\input phyzzx

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

\usechapterlabeltrue
{\unlock \global\cn@@true }
\xdef\chapterlabel{{\noexpand\rm S}}
\def\QED{\hskip 1em $\cal Q.~E.~D.$}

\hsize=6.6in	\hoffset=0in	\vsize=8.5in	\voffset=0in
\noindent PHY--396~K.\quad
Solutions for problems 1 and 2 of set \#5.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\noindent\undertext{Problem {\bf 1}(a)}:\brk
The conjugacy relations $\hat A_\kl^\dagger=-\hat A_{-\kl}$,
$\hat E_\kl^\dagger=-\hat E_{-\kl}$ follow from hermiticity
of the ${\bf\hat A}(\bx)$ and ${\bf\hat E}(\bx)$ quantum fields
and from the third eq.~(6) for the polarization vectors:
$$
\hat A_\kl^\dagger\
=\int\!\!d^3\bx\,e^{+i\bf kx}{\bf e}^{}_\lambda(\bk)
	\cdot {\bf\hat A}^\dagger(\bx)\
=\int\!\!d^3\bx\,e^{-i(-\bk)\bx}(-{\bf e}^*_\lambda(-\bk))
	\cdot {\bf\hat A}\
=\ -\hat A^\pd_{-\kl}\,,
\eqn\Arel
$$
and likewise
$$
\hat E_\kl^\dagger\
=\int\!\!d^3\bx\,e^{+i\bf kx}{\bf e}^{}_\lambda(\bk)
	\cdot {\bf\hat E}^\dagger(\bx)\
=\int\!\!d^3\bx\,e^{-i(-\bk)\bx}(-{\bf e}^*_\lambda(-\bk))
	\cdot {\bf\hat E}\
=\ -\hat E^\pd_{-\kl}\,,
\eqn\Erel
$$
The equal-time commutation relations follow from eqs.~(1):
Obviously,
$$
[\hat A_\kl,\hat A_{\bk',\lambda'}]\ =\ 0,\qquad
[\hat E_\kl,\hat E_{\bk',\lambda'}]\ =\ 0.
\eqn\commAAEE
$$
Less obviously,
$$
\eqalign{
[\hat A^\pd_\kl,\hat E_{\bk',\lambda'}^\dagger]\ &
=\int\!\!d^3\bx\int\!\!d^3\by\,
    e^{-i\bf kx}\bigl({\bf e}^*_\lambda(\bk)\bigr)^i\times
    e^{+i\bf k'y}\bigl({\bf e}^{}_{\lambda'}(\bk')\bigr)^j\times
    [\hat A^i(\bx),\hat E^j(\by)]\cr
&=\int\!\!d^3\bx\int\!\!d^3\by\,
    e^{-i\bf kx}\bigl({\bf e}^*_\lambda(\bk)\bigr)^i\times
    e^{+i\bf k'y}\bigl({\bf e}^{}_{\lambda'}(\bk')\bigr)^j\times
    (-i)\delta^{(3)}(\bx-\by)\delta_{ij}\cr
&=\ -i\!\int\!\!d^3\bx\,e^{-i(\bk-\bk')\bx}\times
    \bigl( {\bf e}^*_\lambda(\bk)\cdot
	{\bf e}^{}_{\lambda'}(\bk')\bigr)\cr
&=\ -i(2\pi)^3\delta^{(3)}(\bk-\bk')\times
    \bigl( {\bf e}^*_\lambda(\bk)\cdot
	{\bf e}^{}_{\lambda'}(\bk)\bigr)\cr
&=\ -i(2\pi)^3\delta^{(3)}(\bk-\bk')\times\delta_{\lambda,\lambda'}\,,\cr
}\eqn\commAEd
$$
or equivalently,
$$
[\hat A_\kl,\hat E_{\bk',\lambda'}]\
=\ +i(2\pi)^3\delta^{(3)}(\bk+\bk')\times\delta_{\lambda,\lambda'}\,.
\eqn\commAE
$$

