\input phyzzx

\def\bx{{\bf x}}
\def\bp{{\bf p}}
\def\bsigma{\setbox0=\hbox{\fourteeni\char27}\setbox2=\hbox{\twelvei\char27}%
        \vbox to \ht0{\vss\hbox to\wd0{\hss\box2\hss}\vss}\kern -\wd0\box0}
\def\D{\!\not\!}
\def\P{\hat{\cal P}}
\def\eqdef{\buildrel\rm def\over=}

\hsize=6.8in	\hoffset=-0.2in
\vsize=8.75in	\voffset=-0.25in
\noindent PHY--396~K.\quad
Problem set \#6.\quad
Due October 12, 2004.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\item{\bf 0.}
First of all, finish problem \#3 from the previous homework set \#5.
It is due by Thursday, October 7.

\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
The first problem of this set concerns finite representations of
the continuous Lorentz symmetry $SO^+(3,1)$, or rather its double cover
${\rm Spin(3,1)}\cong SL(2,{\bf C})$. 
Let us define
$$
{\bf\hat J}_\pm\ =\ \half\bigl({\bf\hat J}\,\pm\,i{\bf\hat K}\bigr).
\eqn\Jpm
$$
where $\hat K^i=\hat J^{0i}=-\hat J^{i0}$ are generators of the Lorentz
boosts and $\hat J^i$ are generators of the space rotations.

\spointbegin
Show that the ${\bf\hat J}_+$ and the ${\bf\hat J}_-$
commute with each other and that each satisfies the commutations relations
of an angular momentum,
$[\hat J^k_\pm,\hat J^\ell_\pm]=i\epsilon^{k\ell m}\hat J^m_\pm$.

\pointcon
The ``angular momentum'' ${\bf\hat J}_+$ is non-hermitian and hence its
finite irreducible representations are non-unitary
analytic continuations of the spin--$j$ representations of
a hermitian $\bf\hat J$.
The same is true for the ${\bf\hat J}_-={\bf\hat J}_+^\dagger$.
Thus altogether,
the finite irreducible representations of the Lorentz algebra are specified
by two integer or half-integer `spins' $j_+$ and $j_-$.

\pointcon
The simplest non-trivial representations of the Lorentz algebra are
$(j_+=\half,j_-=0)$ --- the left-handed Weyl spinor  where ${\bf\hat J}$
acts as $\half\bsigma$ and ${\bf\hat K}$ as $-\coeff{i}{2}\bsigma$,
and  $(j_+=0,j_-=\half)$ --- the right-handed Weyl spinor
where ${\bf\hat J}$ also acts as $\half\bsigma$
but ${\bf\hat K}$ acts as $+\coeff{i}{2}\bsigma$.
Together, the two Weyl spinors  comprise the Dirac spinor.
From the $SL(2,{\bf C})$ point if view, the left-handed Weyl spinor is the
doublet representation $\bf 2$ which defines the $SL(2,{\bf C})$ group
while the right-handed Weyl spinor is the conjugate doublet $\bf\bar 2$.
As discussed in class, the Weyl spinors transform according to
$$
\psi^L_\alpha\ \mapsto\ M_\alpha^{\,\,\beta}\psi^L_\beta
\qquad{\rm and}\quad (\sigma_2\psi^R_{})_{\dot\alpha}\
\mapsto\ M_{\dot\alpha}^{*\dot\beta}(\sigma_2\psi^R_{})_{\dot\beta}
\eqn\SLtwo
$$
where $M\equiv M_L$ and $\sigma_2 M^*\sigma_2=M_R$.
Note the notations: the un-dotted indices from the beginning
of the Greek alphabet for the left-handed spinors,
and the dotted indices for the right-handed spinors.

\pointcon
A generic $(j_+,j_-)$ representation of the Lorentz algebra becomes in the
$SL(2,{\bf C})$ terms a tensor
$\Phi_{\alpha_1\ldots\alpha_{(2j_+)},\dot\gamma_1\ldots \dot\gamma_{( 2j_-)}}$,
totally symmetric in its $2j_+$ un-dotted indices
$\alpha_1,\ldots, \alpha_{(2j_+)}$
and separately totally symmetric in its
$2j_-$ dotted indices $\dot\gamma_1,\ldots, \dot\gamma_{(2j_-)}$;
it transforms according to
$$
\Phi_{\alpha_1\ldots\alpha_{(2j_+)},\dot\gamma_1\ldots \dot\gamma_{(2j_-)}}\
\mapsto\ M_{\alpha_1}^{\;\beta_1}\cdots M_{\alpha_{(2j_+)}}^{\;\beta_{(2j_+)}}\
M_{\dot \gamma_1}^{*\,\dot\delta_1}\cdots
	U_{\dot\gamma_{(2j_-)}}^{*\,\dot\delta_{(2j_-)}}\
\Phi_{\beta_1\ldots \beta_{(2j_+)},\dot\delta_1\ldots \dot\delta_{(2j_-)}}\,.
\eqn\SLTwoCLaw
$$

