Spin Angular Momentum

The goal of this section is to introduce the spin angular momentum $S$, as a generalized angular momentum operator that satisfies the general commutation relations $\boxed{S\times S = i\hbar S}$. The main difference between the angular momenta $S$, and $L$, is that $S$ can have half-integer quantum numbers.

Note: Remember that the quantization rules established by the commutation relations did not rule out the possibility of half-integer values for $j$ (see page 46). However, such possibility was ruled out by the periodicity requirement, $Y(\theta + 2\pi) =Y(\theta)$, associated with the eigenfunctions of $L_z$ and $L^2$. Since the spin eigenfunctions (i.e., the spinors) do not depend on spatial coordinates, they do not have to satisfy any periodicity condition and therefore their eigenvalues can be half-integer.

Electron Spin:

A particular case of half-integer spin is the spin angular momentum of an electron with $l=1/2$ (see http://www.lorentz.leidenuniv.nl/history/spin/goudsmit.html, for Goudsmit's historical recount of the discovery of the electron spin). In discussing the spin properties of a particle we adopt the notation $l=S$, and $m=m_s$.

The spin functions $\alpha$ and $\beta$ are eigenfunctions of $S_z$ with eigenvalues $+\frac{1}{2}\hbar$ and $-\frac{1}{2}\hbar$, respectively. These eigenfunctions are normalized according to,

$$\sum_{m_s=-1/2}^{1/2} \vert\alpha (m_s)\vert^2 =1, \qquad \sum_{m_s=-1/2}^{1/2} \vert\beta (m_s)\vert^2 =1,$$ (36)

since $m_s$ can be either $\frac{1}{2}$, or $-\frac{1}{2}$. Also, since the eigenfunctions $\alpha$ and $\beta$ correspond to different eigenvalues of $S_z$, they must be orthogonal:

$$\sum_{m_s=-1/2}^{1/2} \alpha^* (m_s) \beta (m_s) =0.$$ (37)

In order to satisfy the conditions imposed by Eqs. (36) and (37),

$$\alpha (m_s) = \delta_{m_s, 1/2},$$   and,$$\qquad \beta(m_s)= \delta_{m_s, -1/2}.$$

It is useful to define the spin angular momentum ladder operators, $\boxed{S_+=S_x+iS_y}$ and $\boxed{S_-=S_x-i S_y}$. Here, we prove that the raising operator $S_+$ satisfies the following equation:

$$\boxed{S_+\beta = \hbar \alpha}.$$

Proof:

Using the normalization condition introduced by Eq. (36) we obtain,

$$\sum_{m_s=-1/2}^{1/2} \alpha^*(m_s)\alpha(m_s)=\sum_{m_s=-1/2}^{1/2}(\hat{S}_+\frac{\beta}{c})^*(\hat{S}_+ \frac{\beta}{c})=1,$$

and

$$\vert c\vert^2 = \sum_{m_s} (\hat{S}_+\beta)^*(\hat{S}_x \beta + i\hat{S}_y \beta).$$

Now, using the hermitian property of $S_x$ and $S_y$ we obtain,

$$\sum_{m_s} f^*S_x g = \sum_{m_s} g S_x^*f^*,$$

$\vert c\vert^2 = \sum_{m_s} \beta S_x^* (S_+\beta)^* + i \beta S_y^*(S_+\beta)^*,$

where,

$\vert c\vert^2 =\sum_{m_s} \beta^* S_x S_+\beta - i\beta^* S_y S_+\beta,$

$\vert c\vert^2 =\sum_{m_s} \beta^*S_-S_+\beta,$

$\vert c\vert^2 =\sum_{m_s} \beta^*(S^2 -S_z^2 - \hbar S_z) \beta,$

$\vert c\vert^2 =\sum_{m_s} \beta^*(\frac{3}{4} \hbar^2 - \frac{\hbar^2}{4} +\frac{\hbar^2}{2})\beta,$

$\vert c\vert^2 =\hbar^2.$

Since the phase of c is arbitrary, we can choose c=$\hbar$.

Similarly, we obtain $\boxed{S_-\alpha = \hbar \beta}.$

Since $\alpha$ is the eigenfunction with highest eigenvalue, the operator $S_+$ acting on it must annihilate it as follows,

$$S_+ \alpha=0,$$   and$$\qquad S_-\beta=0.$$

$S_x\alpha =(S_+ + S_-)\frac{\alpha}{2} = \frac{\hbar}{2} \beta, \qquad \Rightarrow \qquad \boxed{S_x\alpha=\frac{1}{2}\hbar \beta.}$

$S_y\beta =(S_+ - S_-)\frac{\beta}{2i} = \frac{\hbar}{2} \alpha, \qquad \Rightarrow \qquad \boxed{S_y\beta=-\frac{1}{2} i \hbar \alpha.}$

Similarly, we find $\boxed{S_x\beta=\frac{1}{2}\hbar \alpha},$ and $\boxed{S_y\alpha=\frac{1}{2} i \hbar \beta}.$

$\boxed{\begin{array}{c\vert cc} <\vert S_x\vert>& \alpha & \beta \\ \alpha & ... ...alpha & \beta \\ \alpha & \hbar /2 & 0 \\ \beta & 0 &-\hbar /2 \end{array}}$