\medskip\goodbreak
\noindent\undertext{Problem {\bf 1}(b)}:\brk
There are four terms in the Hamiltonian density~(2), so let us consider
them one by one.
Combining Fourier transform with decomposition into polarization modes
it is easy to see that in light of eq.~(4),
$$
\int\!\!d^3\bx\,{\bf\hat E}^2(\bx)\
=\int\!\!{d^3\bk\over(2\pi)^3}\sum_\lambda \hat E_\kl^\dagger\hat E_\kl^\pd
\eqno\eq
$$
and likewise
$$
\int\!\!d^3\bx\,{\bf\hat A}^2(\bx)\
=\int\!\!{d^3\bk\over(2\pi)^3}\sum_\lambda \hat A_\kl^\dagger\hat A_\kl^\pd\,.
\eqno\eq
$$
Furthermore, using eq.~(5) we have
$$
\nabla\times{\bf\hat A}(\bx)\
= \int\!\!{d^3{\bf k}\over(2\pi)^3}\sum_\lambda e^{i\bf kx}
\Bigl(i\bk\times{\bf e}^{}_\lambda(\bk)
	=\lambda|\bk|{\bf e}^{}_\lambda(\bk)\Bigr)\,
\hat A_\kl
\eqn\CurlA
$$
and hence
$$
\int\!\!d^3\bx\,\left(\nabla\times{\bf\hat A}(\bx)\right)^2\
=\int\!\!{d^3\bk\over(2\pi)^3}\sum_\lambda\,
\lambda^2\bk^2\, \hat A_\kl^\dagger\hat A_\kl^\pd\,.
\eqno\eq
$$
Finally, the first eq.~(6) gives us
$$
\nabla\cdot{\bf\hat E}(\bx)\
= \int\!\!{d^3{\bf k}\over(2\pi)^3}\sum_\lambda e^{i\bf kx}
\Bigl(i\bk\cdot{\bf e}^{}_\lambda(\bk)=i|\bk|\delta_{\lambda,0}\Bigr)\,
\hat E_\kl
\eqn\divE
$$
and hence
$$
\int\!\!d^3\bx\,\left(\nabla\cdot{\bf\hat E}(\bx)\right)^2\
=\int\!\!{d^3\bk\over(2\pi)^3}\,
	\bk^2 \hat E^\dagger_{\bk,0}\hat E^\pd_{\bk,0}\,.
\eqno\eq
$$

In light of all these formul\ae, we assemble the Hamiltonian~(2) as
$$
\!\hat H\
=\int\!\!{d^3\bk\over(2\pi)^3}\sum_\lambda\,\left(
    \left({1\over2}+{\bk^2\over 2m^2}\,\delta_{\lambda,0}
		={C_\kl\over2}\right)
	\hat E_\kl^\dagger\hat E_\kl^\pd\,
    +\, \left({m^2+\lambda^2\bk^2\over 2}
		={\omega_{\bk}^2\over 2C_\kl}\right)
	\hat A_\kl^\dagger\hat A_\kl^\pd
	\right).
\eqno(8)
$$

