\input phyzzx

\def\bx{{\bf x}}
\def\eqdef{\buildrel\rm def\over=}
\def\a{\hat a^{\vphantom\dagger}}
\def\ad{\hat a^\dagger}
\def\bk{{\bf k}}
\def\kl{{\bk,\lambda}}
\def\klp{{\bk',\lambda'}}


\hsize=6.8in	\hoffset=-0.2in	%\vsize=8.5in	\overfullrule=0pt
\noindent PHY--396~K.\quad
Problem set \#4.\quad
Due October 1, 1998.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
Conside a {\sl massive} real vector field $A^\mu(x)$ with a classical
free Lagrangian
$$
{\cal L}\ =\ -\coeff14\, F_{\mu\nu} F^{\mu\nu}\
+\ \half m^2 A_\mu A^\mu 
\eqn\MassiveL
$$
where --- similar to the massless case ---
$F_{\mu\nu}\eqdef\partial_\mu A_\nu-\partial_\nu A_\mu$.
Unlike its massless counterpart however, the Lagrangian~\MassiveL\
is {\sl not} gauge invariant.

\spointbegin
Let us add a source term $-J^\mu A_\mu$ to the free Lagrangian~\MassiveL.
Show that for any {\sl conserved} current $J^\mu(x)$,
$\partial_\mu J^\mu=0$, the vector field $A^\mu$ satisfies
$$
\displaylines{
\partial_\mu\,A^\mu\ =\ 0 \hfil\eqname\LorentzGauge\hfilneg\cr
{\rm and}\quad \left( \partial^2+m^2\right) A^\mu\  =\ J^\mu.
	\hfil\eqname\Dalambert\hfilneg\cr
}$$
Note that the massless electromagnetic field generally does not satisfy
eq.~\LorentzGauge, although it may be imposed as a Lorentz-invariant
gauge-fixing condition (`the Lorentz gauge').
Likewise, in a general gauge, the field equation for the massless vector
field is $\partial_\mu F^{\mu\nu}\equiv \partial^2 A^\mu
-\partial^\mu(\partial A)= J^\mu$, which reduces to the massless version
of the Dalambert equation~\Dalambert\ {\sl only} in the Lorentz gauge
\LorentzGauge.
By contrast, there are no gauge transformations of the massive field
and both eqs.~\LorentzGauge\ and ~\Dalambert\ are field equations
(for a conserved current $J^\mu$ only).

\pointcon
The Lagrangian~\MassiveL\ does not depend on the $\partial_t A^0$, so the
$A^0$ component of the 4--vector field $A^\mu$ is not dynamical but auxiliary.
In the massive case, the field equation for the $A^0$ does {\sl not} impose a
transversality constraint $\nabla\cdot{\bf E}=0$ but simply a time-independent
(or rather history-independent) equation for the $A^0$ itself,
$A^0=(-1/m^2)\nabla\cdot{\bf E}$.
Consequently, one may eliminate the auxiliary field $A^0(\bx)$ from the Lagrangian
for the dynamical fields $A^i(\bx)$ and their time derivatives by means of
solving the field equation for the $A^0$ in terms of the
$\partial_t{\bf A}$ and substituting the result back into
the Lagrangian~\MassiveL. 

\pointcon
Practically, it is easier to eliminate the auxiliary $A^0$ in terms of the
Fourier transformed fields $A^\mu(\bk)$ and helicity basis ${\bf e}_\kl$
for the 3--vectors:
$$
\displaylines{
i\bk\times{\bf e}_\kl\ =\ \lambda|\bk|\,{\bf e}_\kl\qquad
	{\rm for}\ \lambda=0,\pm1,
	\hfil\eqname\Helicity\hfilneg\cr
{\bf e}^*_\kl \cdot {\bf e}^{\phantom{*}}_{\bk,\lambda'}\
	=\ \delta_{\lambda,\lambda'}\,,\qquad
    {\bf e}_{\bk,0}\ =\ {\bk\over|\bk|}\,,\qquad
    {\bf e}^{\phantom{*}}_{-\bk,\lambda}\ =\ -(-1)^\lambda {\bf e}^*_\kl\,.\cr
}$$

