Setting. We are considering a transformation, which may transform the field variables $\phi^{\alpha}(x)$ and which may transform the space-time points $x^{\mu}$. The transformation in turn apply to
The action $S_V[\phi]=\int_V \! d^nx~{\cal L} $.
The Euler-Lagrange equations = the equations of...
Only one state is actually symmetric under the Poincaré group by the way, and that is the vacuum
I wonder what the Giant Diagram of Groups look like if you include the various structures, extensions, covers, projective and conformal variants, contractions and Whitehead towers of the Poincaré group
Sorry by $SO(3)$ symmetry, I meant that if you actively rotate the point particle around it's center and keep everything else about the system the same, the physics doesn't change
You can't rotate a point, it's a point. The symmetries are symmetries of the Lagrangian or Hamiltonian describing a given physical system, some systems have symmetries, others don't, for example the Kepler problem $L = \frac{1}{2} m \dot{\mathbf{r}}^2 - V(r)$ has spherical symmetry
no one cares if the numeric values of your position vector don't change just because you chose the origin cleverly
physical "symmetries" are symmetries of things like the action or the equations of motion or their solutions - it's pointless to talk about "a point particle", you actually have to talk about a physical system with an action and everything
I mean initial conditions or configurations having certain symmetries certainly has its importance
"In quantum theory, spontaneous symmetry breaking requires the system to be infinite dimensional. When the number of degrees of freedom is finite, spontaneous symmetry breaking does not take place."
we know that $\mathrm{SO}(3)$ is a group under which the theory should transform. The nature of quantum states means that therefore the group is represented projectively on the quantum space of states
that's the argument for spin without ever even mentioning classical mechanics
For a system in eq. it's distribution function \rho(x,p,t) is non-zero constant in some region and zero elsewhere. In the case of MCE this region is practically a line, along the which, \rho is constant?
@imbAF it's not a line, it's a (hyper)surface of codimension 1 - as you know, for the ideal gas it's a cube in the position part and a sphere in the momentum part
but because I draw that sphere in a 2d plane it's a line :P
@ACuriousMind I am familiar with the sphere only, I don't know about the cube in position part. The only cube I know, is the 6N dimensional cube of a microstate in phase space
@ACuriousMind If we have a system AB, made by two sub-systems A and B, and for simplicity the total system (AB) is in a pure state $|\Psi\rangle_{AB}=|\Psi\rangle_{A} \times |\Psi\rangle_{B}$. The density matrix of it would be $\rho_{AB}=|\Psi\rangle_{AB} \langle \Psi|=|\Psi\rangle_{A}\langle \Psi| \times |\Psi\rangle_{B}\langle \Psi|= \rho_A \times \rho_B$
For a particle, if $e(\tau)$ transforms as $e'(\tau') = \frac{d \tau}{d \tau'} e(\tau)$ under a 'gauge' transform, and you 'fix' a gauge by setting $e = 1$, this is supposed to be saying $1 = \frac{d \tau}{d \tau'} 1$ so that there would be no more 'gauge' transformations that preserve this, however we can actually have $c = \frac{d \tau}{d \tau'} 1$ for $c$ a constant, as long as $\frac{d \tau}{d \tau'} = c$, so it's only partial, right
Now if the joint system AB is in a mixed state his density matrix would be: $\rho_{AB} =\Sigma_i P_i |\Psi\rangle_{AB}^i \langle \Psi|^i$, is that right until now?
@ACuriousMind and how would i write that same density matrix,but expressed via the density matrix of the sub systems, just like I did initially for the pure state of the joint system?
@DIRAC1930 that's an ill-defined question - if you just say "I do a rotation", then you have to act with the total angular momentum, i.e. both orbital and spin. But there are situations where it makes sense to only act with spin or only act with orbital angular momentum, since they can be conserved separately
@imbAF there is - from the density matrices you get from the partial trace, you get back the original matrix precisely as $\rho = \rho_A\otimes \rho_B$
if you have two representations $V_1$ and $V_2$ of a group $G$, you can always talk about only transforming $V_1$ while doing nothing on $V_2$ and vice versa
whether that's a useful idea depends on the context, but you can do that
you could think of this as producing "copies" of $G$ where now $V_2$ transforms in the trivial rep of the copy or something, but you also can just...not think like that
@imbAF yes (I see that as the justification for why the partial trace is the correct operation to begin with)
And that's what I was trying to do early on. Find a way to change this $\rho_{AB} =\Sigma_i P_i |\Psi\rangle_{AB}^i \langle \Psi|^i$ into $\rho = \rho_A\otimes \rho_B$
@imbAF I give the way to obtain the probabilities in the subsystem from the $p_{ij}$ in the combined system explicitly in eq. (1) in the answer I linked
and $L^2(\mathbb{R}^3)$ decomposes into the representations $H_\ell$ (spherical harmonics!) with integer spin $\ell$ under the action of orbital angular momentum $x\times p$
@DIRAC1930 you can't "derive" that the electron is a spin-1/2 particle any more than you can derive its mass or charge!
I look at a Stern-Gerlach experiment with electrons, I see spin-1/2 behaviour, so I choose spin-1/2 particles as my electrons
Okay so a theory with a classical vector field that transforms under $SO(3)$ when quantized is essentially just a quantum theory with quantum point particles and no spin
if all your weird questions amount to "why does the particle associated with a quantum field have the same spin as the representation on the target space of the field", it basically once again comes down to the compatibility condition $\rho(g)\phi(x) = U(g)\phi(x)U(g)^\dagger$, but Weinberg's QFT book also has a lengthy technical demonstration of that (and of course also a discussion of massless relativistic non-scalar fields being special)
I do not recommend it for learning QFT, I do recommend it for a) learning at lot of subtle arguments other sources skip over and b) learning a very different perspective on QFT since he starts with the particles and then constructs the fields, while everyone else does it the other way around
the notation is terrible, you just have to suck that up :P
it's more a tool than a subfield on its own, much like "quantum mechanics". The subfields that use it are things like high-energy physics, condensed matter, ...