\medskip\goodbreak
\noindent\undertext{Problem {\bf 1}(c)}:\brk
Given eqs.~\commAAEE\ and \commAE, we have
$$
\eqalign{
[\a_\kl,\a_{\bk',\lambda'}]\ &
=\ -i\omega_\bk\sqrt{C_{\bk',\lambda'}\over C_\kl}\,
    \left( [\hat A_\kl,\hat E_{\bk',\lambda'}]=
    	(+i)(2\pi)^3\delta^{(3)}(\bk+\bk')\delta_{\lambda,\lambda'}\right)\cr
&\quad+\ -i\omega_{\bk'}\sqrt{C_\kl\over C_{\bk',\lambda'}}
    \left( [\hat E_\kl,\hat A_{\bk',\lambda'}]=
    	(-i)(2\pi)^3\delta^{(3)}(\bk+\bk')
	\delta_{\lambda,\lambda'}\right)\cr
&=\ \omega_\bk\times(2\pi)^3\delta^{(3)}(\bk+\bk')\delta_{\lambda,\lambda'}\
    -\ \omega_{\bk'}\times(2\pi)^3\delta^{(3)}(\bk+\bk')
	\delta_{\lambda,\lambda'}\
    =\ 0\cr
}\eqno\eq
$$
and likewise $[\ad_\kl,\ad_{\bk',\lambda'}]=0$.
On the other hand,
$$
\eqalign{
[\a_\kl,\ad_{\bk',\lambda'}]\ &
=\ +i\omega_\bk\sqrt{C_{\bk',\lambda'}\over C_\kl}\,
    \left( [\hat A^\pd_\kl,\hat E^\dagger_{\bk',\lambda'}]=
    	(-i)(2\pi)^3\delta^{(3)}(\bk-\bk')
	\delta_{\lambda,\lambda'}\right)\cr
&\quad+\ -i\omega_{\bk'}\sqrt{C_\kl\over C_{\bk',\lambda'}}
    \left( [\hat E^\pd_\kl,\hat A^\dagger_{\bk',\lambda'}]=
    	(+i)(2\pi)^3\delta^{(3)}(\bk-\bk')
	\delta_{\lambda,\lambda'}\right)\cr
&=\ \omega_\bk\times(2\pi)^3\delta^{(3)}(\bk-\bk')\delta_{\lambda,\lambda'}\
    +\ \omega_{bk'}(2\pi)^3\delta^{(3)}(\bk-\bk')\delta_{\lambda,\lambda'}\cr
&=\ 2\omega_\bk\times(2\pi)^3\delta^{(3)}(\bk-\bk')
	\delta_{\lambda,\lambda'}\,.\cr
}\eqno\eq
$$
\QED

\medskip
\noindent\undertext{Problem {\bf 1}(d)}:\brk
Expanding the operators $\a_\kl$ and $\ad_\kl$ according to eqs.~(9), we have
$$
\ad_\kl\a_\kl\
=\ {\omega^2_{\bk}\over C_\kl}\,\hat A^\dagger_\kl\hat A^\pd_\kl\
+\ C_\kl\,\hat E^\dagger_\kl\hat E^\pd_\kl\
+\ i\omega_{\bk}\,\hat E^\dagger_\kl\hat A^\pd_\kl\
-\ i\omega_{\bk}\,\hat A^\dagger_\kl\hat E^\pd_\kl
\eqn\decomp
$$
and therefore
$$
\hat H^{\rm eq.~(8)}\
=\int\!{d^3\bk\over (2\pi)^3\,2\omega_\bk}\sum_\lambda\omega_\bk\ad_\kl\a_\kl\
+\ \Delta\hat H
\eqno\eq
$$
where
$$
\Delta\hat H\ =\int\!{d^3\bk\over(2\pi)^3}\,{i\omega_\bk\over2}
\sum_\lambda\left(\hat A^\dagger_\kl\hat E^\pd_\kl\
	-\ \hat E^\dagger_\kl\hat A^\pd_\kl\right)
\eqn\DH
$$
Thus, to prove eq.~(10) we need to show that $\Delta\hat H$
is a c-number constant.

