The h Bar

1:24 AM

@Obliv The issue is that the B field is already determined by the original flat surface loop, so the same B field has to be found from deforming the flat surface into a balloon surface that doesn't pass through the wire, and instead passes through the gap between the capacitor plates.

@SillyGoose yes, this is covered in basically every treatment of EM that has this level of maths. Not that they would say a lot about it, but they would note that two of them are mathematical requirements of consistency, and two are not. And often they would also cover the magnetic monopoles, in which case the "always zero" pair is also possibly circumvented.

Relativisticcucumber

1:59 AM

the more i study physics the more i get confused about things i once thought i understood

onto my 50th revisit of the pauli exclusion principle

i understand that on physical ground, it should be that probability amplitudes dont change upon particle exchange. this would mean that if we do exchange particles, we should just pick up a phase. is there a reason that we only have the case of symmetry and antisymmetry as opposed to a range of phases?

Obliv

2:14 AM

what do u guys think

"from the ground" means

nvm I see it now

naturallyInconsistent

3:28 AM

@Relativisticcucumber That goes all the way up to spin statistics theorem, but at your level maybe you could just see that, if you apply the same exchange operator twice, it needs to go back to itself. i.e. the eigenvalue is only $\pm1$ because only $(\pm1)^2=1$. That there are no other possible phase relationships is due to spin statistics theorem.

Relativisticcucumber

how does your whole state/system picture look for e&m? @SillyGoose

naturallyInconsistent

@Obliv SR requires specification of which frame you are measuring things by. The question is measuring everything in terms of the ground, i.e. takes the ground as stationary, and then asks you what the car will think the train's length would be.

Relativisticcucumber

@naturallyInconsistent I get that if I have a ground state and excited state, for example. Then if I have two fermions. If they were both in the ground state, I would need to swap them and pick up a negative sign. But I cannot do that because they would be indistinguishable, so I should really just get the same state back when I swap. This means that fermions cannot be in the same state. [...]

[...] But this is quite confusing to me because how do they just distribute in the proper way — I asked this on the main site awhile back and it seems fermions naturally experience repulsion, but I dont understand how that happens physically

naturallyInconsistent

@Relativisticcucumber no, when you swap, including indistinguishability, you still get a negative sign.

Relativisticcucumber

and iiuc, this q is purely about the stat mech side of things. i dont care so much about relating this to spin. but when i ask this to many people, they always answer w SST. im not sure why though because iiuc, all SST tells us is the spin value of the particles we classify in stat mech

@naturallyInconsistent right so isnt that why the fermions cant be in the same state -- because we cannot have this happen?

or am i misunderstanding

it seems a fermion must know somehow that another fermion is already in that state but this seems absurd

naturallyInconsistent

3:32 AM

@Relativisticcucumber The "repulsion" from wavefunction needing to obey Fermi-Dirac statistics, which is the core underlying principle that enforces Pauli Exclusion Principle, is not a repulsion at all. There is no force involved. It is purely that the wavefunction has a certain antisymmetry behaviour in there.

Relativisticcucumber

i dont understand how to develop intuition for this. i worked through the calculations for the BE stats and FD stats but i am left just puzzled

i feel it just tells me the answer but gives no mechanism

naturallyInconsistent

@Relativisticcucumber It is not absurd. If you think of it as fermions running around, then you will feel it is absurd for one to know another. But if you think of it as that there is one single underlying field, and what we call as particles are merely excitations on top of that one field, then it makes much more sense that all excitations on the field will know that there are other excitations on the field.

Relativisticcucumber

@naturallyInconsistent really? :o

hm so the information is somehow engrained in the field?

naturallyInconsistent

@Relativisticcucumber Yes. And so there is a need, say, for the spin-half portion to be antisymmetric, so that when the orbital part is symmetric "due to being in the same orbital state", the fermion system is still acceptable.

@Relativisticcucumber only in the form of "these states have already been occupied", which is not possible to communicate faster than light. The proofs surrounding these are quite intricate and Feynman famously asserted that he could not explain this to a child.

@Relativisticcucumber the spin statistics theorem is the only thing that enforces which set of particles obey BE stats and which set of particles obey FD. Trying to separate the behaviour (of FD v.s. BE stats) away from spin is thus somewhat futile. But yes, you can easily consider some spinless yet FD stats obeying nonsense, if you want to consider them theoretically.

Anyway, this is also why a popular modern thinking is that all these quantum weirdness are due to quantum fields. It is not waves and/or particles. It is that the quantum fields underlying them are the real thing, and excitations on these quantum fields can look like waves or look like particles. No need to keep the outdated dichotomy of classical waves v.s classical particles

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