@Obliv i think whether one can get "core ideas of a physical theory" has quite a social dependence. electricity and magnetism has sufficiently infiltrated colloquial speech such that the supposed mechanisms are just accepted without thought

i imagine in 50 or 100 years (if humans still exist) quantum mechanics as it is written in textbooks now may no longer be considered "unintuitive"

I still don't think some of its basic notions are that unintuitive even now. one probably has wondered before when a coin has been flipped and is out of sight, whether the coin really, truly occupies a state of heads or tails.

if one did not know classical physics why should one think that it does

@naturallyInconsistent Ah that was simple. I solved that using a regularization technique earlier though :p

@Sanjana miao miao hath tonnes of quick tricks miehehehe~

one thing that is quite sad of a lot of people who were shortchanged by their teachers to not learn anything outside of the basic syllabus in maths, is that they never get to learn quick tricks. Of course, quick tricks are completely unnecessary to learn physics, but that makes learning physics arduous. Miao miao can always pick the shortest maths path for finding solutions to quite a lot of physics problems.

@SillyGoose doubt. The way it is taught is just appalling.

It takes a while for people to clear out the obstacles in pedagogy and find out the best ways to teach things

That is also why it is not always the best idea to stick to the originals.

@SillyGoose the reason quantum is said to be unintuitive has nothing to do with how its taught

its about the fact that u live ur life in a classical world, so your expectations about the behaviors is necessarily classical

at least this is what people mean when they say qm is unintuitive

A bartender walks into a bar and ...?

Oh sorry, I mixed some things up. Let's try again: knock, knock ...?

Knock knock ...?

@Relativisticcucumber i'd be much more specific than this. I mean, there are plenty of QM behaviours that are completely intuitive. For example, the superposition of horizontal polarisation and vertical polarisation can make one of the two diagonal polarisations, is completely intuitive even in classical physics and nobody complains about it. However, that this is the exact same mathematics that implies that atoms can be in two places at the same time is unintuitive.

@SillyGoose That is, it is that it is clashing with the classical notions of what are fundamental independent underlying quantities, guaranteed to be good reasoning when used in classical settings, is what is being complained as unintuitive, and this seems to be hardwired in the brain. We can do experiments to illicit responses in babies and animals and whatnot, when we violate brains' expectation for, e.g. things to stay in the same place, thereby taking as conclusion that it is hardwired.

hi

@naturallyInconsistent yes i agree

B A H ~

how can this q be more focused? seems p focused to me? physics.stackexchange.com/questions/818644/…

closed in like one second kms

god i hate the mainsite

i jsut want one example tho

Also, entanglement is mostly in the lab. It is difficult to say "Nature outside of lab" because if it were out in Nature, we would not have been so shocked by it existing at all.

i thought "a" means one

hm okay ill clarify that i want one example

Wait,

you should start by clarifying why it is it has to be in Nature and outside of lab

what is the purpose of having such an example?

okay ill clarify those

bc one of my students asked me and i could only think of bbo crystal entanglement but she asked if entanglement happens in nature

i said well surely yes

she said where

i said um XD

hahahahahahah

then i looked online and ppl were all like "things interact and thus entangle"

but nowhere did i find an example to provide her with

Well, this question would somewhat be interpretation dependent. If you accept Many Worlds, e.g. anything that accepts decoherence, then the very act of measurement itself is entanglement in Nature

@naturallyInconsistent hm can you elaborate

However, if you accept collapse, Copenhagen, then I don't think we have natural examples of entanglement. There are natural examples of superposition, of resonances, and so forth.

@Relativisticcucumber In decoherence, the evolution of a system + measurement apparatus (and environment if you want) is modelled under a gigantic Hamiltonian, and what happens is that each branch of the measurement apparatus's possible outcomes are entangled with a different set of the system's possible wavefunction branch. The multiverse continues to see that every possibility exists, but each branch sees just their specific one.

Also, do you consider LCAO's "electron is in two places at once" as entanglement, or do you insist to use stationary molecular orbitals, which is already spread throughout the molecule?

i.e. what is the specificity of how you define entanglement also impacts the selectivity of the result.

@naturallyInconsistent oh no i thought lcao was superposition wait lemme read

It is superposition, definitely, but do you also consider it to be entanglement enough? I mean, you can have the orbital and its spin part entangled. Does that matter to you?

