\input phyzzx

\def\pd{{\vphantom{\dagger}}}
\def\a{\hat a^\pd}
\def\ad{\hat a^\dagger}
\def\b{\hat b^\pd}
\def\bd{\hat b^\dagger}
\def\bx{{\bf x}}
\def\bp{{\bf p}}
\def\eqdef{\buildrel\rm def\over=}

\hsize=6.8in	\hoffset=-0.2in	\vsize=8.5in	\overfullrule=0pt
\noindent PHY--396~K.\quad
Problem set \#6.\quad
Due October 23, 2000.
\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
Verify $M^{-1}(L)\gamma^\mu M(L)= L^\mu_{\,\,\nu}\gamma^\nu$
for $L=\exp(\theta)$ (\ie, $L^\mu_{\,\,\nu}=\delta^\mu_\nu+
 \theta^\mu_{\,\,\nu}+\half\theta^\mu_{\,\,\lambda}\theta^\lambda_{\,\,\nu}+\cdots$)
and $M(L)=\exp\bigl(-\coeff{i}{2}\theta_{\alpha\beta}S^{\alpha\beta}\bigr)$

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

\spoint %e
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
$\bigl( m^2+D^2+qF_{\mu\nu}S^{\mu\nu}\bigr)\Psi(x)=0$.

\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
$\half\vec\sigma$ and ${\bf\hat K}$ as $-\coeff{i}{2}\vec\sigma$
and the congugate Weyl `anti-spinor' $(j_+=0,j_-=\half)$
where ${\bf\hat J}$ also acts as $\half\vec\sigma$
but ${\bf\hat K}$ acts as $+\coeff{i}{2}\vec\sigma$.
Together the Weyl spinor and the Weyl antispinor comprise the Dirac spinor.

\spoint %b
Show that for any infinitesimal combination of a Lorentz boost $\vec b$
and rotation $\vec\theta\equiv\theta\bf n$,
$$
\Psi'(x')\ =\ \Psi(x)\
+\,\pmatrix{ -\coeff{i}{2}(\vec\theta-i\vec b)\cdot\vec\sigma & 0\cr
	0 & -\coeff{i}{2}(\vec\theta+i\vec b)\cdot\vec\sigma\cr }\,
\Psi(x),
\eqn\DiracJ
$$
which means that a Dirac spinor indeed decomposes into a Weyl spinor
and a Weyl antispinor.

\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 $X^\mu\sigma_\mu=T-{\bf X}\cdot\vec\sigma$.
(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
$U=\exp\bigl(-\coeff{i}{2}\theta{\bf n}\cdot\vec\sigma\bigr)$,
$L(U)$ is a rotation by angle $\theta$ around axis $\bf n$
while for $U=\exp\bigl(-\coeff{1}{2} r {\bf n}\cdot\vec\sigma\bigr)$,
$L(U)$ is a boost of rapidity $r$ ($\beta=\tanh r$, $\gamma=\cosh r$)
in the direction $\bf n$.

\bigskip
\point %3
Finally, consider the relation between Lorentz transformations of the fields
and of the particles.
In mechanics (classical or quantum), one must distinguish between two
opposite kinds of rotations, namely coordinate-frame rotations of bodies
and body-frame rotations of coordinate systems.
For the Lorentz transformations of fields and particles, there is a
similar distinction between the particle-frame and field-frame Lorentz transforms.

\pointcon
For example, consider a real (hermitian) scalar quantum 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 field-frame Lorentz transform $L$ acts on this field according to
$$
\hat\Phi'(x')\ 
\equiv\ \hat{\cal D}^\dagger(L)\,\hat\Phi(x')\,\hat{\cal D}^\pd(L)\
=\ \hat\Phi(x=L^{-1}x')
\eqn\LSF
$$
while the corresponding particle-frame transform acts precisely in reverse:
$$
\hat{\cal D}^\pd(L)\,\hat\Phi(x)\,\hat{\cal D}^\dagger(L)\
=\ \hat\Phi(Lx) .
\eqn\LSP
$$
In both cases $\hat{\cal D}(L)=\exp\bigl(\coeff{i}{2}\theta_{\alpha\beta}
	\hat J^{\alpha\beta}\bigr)$ is a unitary operator representing
the lorentz transform $L$ in the Fock space of the quantum field theory.

\spointbegin
Show that \LSP\ implies
$$
\eqalign{
\hat{\cal D}(L)\bigl(\sqrt{2p^0}\,\a(p)\bigr)\hat{\cal D}^\dagger(L)\ &
=\ \sqrt{2(Lp)^0}\,\a(Lp),\cr
\hat{\cal D}(L)\bigl(\sqrt{2p^0}\,\ad(p)\bigr)\hat{\cal D}^\dagger(L)\ &
=\ \sqrt{2(Lp)^0}\,\ad(Lp),\cr
\noalign{\hbox{and hence}}
\hat{\cal D}(L)\ket{p}\ &=\,\ket{Lp},\cr
\hat{\cal D}(L)\ket{p_1,p_2}\ &=\,\ket{Lp_1,Lp_2},\cr
\omit\span\omit \dotfill\cr
}\eqn\LAD
$$
(Thus {\sl particle}-frame Lorentz transform.)


\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, thus
$$
\hat\phi'_a(x')\ 
\equiv\ \hat{\cal D}^\dagger(L)\,\hat\phi_a(x')\,\hat{\cal D}^\pd(L)\
= \sum_b\,M_a^{\,\,b}(L)\, \hat\phi_b(x=L^{-1}x') 
\eqn\LGenF
$$
in the field frame and
$$
\hat{\cal D}^\pd(L)\,\hat\phi_a(x)\,\hat{\cal D}^\dagger(L)\
= \sum_b\,M_a^{\,\,b}(L^{-1})\, \hat\phi_b(Lx) 
\eqn\LGenP
$$
in the particle frame.

\spoint %b
Verify that formula \LGenP\ 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)$.

\spoint %c
A particle-frame Lorentz transform should act on particle --- or antiparticle ---
quantum numbers  according to
$$
\hat{\cal D}(L)\,\ket{p,\pm,s}\
=\sum_{s'}\,C_{s,s'}(L,p)\,\ket{Lp,\pm,s'} .
\eqn\LGRule
$$
Show that eqs.\ \LGenP\ and \LGRule\ are consistent with each other
if and only if
$$
\eqalign{
f_a(Lp,s')\ &
=\sum_b\sum_s\,M_a^{\,\,b}(L)\,C^*_{s,s'}(L,p)\,f_b(p,s),\cr
h_a(Lp,s')\ &
=\sum_b\sum_s\,M_a^{\,\,b}(L)\,C^{}_{s,s'}(L,p)\,h_b(p,s).\cr
}\eqno\eq
$$

\bye
