Consider the Goldstone model of a complex scalar field ΦΦ. It has U(1)U(1) global symmetry, so if we apply the transformation Φ→eiαΦΦ→eiαΦ the Lagrangian is left invariant
Dear Islam, I am confused about what your asking. If you have a single complex scalar, and I can't tell from the question, but that theory usually has a single U(1) symmetry...in the typsetting it looked maybe like you wanted a theory that has U(1)xU(1)? I can't quite tell from the formatting of the question. For a single complex scalar with a single U(1) global phase symmetry then if we quantize the system in a big box that respects the global symmetry (for example a box of volume V that has the field vanishing at the surface of the box) then whenever states have a different "charge" of some type (i.e. value of an observable, like energy, or U(1) charge or momentum if you are in periodic boundary conditions...etc) then they are physically distinct. Being physically distinct is not the same as being orthogonal in the Hilbert space, as clearly mixed states can be physically distinct, for example. Finally, some mixed states may end up becoming orthogonal in the V-> infinity limit...a special limit in which the Hilbert space breaks into (a generally infinite) direct sum of orthogonal Hilbert spaces called "superselection sectors". States in different superselection sectors are also physically distinct. Hope that helps and that I've understood the question correctly! Peace, -Mike C.
In a U(1) x U(1) theory we have to assign two different hypercharges to matter multiplets (fermions and scalars). For example a quark doublet will look like (3,2,1/6,x) and a lepton doublet will look like (1,2,1/2,y) and so on where x,y, etc. are extra U(1) charges. Here I have assumed that the extra charge does not contribute to U(1)_em charge, which is electric charge. But in a more general case both the U(1) charges should contribute to electric charge after symmetry breaking and mixing, and there should exist a relation like Q=T3L + c1 q1 + c2 q2 where the c coefficients should be obtained from normalization.
The statements about the Higgs or the Goldstone mode aren't relevant for answering the question of the main text. Nor does it matter that the group is U(1) or U(1) x U(1) or any other.
States under transformations are physically different when the invariants of the transformations take different values in each state.
Take a spin. The group is su(2) which has only a single algebraically independent invariant, the quadratic casimir. That has the same value on both spin up and spin down, so by your logic, spin up and spin down are not physically different. Huh? What am I missing about your argument. Point is, in quantum mechanics, as in GR,physically distinct states do not have to be invariant, they have to be covariant. World of difference there....
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