The trick here is to change the integration variable $\bk\to -\bk$ in the
the first term in the integrand of eq.~\DH\
and then apply eqs.~\Arel\ and~\Erel:
$$
\int\!\!{d^3\bk\over(2\pi)^3}\sum_\lambda\,
    {i\omega_{\bk}\over2}\,\hat A^\dagger_\kl\hat E^\pd_\kl\
= \int\!\!{d^3\bk\over(2\pi)^3}\sum_\lambda\,
    {i\omega_{\bk}\over2}\,\hat A^\dagger_{-\kl}\hat E^\pd_{-\kl}\
= \int\!\!{d^3\bk\over(2\pi)^3}\sum_\lambda\,
    {i\omega_{\bk}\over2}\,\hat A^\pd_{+\kl}\hat E^\dagger_{+\kl}\,.
\eqno\eq
$$
Consequently
$$
\eqalign{
\Delta\hat H\  &
=\int\!\!{d^3\bk\over(2\pi)^3}\sum_\lambda\,{i\omega_{\bk}\over2}
\left(\hat A^\pd_\kl\hat E^\dagger_\kl\,-\, \hat E^\dagger_\kl\hat A^\pd_\kl\
	\equiv\ [\hat A^\pd_\kl,\hat E^\dagger_\kl]\right)\cr
&= \int\!\!{d^3\bk\over(2\pi)^3}\sum_\lambda\,
    {\omega_{\bk}\over2}\,(2\pi)^3\delta^{(3)}(\bk-\bk={\bf 0})\cr
&\equiv\ E_{\rm vacuum}\cr
}\eqn\Evacuum
$$
which is indeed a c-number constant, albeit divergent.
\QED

Physically, the vacuum energy~\Evacuum\ is the net zero-point energy of all
the oscillatory modes of the vector field theory.
This energy is infinite for two reasons, one having do do with the infinite
volume of space and the other with its perfect continuity.
The infinite-volume divergence of $\int\!d^3\bx$ of a constant vacuum
energy {\sl density} manifest itself via the
$(2\pi)^3\delta^{(3)}({\bf 0})$ factor, which is simply the Fourier transform
of $\int\!d^3\bx(1)$.
Indeed, had we quantized the theory in a very large but finite box,
we would have obtained the $L^3$ volume factor in eq.~\Evacuum\
instead of the delta function.
In other words, the vacuum has energy density
$$
{\rm \left.{Energy\over Volume}\right|_{Vacuum}}\
=\int\!\!{d^3\bk\over(2\pi)^3}\,3\times{\omega_{\bk}\over2}\,.
\eqn\Edensity
$$
Alas, this integral diverges at large momenta so the vacuum energy
{\sl density} is also infinite.
This is a generic problem of all Quantum Field Theories in a perfectly
continuous space (and hence unlimitedly high momenta).
Ultimately, this problem should be resolved by the fundamental theory of
physics at ultra-short distances, whatever such theory might be.

Fortunately, for all practical purposes, we may safely disregard any
c-number constant term in the Hamiltonian, even if such term is infinite
--- and that is exactly what we shall do in this course!


\medskip
\noindent\undertext{Problem {\bf 1}(e)}:\brk
Reversing eqs.~(9), we have
$$
\hat A_\kl\ =\ {\sqrt{C_\kl}\over 2\omega_{\bk}}
\left(\a_\kl-\ad_{-\kl}\right)
\eqn\AklReversal
$$
and therefore
$$
\eqalign{
{\bf\hat A}(\bx)\ &
=\int\!\!{d^3{\bf k}\over(2\pi)^3}\,{e^{i\bk\bx}\over 2\omega_\bk}
	\sum_\lambda\sqrt{C_\kl}\,{\bf e}^{}_\lambda(\bk)
	\left(\a_\kl-\ad_{-\kl}\right)\cr
&=\int\!\!{d^3{\bf k}\over(2\pi)^3}\,{e^{+i\bk\bx}\over 2\omega_\bk}
	\sum_\lambda\sqrt{C_\kl}\,{\bf e}^{}_\lambda(\bk)\,\a_\kl\
    -\int\!\!{d^3{\bf k}\over(2\pi)^3}\,{e^{-i\bk\bx}\over 2\omega_\bk}
	\sum_\lambda\sqrt{C_\kl}\,{\bf e}^{}_\lambda(-\bk)\,\a_{+\kl}\cr
&=\int\!\!{d^3{\bf k}\over(2\pi)^3\,2\omega_\bk}\sum_\lambda\sqrt{C_\kl}
    \left( e^{+i\bk\bx}{\bf e}^{}_\lambda({\bf k})\,\a_\kl\
	+\ e^{-i\bk\bx}{\bf e}^*_\lambda({\bf k})\,\ad_\kl \right).\cr
}\eqn\Avec
$$

