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Next: Wavefunction of Spin One-Half Up: Spin Angular Momentum Previous: Introduction

Properties of Spin Angular Momentum

Let us denote the three components of the spin angular momentum of a particle by the Hermitian operators $ (S_x, S_y, S_z)\equiv {\bf S}$ . We assume that these operators obey the fundamental commutation relations (297)-(299) for the components of an angular momentum. Thus, we can write

$\displaystyle {\bf S} \times {\bf S} = {\rm i}\,\hbar \, {\bf S}.$ (417)

We can also define the operator

$\displaystyle S^2 = S_x^{\,2}+S_y^{\,2} + S_z^{\,2}.$ (418)

According to the quite general analysis of Section 4.1,

$\displaystyle [{\bf S}, S^2] = 0.$ (419)

Thus, it is possible to find simultaneous eigenstates of $ S^2$ and $ S_z$ . These are denoted $ \vert s, s_z\rangle$ , where

$\displaystyle S_z \,\vert s, s_z\rangle$ $\displaystyle = s_z \,\hbar \,\vert s, s_z\rangle,$ (420)
$\displaystyle S^2 \,\vert s, s_z\rangle$ $\displaystyle = s\,(s+1)\,\hbar^2\, \vert s, s_z\rangle.$ (421)

According to the equally general analysis of Section 4.2, the quantum number $ s$ can, in principle, take integer or half-integer values, and the quantum number $ s_z$ can only take the values $ s, s-1 \cdots -s+1, -s$ .

Spin angular momentum clearly has many properties in common with orbital angular momentum. However, there is one vitally important difference. Spin angular momentum operators cannot be expressed in terms of position and momentum operators, like in Equations (290)-(292), because this identification depends on an analogy with classical mechanics, and the concept of spin is purely quantum mechanical: i.e., it has no analogy in classical physics. Consequently, the restriction that the quantum number of the overall angular momentum must take integer values is lifted for spin angular momentum, since this restriction (found in Sections 4.3 and 4.4) depends on Equations (290)-(292). In other words, the spin quantum number $ s$ is allowed to take half-integer values.

Consider a spin one-half particle, for which

$\displaystyle S_z \,\vert\pm \rangle$ $\displaystyle =\pm \frac{\hbar}{2} \,\vert\pm \rangle,$ (422)
$\displaystyle S^2\, \vert\pm\rangle$ $\displaystyle = \frac{3 \,\hbar^2}{4}\,\vert\pm \rangle.$ (423)

Here, the $ \vert\pm \rangle$ denote eigenkets of the $ S_z$ operator corresponding to the eigenvalues $ \pm \hbar/2$ . These kets are mutually orthogonal (since $ S_z$ is an Hermitian operator), so

$\displaystyle \langle +\vert -\rangle = 0.$ (424)

They are also properly normalized and complete, so that

$\displaystyle \langle +\vert + \rangle=\langle -\vert - \rangle = 1,$ (425)


$\displaystyle \vert+\rangle \langle +\vert + \vert-\rangle \langle -\vert = 1.$ (426)

It is easily verified that the Hermitian operators defined by

$\displaystyle S_x$ $\displaystyle = \frac{\hbar}{2} \left(\, \vert+\rangle \langle -\vert + \vert-\rangle \langle +\vert\, \right),$ (427)
$\displaystyle S_y$ $\displaystyle = \frac{{\rm i}\,\hbar}{2}\left(\, -\,\vert+\rangle \langle -\vert + \vert-\rangle \langle +\vert\,\right),$ (428)
$\displaystyle S_z$ $\displaystyle = \frac{\hbar}{2}\left(\, \vert+\rangle \langle +\vert - \vert-\rangle \langle -\vert\,\right),$ (429)

satisfy the commutation relations (297)-(299) (with the $ L_j$ replaced by the $ S_j$ ). The operator $ S^2$ takes the form

$\displaystyle S^2 = \frac{3\,\hbar^2}{4}.$ (430)

It is also easily demonstrated that $ S^2$ and $ S_z$ , defined in this manner, satisfy the eigenvalue relations (422)-(423). Equations (427)-(430) constitute a realization of the spin operators $ {\bf S}$ and $ S^2$ (for a spin one-half particle) in spin space (i.e., the Hilbert sub-space consisting of kets which correspond to the different spin states of the particle).

next up previous
Next: Wavefunction of Spin One-Half Up: Spin Angular Momentum Previous: Introduction
Richard Fitzpatrick 2013-04-08