\input phyzzx

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

\hsize=6.8in	\hoffset=-0.2in	\vsize=8.75in \voffset=-0.25in
%\overfullrule=0pt
\noindent PHY--396~K.\quad
Problem set \#7.\quad
Due October 19, 2004.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
Consider the matrix $\gamma^5\eqdef i\gamma^0\gamma^1\gamma^2\gamma^3$.

\spointbegin
Show that $\gamma^5$ anticommutes with each of the $\gamma^\mu$ matrices,
$\gamma^5\gamma^\mu=-\gamma^\mu\gamma^5$.

\spoint %b
Show that $\gamma^5$ is hermitian and that $(\gamma^5)^2=1$.

\spoint %c
Show that $\gamma^5=(-i/24)\epsilon_{\kappa\lambda\mu\nu}
\gamma^\kappa\gamma^\lambda\gamma^\mu\gamma^\nu$ and
$\gamma^{[\kappa}\gamma^\lambda\gamma^\mu\gamma^{\nu]}=
-i\epsilon^{\kappa\lambda\mu\nu}\,\gamma^5$.

\spoint %d
Show that $\gamma^{[\lambda}\gamma^\mu\gamma^{\nu]}=
i\epsilon^{\kappa\lambda\mu\nu}\,\gamma_\kappa\gamma^5$.

\spoint %e
Show that any $4\times4$ matrix $\Gamma$ is a unique linear combination
of the following 16 matrices:
1, $\gamma^\mu$, $\gamma^{[\mu}\gamma^{\nu]}$, $\gamma^5\gamma^\mu$
and $\gamma^5$.

\pointcon
Conventions: $\epsilon^{0123}=+1$, $\epsilon_{0123}=-1$,
$\gamma^{[\mu}\gamma^{\nu]}=
{1\over2}(\gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu)$,\brk
$\gamma^{[\lambda}\gamma^\mu\gamma^{\nu]}={1\over6}(
\gamma^\lambda\gamma^\mu\gamma^\nu-\gamma^\lambda\gamma^\nu\gamma^\mu
+\gamma^\mu\gamma^\nu\gamma^\lambda-\gamma^\mu\gamma^\lambda\gamma^\nu
+\gamma^\nu\gamma^\lambda\gamma^\mu-\gamma^\nu\gamma^\mu\gamma^\lambda)$,\brk
and ditto for the $\gamma^{[\kappa}\gamma^\lambda\gamma^\mu\gamma^{\nu]}$.

\bigskip
\point %2
Consider bilinear products of a Dirac field $\Psi(x)$ and its
conjugate $\overline\Psi(x)$.
Generally, such products have form $\overline\Psi\Gamma\Psi$ where
$\Gamma$ is one of 16 matrices discussed in {\bf1}.(e); altogether, we have
$$
S\ =\ \overline\Psi\Psi,\quad
V^\mu\ =\ \overline\Psi\gamma^\mu\Psi,\quad
T^{\mu\nu}\ =\ \overline\Psi i\gamma^{[\mu}\gamma^{\nu]}\Psi,\quad
A^\mu\ =\ \overline\Psi\gamma^5\gamma^\mu\Psi\,\quad{\rm and}\quad
P\ =\ \overline\Psi i\gamma^5\Psi.
\eqn\BiLinears
$$

\spointbegin
Show that all the bilinears~\BiLinears\ are Hermitian.\brk
Hint: First, show that $\left(\overline\Psi\Gamma\Psi\right)^\dagger=
\overline\Psi\overline\Gamma\Psi$

\spoint %b
Show that under {\sl continuous} Lorentz symmetries, the $S$ and the $P$
transform as  scalars, the $V^\mu$ and the $A^\mu$ as vectors and the
$T^{\mu\nu}$ as an antisymmetric tensor.

\spoint %c
Find the transformation rules of the bilinears~\BiLinears\ under parity
(\cf\ problem~2 of the previous set)
and show that while $S$ is a true scalar and $V$ is a true (polar) vector,
$P$ is a pseudoscalar and $A$ is an axial vector.

