2.4: Some group theoretic concepts

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The spin-1/2 rotation group has a special name. It is known as SU(2). SU(2) is the group of 2$\times$ 2 unitary matrices with unit determinant. The representation of such a matrix as

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

shows that there are an infinite number of such matrices, since the parameters $\alpha$ and ${\hat{\bf n}}$, which constitutes three parameters (remember ${\hat{\bf n}}$, being a unit vector, has only two independent components), and thus, SU(2), is an example of a continuous Lie group (because the generators satisfy a Lie algebra).

In general, SU($n$) is the group of $n\times n$ unitary matrices with unit determinant. The number of generators belonging to SU($n$) is $n^2-1$. Thus, for SU(2), there should be $2^2-1=3$ generators, which is, indeed, the number of Pauli matrices. SU(3), for example, should have $3^2-1=8$ generators. (Since SU(3) is the group in terms of which quantum chromodynamics, the theory of quarks, is formulated, the eight generators correspond to the eight gluons in the theory.)

Note that it is possible to represent the group in terms of matrices of higher dimension then $n$, so long as the number of generators and independent parameters remains the same. For example, the group SU(2) and the group SO(3) (SO($n$) is the group of $n\times n$ orthogonal matrices with unit determinant), which is used to generate rotations of vectors in ordinary Cartesian space, have the same number of generators and independent parameters. Thus, SU(2) is said to be isomorphic to SO(3), and, therefore, there should be a representation of SU(2) in terms of 3$\times$ 3 matrices. This will be true of any group to which SU(2) is isomorphic.

In order to generate a representation of SU(2), we need to determine the generators of that representation. This can be accomplished by knowing the action of the raising and lowering operators and the operator $S_z$ on the spin states. The general relations are:

$\displaystyle S_z\vert s\;\;m_s\rangle$	$\textstyle =$	$\displaystyle m_s\hbar\vert s\;\;m_s\rangle$
$\displaystyle S_+\vert s\;\;m_s\rangle$	$\textstyle =$	$\displaystyle \sqrt{(s-m_s)(s+m_s+1)}\vert s\;\;m_s+1\rangle$
$\displaystyle S_-\vert s\;\;m_s\rangle$	$\textstyle =$	$\displaystyle \sqrt{(s+m_s)(s-m_s+1)}\vert s\;\;m_s-1\rangle$

Note that for spin-1/2, this reduces to the relations we wrote down before. This are general relations that we will need later when we consider addition of angular momentum.

From these relations, we can construct a representation of SU(2). Consider the case of a spin-1/2 particle. It is clear that the operator $S_z$ is diagonal, and its eigenvalues must be $\pm \hbar/2$, so we can write down the form of $S_z$ immediately, using the fact that it is diagonal in the basis we are working with:

$\begin{displaymath} S_z = {\left(\matrix{{\hbar \over 2} & 0 \cr 0 & -{\hbar \over 2}}\right)} \end{displaymath}$

In order to get $S_x$ and $S_y$, note that the raising and lowering operators must satisfy

$\displaystyle S_+{\left(\matrix{0 \cr 1}\right)}$	$\textstyle =$	$\displaystyle \hbar {\left(\matrix{1 \cr 0}\right)}$
$\displaystyle S_-{\left(\matrix{1 \cr 0}\right)}$	$\textstyle =$	$\displaystyle \hbar {\left(\matrix{0 \cr 1}\right)} \nonumber$

The matrix forms for $S_+$ and $S_-$ that produce this action on the spin states must be

$\displaystyle S_+$	$\textstyle =$	$\displaystyle {\hbar\left(\matrix{0$
$\displaystyle S_-$	$\textstyle =$	$\displaystyle {\hbar\left(\matrix{0$

Then, $S_x$ and $S_y$ are given by

$\displaystyle S_x$	$\textstyle =$	$\displaystyle {1 \over 2}\left(S_+ + S_-\right) = {\left(\matrix{0$
$\displaystyle S_y$	$\textstyle =$	$\displaystyle {1 \over 2i}\left(S_+-S_-\right) ={\left(\matrix{0$

The same can be done, for example, for a spin-1 particle, which will yield the 3$\times$ 3 representation of the group generators.

\(\displaystyle S_z\vert s\;\;m_s\rangle\)	\(\textstyle =\)	\(\displaystyle m_s\hbar\vert s\;\;m_s\rangle\)
\(\displaystyle S_+\vert s\;\;m_s\rangle\)	\(\textstyle =\)	\(\displaystyle \sqrt{(s-m_s)(s+m_s+1)}\vert s\;\;m_s+1\rangle\)
\(\displaystyle S_-\vert s\;\;m_s\rangle\)	\(\textstyle =\)	\(\displaystyle \sqrt{(s+m_s)(s-m_s+1)}\vert s\;\;m_s-1\rangle\)

\(\displaystyle S_+{\left(\matrix{0 \cr 1}\right)}\)	\(\textstyle =\)	\(\displaystyle \hbar {\left(\matrix{1 \cr 0}\right)}\)
\(\displaystyle S_-{\left(\matrix{1 \cr 0}\right)}\)	\(\textstyle =\)	$\displaystyle \hbar {\left(\matrix{0 \cr 1}\right)} \nonumber$

\(\displaystyle S_+\)	\(\textstyle =\)	\(\displaystyle {\hbar\left(\matrix{0\)
\(\displaystyle S_-\)	\(\textstyle =\)	\(\displaystyle {\hbar\left(\matrix{0\)

\(\displaystyle S_x\)	\(\textstyle =\)	\(\displaystyle {1 \over 2}\left(S_+ + S_-\right) = {\left(\matrix{0\)
\(\displaystyle S_y\)	\(\textstyle =\)	\(\displaystyle {1 \over 2i}\left(S_+-S_-\right) ={\left(\matrix{0\)