- #1
Zoot
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The Pauli-Lubanski spin 4-vector for a particle in its rest frame is given by Sμ = (0, Sx, Sy, Sz). (Note this is the version from Rindler, which defines S in terms of the 4-velocity instead of 4-momentum) From the 3 spatial components, it is easy to construct a unit vector that points along the direction of spin. (I realize that since quantum mechanics doesn't allow us to simultaneously know Sx, Sy, and Sz these represent expectation values). Now suppose we boost this particle to another frame using a standard Lorentz transformation. We pick up a time component S0, and the spatial components change also. My question is, can we still use the 3 boosted spatial components Sx', Sy', Sz' to determine the 3-direction that the spin points in the boosted frame?
As an example, suppose an electron is "spin-up" along the z-axis in its rest frame (meaning 100% chance of measuring spin up along z). Would its spin vector be S = (0, 0, 0, h/2) where h = h bar?
As another example, suppose the electron's spin in the rest frame points in some arbitrary direction such that S = (0, Sx, Sy, Sz) and we do a boost along the x-axis to get
S = (γβSx, γSx, Sy, Sz). If the boost is LARGE, so the x component dominates, does this indicate that the spin now points along the x direction?
As an example, suppose an electron is "spin-up" along the z-axis in its rest frame (meaning 100% chance of measuring spin up along z). Would its spin vector be S = (0, 0, 0, h/2) where h = h bar?
As another example, suppose the electron's spin in the rest frame points in some arbitrary direction such that S = (0, Sx, Sy, Sz) and we do a boost along the x-axis to get
S = (γβSx, γSx, Sy, Sz). If the boost is LARGE, so the x component dominates, does this indicate that the spin now points along the x direction?