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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.