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 $m_s=\pm 1/2$.

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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