Obtaining the Quantum exp(ipx) From Pauli Matrices?
Francesco R. Ruggeri
We have pointed out in previous notes that the free particle wavefunction exp(-iEt+ipx) holds for both a photon and particle with rest mass, while the latter has spin ½ and the former, spin 1. We thus suggested that spin is obtained in a very different manner from exp(-iEt+ipx). Here, we start with the Lorentz invariant -EE+ p dot p cc = -momocc ((1)) and argue, as in (1), that it must be linearized to obtain a linear equation in the intrinsic force, namely mo which is linked to inertia, because force acts in a linear form in nature (e.g. F=dp/dt). At this point, however, p is a vector and there is no concept of -ihbar grad. One may mathematically linearize ((1)) using the approach of Dirac (except that now p is a vector of numbers not an operator). This leads to 4x4 matrices with off diagonal Pauli matrices (multiplied by 1 and -1). One may see that the rotational invariant p dot p may be linearized to s dot p (s = three 2x2 Pauli matrices) because these matrices anticommute. Thus, a 2x2 Pauli matrix is associated with rotational invariance. At this point, this Pauli matrix is simply a math entity. In terms of measurement, given that one has matrices, there are also commutation relations which are important. One may consider another math operator linked with rotational invariance, namely d/dphi for a rotation about a z-axis. In classical physics, one does not have such a derivative appearing as one deals with functions. If d/dphi is an operator generating rotations, then d/dx should generate translations, but this is simply math. The “catch” seems to be that if one introduces a probability f(x,p) (which is unusual because P(x)dx = dx/L, where L is an arbitrary length), then f(p1,x)f(p2,x)=f(p3,x)f(p4,x) if p1+p2=p3+p4 (we use 1-dimension for simplicity). One may, however, shift this equation to x+dx which implies that df(p1)/dx f(p2) + f(p1)df(p2)/dx = df(p3)/dx f(p4) +f(p3) df(p4)/dx. A solution which keeps p present is: exp(i C p x). We use i because we want to have equal modulus at all x points. If C=1, the one has exp(i px) with -id/dx as the operator which retrieves p. One may then apply this idea to L = r x p and find that this new operator commutes like the Pauli matrices and gives rise to d/dphi, the rotational operator postulated above. Thus, it seems that one could start with the notion of spin and obtain the notion of exp(-iEt+ipx), we argue and spin being linked to angular momentum, we argue.
•Feb 25, 2026•0 citations