The symmetry group underlying a de Sitter SR is certainly not the same as the one underlying Minkowski SR. The symmetry group of the latter is the Poincaré group P, and the Minkowski spacetime is defined as the quotient group P/SO(1,3) such that it is transitive under ordinary translations. The...
The difference between a real gravitational force and a fictious inertial force is that the Riemann tensor does not vanish in a region of space-time when there is a real force, as opposite to a fictious force. The reason is that when there is curvature it is impossible to find a GCT such that...
You should be careful in considering the "mass" of an object in a relativistic theory. In special relativity the rest-mass is a form of energy but not all energy is rest-mass. Basically particles can be divided in two subclasses: those with a rest-mass (e.g. the electron) and those without (e.g...
Ok I solved it. For those interested, use the defining equation for the clifford algebra and the fact that the antisymmetric part in spacetime indices is symmetric in its weyl indices, such that this term is zero when combined with the antisymmetry of the theta's. This gives a kronecker delta...
Hello,
I'm trying to deduce the suspersymmetry transformation of the chiral multiplet out of the superfield formalism. In doing this I got stuck with this...
I think that in the first place one has to define what is meant with something physical. In my opinion I would define it as how we observe the universe in a real sense, that is by experiment. So in this definition one eliminates the question whether the universe is physical or not: it is...
A wavefunction is non-physical in the sense that you'll never encounter a wavefunction as such (a complex valued function) in experiments. The only thing you measure upon are amplitudes (which are real).
You have to be careful with what you call non-physical and what not. Physics as such tries...
Why is that? Because for these states not all Pij commute with each other?
So if I understand you well: it is within the framework of QM that we can conclude that only symmetric and antisymmetric states are physical. But to know which particles belong to the symmetrical part and which to the...
Ok, maybe this is the explanation: because of the very nature of indistinguishability every operator Q must commute with every element of the permutation group. But this can only be true for symmetric or antisymmetric states, which form invariant subspaces of the product space of the one...
Alright, this means that once your system is (anti)symmetric it stays that way because every observable commutes with P. But this does not explain why a many particle state is (anti)symmetric in the first place (or does it, and if so could you please clearify this)?
For the composite system of identical particles only symmetric and antisymmetric states in the tensor-product (from the one-particle spaces) space are allowed to represent particles in nature. Why is that?
Is it an experimental fact which is used as an input in the theory of many particle QM...
The http://en.wikipedia.org/wiki/Dirac_equation" , which itself is based upon the relativistic energy-momentum relation E^2 = p^2 + m^2 (natural units). And here comes my question then:
Why do we throw away the negative energy solutions in relativity but do we keep them when we combine it...