\pointcon
The vector representation of the Lorentz group has $j_+=j_-=\half$.
To cast the action of the Lorentz group in $SL(2,{\bf C})$ terms~\SLTwoCLaw,
we define a $2\times2$ matrix
$$
X_\mu\sigma^\mu\ \equiv\ X_0\ -\ {\bf X}\cdot\bsigma
\eqno\eq
$$
where $\sigma^0$ is a unit $2\times2$ matrix while $\sigma^1$, $\sigma^2$ and
$\sigma^3$ are the Pauli matrices.
In $SL(2,{\bf C})$ terms, this matrix has one dotted and one un-dotted index,
$$
X_{\alpha\dot\gamma}\ =\ X_\mu\sigma^\mu_{\alpha\dot\gamma}\
=\ X_0\delta_{\alpha\dot\gamma}\ -\ {\bf X}\cdot\bsigma_{\alpha\dot\gamma}\,,
\eqno\eq
$$
thus it transforms under the $SL(2,{\bf C})$ as a $(\half,\half)$ bi-spinor,
$$
X'_{\alpha\dot\gamma}\
=\ M_\alpha^{\,\,\beta}M_{\dot\gamma}^{*\dot\delta}X_{\gamma\dot\delta}\,,
\eqn\BiSpinor
$$
or in matrix form
$$
X'_\mu\sigma^\mu\ =\ M(X_\mu\sigma^\mu)M^\dagger.
\eqn\LorentzM
$$

\spoint %b
Show that for any $SL(2,{\bf C})$ matrix $M$, eq.~\LorentzM\ defines
an orthochronous Lorentz transform $X'_\mu=L_\mu^{\,\,\nu}(M)X_\nu$.
(Hint: prove and use $\det(X_\mu\sigma^\mu)=X^2\equiv X_\mu X^\mu$).
\subsubitem\ast
For extra challenge, show that $L$ is proper, \ie\ $\det(L)=+1$.

\spoint %c
Verify the group law, $L(M_2 M_1)=L(M_2) L(M_1)$.

\spoint %d
Verify explicitly that for
$M=\exp\bigl(-\coeff{i}{2}\theta\,{\bf n}\cdot\bsigma\bigr)$,
$L(M)$ is a rotation by angle $\theta$ around axis $\bf n$,
while for $M=\exp\bigl(-\coeff{1}{2} r\, {\bf n}\cdot\bsigma\bigr)$,
$L(M)$ is a boost of rapidity $r$ ($\beta=\tanh r$, $\gamma=\cosh r$)
in the direction $\bf n$.

\pointcon
In general, any $(j_+,j_-)$ multiplet of the $SL(2,{\bf C})$
with integer net spin $j_++j_-$ is equivalent to some kind of a Lorentz
tensor.
(Here, we include the scalar and the vector among the tensors.)
For example, the $(1,1)$ multiplet is equivalent to a symmetric,
traceless 2--index tensor $T^{\mu\nu}=T^{\nu\mu}$, $T^\mu_\mu=0$.
For $j_+\neq j_-$ the representation is complex, but one can make
a real tensor by combining two multiplets with opposite
$j_+$ and $j_-$, for example the $(1,0)$ and $(0,1)$ multiplets
are together equivalent to an antisymmetric 2--index tensor
$F^{\mu\nu}=-F^{\nu\mu}$.

\spoint %e
Verify the above examples.\brk
Hint: For any angular momentum $(j=\half)\otimes(j=\half)=(j=1)\oplus(j=0)$.

