9.3: Eigenstates of Sz and S²
( \newcommand{\kernel}{\mathrm{null}\,}\)
Because the operators Sz and S2 commute, they must possess simultaneous eigenstates. (See Section [smeas].) Let these eigenstates take the form [see Equations ([e8.29]) and ([e8.30])]: Szχs,ms=msℏχs,ms,S2χs,ms=s(s+1)ℏ2χs,ms.
Now, it is easily demonstrated, from the commutation relations ([e10.9]) and ([e10.10]), that Sz(S+χs,ms)=(ms+1)ℏ(S+χs,ms),
Consider the wavefunction χ=S+χs,ms. Because we know, from Equation ([e10.11]), that χ†χ≥0, it follows that (S+χs,ms)†(S+χs,ms)=χ†s,msS†+S+χs,ms=χ†s,msS−S+χs,ms≥0,
Let msmin be the minimum possible value of ms. It follows that (see Section [slsq]) S−χs,msmin=0.
Now, the raising operator S+, acting upon χs,−s, converts it into some multiple of χs,−s+1. Employing the raising operator a second time, we obtain a multiple of χs,−s+2. However, this process cannot continue indefinitely, because there is a maximum possible value of ms. Indeed, after acting upon χs,−s a sufficient number of times with the raising operator S+, we must obtain a multiple of χs,s, so that employing the raising operator one more time leads to the null state. [See Equation ([e10.31]).] If this is not the case then we will inevitably obtain eigenstates of Sz corresponding to ms>s, which we have already demonstrated is impossible.
It follows, from the previous argument, that msmax−msmin=2s=k,
In 1940, Wolfgang Pauli proved the so-called spin-statistics theorem using relativistic quantum mechanics . According to this theorem, all fermions possess half-integer spin (i.e., a half-integer value of s), whereas all bosons possess integer spin (i.e., an integer value of s). In fact, all presently known fermions, including electrons and protons, possess spin one-half. In other words, electrons and protons are characterized by s=1/2 and ms=±1/2.
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)