\input phyzzx

\def\bx{{\bf x}}
\def\eqdef{\buildrel\rm def\over=}
\def\a{\hat a^{\vphantom\dagger}}
\def\ad{\hat a^\dagger}
\def\bd{\hat b^\dagger}
\def\bp{{\bf p}}
%\def\kl{{\bk,\lambda}}
%\def\klp{{\bk',\lambda'}}
\def\PMBOLD#1#2{\setbox0=\hbox{$#1{#2}$}\setbox2=\hbox{$#1{0}$}%
	\copy0\kern-\wd0\kern 0.1\wd2\box0 }
\def\pmbold#1{\mathpalette\PMBOLD{#1}}


\hsize=6.8in	\hoffset=-0.2in	%\vsize=8.5in	\overfullrule=0pt
\noindent PHY--396~K.\quad
Problem set \#5.\quad
Due October 8, 1998.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
The first problem is about Dirac's $\gamma$ matrices.
\spointbegin
Verify
$\bigl[S^{\kappa\lambda},S^{\mu\nu}\bigr]=i\bigl(
g^{\lambda\mu}S^{\kappa\nu}-g^{\lambda\nu}S^{\kappa\mu}
-g^{\kappa\mu}S^{\lambda\nu}+g^{\kappa\nu}S^{\lambda\mu}\bigr)$.

\spoint %b
Calculate $\{\gamma^\rho,\gamma^\lambda\gamma^\mu\gamma^\nu\}$,
$[\gamma^\rho,\gamma^\kappa\gamma^\lambda\gamma^\mu\gamma^\nu]$ and
$[S^{\rho\sigma},\gamma^\lambda\gamma^\mu\gamma^\nu]$.

\spoint %c
Show that
$\gamma^\alpha\gamma_\alpha=4$,
$\gamma^\alpha\gamma^\nu\gamma_\alpha=-2\gamma^\nu$,
$\gamma^\alpha\gamma^\mu\gamma^\nu\gamma_\alpha=4g^{\mu\nu}$ and
$\gamma^\alpha\lambda\gamma^\mu\gamma^\nu\gamma_\alpha=
	-2\gamma^\nu\gamma^\mu\gamma^\lambda$.\brk
Hint: use $\gamma^\alpha\gamma^\nu=2g^{\nu\alpha}-\gamma^\nu\gamma^\alpha$
repeatedly.

\spoint %d
Consider the electron's spinor field $\Psi(x)$
in an electomagnetic background.
Show that the gauge-covariant Dirac equation
$\bigl(i\gamma^\mu D_\mu+m\bigr)\Psi(x)=0$ implies\brk
$\bigl( m^2+D^2+eF_{\mu\nu}S^{\mu\nu}\bigr)\Psi(x)=0$.

\medskip
\point %2
The second problem is about the Lorentz group and its generators $\hat J^{\mu\nu}$.
In 3--index notations, $\hat J^{ij}=\epsilon^{ij\ell}\hat J^\ell$ generate
ordinary rotations while $\hat J^{0i}= -\hat J^{i0}=\hat K^i$ generate the
Lorentz boosts.
Let
$$
{\bf\hat J}_\pm\ =\ \half\bigl({\bf\hat J}\ \pm\ i{\bf\hat K}\bigr).
\eqn\Jpm
$$
\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 ${\bf\hat J}_-={\bf\hat J}_+^\dagger$, so 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 the
Weyl spinor $(j_+=\half,j_-=0)$ --- a doublet where ${\bf\hat J}$ acts as
$\half\pmbold\sigma$ and ${\bf\hat K}$ as $-\coeff{i}{2}\pmbold\sigma$
and the congugate Weyl `anti-spinor' $(j_+=0,j_-=\half)$
where ${\bf\hat J}$ also acts as $\half\pmbold\sigma$
but ${\bf\hat K}$ acts as $+\coeff{i}{2}\pmbold\sigma$.
Together the Weyl spinor and the Weyl antispinor comprise the Dirac spinor:
$$
\half\theta_{\mu\nu}\hat S^{\mu\nu}\
=\ \pmatrix{ \half(\pmbold\theta-i\pmbold\beta)\cdot\pmbold\sigma & 0\cr
	0 & \half(\pmbold\theta+i\pmbold\beta)\cdot\pmbold\sigma\cr }
\eqn\DiracJ
$$
for any infinitesimal Lorentz boost $\pmbold\beta$ and rotation $\pmbold\theta$.

