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 Jul 21 comment How to prove that a spacetime is maximally symmetric? Ok. Thanks for the clarification. Jul 21 comment How to prove that a spacetime is maximally symmetric? Sorry if its a trivial question but I don't understand why you didn't consider the possibilities $x^2+y^2+z^2+w^2-t^2=-r_0^2$ and $x^2+y^2+z^2-t_1^2-t_2^2=r_0^2$? Jun 20 awarded Yearling Apr 5 answered How to derive the form of the invariant spinor inner product? Jan 30 revised Irreducible Representations Of Lorentz Group added 294 characters in body Jan 30 revised Irreducible Representations Of Lorentz Group added 43 characters in body Nov 15 comment Path integral derivation of the state-operator correspondence in a CFT The space of boundary conditions at the origin is anyway much smaller than on a circle of nonzero radius and hence can't be the whole configuration space. Nov 15 comment Path integral derivation of the state-operator correspondence in a CFT @Prahar ya you are right. I read that in a hurry. Perhaps the argument is that to each primary state we can assign a local field (through some algorithm that i don't remember) and then fields corresponding to other states can be generated by applying differential operators (Ln's) to the primary fields. But, I have never encountered any rigorous proof of these statements. Nov 15 comment Path integral derivation of the state-operator correspondence in a CFT I mean the map may not be surjective Nov 15 comment Path integral derivation of the state-operator correspondence in a CFT Also, the correspondence is 1-1 in one direction i.e. to each local functional we can assign a state on the boundary. However i am not sure if to each state specified on the boundary we can construct a local functional or not. Nov 15 comment Path integral derivation of the state-operator correspondence in a CFT @Prahar the wavefunctions are of the form $\psi(\phi_i(\sigma))$ where $\phi_i(\sigma)$ is field specified on a circle of nonzero radius. On a circle of zero radius (i.e. a point) there are no (nontrivial) boundary conditions to be specified and we can at most associate a local functional depending upon the value of the field and its derivatives at that point. So by taking the r->0 limit of a wave function assigned to the inner boundary of an annulus we may only get a local functional at the origin and not a wave function. Nov 15 comment Path integral derivation of the state-operator correspondence in a CFT @Prahar At the origin $\psi(\phi_i)$ is not a wave function. Its rather a local functional of the field. Wave functions are assigned to proper boundaries. However, I am not sure if the above map $T_D$ is invertible. Nov 15 revised Path integral derivation of the state-operator correspondence in a CFT added 103 characters in body Nov 15 revised Path integral derivation of the state-operator correspondence in a CFT added 103 characters in body Nov 15 answered Path integral derivation of the state-operator correspondence in a CFT Sep 30 awarded Explainer Aug 20 comment The states of the adjoint representation correspond to the generators The matrices $T_a$ span a Lie algebra $g$ which, in particular, is also a vector space(or state space). The adjoint representation is a representation of $g$ on itself. That is, the states are again the matrices $T_b$ (or in bra notation $|T_b>$), and a matrix $T_a$ acts on state $|T_b>$ as $T_a|T_b>=|[T_a,T_b]>=if_{abc}|T_c>$. Aug 13 revised What is the physical meaning of commutators in quantum mechanics? added 838 characters in body Aug 13 answered What is the physical meaning of commutators in quantum mechanics? Jul 16 comment Is there a physical system whose phase space is the torus? +1: Just a minor point: mathematically one may consider non-relativistic massless particles. But no such particles are found in nature.