In central potential problems, typically we have non-trivial $\phi$ dependence, but during phase shift analysis for some reason, the expansions contain only Legendre polynomials not spherical harmonics/associated Legendre Polynomials. Why? Books say it is because the potential is "central". But central potential means independence from both $\theta$ and $\phi$. Why is $\phi$ special?
@ACuriousMind Thank u so much. But one last question. Where does the extra $U(\Lambda)$s come from? I thought covector fields transform as $A_\mu'(x')=\Lambda_\mu^\nu A_\nu(x)$. Is the $U(\Lambda)$ implementing the primes on the left on $A_\mu$?
So $\langle 0 | A_\mu \Lambda | 0 \rangle=\langle 0| A_\mu |0 \rangle$. If the RHS is non-zero, then the LHS should be non-zero too. Is it zero in general for some reason?
People say because the VEV $\langle A_\mu \rangle=v_\mu$ points out in a particular direction: that breaks the isotropy of spacetime. Well, doesn't the field $A_\mu$ do the same, then?
Usually people say that you can't have a VEV for non-scalar fields because it violates the translation/rotation symmetries, but in equations why does it do so? I mean why won't the field itself $A_\mu$ itself then "violate" Lorentz symmetry?
I have another unrelated question: how do you define action principles on non-orientable surfaces? E.g. I know people do string theory with let's say the Mobius strip as the worldsheet, but how do you even define the Polyakov path integral which has a volume form in it's very definition?
This wiki page says that $\mathrm{Spin}^{\mathbb{C}}(n)$ is a central extension of $\mathrm{SO}(n)$ by $U(1)$. But $U(1)$ does not even lie in the center of $SO(n)$ for generic $n$. The center is $\mathbb{Z}_2$ $\forall n \ne 2$.
Lubos Motl's answer here supports my claim that asymptotic freedom implies UV finiteness. So what could the authors of the paper could have possibly meant?
> "physicists have constructed many, many variations of Yang-Mills theory in the search for regulated (UV finite) versions as well as more tractable analogs of the theory"
A basic question: I know that finite dimensional vector spaces of the same dimension are isomorphic to each other. So why/when do people want to distinguish between a space and its dual. E.g. Why/When do people go crazy over $T_p M$ vs $T_p^* M$. When does this distinction actually matter?
@ACuriousMind But in the definition we have maps to local subsets $U_i$ of $\mathbb{R}^2$. Here we have a map from the manifold to all of $\mathbb{R}^2$. So I have a $K=+1$ map not only on the UHP but all of $\mathbb{R}^2$?
@ACuriousMind The stereographic projection: there's also a point on the sphere which is not mapped to the plane. But that's not a problem, right? Because then it will still be a map just not surjective
Can u put a metric of positive definite curvature $K=1$ say on the upper half plane instead of the Poincare metric? I know you can use the flat metric, but is the $K=+1$ allowed too?
In books like Nakahara the gauge transformation of connection is given by $A'=g^{-1}(A+d)g$. So there are two "Jacobians" : the two $g$s. Why do we have three Jacobians in the homogeneous part of the transformation of Christoffel symbols then?
@ACuriousMind But the current density is sometimes thought as the Noether current of this global $U(1)$, but then yeah you consider only rays it is not a real global symmetry. So how is there a Noether current?!
@ACuriousMind niceee. Thanks. But one last question: this is a gauge symmetry. If it was global only, then it would also have a conserved charge... What is the conserved charge due to this? At first I thought it is the number operator but that is due to a $U(1)$ probability conservation. This is different
@ACuriousMind It's just strange to me that the Hamiltonian is a connection. Usually in physics (atleast classically), connections are something which is a bit "unphysical": and we look at the curvature... but the defining equation is not based on the curvature but the connection itself?
I mean not that new :( pleeeeease I have looked into YM stuff, instantons, even read your answers to make the notion of diffs in GR precise using frame bundles and solder forms. It is only that I didn't find discussions on this which is why I am asking you.
@ACuriousMind Thing is I can't know by myself that I know "enough", right? One of the ways to figure out is try to answer questions that come up while studying something else and when I have tried for long enough I come to ask here :)
@ACuriousMind Because that's how I usually define my gauge fields in physics? I look at the tangent space of the $G$- bundle and then use a connection form to project down to the vertical subspace. That's my Ehresmann connection. And then I use a local section of the $G$-bundle to pullback to one of the covers of my base to define a connection there: and that's my gauge field... This is what I learnt. So this is what I said
@ACuriousMind And all of these are trivial bundles... ok. Which one of these do I want? I mean if I want to look at the Hamiltonian as a pullback of the Ehresmann connection by a section of the bundle to the base, then which of these bundles should I use for defining the Ehresmann connection?
@ACuriousMind I want to talk about bundles here, because I saw that the Hamiltonian transforms as a connection. And I want to know connection in which bundle? I think $\mathbb{R}$ which parameterizes the time parameter $t$ forms the base space?
@ACuriousMind So is the $G$ here just $U(\mathcal{H})$? This $U(\mathcal{H})$ is infinite dimensional so the $G$-bundle is infinite dimensional too, right?
@ACuriousMind What about infinite dimensional? So: My original question was that since $\psi \to U \psi$ implies $H \to UH U^\dagger+ i \dot{U} U^\dagger$ that is $H$ transforms as a connection. What is the $G$ in the $G$-bundle which is associated with this connection?