It remains to work out the time dependence in the Heisenberg picture.
For the free field governed by Hamiltonian~(10),
$\a_\kl(t)=e^{-i\omega t}\a_\kl(0)$
and $\ad_\kl(t)=e^{+i\omega t}\ad_\kl(0)$ where $\omega\equiv\omega_{\bf k}$.
Substituting this time dependence into eq.~\Avec\
and switching to relativistic notations, we immediately arrive at
$$
{\bf\hat A}(x)\
=\int\!\!{d^3{\bf k}\over(2\pi)^3\,2\omega_\bk}
\sum_\lambda\sqrt{C_\kl}\left(
	e^{-ikx}{\bf e}^{}_\lambda({\bf k})\,\a_\kl(0)\
	+\ e^{+ikx}{\bf e}^*_\lambda({\bf k})\,\ad_\kl(0)
	\right)_{k^0=+\omega_{\bf k}}\, .
\eqno(11)
$$
\QED

\medskip
\noindent\undertext{Problem {\bf 1}(f)}:\brk
The 3--scalar field $\hat A^0(x)$ is governed by eqs.~(3) and \divE,
$$
\hat A^0(\bx,t)\
=\int\!\!{d^3{\bf k}\over(2\pi)^3}\, 
	{-i|\bk|\over m^2}\,e^{i\bf kx}\, \hat E_{\bk,0}(t) .
\eqno\eq
$$
Reversing eqs.~(9) for the $\hat E_\kl$ operator, we have
$$
\hat E_\kl\ =\ {i/2\over\sqrt{C_\kl}}\left(\a_\kl+\ad_{-\kl}\right),
\eqn\EklReversal
$$
and in particular
$$
\hat E_{\bk,0}\ =\ {im\over 2\omega_\bk}
\left(\a_{\bk,0}+\ad_{-\bk,0}\right).
$$
Hence,
$$
\eqalign{
\hat A^0(\bx)\ &
=\int\!\!{d^3{\bf k}\over(2\pi)^3\,2\omega_\bk}\,{|\bk|\over m}\,
	e^{+i\bk\bx}\left(\a_{\bk,0}+\ad_{-\bk,0}\right)\cr
&=\int\!\!{d^3{\bf k}\over(2\pi)^3\,2\omega_\bk}\,{|\bk|\over m}\,
	e^{+i\bk\bx}\,\a_{\bk,0}\
    +\int\!\!{d^3{\bf k}\over(2\pi)^3\,2\omega_\bk}\,{|\bk|\over m}\,
	e^{-i\bk\bx}\,\ad_{+\bk,0}\cr
&=\int\!\!{d^3{\bf k}\over(2\pi)^3\,2\omega_\bk}\,
    {|\bk|\over m}\left( e^{+i\bk\bx}\,\a_{\bk,0}\
	+\ e^{-i\bk\bx}\,\ad_{\bk,0} \right).\cr
}\eqn\Ascalar
$$
As to the time dependence, it works exactly as in eq.~(11)
for the vector field, thus
$$
\hat A^0(x)\
=\int\!\!{d^3{\bf k}\over(2\pi)^3\,2\omega_\bk}\,
{|\bk|\over m}\left( e^{-ikx}\,\a_{\bk,0}(0)\ +\ e^{+ikx}\,\ad_{\bk,0}(0)
	\right)_{k^0=+\omega_{\bf k}}\, .
\eqn\AzeroTD
$$
In light of similarity between eqs.~(11) and \AzeroTD,
we may combine them into a single eq.~(12) where
$$
{\bf f}(\bk,\lambda)\ =\ C_\kl\,{\bf e}_\lambda(\bk)\quad
{\rm and}\quad
f^0(\bk,\lambda)\ =\ {|\bk|\over m}\,\delta_{\lambda,0}
\eqno\eq
$$
or equivalently,~(13).