\spoint %b
Show that solving for the $A^0_\bk$ yields
$$
\displaylines{
A^0_\bk\ =\ {-i|\bk|\over m^2}\,E_{\bk,0}\
	=\ {i|\bk|\over\omega_\bk^2}\,\dot A_{\bk,0}\,,\qquad
   E_{\bk,0}\ =\ {-m^2\over\omega_\bk^2}\,\dot A_{\bk,0}
   \hfil\eqname\Auxilrel\hfilneg\cr
{\rm and}\quad L\ \equiv\int\!\!d^3\bx\,{\cal L}\
    = \int\!{d^3\bk\over(2\pi)^3} \sum_\lambda\,\half C_\kl
	\left( |\dot A_\kl|^2\ -\ \omega_\bk^2\,|A_\kl|^2\right) ,
   \hfil\eqname\KspaceL\hfilneg\cr
{\rm where}\quad \omega_\bk^2\ =\ \bk^2\ +\ m^2,\qquad
    C_{\bk,\pm1}\ =\ 1\quad {\rm and}\quad C_{\bk,0}\ =\ {m^2\over \omega^2}\,.
   \hfil\eqname\DEFS\hfilneg\cr
}$$
Note that in the massless limit $m^2=0$, $C_0$ vanishes and the longitudinal
polarization decouples from the Lagrangian~\KspaceL\ and becomes unphysical;
in contrast, all three polarizations are physical
in the massive case $m^2>0$.

\spoint %c
Next, let us go over to the Hamiltonian formalism.
Show that the canonically conjugate fields are $\Pi_\kl=-E^*_\kl$ and
write down the Hamiltonian in terms of the $A_\kl$ and the $E_\kl$.

\pointcon
Now we are ready to quantize the field modes $A_\kl$ and $E_\kl$ and to
re-express everything in term of the bosonic creation and annihilation operators
$\ad_\kl$ and $\a_\kl$ satisfying the standard equal-time commutation relations
$$
\bigl[\a_\kl,\a_\klp\bigr]\ =\ \bigl[\ad_\kl,\ad_\klp\bigr]\ =0,\qquad
\bigl[\a_\kl,\ad_\klp\bigr]\ =\ (2\pi)^3\delta^{(3)}(\bk-\bk')\,\delta_{\lambda,\lambda'}\,.
\eqn\BCR
$$

\goodbreak
\spoint %d
Perform the quantization and show that
$$
\eqalign{
\hat A_\kl\ =\,\hphantom{\half} {1\over\sqrt{2\omega_\bk C_\kl}}\, &
\left(\a_\kl\ -\ (-1)^\lambda\,\ad_{-\bk,\lambda}\right)\cr
{\rm and}\quad \hat E_\kl\ =\; \coeff{i}{2}\sqrt{2\omega_\bk C_\kl}\,&
\left(\a_\kl\ +\ (-1)^\lambda\,\ad_{-\bk,\lambda}\right)\cr
}\eqn\FieldExp
$$
satisfy the correct commutation relations and that
$$
\hat H\ = \int\!{d^3\bk\over(2\pi)^3} \sum_\lambda\,\omega_\bk\,\ad_\kl\a_\kl\
+\ \rm const.
\eqn\Hamiltonian
$$

\bigskip
\point %2
The second problem is about the spacetime properties of the massive vector field.
\spointbegin
Use the results of problem 1 to show that
$$
\hskip -\itemsize
\hat A^\mu(x)\ = \int\!{d^3\bk\over(2\pi)^3}\,{1\over\sqrt{2\omega_\bk}}
\sum_\lambda\,\left.\left( e^{-ikx}\, f^\mu_\kl\,\a_\kl(t=0)\
    +\ e^{+ikx}\, (f^\mu_\kl)^*\,\ad_\kl(t=0)\right)\right|_{k^0=\omega_\bk}
\eqn\SpaceTimeA
$$
where
$$
f^\mu_{\bk,\pm1}\ =\ \left(0,{\bf e}_{\bk,\pm1}^{\phantom{*}}\right)\quad
{\rm and}\quad f^\mu_{\bk,0}\
=\ \left( {|\bk|\over m},{\omega_\bk\,\bk\over m|\bk|} \right) .
\eqn\FourVectors
$$

\spoint %b
Finally,
calculate the Feynman propagator of the massive vector field and show that
$$
\def\0{0\vphantom{\hat A^\mu}}
G_F^{\mu\nu}(x-y)\
\eqdef\ \bra{\0} T\left(\hat A^\mu(x),\hat A^\nu(y)\right)\ket{\0}\
= \int\!{d^4k\over(2\pi)^4}\,{ie^{-ik(x-y)}\over k^2 -m^2 +i0}\,
\left( -g^{\mu\nu}\, +\,{k^\mu k^\nu\over m^2}\right) .
\eqn\Feynman
$$


\bye
