The short answer is that relativistic quantum fields correspond toTh[...] is exactly my concern: The relativistic E^2=p.p+m^2 only comes from the field satisfying Klein Gordon which is the case when we have a scalar field or a spinor field but for an arbitrary theory I do not see why the e-values associated with the Casimir operator p.p (4-vectors) should be the mass^2.
Most physicists define the mass of an object as the invariant length of its four-momentum. This turns out to be a useful definition. It has nothing to do with a wave equation.Thanks for the reply, but what you are pointing out is exactly my concern: The relativistic E^2=p.p+m^2 only comes from the field satisfying Klein Gordon which is the case when we have a scalar field or a spinor field but for an arbitrary theory I do not see why the e-values associated with the Casimir operator p.p (4-vectors) should be the mass^2.
Yes, it is. What is the purpose of builing a lagrangian for an arbitrary classical field ? To use it in the quantization formalism. Therefore, we need a mass term, since the mass of the quanta of the field is essential in the quantization...I agree with the earlier posts and should rephrase my question: Is the coefficient of the quadratic term of a fundamental field in an arbitrary Lagrangian density always related to the eigenvalue of the Casimir operator p.p (4-momentum squared)? If this is the case, could you pleae tell me how and why it should be related? Thanks in advance
That's how we define the mass of the quanta of the field.Could somebody please remind me why the e-values of the Casimir operator p.p (4-vector product) are the mass^2.
A (relativistic) Lagrangian represents a particular physical theory[...] rephrase my question: Is the coefficient of the quadratic term of a fundamental field in an arbitrary Lagrangian density always related to the eigenvalue of the Casimir operator p.p (4-momentum squared)? If this is the case, could you pleae tell me how and why it should be related?
This is the one.the operator commuting with all the infinitesimal generators of the Group Algebra?
Provided thatFor example: If I take an Algebra [tex]\mbox{so(3)}[/tex], identify its infinitesimal generators, [tex]\left{ \tau^i \right}[/tex] and take an operator, call it, [tex]L^2[/tex] and find that
[tex]\left[ L^2, \tau^i \right]=0[/tex]
for each [tex]i=1,2,3[/tex].
Then this is sufficient to claim [tex]L^2[/tex] is a Casimir of [tex]\mbox{SO(3)}[/tex]?
1) [itex]L^{2}[/itex] is invariant under SO(3) transformations.
2) it is constructed out of the group generators.
regards
sam