\spoint %b
Verify \DiracJ.

\pointcon
Finite Lorentz transformations act on Weyl spinors as complex, unimodular
($\det=1$) but non-unitary two-by-two matrices.
The group $SL(2,{\bf C})$ of such matrices is actually isomorphic to the
${\rm Spin}\)(3,1)$ --- the double cover of the continuous Lorentz group.
(This is similar to ${\rm Spin}\)(3)\cong SU(2)$.)
Any $(j_+,j_-)$ representation of the ${\rm Spin}\)(3,1)$ becomes in the
$SL(2,{\bf C})$ terms a tensor
$\Phi_{a_1\ldots a_{(2j_+)},\dot a_1\ldots \dot a_{( 2j_-)}}$, totally
symmetric in its $2j_+$ un-dotted indices $a_1,\ldots, a_{(2j_+)}$
and separately totally symmetric in its
$2j_-$ dotted indices $\dot a_1,\ldots, \dot a_{(2j_-)}$,
transforming according to
$$
\Phi'_{a_1\ldots a_{(2j_+)},\dot a_1\ldots \dot a_{(2j_-)}}\
=\ U_{a_1}^{\;b_1}\cdots U_{a_{(2j_+)}}^{\;b_{(2j_+)}}\
U_{\dot a_1}^{*\,\dot b_1}\cdots U_{\dot a_{(2j_-)}}^{*\,\dot b_{(2j_-)}}\
\Phi_{b_1\ldots b_{(2j_+)},\dot b_1\ldots \dot b_{(2j_-)}}\,.
\eqn\SLTwoCLaw
$$

\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,
consider $X^\mu\sigma_\mu=T-{\bf X}\cdot\pmbold\sigma$.
(Here $\sigma^0=1$ while $\sigma^1$, $\sigma^2$ and $\sigma^3$ are the Pauli
matrices.)
Let
$$
X^{\prime\mu}\sigma_\mu\ \equiv\ L^\mu_{\,\,\nu}(U)\,X^\nu\sigma_\mu\
=\ U\bigl(X^\mu\sigma_\mu\bigr)U^\dagger .
\eqn\LorentzU
$$
\spoint %c
Show that for any $SL(2,{\bf C})$ matrix $U$, eq.~\LorentzU\ indeed defines
a Lorentz transform.
(Hint: prove and use $\det(X^\mu\sigma_\mu)=X^2\equiv X_\mu X^\mu$).\brk
Also verify the group law, $L(U_2 U_1)=L(U_2) L(U_1)$.

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

\medskip
\point %3
Finally, consider the relation between Lorentz transformations of the fields
and of the particles.
For example, consider a real scalar field
$$
\hat\Phi(x)\ = \int\!{d^3\bp\over(2\pi)^3}\,{1\over\sqrt{2E_\bp}}\,
\left[ e^{-ipx}\,\a(p)\ +\ e^{+ipx}\,\ad(p) \right]_{p^0\equiv E_\bp}
\eqn\Scalar
$$
(where $\a(p)$ stands for the $\a_\bp(t=0)$ and ditto for the $\ad(p)$).
A particle-frame Lorentz transform $L$ acts on $\hat\Phi$ according to
$$
\hat\Phi'(x)\ 
\equiv\ \hat{\cal D}(L)\,\hat\Phi(x)\,\hat{\cal D}^\dagger(L)\
=\ \hat\Phi(Lx) .
\eqn\LScalar
$$
\spointbegin
Show that \LScalar\ implies
$$
\hat{\cal D}(L)\,\ad(p)\,\hat{\cal D}^\dagger(L)\
=\ \sqrt{(Lp)^0\over p^0}\,\ad(Lp)
\eqn\LAD
$$
and ditto for the annihilation operators $\a(p)$.