@naturallyInconsistent also about this, am i correct in thinking an entanglement operator should take in two systems and return an entangled system? like it can take in $\vert \psi_1 \rangle$, $\vert \psi_2 \rangle$ $\vert \phi_1 \rangle$, $\vert \phi_2 \rangle$ and return something like $\vert \psi_1 \rangle \vert \phi_1 \rangle + \vert \psi_2 \rangle \vert \phi_2 \rangle$?

@Relativisticcucumber I don't think we should be talking about "entanglement operator" at all, especially if you understand the no-cloning theorem. However, the latter part is a yes; entangled stuff look like that.

hm so im trying to understand how i can think of making a measurement as producing entanglement in many worlds picture

i think i see why copenhagen interpretation would make entanglement not possible since it just collapses onto an item in the superposition

@naturallyInconsistent hm im not sure ive ever seen this. all i recall is that to get the canonical images of how atoms look, you have to take superpositions of the actual wave function solutions. i dont recall how spin plays into things

(I am not talking about Many Worlds in the following) Consider a Stern-Gerlach system. Initially, a beam of silver atoms flies into a SG splitter; you pick the $\left|+x\right>$ state, and it flies in a specific direction that we now call $+y$. Later, you let it pass another SG splitter so that the initial state of $\left|+y\right>\frac{\left|\uparrow\right>+\left|\downarrow\right>}{\sqrt2}$ evolves to $\frac{\left|u\right>\left|\uparrow\right>+\left|d\right>\left|\downarrow\right>}{\sqrt2}$

where u and d refers to whether the spot it up or down in z position. Now, do you consider such a thing as entanglement or not? It certainly mathematically looks so, but very few physicists would consider this as entanglement in the usual way we mean it.

wait im confused about the notation

i know that if we pick one state, say $\vert + x\rangle$ then this is still a superposition of, say $\vert + y \rangle$ and $\vert - y \rangle$ -- is this related to what you are referring to?

Yes and no. The $\left|+x\right>$ state is a superposition of z eigenstates (and also y), but my $\left|+y\right>$ is meant to represent position, not spin

Or you can consider it as momentum, the propagation direction. The SG apparatus nudges the atoms into flying at different directions.

ok i think i see

no i do not consider that entanglement

Well, you need to be aware that all entanglement stuff look just like this. $\dfrac{\left|\uparrow\right>\left|\downarrow\right>+\left|\downarrow\right>\left|\uparrow\right>}{\sqrt2}$ is no different from what I gave you earlier.

it is just that we often have the thing on the left and on the right be separated by kilometres and suddenly the entanglement result is in conflict with realism+SR

@naturallyInconsistent thats true but it disturbs me to call this entanglement xD

the SG eg i mean

I know it might feel alien to you, but it is a very standard thing. actual entanglement experiments dont care about how you see it. The example I gave is directly analogous to actual experiments. You have a particular spin system, you do something, and then they fly apart in different directions, and suddenly you have a spatially separated entangled system that becomes macroscopically obviously far apart.

Similarly, the classically un-explain-able quantum correlation does not care if it is anti-correlated (as is the usual exposition) or correlated.

Anyway, if you do not accept this type of entanglement, then the answer to your question is no. Your strict definition of entanglement will thus only happen in lab.

Hi

After being struck by Shelob, Frodo is bundled into the Spider's web and he is carried by an orc towards the watchtower in a long dark path 700 m, pulling Frodo at constant speed with a rope inclined at 30°; how much work does the ogre perform? N.B. The coefficient of dynamic friction between the spider's web and the stone humid is 0.3. Frodo has a mass of 35 kg

What am I doing wrong?

The solution is W = 61.4 kJ

For me it's 57.75 kJ (but I think I'm doing the exercise wrong)

Could someone take a look pls:)

W_attr: Friction Work, W_g: Gravitational Work, Ip: hypotenuse

@Pizza your conception of the problem is wrong. The path is horizontal.

@naturallyInconsistent oh ok, thanks!

@ACuriousMind says the guy who lectured me on Latin twice :P

@Relativisticcucumber "a" means "any" is how I remember it

Guys I'm having trouble solving an exercise: I have a particle of mass M and spin 3/2 is subjected to the following Hamiltonian: $$ H = \frac{p^2}{2M} + \frac{k}{r}(\mathbf{J}^2-\mathbf{S}^2), k >0 $$

I need to look for bound states

now I see the hydrogen-like structure of the Hamiltonian, I see its rotational invariance

but how do I look for bound states?

ok looking at Cohen-T I probably just need to require that $(J^2-S^2)$ be positive so that I have the usual behavior of the centrifugal barrier($\frac{1}{r^2}$) and the coulomb potential $1/r$, and then I'm sure that the motion is bounded for negative energies