2.3: Rotations in spin space

Last updated
Save as PDF

Page ID: 20879

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

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

$\begin{displaymath} {\bf J}= {\bf L}+ {\bf S} \end{displaymath}$

${\bf J}$ 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

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

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

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

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

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

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

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

This can be evaluated by a Taylor series:

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

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:

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

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

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

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

$\begin{displaymath} \vert\psi\rangle = \vert\phi\rangle \bigotimes \vert\chi\rangle \end{displaymath}$

will be transformed according to

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

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

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

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

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

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

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

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

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

Search

Text Color

Text Size

Margin Size

Font Type