\pointcon
Next consider the spinless particles created by the $\ad(p)$ and annihilated
by the $\a(p)$.
It is convenient to have a simple rule for the Lorentz transformations of
their quantum states in the momentum basis, namely
$$
\hat{\cal D}(L)\ket{p}\ =\ \ket{Lp},\quad
\hat{\cal D}(L)\ket{p_1,p_2}\ =\ \ket{Lp_1,Lp_2},\quad\etc
\eqn\LSRule
$$
In light of eq.~\LAD, this rule requires us to normalize the momentum
eigenstates states  according to the Lorentz invariant measure
$$
\vev{p|p'}\ =\ 2E_\bp\,(2\pi)^3\delta^{(3)}(\bp-\bp'),
\eqn\Measure
$$
thus
$$
\ket{p}\ =\ \sqrt{2E_\bp}\,\ad_\bp\ket{0},\quad
\ket{p_1,p_2}\ =\ \bigl(\sqrt{2E_{\bp_1}}\,\ad_{\bp_1}\bigr)
	\bigl(\sqrt{2E_{\bp_2}}\,\ad_{\bp_2}\bigr)\ket{0},\quad\etc
\eqn\States
$$

\pointcon
Now consider a generic quantum field
$$
\hat\phi_a(x)\ = \int\!{d^3\bp\over(2\pi)^3}\,{1\over\sqrt{2E_\bp}}\sum_s\,
\left[ e^{-ipx}f_a(p,s)\,\a(p,s)\
	+\ e^{+ipx}h_a(p,s)\,\bd(p,s) \right]_{p^0\equiv E_\bp}
\eqn\Generic
$$
where $e^{-ipx}f_a(p,s)$ and $e^{+ipx}h_a(p,s)$
are independent plane-wave solutions of the free field equation for the $\phi_a$,
whatever that might be.
We assume complex (\ie, non-hermitian) $\hat\phi_a(x)$; otherwise
we would have $\bd(p,s)=\ad(p,s)$ and $h_a(p,s)=f_a^*(p,s)$.
\subpar
The field $\hat\phi_a(x)$ transforms
according to some representation $M_a^{\,\,b}(L)$ of the
Lorentz symmetry,
$$
\hat\phi'_a(x)\ 
\equiv\ \hat{\cal D}(L)\,\hat\phi_a(x)\,\hat{\cal D}^\dagger(L)\
= \sum_b\,M_a^{\,\,b}(L^{-1})\, \hat\phi_b(Lx) .
\eqn\LGeneric
$$
\spoint %b
Verify that formula \LGeneric\ is consistent with the group Law for the
Lorentz symmetry, $\hat{\cal D}(L_2L_1)=\hat{\cal D}(L_2)\)\hat{\cal D}(L_1)$.

\pointcon
The particles created by the $\ad(p,s)$ operators are normalized according to
$$
\ket{p,s}\ =\ \sqrt{2E_\bp}\,\ad(p,s)\ket{0},\qquad
\vev{p,s|p',s'}\ =\ 2E_\bp\,(2\pi)^3\delta^{(3)}(\bp-\bp')\,\delta_{s,s'}
\eqn\GMeasure
$$
and ditto for the antiparticles created by the $\bd(p,s)$.
The Lorentz transformations act on the particle --- or antiparticle ---
 states according to
$$
\hat{\cal D}(L)\,\ket{p,s}\
=\sum_r\,C_{s,r}(L,p)\,\ket{Lp,r} .
\eqn\LGRule
$$

\spoint %c
Show that eqs.\ \LGeneric\ and \LGRule\ are consistent with each other
if and only if
$$
\eqalign{
\sum_s\,f_a(p,s)\,C^*_{s,r}(L,p)\ &
= \sum_b\,M_a^{\,\,b}(L^{-1})\,f_b(Lp,r),\cr
\sum_s\,h_a(p,s)\,C^{\phantom*}_{s,r}(L,p)\ &
= \sum_b\,M_a^{\,\,b}(L^{-1})\,h_b(Lp,r).\cr
}\eqno\eq
$$

\bye