\medskip
\noindent\undertext{Problem {\bf 1}(g)}:\brk
According to eq.~(12),
$$
\eqalignno{
(\partial^2+m^2)\hat A^\mu(x)\
    =\int\!\!{d^3{\bf k}\over(2\pi)^3\,2\omega_\bk}\sum_\lambda\Bigl( &
(-k^2+m^2)e^{-ikx}f^\mu(\kl)\,\a_\kl(0) &
\eq\cr
&+\ (-k^2+m^2)e^{+ikx}f^{*\mu}(\kl)\,\ad_\kl(0)
	\Bigr)_{k^0=+\omega_{\bf k}}\,,\cr
}$$
which vanishes because $(-k^2+m^2)=0$ for $k^0=\omega_{\bk}$.
Likewise,
$$
\eqalignno{
\partial_\mu\hat A^\mu(x)\
    =\int\!\!{d^3{\bf k}\over(2\pi)^3\,2\omega_\bk}\sum_\lambda\Bigl( &
e^{-ikx}\bigl(ik_\mu f^\mu(\kl)\bigr)\a_\kl(0) &
\eq\cr
&+\ e^{+ikx}\bigl(-ik_\mu f{*^\mu}(\kl)\bigr)\ad_\kl(0)
	\Bigr)_{k^0=+\omega_{\bf k}}\,,\cr
}$$
which vanishes because $k_\mu f^\mu(\kl)=0$ for all polarizations~$\lambda$.
\QED

\bigskip
\noindent\undertext{Problem {\bf 2}(a)}:\brk
The simplest way to prove this lemma is by direct inspection, component by
component:
$$
\eqalign{
\sum_\lambda f^i(\kl)f^{*j}(\kl)\ &
=\sum_\lambda e^i_\lambda(\bk)e^{*j}_\lambda(\bk)\
    +\ {\bk^2\over m^2}\,e^i_0(\bk)e^{*j}_0(\bk)\
    =\ \delta^{ij}\ +\ {k^i k^j\over m^2}\,;\cr
\sum_\lambda f^i(\kl)f^{*0}(\kl)\ &
=\ f^i(\bk,0)f^{*0}(\bk,0)\ =\ {k^i\omega_{\bk}\over m^2}\,;\cr
\sum_\lambda f^0(\kl)f^{*0}(\kl)\ &
=\ \left|f^0(\bk,0)\right|^0\ =\ {\bk^2\over m^2}\
    =\ -1\ + {\omega_{\bk}^2\over m^2}\,.\cr
}\eqno\eq
$$
Alternatively, we may use the fact that the three four-vectors
$f^\mu(\kl)$ (fixed $\bk$, $\lambda=-1,0,+1$) are orthogonal to each other
and also to the $k^\mu=(\omega_\bk,\bk)$.
Furthermore, each $\bigl(f(\kl)\bigr)^2=-1$.
Consequently, the symmetric matrix (in Lorentz indices $\mu,\nu$) on the left
hand side of eq.~(14) has to be (minus) the projection matrix onto four-vectors
orthogonal to the $k^\mu$, and that is precisely the matrix appearing
on the right hand side of eq.~(14) (note $k^2=m^2$).

