@Slereah obviously I'd prefer to forget that I've been in labs ;)

The man is a genius

Look, if a asystem is in a state,which is superposition of eigenstates, and expressed as a linear combination of basis kets multiplied with some coef. the probability for you to measure an eigenvalue, is equal to the square of coef. of the corresponding eigenstate, this is the simple case of non degenerative and discrete basis

Some 'ugly' things he does in his book are actually simply brilliant

@imbAF correct, so what issues do you have applying that here?

the space of spin-states is two-dimensional, it doesn't get simpler than that

$\mathbb{C}$ is a zero-dimensional projective Hilbert space

It's even better

But in my example, I am having particles with pozitive z-spin direction and I want to measure the x-spin. ANd I am asked about the result of this experiment, which are the states present

One state, one operator

@imbAF you can express the state of definite z-spin as a superposition of states of definite x-spin

we usually don't do the reverse? use the eigenstates of the z-component

to express x and y-spin

The $z$ matrix is diagonal so it's usually the one people pick

But they're all pretty equivalent

I don't know what "usually" means - you just told me that if you want to measure an operator, you need to figure out how to express your state as a superposition of eigenstates of that operator

which x-spin = $\frac 1 {\sqrt(2)} [|z,+> + |z,->]$

so if you want to measure x-spin, you should express everything in the basis of x-spin eigenstates

Projection to the zero dimensional Hilbert space is used to determine if something is going on

probability of something going on is $1$

idk honestly, I mostly don't remember these stuff well. Like I know that spin components do not commute

but that does;t mean that a system

forget it, I can't even articulate what I want

Apparently you can decompose the $so$ algebra into a graded algebra???

part of which being the conformal orthogonal group

Man I really don't know nothing about groups

What is the conformal orthogonal group

good question

And the best part is when the professor hits you with that : You don't know about groups? SO3 symmetry and Clifford algerbra? ooohhh you should have done that in high school

There's apparently a million variants of the orthogonal group

@ACuriousMind later on in the same sub section of the example, I am asked :Is the state of the system after the measurement a pure state or a mixed state and why?

And it's a pure one

we know that from the QM postulates

You can re-write the conformal group in $D$ dimensions as $SO(1,D+1)$, maybe that's what is meant

@imbAF that's a bit of an ambiguous question

well

which implies that the description of the problem early on

has to do with mixed states

it depends on a) whether you know the result of the measurement or not and b) your quantum interpretation :P

well I am asked about the result

for which I am unsure as to what to say

"In the Lorentzian case the space $S^{1,n−1}$ is the compactification of the Minkowski space $R^{1,n−1}$ by a light cone at infinity."

Whaaat

by "result" I mean the value which the measurement device shows for the spin

that is something you should deduce

from the set up described to us

z-spin positive and find x-spin, what are the states, what is the result of the experiment

this is exactly how it's framed

sure, but if you just know "a x-spin measurement has occured, but I don't know what the result was", then it's perfectly reasonable to say that the state is currently a mixed state with 50% x-spin up and 50% x-spin down. if you know the result, then it's just the pure state corresponding to that result

as I said, that's just an ambiguous question

if that's the case

then why is it give to us also that the system, particles have a postiive z-spin

because the input matters?

what does this information does to us?

if you had sent in a state e.g .with x-spin up, the result would just always be "x-spin up" and the resultant state 100% x-spin up

So, the input are particles with z-spin positive, and the magnetic fields are in the x direction?

...why are we back to magnetic fields? I thought we had agreed the exact method of measurement doesn't matter?

yes

The thing is. I am confused about one thing,

If the input are z-spin pozitive particles

for which we want to measure the x-spin

I think it's just saying you treat infinity as a point of space-time and you get $S^{1,n-1}$

@bolbteppa I guess maybe it's like identifying the two conformal boundaries of a Penrose diagram?

idk

@ACuriousMind Usually when you express the density matrix for a system, that is an indicator of whether the system is in a statistical mixture of states, or in a pure state. But then if you have to give the density matrix of N-particles, N system, how do you do it?

Usually in my class we mostly confine ourselves to one system

and express the density matrix

I'm afraid I don't understand the question

I think the exercise at hand, is TOTALLY unrelated to what we do in the lectures, since the person who does the lectures and the one who does homework exercises are two different people, that do not interact at all

that's why

right now, I feel like I am blindfolded and asked to walk over a narrow bridge

but out of curiosity in the same example that we were discussing, I am asked to give the density matrix of a system of N particles, from which n are in the |z,-> state and the rest in the |x,+> state

ofc that would mean

n/N |z,-> <z,-| + (N-n)/N |x,+> <x,+|

to me at least

this makes sense

"The space $S^{1, n-1}$ is diffeomorphic to $(S^1 \times S^{n-1})$"

Is that kosher

I know that it's not necessarily a spacetime but it is weird to have spherical spaces for that

Although I guess that since one point is supposed to be the point at infinity maybe it's allowed not to have a non-zero timelike vector on it

Also how am I supposed to take this seriously when one object is called the tractor bundle

"The term tractor is a portmanteau of "Tracy Thomas" and "twistor", the bundle having been introduced first by T. Y. Thomas as an alternative formulation of the Cartan conformal connection"

I fear if I want to do some kind of hierarchy of spacetime structures I'm gonna have to do a big old diagram

@ACuriousMind In the micro canonical ensemble, the microstates are eigenstates of the Hamiltonian. But what about in the case of a canonical ensemble? They shouldn't,right? Since energy changes

Congratulations to @DanielSank who made the 2021 list of highly cited researchers: recognition.webofscience.com/awards/highly-cited/2021/…

what's cross field?

aka multidisciplinary . And in itself it’s a bit of a hodge podge.

biomolecular engineering and the likes…