\input phyzzx

\def\dim{\mathop{\rm dim}\nolimits}

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\noindent PHY--396~K.\quad
Problem set \#12 (last!).\quad
Due May 6, 1999.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
Consider a classical Yang--Mills theory
$$
{\cal L}\ =\ {-1\over 2g^2}\,\tr\bigl(F_{\mu\nu}F^{\mu\nu}\bigr)\
+\ \overline\Psi(i\!\not\!\! D-m)\Psi .
\eqn\Lagrange
$$
\spointbegin
Verify explicitly that $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu
-i[A_\mu,A_\nu]$ transforms covariantly under the gauge transformations,
$F_{\mu\nu}(x)\mapsto U(x)F_{\mu\nu}U^\dagger(x)$.

\spoint
Verify the Bianchi identity
$$
D_\lambda F_{\mu\nu}\ +\ D_\mu F_{\nu\lambda}\ +\ D_\nu F_{\lambda\mu}\
=\ 0.
\eqn\Bianchi
$$

\spoint
Show that the classical equation for the non-abelian vector field
is simply 
$$
D^\mu F_{\mu\nu}\ =\ -g^2 J_\nu
\eqn\EoM
$$
where
$$
J_\nu^a\ =\ \overline\Psi\gamma_\nu{\tau^a\over2}\Psi .
\eqn\Current
$$
Hint: prove and use $\delta F_{\mu\nu}=\partial_\mu\delta A_\nu
-\partial_\nu\delta A_\mu$.

\spoint
Show that eq.~\EoM\ requires the fermionic current $J_\nu$ to be
{\sl covariantly conserved}, $D^\nu J_\nu=0$.

\spoint
Verify that the current~\Current\ is indeed covariantly conserved.


\bigskip
\point
Consider a generic simple non-abelian Lie group $G$ and its
generators $T^a$.
The (quadratic)
 Casimir operator $C_2=\sum_a T^aT^a$ commutes with all the generators
and hence for any irreducible representation $(r)$ of the group,
$C_2$ restricted to $(r)$ is simply a unit matrix times a number $C(r)$.
In other words, if $T^a_{(r)}$ is a matrix of the generator $T^a$ in the
representation $(r)$, then $\sum_a T^a_{(r)}T^a_{(r)}=C(r)
\times\bf 1$.
For example, for the isospin group $SU(2)$, the irreps are
characterized by the isospin $I$ and $C(I)=I(I+1)$.

\spointbegin
By symmetry, for any complete representation $(r)$ of the group,
$$
\tr_{(r)}(T^aT^a)\ \equiv\ \tr\left(T^a_{(r)}T^b_{(r)}\right)\
=\ R(r)\delta^{ab}
\eqno\eq
$$
for some coefficient $R(r)$.
Show that for any irreducible representation,
$$
{R(r)\over C(r)}\ =\ {\dim(r)\over\dim(G)}\,.
\eqn\RC
$$
In particular, for the $SU(2)$ group, this formula gives
$R(I)={1\over3}I(I+1)(2I+1)$.

\spoint
Suppose the first three generators of $G$ generate an $SU(2)$ subgroup.
Show that if a representation $(r)$ of $G$ decomposes into several
$SU(2)$ multiplets of isospins $I_1,I_2,\ldots,I_n$, then
$$
R(r)\ =\sum_{i=1}^n\,\coeff13 I_i(I_i+1)(2I_i+1).
\eqn\decomp
$$

\spoint
Now consider the $SU(N)$ group with an obvious $SU(2)$ subgroup
of matrices acting on the first two components of a complex $N$-vector.
The fundamental representation $(N)$ of the $SU(N)$ decomposes into
one doublet and $(N-2)$ singlets of the $SU(2)$ subgroup,
hence
$$
R(N)\ =\ \half\qquad{\rm and}\quad C(N)\ =\ {N^2-1\over 2N}\,.
\eqno\eq
$$
Show that the adjoint representation of the $SU(N)$ decomposes into
one $SU(2)$ triplet, $2(N-2)$ doublets and $(N-2)^2$ singlets and hence
$$
R({\rm adj})\ =\ C({\rm adj})\ \equiv\ C(G)\ =\ N.
\eqno\eq
$$
Hint: $(N)\times(\overline N)=({\rm adj})+(1)$.

\spoint
The symmetric and the anti-symmetric 2--index tensors from irreducible
representations of the $SU(N)$ group.
Find out the decomposition of these irreps under an $SU(2)\subset SU(N)$
and calculate their respective $R$ factors.

\bye
