Lagrangian/Action Formalism As a Flow Scheme and Spin Part II
Francesco R. Ruggeri · Zenodo (CERN European Organization for Nuclear Research) · 2022
In Part I of this note, we argued that the exp(ipx) portion of a free quantum particle’s wavefunction follows from d/dx A(x,t) = p extended to an eigenvalue equation: -id/dx exp(iA) = p exp(iAc) “p” appears as a scalar because it is the momentum magnitude, but one may consider flow in 3-space i.e. px,py,pz. In such a case, one might think of using basis vectors ei such that ei dot ei =1 (i=1,2,3). Given that one has an operator equation, one uses basis operators or matrices instead leading to the Pauli matrices or 4x4 matrices as found by Dirac if one imposes the restriction that upon “squaring” the operators disappear because one only has pp. As a result, we argued that intrinsic spin is linked to 3-space recognition for a free particle. The resulting spin, however, is spin ½ so what about other spins say 0 or 1? This seems to imply that free recognition of 3-space does not apply to other spin cases as it does to spin ½. The Stern-Gerlach experiment already shows that spin ½ is interesting in that one may have spin up or down project along any 3-space axis. In Part I, we showed that a wave-equation developed for the photon from d/dt (partial) b (ElEl + BB) + grad dot ElxB = 0 contains an eijk (-i d/dxi) term with eijk being the Levi-Civita symbol as well as the spin matrix. This is not the Pauli matrix (or matrix containing it) and suggests a different geometry or basis scheme. Thus, there is physical angular momentum (spin) in a plane perpendicular to the momentum and the (1,0) spin component is “hidden”, but associated with angular momentum in another spatial direction. As a result, 3-space is not treated in an independent way as in the spin ½ case. In this note, we investigate this idea further by considering the deuteron, He4 and He3 and try to argue that intrinsic spin still seems to have a connection to 3-space recognition, but that the fully independent 3-space recognition corresponds to spin ½. We only consider spin 0, ½ and 1 in this note.