\pointcon
Next, consider the charge-conjugation properties of Dirac bilinears.
%${\overline\Psi}\Gamma\Psi$.
To avoid operator ordering problems, take $\Psi(x)$ and $\Psi^\dagger(x)$
to be ``classical'' fermionic fields which {\it anticommute} with
each other, $\Psi_\alpha^{}\Psi^\dagger_\beta=
-\Psi^\dagger_\beta\Psi_\alpha^{}$.

\spoint %d
In the Weyl convention, $\hat{\cal C}\hat\Psi(x)\hat{\cal C}=
\pm\gamma^2\hat\Psi^*(x)$.
Show that $\hat{\cal C}\hat{\overline\Psi}\Gamma\hat\Psi\hat{\cal C}=
\hat{\overline\Psi}\Gamma^c\hat\Psi$ where
$\Gamma^c=\gamma^0\gamma^2\Gamma^\top\gamma^0\gamma^2$.

\spoint %e
Calculate $\Gamma^c$ for all 16 independent matrices $\Gamma$
and find out which Dirac bilinears are $\cal C$--even and which
are $\cal C$--odd.


\bigskip
\point %3
Next, an exercise in fermionic creation and annihilation operators and their
anticommutation relations,
$$
\{\a_\alpha,\a_\beta\}\ =\ \{\ad_\alpha,\ad_\beta\}\ =\ 0,\quad
\{\a_\alpha,\ad_\beta\}\ =\ \delta_{\alpha,\beta}\,.
\eqn\CCR
$$

\spointbegin
Calculate the commutators $[\ad_\alpha\a_\beta,\ad_\gamma]$,
$[\ad_\alpha\a_\beta,\a_\delta]$
and $[\ad_\alpha\a_\beta,\ad_\gamma\a_\delta]$.

\spoint %b
Consider two one-body operators $\hat A_1$ and $\hat B_1$
and let $\hat C_1$ be their commutator, $\hat C_1=[\hat A_1,\hat B_1]$.
Let $\hat A$ be the second-quantized forms of $\hat A_{\rm tot}$,
$$
\hat A\ =\sum_{\alpha,\beta}\bra\alpha\hat A_1\ket\beta\,\ad_\alpha\a_\beta\,,
\eqn\OneParticle
$$
and ditto for the second-quantized $\hat B$ and $\hat C$.
\subpar
Verify that $[\hat A,\hat B]=\hat C$.

\spoint %c
Calculate the commutator $[\ad_\mu\a_\nu,\ad_\alpha\ad_\beta\a_\gamma\a_\delta]$.

\spoint %d
The second quantized form of a two-body additive operator
$$
\hat B_{\rm tot}\ =\ \half\sum_{i\neq j}\,
\hat B_2( i{\rm th\atop{}}\ {\rm and}\,j{\rm th\atop{}}\,{\rm particles})
$$
acting on identical fermions is
$$
\hat B\ =\ \half\sum_{\alpha,\beta,\gamma,\delta}
\bra{\alpha\otimes\beta}\hat B_2\ket{\gamma\otimes\delta}\,
	\ad_\alpha\ad_\beta\a_\delta\a_\gamma\,.
\eqn\TwoParticles
$$
This expression is similar to its bosonic counterpart,
but note the reversed order
of the annihilation operators $\a_\delta$ and $\a_\gamma$.
\subpar
Consider a one-body operator $\hat A_1$ and two two-body operators $\hat B_2$
and $\hat C_2$.
Show that if
$\hat C_2=\left[\bigl(\hat A_1(1\coeff{\rm st}{})
 +\hat A_1(2\coeff{\rm nd}{})\bigr),\hat B_2\right]$,
then the respective second-quantized operators in the fermionic Fock space
satisfy $\hat C=[\hat A,\hat B]$.

