm
never been able to understand that section
@Relativisticcucumber The question of in how far science includes necessarily the notions of cause and effect has a long history (cf. "correlation is not causation", Hume's radical skepticism or Norton's Causation as Folk Science), but I don't see how it relates to the questions about the classical limit at hand :P
09:44
just think of it as cartesian space with a scalar function $\rho$ that can be 0 here and there, a vector function $J$, and 2 vector functions $E$ and $B$ that depend on each other and the previous 2 functions (from jefimenkos, it only depends on the previous 2 though). To rigorously do boundary cases one may need some pde knowledge i guess, i never bothered to do it rigorously, im happy with handwave
@ACuriousMind oh because earlier when i invoked cause and effect to describe my model of the universe, silly goose claimed something akin to we dont have a reason to believe cause and effect is true. which is kind of confusing to me because if we invoke skepticism on cause and effect then can we even say or do anything.... but it was similar to me about being sus about classical physics at a large level
cause its like i see where the thoughts are coming from but that level of skepticism/abstraction pains me
but i guess this is the weird dilemma of a scientist. like what level of skepticism is proper. i guess we choose what is practical but in what sense that is reasonable im not really sure.
er wait
i guess the level that is proper depends on when the model starts to work well enough?
being sus is fine if youre right enough? sounds bad but sufficient.
all models are wrong, some are useful
10:26
what structure cud make spacetime emergent
we want all the components of a manifold to emerge: the set, the topology, the smooth structure, and the metric
 
5 hours later…
14:59
@naturallyInconsistent why so long?
we don't see Veritasium making videos like this^
15:16
@user85795 Veritasium is more for pop sci stuff
but great quality and informative
He puts a lot of effort into his work.
yes. he meets with the experts and reads the relevant books
The solution one gets for the TDSE via the separation of variables is considered to be a stationary state. For a state $\Psi$, which can be a linear combination of stationary states, can one say that this state is also a stationary state?
I ask this because, if one was to perform $\lang\psi|\psi\rangle$, the time dependent complex exponentials wouldn't vanish, but instead one would have phase differences
@imbAF What problem do you have in figuring this out yourself? You know the definition of a stationary state, what's the problem in looking at the sum of stationary states and seeing whether they also fulfill that definition? (also, we've been over this before)
I did figure it out, I did the calculations, and because I have phase differences, I believe that's the case
15:30
you believe that what is the case?
That it doesn't represent a stationary state
Then what is your question?
Whether I stand correct or not
you're right, the sum of stationary states is not necessarily a stationary state
you should recognize that this is just an application of the linear-algebraic fact that the sum of eigenstates (of different eigenvalues) is not an eigenstate
yes, I am aware
And an additional reason I asked was because we were considering
a state $\psi(t_0)=\sum_\tau c_{n,\tau}|\phi_{n,\tau}\rangle$

And after some time t, one would have
$|\psi(t)\rangle=e^{-iE_n(t-t_0)/\hbar}|\psi(t_0)\rangle$
And with this one can show both states are physically the same
And I thought, this is true, but this is only valid when the system is initially at a stationary state. Instead, if it was at a superposition state of them, the above equations wouldn't hold
Hence why I asked to confirm my suspicion regarding a superposition state
15:43
Your equations make no sense - what is the index $n$ and $r$ here, what are the $\phi_{n,r}$? You need to consider that other people are not looking at the exact same text as you and have only the context you give them
you mean $\tau$
not r
And that is just a representation of degeneracy of the considered discrete spectrum
@imbAF do u also have a question about $\langle \psi |\psi \rangle$?
What?
u wrote this in ur first comment
maybe i misinterpreted
@imbAF what "considered discrete spectrum"? Why is there a sum over $\tau$ but not $n$? That's what I mean by the missing context - if you want good answers to your questions, you need to explain them properly
with the information you have given, it's completely impossible for me to be able to say whether what you say about these equations is correct
15:56
In my notes I have the following
The system is initially, at t=t_0 in a stationary state:
$|\psi(t_0)\rangle=\psi_0e^{i\frac{E_n}{\hbar}t_0}\phi_n$ which can be written also as:

