\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\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\A{{\bar A}}
\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 14, 2003.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
First, 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 $\bar v'\gamma^\mu v$.

\bigskip
\point %2
The second problem concerns finite representations of the Lorentz symmetry,
or rather ${\rm Spin(3,1)}\cong SL(2,{\bf C})$. 
Consider the Lorentz generators $\hat J^{\mu\nu}$:
In 3--index notations, the $\hat J^{ij}=\epsilon^{ij\ell}\hat J^\ell$ generate
ordinary rotations while the $\hat J^{0i}= -\hat J^{i0}=\hat K^i$ 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 ${\bf\hat J}_-={\bf\hat J}_+^\dagger$.
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 spinirs, 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
$$
X_{\alpha\dot\gamma}\ =\ X_\mu\sigma^\mu_{\alpha\dot\gamma}\
=\ X_0\delta_{\alpha\dot\gamma}\ -\ {\bf X}\cdot\bsigma_{\alpha\dot\gamma}\,,
\eqno\eq
$$
or in $2\times2$ matrix notations, $X_\mu\sigma^\mu=X_0-{\bf X}\cdot\bsigma$
where $\sigma^0$ is the unit matrix while $\sigma^1$, $\sigma^2$ and $\sigma^3$
are the Pauli matrices.
In the $SL(2,{\bf C})$ terms, we have
$$
X'_{\alpha\dot\gamma}\
=\ M_\alpha^{\,\,\beta}M_{\dot\gamma}^{*\dot\delta}X_{\gamma\dot\delta}\qquad
i.\,e.,\quad X'_\mu\sigma^\mu\ =\ M(X_\mu\sigma^\mu)M^\dagger.
\eqn\LorentzM
$$

\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 %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
$$
\hat\Phi(x)\ = \int\!{d^3\bp\over(2\pi)^3\,2E_\bp}\,
\left[ e^{-ipx}\,\a(p)\ +\ e^{+ipx}\,\ad(p) \right]_{p^0\equiv E_\bp}
\eqn\Scalar
$$
(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
$$
\hat{\cal D}(L)\,\a(p)\,\hat{\cal D}^\dagger(L)\ =\ \a(Lp),\qquad
\hat{\cal D}(L)\,\ad(p)\,\hat{\cal D}^\dagger(L)\ =\ \ad(Lp),
\eqn\LAD
$$
and therefore
$$
\hat{\cal D}(L)\ket{p}\ =\,\ket{Lp},\quad
\hat{\cal D}(L)\ket{p_1,p_2}\ =\,\ket{Lp_1,Lp_2},\quad
\etc,\ \etc
\eqn\LAP
$$
thus {\sl particle}-frame Lorentz transform.

\pointcon
Now consider a generic Lorentz multiplet of quantum fields
$\hat\phi_A(x)$ which transform into each other according to
$$
\hat\phi'_A(x')\ 
\equiv\ \hat{\cal D}^\dagger(L)\,\hat\phi_A(x')\,\hat{\cal D}^\pd(L)\
= \sum_B\,M_A^{\,\,B}(L)\, \hat\phi_B(x=L^{-1}x') 
\eqn\LGenF
$$
in the field frame, or
$$
\hat{\cal D}^\pd(L)\,\hat\phi_A(x)\,\hat{\cal D}^\dagger(L)\
= \sum_B\,M_A^{\,\,B}(L^{-1})\, \hat\phi_B(Lx) 
\eqn\LGenP
$$
in the particle frame.
In both frames, the matrices $M_A^{\,\,B}(L)$ form a finite but non-unitary
representation of the Lorentz group while the Fock-space operators ${\cal D}(L)$
form a unitary but infinite representation.

\spoint %b
Verify that formula \LGenP\ is consistent with the same group law for both
the field-multiplet and the Fock--space representations,
$M_A^{\,\,C}(L_1L_2)=\sum_B M_A^{\,\,B}(L_1)M_B^{\,\,C}(L_2)$ while
 $\hat{\cal D}(L_2L_1)=\hat{\cal D}(L_2)\)\hat{\cal D}(L_1)$.

\pointcon
A free (complex) quantum field comprises particle and antiparticle
creation and annihilation operators according to
$$
\eqalign{
\hat\phi^{}_A(x)\ &
= \int\!{d^3\bp\over(2\pi)^3\,2E_\bp}\sum_s\,
    \left[ e^{-ipx}f_A(p,s)\,\a(p,s)\
	+\ e^{+ipx}h_A(p,s)\,\bd(p,s) \right]_{p^0\equiv E_\bp}\cr
\hat\phi^\dagger_\A(x)\ &
= \int\!{d^3\bp\over(2\pi)^3\,2E_\bp}\sum_s\,
    \left[ e^{-ipx}h^*_\A(p,s)\,\b(p,s)\
	+\ e^{+ipx}f^*_\A(p,s)\,\ad(p,s) \right]_{p^0\equiv E_\bp}\cr
}\eqn\Generic
$$
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.
For the real (\ie, non-hermitian) fields, there are similar formulae where
$h_A^{}(p,s)=f^*_\A(p,s)$, $\b(p,s)=\a(p,s)$ and $\bd(p,s)=\ad(p,s)$, \ie,
the particles are their own antiparticles.

\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
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pointcon
Finally, let us define matrices
$$
{\cal F}_a^{\,\,b}(p)\ =\ \sum_s f_a(p,s)f^{*b}(p,s)
\quad{\rm and}\quad
{\cal H}_a^{\,\,b}(p)=\sum_s h_a(p,s)h^{*b}(p,s).
\eqno\eq
$$
In the next homework, we shall use them to prove the spin-statistics theorem
of the Quantum Field Theory.
But right now, let us simply show that they transform under (proper, orthochronous)
Lorentz transformations according to
$$
{\cal F}_a^{\,\,b}(Lp)\
=\ \sum_{c,d} M_a^{\,\,c}(L)\,{\cal F}_c^{\,\,d}(p)\,M^{*b}_{\,\,\,\,\,d}(L)
\quad{\rm and}\quad {\cal H}_a^{\,\,b}(Lp)\
=\ \sum_{c,d} M_a^{\,\,c}(L)\,{\cal H}_c^{\,\,d}(p)\,M^{*b}_{\,\,\,\,\,d}(L),
\eqn\FH
$$
or in matrix notations, ${\cal F}(Lp)=M(L){\cal F}(p)M^\dagger(L)$ and
${\cal H}(Lp)=M(L){\cal H}(p)M^\dagger(L)$.
%
\spoint %d
Verify eq.~\FH.
\bye