\bigskip\goodbreak
\point %4
Finally, consider the quantum Dirac fields
$$
\eqalign{
\hat\Psi(x)\ &
=\int\!{d^3\bp\over(2\pi)^3}\,{1\over2E_\bp}\sum_s\left(
	e^{-ipx}u(\bp,s)\a_{\bp,s}\,+\, e^{+ipx}v(\bp,s)\bd_{\bp,s}
	\right)_{p^0=+E_\bp}\,,\cr
\hat{\overline\Psi}(x)\ &
=\int\!{d^3\bp\over(2\pi)^3}\,{1\over2E_\bp}\sum_s\left(
	e^{-ipx}\bar v(\bp,s)\b_{\bp,s}\,+\,e^{+ipx}\bar u(\bp,s)\ad_{\bp,s}
	\right)_{p^0=+E_\bp}\,,\cr
}\eqn\DiracFields
$$
where $\a$, $\b$, $\ad$, and $\bd$ are relativistically normalized
fermionic annihilation and creation operators, thus
$$
\{\a_{\bp,s},\ad_{\bp',s'}\}\ =\ \{\b_{\bp,s},\bd_{\bp',s'}\}\
=\ \delta_{s,s'}\times 2E_\bp(2\pi)^3\delta^{(3)}(\bp-\bp')
\eqn\RelAC
$$
while all other anticommutators vanish,
$$
\def\or{\,{\rm or}\,}
\{\a\or\b\,,\,\a\or\b\}\ =\ 0,\quad
\{\ad\or\bd\,,\,\ad\or\bd\}\ =\ 0,\quad
\{\a,\bd\}\ =\ \{\b,\ad\}\ =\ 0.
\eqn\RelACZ
$$
As discussed in class, the free Dirac Hamiltonian is
$$
\hat H\
=\int\!d^3\bx\hat{\overline\Psi}(-i\vec\gamma\cdot\nabla\,+\,m)\hat\Psi\
=\int\!{d^3\bp\over(2\pi)^3}\,{1\over2E_\bp}\sum_s\left(
	E_\bp\,\ad_{\bp,s}\a_{\bp,s}\,+\,E_\bp\,\bd_{\bp,s}\b_{\bp,s}\right)\
+\ {\rm const}.
\eqn\DiracHam
$$

\spointbegin
Derive Dirac field's stress-energy tensor (use Noether theorem) and show that
the net mechanical momentum is
$$
{\bf\hat P}_{\rm mech}\
=\int\!d^3\bx\,\hat\Psi^\dagger(-i\nabla)\hat\Psi\
=\int\!{d^3\bp\over(2\pi)^3}\,{1\over2E_\bp}\sum_s\left(
	\bp\,\ad_{\bp,s}\a_{\bp,s}\,+\,\bp\,\bd_{\bp,s}\b_{\bp,s}\right).
\eqn\Momentum
$$

\spoint %b
Show that the electric 4-current of the electron field is
$J^\mu(x)=-e\overline\Psi(x)\gamma^\mu\Psi(x)$
and that the net electric charge operator is
$$
\eqalign{
\hat Q\ &
=\ -e\!\int\!d^3\bx\,\hat\Psi^\dagger(x)\hat\Psi(x)\
	+\rm constant\cr
&=\int\!{d^3\bp\over(2\pi)^3}\,{1\over2E_\bp}\sum_s\left(
	-e\,\ad_{\bp,s}\a_{\bp,s}\,+\,e\,\bd_{\bp,s}\b_{\bp,s}\right).\cr
}\eqn\Charge
$$
Note: The constant term in the first line arises from the
operator ordering ambiguity when the classical electron field is quantized.
It's actual value --- which happens to be infinite --- is determined
by demanding that the vacuum state has zero electric charge.

\spoint %c
Finally, consider the net spin of electrons and positrons,
$$
{\bf\hat S}_{\rm net}\ =\int\!d^3\bx\,\hat\Psi^\dagger{\bf S}\hat\Psi .
\eqn\Spin
$$
Expand this operator into momentum modes
$$
{\bf\hat S}_{\rm net}\
=\int\!{d^3\bp\over(2\pi)^3}\,{1\over2E_\bp}\,{\bf\hat S}_\bp
\eqno\eq
$$
and show that for the non-relativistic modes ($|\bp|\ll m$)
$$
{\bf\hat S}_\bp\
=\,\sum_{s,s'}\xi^\dagger_s{\bsigma\over2}\xi_{s'}^{}\times\left(
	\ad_{\bp,s}\a_{\bp,s'}\,+\,\bd_{\bp,s}\b_{\bp,s'}\right)\
+\ O(|\bp|/m).
\eqn\SpinMode
$$
The relativistic modes with $|\bp|\gsim O(m)$ are more complicated
because of mixing between the spin and the orbital angular momentum.
\subpar
Hint: Approximate $u(\bp,s)\approx u({\bf 0},s)$ and
$v(-\bp,s)\approx v({\bf0},s)$ for small $|\bp|\ll m$,
and use $\eta_s=\sigma_2\xi_s^*$.

