Rotations in spin space

Given two types of angular momentum, orbital and spin, it is possible to define a total angular momantum

$${\bf J}= {\bf L}+ {\bf S}$$

${\bf J}$ plays a special role in quantum mechanics. Not only is it often a constant of the motion even when $H$ is spin-dependent, but it is the generator of rotations in the Hilbert space.

To see what this means, consider a simpler situation with the total linear momentum $P$. The linear momentum is known as the generator of translations in the Hilbert space. By this, we mean that the operator

$$T_{{\bf a}} = e^{-i{\bf P}\cdot{\bf a}/\hbar}$$

which is a function of $P$ produces translations in space by an amount ${\bf a}$. Thus, its action on an arbitrary function of ${\bf r}$ is

$$T_{{\bf a}}\psi({\bf r}) = \psi({\bf r}-{\bf a})$$

To see that this is true, consider the one-dimensional version of this operator

$$T_a = e^{-iPa/\hbar}$$

Using the fact that $P=(\hbar/i)(d/dx)$, the action of $T_a$ on an arbitrary function $\psi(x)$ is

$$T_a \psi(x) = e^{-ad/dx}\psi(x)$$

This can be evaluated by a Taylor series:

$$e^{-ad/dx}\psi(x) = \left[1-a{d \over dx} + {1 \over 2!}a^2{d^2 \over dx^2} - {1 \over 3!}a^3{d^3 \over dx^3} + \cdots\right]\psi(x)$$

$$= \psi(x)-a\psi'(x) + {1 \over 2!}a^2 \psi''(x) - \cdots$$

$$= \psi(x-a)$$

That is, the next to last line is just the Taylor expansion of $\psi(x-a)$ about $a=0$. ${\bf P}$ is, therefore, called the generator of the translation group.

By analogy and by similar reasoning, it can be shown that ${\bf J}$ is the generator of rotations of vectors in the Hilbert space via the operator:

$${R_{\alpha}({\bf n})}= \exp\left[-{i \over \hbar}\alpha{\bf J}\cdot{\hat{\bf n}}\right]$$

which produces rotations of a vector by an angle $\alpha$ about an axis defined by the unit vector ${\hat{\bf n}}$. ${\bf J}$ is called the generator of the rotation group.

Since ${\bf L}$ and ${\bf S}$ commute (they act in different Hilbert spaces), the rotation operator can be written as

$${R_{\alpha}({\bf n})} = \exp\left[-{i \over \hbar}\alpha({\bf L}+{\bf S})\cdot{\hat{\bf n}}\right]$$

$$= \exp\left[-{i \over \hbar}\alpha{\bf L}\cdot{\hat{\bf n}}\right] \exp\left[-{i \over \hbar}\alpha{\bf S}\cdot{\hat{\bf n}}\right]$$

$$= {R_{\alpha}^{(r)}({\bf n})}{R_{\alpha}^{(s)}({\bf n})}$$

Thus, a particle whose state vector is separable into spatial and spin components according to

$$\vert\psi\rangle = \vert\phi\rangle \bigotimes \vert\chi\rangle$$

will be transformed according to

$$\vert\psi'\rangle = {R_{\alpha}({\bf n})}\vert\psi\rangle = \left[\exp\left(-{i \over... ...-{i \over \hbar}\alpha{\bf S}\cdot{\hat{\bf n}}\right) \vert\chi\rangle \right]$$

$$= \vert\phi'\rangle \bigotimes \vert\chi'\rangle$$

Let us focus on the spin part of this equation, which transform $\vert\chi\rangle\longrightarrow \vert\chi'\rangle$ by

$$\vert\chi'\rangle = \exp\left(-{i \over \hbar}\alpha{\bf S}\cdot{\hat{\bf n}}\right) \vert\chi\rangle$$

Since ${\bf S}= (\hbar/2)\stackrel{\rightarrow}{\sigma}$, where $\stackrel{\rightarrow}{\sigma}$ is the vector of Pauli matrices, the spin rotation operator becomes

$${R_{\alpha}^{(s)}({\bf n})}= \exp\left[-i{\alpha \over 2}\stackrel{\rightarrow}{\sigma}\cdot{\hat{\bf n}}\right]$$

Thus, the generators of the spin-1/2 rotation group are just the 2$\times$ 2 Pauli matrices.