Therefore, $S=\frac{1}{2}\hbar \sigma$, where $\sigma$ are the Pauli matrices defined as follows,

$$\sigma_x = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{a... ...\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right),$$

where, $\sigma_x^2=\sigma_y^2=\sigma_z^2 = 1.$

Exercise 25: Prove that the Pauli matrices anti-commute with each other, i.e.,

$$\sigma_i\sigma_j +\sigma_j\sigma_i =0,$$

where $i \neq j$, and $i,j=(x,y,z).$ In order to find the eigenfunctions of $S_z$, called eigenspinors, consider the following eigenvalue problem:

$$S_z \left ( \begin{array}{c} u_{\pm} \\ v_{\pm} \end{array} \righ... ...rac{\hbar}{2} \left ( \begin{array}{c} u_{\pm} \\ v_{\pm} \end{array} \right ),$$

$$\left ( \begin{array}{cc} 1 &0 \\ 0 & -1 \end{array} \right )\lef... ...right ) = \pm \left ( \begin{array}{c} u_{\pm} \\ v_{\pm} \end{array} \right ),$$

$$\left ( \begin{array}{c} u_{\pm} \\ -v_{\pm} \end{array} \right )... ...rray} \right ), \qquad \Rightarrow \qquad \boxed{v_+=0}, \qquad \boxed{u_+ =1}.$$

Similarly we obtain, $\boxed{u_-=0},$   and$\hspace{.20cm} \boxed{v_- =1}$. Therefore, electron eigenspinors satisfy the eigenvalue problem,

$$S_z \chi_{\pm} = \pm \frac{\hbar}{2} \chi_{\pm},$$

with,

$$\chi_-=\left ( \begin{array}{c} 0 \\ 1 \end{array} \right ),$$   and$$\hspace{.3cm} \chi_+=\left ( \begin{array}{c} 1 \\ 0 \end{array} \right ).$$

Any spinor can be expanded in the complete set of eigenspinors as follows,

$$\left ( \begin{array}{c} \alpha_+ \\ \alpha_- \end{array} \right ... ...rray} \right ) + \alpha_- \left ( \begin{array}{c} 0 \\ 1 \end{array} \right ),$$

where $\vert\alpha_+\vert^2$, and $\vert\alpha_-\vert^2$, are the probabilities that a measurement of $S_z$ yields the value $+\frac{1}{2}\hbar$, and $-\frac{1}{2}\hbar$, respectively, when the system is described by state $\left ( \begin{array}{c} \alpha_+ \\ \alpha_- \end{array} \right )$.

Exercise 26: Prove that, $S^2 \chi_+ = \frac{\hbar^2}{2} (\frac{1}{2} +1) \chi_+$.

Exercise 27: Consider an electron localized at a crystal site. Assume that the spin is the only degree of freedom of the system and that due to the spin the electron has a magnetic moment,

$$M =- \frac{eg}{2mc} S,$$

where $g\approx2$, $m$ is the electron mass, $e$ is the electric charge and $c$ is the speed of light. Therefore, in the presence of an external magnetic field $B$ the Hamiltonian of the system is,

$$H = - M \cdot B.$$

Assume that $B$ points in the $z$ direction and that the state of the system is,

$$\psi(\epsilon) = e^{i \omega t} \left ( \begin{array}{c} \alpha_+ \\ \alpha_- \end{array} \right ).$$

Consider that initially (i.e., at time $t=0$) the spin points in the $x$ direction (i.e., the spinor is an eigenstate of $\sigma_x$ with eigenvalue $\frac{1}{2}\hbar$).

Compute the expectation values of $S_x$ and $S_y$ at time $t$.

Addition of Angular Momenta

Since $L$ depends on spatial coordinates and $S$ does not, then the two operators commute (i.e., $[L, S] =0$). It is, therefore, evident that the components of the total angular momentum,

$$J = L+S,$$

satisfy the commutation relations,

$$J\times J= i \hbar J.$$

Eigenfunctions of $J^2$ and $J_z$ are obtained from the individual eigenfunctions of two angular momentum operators $L_1$ and $L_2$ with quantum numbers ($l_1$, $m_1$) and ($l_2$, $m_2$), respectively, as follows:

$$\psi_{j}^m=\sum_{l_1,m_1,l_2,m_2} \underbrace{ C(jm, l_1m_1 \ l_2m_2) }\phi_{l_1}^{m_1}\phi_{l_2}^{m_2},$$

Clebsch-Gordan Coefficients

where,

$$J^2 \psi_{jm}= \hbar^2 j (j+1) \psi_{jm},$$

$$J_z \psi_{jm}= \hbar m \psi_{jm}.$$

Exercise 28: Show that, $\psi_j^{m+1/2} = C_1 Y_l^m \chi_+ +C_2 Y_l^{m+1} \chi_-$, is a common eigenfunction of $J^2$ and $J_z$ when, $C_1=\sqrt{\frac{l+m+1}{2l+1}}$, and $C_2=\sqrt{\frac{l-m}{2l+1}}$, or when, $C_1=\sqrt{\frac{l-m}{2l+1}}$, and $C_2= - \sqrt{\frac{l+m+1}{2l+1}}$.

Hint: Analyze the particular case $j=l - 1/2$, and $j=l+1/2$. Note that,

$J^2=L^2 + S^2 + 2LS= L^2 +S^2 +2 L_zS_z + L_+S_- + L_-S_+,$

$J_z= L_z + S_z,$