The symmetry depends on the Lagrangian.
Let γμ be a real representation of the Dirac matrices for 4d spacetime, and define Γ:=γ0γ1γ2γ3. Then Γ is also a real matrix, and Γ2=−1.
If ψ is a Majorana spinor field (with self-adjoint components), then the corresponding left-handed Weyl spinor is
ψL:=1+iΓ2ψ.
So yes, the Weyl spinor ψL may be written in terms of the Majorana spinor ψ and conversely, but the two Lagragnians
L∝¯ψLγμ∂μψLL′∝¯ψγμ∂μψ
are not the same. (I'm suppressing the flavor index.) In particular, they have different flavor symmetries. If we start with L and rewrite it in terms of the Majorana spinor, we get
L∝¯ψγμ∂μ1+iΓ2ψ,
which is different than L′. The flavor symmetry of L is still U(N). To see this, use the identity
iψL=−ΓψL
to see that multiplying the Weyl spinor ψL by i is the same as multiplying the Majorana spinor ψ by −Γ, after which its components are still self-adjoint. This shows that every U(N) flavor transformation of the original version of L can be re-written as an equivalent flavor transformation of the new version of L, using −Γ in place of i, so the flavor symmetry group is still (isomorphic to) U(N).
This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Chiral Anomaly