$|\psi(t_0)\rangle=\psi_0e^{i\frac{E_n}{\hbar}t_0}|\phi_n\rangle$

$|\psi(t_0)\rangle=c_n(t_0)|\phi_n\rangle$

But if we consider that the spectrum is degenerate then:


a state $\psi(t_0)=\sum_\tau c_{n,\tau}(t_0)|\phi_{n,\tau}\rangle$
Ok, so you have an Hamiltonian $H$ with discrete and degenerate spectrum, namely $$H|\psi_{n}^{i}\rangle = E_n|\psi_{n}^{i}\rangle, n \text{ fixed }, i = 1,...g_n $$. You now take a generic state belonging to the subspace $\mathcal{E}$ corresponding to the energy eigenvalue $E_n$. It's time evolution is therefore trivial
where $dim(\mathcal{E}) = g_n$
is this what your notes are trying to say?
no wait, your first line is actually cursed lol, I humbly step down from this conversation
lol
16:22
i feel u r constructing a superposition which is still a stationary state becuz the eigenvalues match
if u sum over $n$, u get a non stationary state like ACuriousMind said @imbAF
yeah that's exactly it
Anyways, I suggest reading Cohen-Tannoudji. The section on this particular topic is extremely clear, I studied from there
@imbAF I feel you still don't understand what I mean by "context". Now I can guess what's going on, but what would have been helpful up-front would have been clarifications such as "The $\lvert \phi_n\rangle$ are energy eigenstates", "The index $r$ runs over states with the same energy eigenvalue $E_n$"
can you see how much more specific especially the second one is than your vague "that is just a representation of degeneracy of the considered discrete spectrum"?
And now I am wondering whether you weren't wrong after all, since the sum of eigenstates with the same eigenvalue is again an eigenstate with that eigenvalue
I keep repeating this, but this part of QM is really not that mysterious: It is just linear algebra. If you just use "stationary state" = "energy eigenstate", all these questions have clear and straightforward answers
Found it. Page 245 of CT vol 1, section is called "Solution of the Schrodinger equation". Interestingly, I think your teacher's notes are taken directly from there :P The notation perfectly matches yours and even the way conclusions are drawn @imbAF. Please take a look at it
nice detective work :P
+1
I know I sound stupid but take a look at this and compare
I assumed those were notes he took in class, but maybe he's just studying from the book lol. Nonetheless, there are undoubtedly some stricking similarities :P
@user85795 thanks lol
16:44
@user85795 such horribleness have a tendency to not be resolved / resolvable. And yet it will keep haunting
@imbAF there is no universe whereby this can be correct. You cannot have $\left|\psi(t_0)\right>$ and $\psi_0$ and $\phi_n$ all appearing at once; that is, it is absolutely necessary that you copied this wrong.
what's that video about? I'm doing some math exercises and I don't have the time to watch it, so can someone explain to me why it has become a meme hahaha
@Claudio it is definitely not a meme. But he worked out an annoying discrepancy in the units of the Planck's constant as used in the various senses in physics
I thought It was a meme. It was brought some days ago too lol. My bad
16:51
the title is self explanatory
Alright, I'll save it in my for later playlist
@Claudio Thanks, I have seen it. And I am clear on everything
it is a meme in the sense that it makes a conundrum out of the units for the title of this chat room
@imbAF you are welcome
@user85795 lol I see
 
