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# 9.3: Eigenstates of Sz and S²

Because the operators $$S_z$$ and $$S^2$$ commute, they must possess simultaneous eigenstates. (See Section [smeas].) Let these eigenstates take the form [see Equations ([e8.29]) and ([e8.30])]: \begin{aligned} S_z\,\chi_{s,m_s}&= m_s\,\hbar\,\chi_{s,m_s},\label{e10.16}\\[0.5ex] S^2\,\chi_{s,m_s} &= s\,(s+1)\,\hbar^{\,2}\,\chi_{s,m_s}.\label{e10.17}\end{aligned}

Now, it is easily demonstrated, from the commutation relations ([e10.9]) and ([e10.10]), that $S_z\,(S_+\,\chi_{s,m_s}) = (m_s+1)\,\hbar\,(S_+\,\chi_{s,m_s}),$ and $S_z\,(S_-\,\chi_{s,m_s}) = (m_s-1)\,\hbar\,(S_-\,\chi_{s,m_s}).$ Thus, $$S_+$$ and $$S_-$$ are indeed the raising and lowering operators, respectively, for spin angular momentum. (See Section [seian].) The eigenstates of $$S_z$$ and $$S^2$$ are assumed to be orthonormal: that is, $\label{e10.20} \chi^\dagger_{s,m_s}\,\chi_{s',m_s'} =\delta_{ss'}\,\delta_{m_s m_s'}.$

Consider the wavefunction $$\chi=S_+\,\chi_{s,m_s}$$. Because we know, from Equation ([e10.11]), that $$\chi^\dagger\,\chi\geq 0$$, it follows that $(S_+\,\chi_{s,m_s})^\dagger\,(S_+\,\chi_{s,m_s}) = \chi_{s,m_s}^\dagger\, S_+^\dagger\,S_+\,\chi_{s,m_s} = \chi_{s,m_s}^\dagger\,S_-\,S_+\,\chi_{s,m_s}\geq 0,$ where use has been made of Equation ([e10.7]). Equations ([e10.8]), ([e10.16]), ([e10.17]), and ([e10.20]) yield $s\,(s+1) \geq m_s\,(m_s+1).$ Likewise, if $$\chi=S_-\,\chi_{s,m_s}$$ then we obtain $s\,(s+1)\geq m_s\,(m_s-1).$ Assuming that $$s\geq 0$$, the previous two inequalities imply that $-s \leq m_s\leq s.$ Hence, at fixed $$s$$, there is both a maximum and a minimum possible value that $$m_s$$ can take.

Let $$m_{s\,{\rm min}}$$ be the minimum possible value of $$m_s$$. It follows that (see Section [slsq]) $S_-\,\chi_{s,m_{s\,{\rm min}}}= 0.$ Now, from Equation ([e10.7a]), $S^2 = S_+\,S_-+S_z^{\,2}-\hbar\,S_z.$ Hence, $S^2\,\chi_{s,m_{s\,{\rm min}}} = (S_+\,S_- +S_z^{\,2}-\hbar\,S_z)\,\chi_{s,m_{s\,{\rm min}}},$ giving $s\,(s+1) = m_{s\,{\rm min}}\,(m_{s\,{\rm min}}-1).$ Assuming that $$m_{s\,{\rm min}}<0$$, this equation yields $m_{s\,{\rm min}} = -s.$ Likewise, it is easily demonstrated that $m_{s\,{\rm max}} = +s.$ Moreover,

$\label{e10.31} S_-\,\chi_{s,-s} = S_+\,\chi_{s,s} = 0.$

Now, the raising operator $$S_+$$, acting upon $$\chi_{s,-s}$$, converts it into some multiple of $$\chi_{s,-s+1}$$. Employing the raising operator a second time, we obtain a multiple of $$\chi_{s,-s+2}$$. However, this process cannot continue indefinitely, because there is a maximum possible value of $$m_s$$. Indeed, after acting upon $$\chi_{s,-s}$$ a sufficient number of times with the raising operator $$S_+$$, we must obtain a multiple of $$\chi_{s,s}$$, 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 $$S_z$$ corresponding to $$m_s>s$$, which we have already demonstrated is impossible.

It follows, from the previous argument, that $m_{s\,{\rm max}}-m_{s\,{\rm min}} = 2\,s = k,$ where $$k$$ is a positive integer. Hence, the quantum number $$s$$ can either take positive integer or positive half-integer values. Up to now, our analysis has been very similar to that which we used earlier to investigate orbital angular momentum. (See Section [sorb].) Recall, that for orbital angular momentum the quantum number $$m$$, which is analogous to $$m_s$$, is restricted to take integer values. (See Section [slz].) This implies that the quantum number $$l$$, which is analogous to $$s$$, is also restricted to take integer values. However, the origin of these restrictions is the representation of the orbital angular momentum operators as differential operators in real space. (See Section [s8.3].) There is no equivalent representation of the corresponding spin angular momentum operators. Hence, we conclude that there is no reason why the quantum number $$s$$ cannot take half-integer, as well as integer, values.

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 $$m_s=\pm 1/2$$.

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