\pointcon
In particle terms, eqs.~\DiracHam--\SpinMode\ mean that
the fermionic operator $\ad_{\bp,s}$ creates and $\a_{\bp,s}$ annihilates
an electron with momentum $\bp$, energy $E_\bp=+\sqrt{m^2+\bp^2}$,
spin${}=\half$ and spin state $\xi_s$, and electric charge${}=-e$, 
while operator $\bd_{\bp,s}$ creates and
$\b_{\bp,s}$ annihilates a positron with exactly the same momentum,
energy, spin and spin state, but electric charge${}=+e$.


\bye
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\bigskip
\point %5
In theories involving both bosons and fermions,
one often has to combine commutation and
anti-commutation relations of various operators, depending on the overall
statistics of the operators involved.
For that purpose, it is useful to define a `mixed' commutator bracket
$$
[\hat A,\hat B\}\ \buildrel \scriptscriptstyle{\rm def}\over= \
\hat A\hat B\ -\ (-1)^{AB}\hat B\hat A
\eqn\MixedDef
$$
where $(-1)^{AB}$ is $-1$ if both $\hat A$ and $\hat B$ have overall Fermi
statistics (\ie, each comprises an odd number of {\sl fermionic}
creation/annihilation operators --- the number of bosonic creation/annihilation
operators does not matter) and $+1$ in all other cases.

\spointbegin
Verify the Leibniz rule for the mixed brackets:
$[\hat A,\hat B\hat C\} = [\hat A,\hat B\}\hat C
+ (-1)^{AB}\hat B[\hat A,\hat C\}$
and write down a similar rule for the
$[\hat A\hat B,\hat C\}$.

\spoint
Similarly, express $[\hat A\hat B,\hat C\hat D\}$ in terms of appropriate
mixed brackets of $\hat A$ or $\hat B$ with $\hat C$ or $\hat D$.

\spoint
Prove the `mixed' Jacobi identity
$$
(-1)^{CA}[\hat A,[\hat B,\hat C\}\}\ +\ (-1)^{AB}[\hat B,[\hat C,\hat A\}\}\
+\ (-1)^{BC}[\hat C,[\hat A,\hat B\}\}\ =\ 0.
\eqn\MixedJacobi
$$
In other words (and notations),
$$
\vcenter{\openup 1\jot \ialign{&$\displaystyle{{}#}$\enspace\hfil\cr
	   [\hat B_1,[\hat B_2,\hat B_3]] &
	+\ [\hat B_2,[\hat B_3,\hat B_1]] &
	+\ [\hat B_3,[\hat B_1,\hat B_2]] &
	=\ 0,\cr
	   [\hat B_1,[\hat B_2,\hat F]] &
	+\ [\hat B_2,[\hat F,\hat B_1]] &
	+\ [\hat F,[\hat B_1,\hat B_2]] &
	=\ 0,\cr
	   \{\hat F_1,[\hat F_2,\hat B]\} &
	-\ \{\hat F_2,[\hat B,\hat F_1]\} &
	+\ [\hat B,\{\hat F_1,\hat F_2\}] &
	=\ 0,\cr
	   [\hat F_1,\{\hat F_2,\hat F_3\}] &
	+\ [\hat F_2,\{\hat F_3,\hat F_1\}] &
	+\ [\hat F_3,\{\hat F_1,\hat F_2\}] &
	=\ 0,\cr
}}\eqn\MixedJacobiSpelledOut
$$
where `$B$' and `$F$' indicate the overall statistics of the operator involved.