1 hour later…
18:12
Why is a projective null cone topologically a sphere?
18:23
you ask these things as if they're widespread notions :P
but assuming that by a projective cone you mean the space you get when you identify all points along a ray from the tip of the cone, is this not obvious (at least in 3d)?
@ACuriousMind I did it on purpose, and it just got fulfilled ;)
@ACuriousMind See this is precisely what they mean by projective null cone. They cut it with a plane, and the intersection is a $S^d$, why should that give me the topology of what is being cut?
yeah, so it's what I just said
the 'null' part is completely irrelevant, this is just geometry of cones
I'm not sure what part exactly is unclear to you
@ACuriousMind Hm. But even in 3D I can't visualize that... I tried for a while, but failed. But what about the thing with intersecting with planes. Why intersecting a cone with plane give you the topology of the cone?
@Sanjana you can't visualize that cutting an ordinary cone with a plane produces an $S^1$?
it's famously one of the conic sections!
@ACuriousMind No that I can, but why would I cut a cone?
I mean (why) projectivising would be equivalent to cutting a cone?
18:32
@Sanjana it gives you the topology of the projective cone, because one way to present projective spaces is to choose one representant from each of the equivalence classes and the plane intersects the cone such that it intersects each ray exactly once
this is the "double cover" logic from recently just for a single intersection
@ACuriousMind Ok so like that, here also plane is just a choice and any surface which would cut through each ray once and only once, would work?
@ACuriousMind Oh ok. that was easy.
I had another question intersection of the 5-cone and 5-sphere gave me a double cover of the cone is given in the book as $S^1 \times S^3$. This I could verify by simply writing the equations of these objects as embeddings in a 6 dim. space. From this double cover we want to get to the original space... and the book says that the original space is same as the double cover...
Why is that? Why is $S^1 \times S^3$ same as its double cover?
it's not unusual - the circle $S^1$ is its own n-fold cover for each $n\in\mathbb{N}$ via $\mathrm{e}^{i\phi} \mapsto \mathrm{e}^{\mathrm{i}n\phi}$
@ACuriousMind Isn't the universal cover of $S^1$ just $\mathbb{R}$?
18:41
that is correct but does not contradict what I said :P
Oh you said nothing about the universal cover. That's why?
@ACuriousMind Also another thing this $S^1 \times S^3$ is just $U(2)$ as a Lie group right?
@Sanjana no, why would it be?
@Sanjana The cover $\mathbb{R}\to S^1$ is infinite - the fibers are not finite
@ACuriousMind oh
@ACuriousMind I was thinking $S^1$ is $U(1)$ and $S^3$ is $SU(2)$ and $SU(2) \times U(1)$ is $U(2)$, that's why?
@Sanjana ah, but you need to pay attention to the group operation - just because two Lie groups are diffeomorphic does not mean they're isomorphic
as a Lie group, $\mathrm{U}(2)$ is the semi-direct product of $S^3$ and $S^1$, not the direct product
@ACuriousMind oh ok, thanks for reminding me this
Oh look ACM! CuriousKid asked this question :p
18:55
no relation
@ACuriousMind But one thing still confuses me. When we cut the 5-cone with a plane we get a 4-sphere. When we cut the 5-cone with a 5-sphere we get the double cover $S^1 \times S^3$ (equivalent to the original space in this case due to some reason). But this is not a 4-sphere.
The text which cuts the cone with a sphere and the one which cuts the same with a plane are different...
@Sanjana your 5-cone wasn't a cone, it was a null-cone for a metric of signature (4,2)
the null cones for metrics of signature (1,d-1) are actual cones, but that one wasn't one
Which one of these 1 or 2 is an actual cone?
By your definitions of an actual cone, none of them actually seem like actual cones :|
@Sanjana nono, this one is an actual cone
because it's a null cone in a spacetime with 1 time dimension, these are actual cones (i.e. $S^{d-2}\times \mathbb{R}$ except the tip)
it's just that your "5-cone" is in a manifold with different signature, and there null cones aren't cones
Oh great...
@ACuriousMind Is there any intuitive reason as to why when we include two timelike coordinates i.e. change the signature the 4-sphere "breaks" off to a torus $S^3 \times S^1$?
19:08
I have no intuition for more than one timelike dimension at all :P
By intuitive reason, I mean sometimes you say cool stuff, you know?? Without solving any equations or so, you just say "oh this was expected because <blah blah blah>"
I understand but I stand by my statement :P
@ACuriousMind So is there any way we could have got the $S^1 \times S^3$ directly for the "5-cone" like thing, without going to its double cover... I mean by intersecting with a plane or something?
in principle, sure
whether it would have been easier I don't know

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