\pointcon
The $SL(2,{\bf C})$ multiplets with half-integer $j_++j_-$ are equivalent
to Lorentz spinors or spin-tensors which carry one Weyl index as well as
0, 1 or more 4--vector indices and transform according to
$$
\psi_\alpha^{\mu,\ldots,\nu}\ \mapsto\
M_\alpha^{\,\,\beta}(L)L^\mu_{\,\,\kappa}\cdots L^\nu_{\,\,\lambda}\,
\psi_\beta^{\kappa,\ldots,\lambda}\qquad
{\rm or}\qquad \psi_{\dot\alpha}^{\mu,\ldots,\nu}\ \mapsto\
M_{\dot\alpha}^{*\dot\beta}(L)L^\mu_{\,\,\kappa}\cdots L^\nu_{\,\,\lambda}\,
\psi_{\dot\beta}^{\kappa,\ldots,\lambda}.
\eqno\eq
$$

\spoint %f
Show that the $(1,\half)$ and $(\half,1)$ multiplets are together equivalent
to the Rarita--Schwinger spin-vector $\Psi_a^\mu$ which has one Dirac index~$a$
and one 4--vector index~$\mu$ and satisfies a Lorentz-covariant constraint
$\gamma_\mu\Psi^\mu=0$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\bigskip
\point %2
Now consider an im-proper Lorentz symmetry, namely
the {\sl parity} symmetry AKA reflection of space
$$
P^\mu_{\,\,\nu}\ 
%=\ {\Tenpoint\pmatrix{
%	+1&0&0&0\cr 0&-1&0&0\cr 0&0&-1&0\cr 0&0&0&-1\cr }}\qquad
=\,\left.\cases{
	+1 & for $\mu=\nu=0$,\cr
	-1 & for $\mu=\nu=1,2,3$,\cr
	0 & for $\mu\neq\nu$,\cr
	}\right\}\qquad	
{\rm thus}\quad P(t,\bx)\ =\ (+t,-\bx);
\eqn\ParityDef
$$
note $P^2=1$.
\subpar
In the Fock space, parity is represented by a unitary operator $\P$;
by the group law, $\P^2=1$ and hence $\P^\dagger=\P^{-1}=\P$.

\spointbegin
Use group law to show that $\P$ commutes with the angular momenta
$\hat J^i$ but anticommutes with the boost generators $\hat K^i$,
then use these commutation relations to show that $\P$
acting on the quantum fields must interchange the $j_+$ and $j_-$
quantum numbers of the field components.
%\subpar
For example, parity must turn left-handed Weyl spinors into right-handed
Weyl spinors and vice verse, thus
$\P(\half,0)\P=(0,\half)$ and $\P(0,\half)\P=(\half,0)$.

\spoint %b
A Dirac spinor field transforms under parity according to
$$
\hat{\cal P}\,\hat\Psi(\bx,t)\,\hat{\cal P}\
\equiv\ \hat\Psi'(\bx,t)\ =\ \pm\gamma^0\,\hat\Psi(-\bx,t)
\eqn\Parity
$$
where the overall $\pm$ sign is the  {\it intrinsic parity}
of a particular Dirac field.
\subpar
Verify that the Dirac equation is covariant under this transformation
and that the Dirac action $\int\!d^4x{\cal L}_{\rm Dirac}$ is invariant.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\bigskip\goodbreak
\point %3
Finally, a few exercises concerning the plane-wave solutions $e^{-ipx}u(p,s)$
and $e^{+ipx}v(p,x)$ of the Dirac equation.
\spointbegin
Show that
$$
\sum_{s=1,2}u_a(p,s)\bar u_b(p,s)\ =\ (\D p+m)_{ab}\quad
{\rm and}\quad\sum_{s=1,2}v_a(p,s)\bar v_b(p,s)\ =\ (\D p-m)_{ab}\,.
\eqn\SpinSum
$$

\spoint %b
Prove the Gordon identity
$$
\bar u(p',s')\gamma^\mu u(p.s)\
=\ {(p'+p)^\mu\over 2m}\,\bar u(p's')u(p,s)\
+\ {i(p'-p)_\nu\over m}\,\bar u(p's') S^{\mu\nu}u(p,s).
\eqn\Gordon
$$
Hint: First, use Dirac equations for the $u$ and the $\bar u'$ to show that\brk
$2m\bar u'\gamma^\mu u=\bar u'({\D p'\gamma^\mu}+{\gamma^\mu\D p})u$.

\spoint %c
Generalize the Gordon identity to $\bar u'\gamma^\mu v$, $\bar v'\gamma^\mu u$
and $\bar v'\gamma^\mu v$.

\spoint %d
[Added on 10/07]
Verify that for $\bp'=-\bp$, $u^\dagger(p,s)v(p',s')=0$.
\bye