\medskip
\noindent\undertext{Problem {\bf 2}(b)}:\brk
The operator product $\hat A^\mu(x)\hat A^\nu(y)$ comprises
$\a\a$, $\ad\ad$, $\ad\a$ and $\a\ad$ terms.
The first three kinds of terms have zero matrix elements between vacuum states
while $\bra0 \a_\kl \ad_{\bk',\lambda'}\ket0
= 2\omega_\bk(2\pi)^3\delta^{(3)}(\bk-\bk')\delta_{\lambda,\lambda'}$.
Consequently,
$$
\eqalign{
\bra{0}\hat A^\mu(x)\hat A^\nu(y) \ket{0}\ &
=\int\!\!{d^3{\bf k}\over(2\pi)^3}\,{1\over{2\omega_{\bf k}}}\sum_\lambda
    \left[e^{-ik(x-y)}\,f^\mu(\kl)f^{*\nu}(\kl)\right]_{k^0=+\omega_{\bf k}}\cr
&=\int\!\!{d^3{\bf k}\over(2\pi)^3}\,{1\over{2\omega_{\bf k}}}
    \left[\left(-g^{\mu\nu}+{k^\mu k^\nu\over m^2}\right)e^{-ik(x-y)}\right]
	_{k^0=+\omega_{\bf k}}\cr
&=\ \left(-g^{\mu\nu}-{1\over m^2}\partder{}{x^\mu}\partder{}{x^\nu}\right)
    \int\!\!{d^3{\bf k}\over(2\pi)^3}\,{1\over{2\omega_{\bf k}}}
    \left[e^{-ik(x-y)}\right]_{k^0=+\omega_{\bf k}}\cr
&\equiv\ \left(-g^{\mu\nu}-{1\over m^2}\partder{}{x^\mu}\partder{}{x^\nu}\right)
	D(x-y).\cr
}\eqno(15)
$$

\medskip\goodbreak
\noindent\undertext{Problem {\bf 2}(c)}:\brk
Starting with eq.~(15), we immediately see that
for the un-modified time-ordering,
$$\eqalign{
\bra{0}{\bf T}\hat A^\mu(x)\hat A^\nu(y) \ket{0}\ &
=\ \theta(x^0-y^0)\bra{0}\hat A^\mu(x)\hat A^\nu(y) \ket{0}\
    +\ \theta(y^0-x^0)\bra{0}\hat A^\nu(y)\hat A^\mu(x) \ket{0}\crr
&=\ \theta(x^0-y^0) \left(-g^{\mu\nu}\,
	-\,{1\over m^2}\partder{}{x_\mu} \partder{}{x_\nu}\right) D(x-y)\cr
&\qquad+\ \theta(y^0-x^0) \left(-g^{\mu\nu}\,
	-\,{1\over m^2}\partder{}{x_\mu} \partder{}{x_\nu}\right) D(y-x).\cr
}\eqn\Tprop
$$
On the other hand,
$$
G_F(x-y)\ =\ \theta(x^0-y^0)D(x-y)\ +\ \theta(y^0-x^0)D(y-x),
\eqno\eq
$$
and hence
$$
\eqalign{
\left(-g^{\mu\nu}\,-\,{1\over m^2}\partder{}{x_\mu} \partder{}{x_\nu}\right)
	G_F(x-y)\ &
=\ \theta(x^0-y^0) \left(-g^{\mu\nu}\,
	-\,{1\over m^2}\partder{}{x_\mu} \partder{}{x_\nu}\right) D(x-y)\cr
&\qquad+\ \theta(y^0-x^0) \left(-g^{\mu\nu}\,
	-\,{1\over m^2}\partder{}{x_\mu} \partder{}{x_\nu}\right) D(y-x)\cr
&\qquad\qquad-\ i\delta^{\mu0}\delta^{\nu0}\delta^{(4)}(x-y)\cr
}\eqn\derivDF
$$
where the $\delta$-function term arises from taking time derivatives of the
$\theta$-functions.
(\cf\ explanation of $(\partial^2+m^2)G_F(x-y)=-i\delta^{(4)}(x-y)$ in class.)
Comparing eqs.~\Tprop\ and \derivDF, we obtain
$$
\bra{0}{\bf T}\hat A^\mu(x)\hat A^\nu(y) \ket{0}\
=\ \left(-g^{\mu\nu}\,
	-\,{1\over m^2}\partder{}{x_\mu} \partder{}{x_\nu}\right)G_F(x-y)\
-\ i\delta^{\mu0}\delta^{\nu0}\delta^{(4)}(x-y)
\eqno\eq
$$
and hence
$$
\bra{0}{\bf T^*}\hat A^\mu(x)\hat A^\nu(y) \ket{0}\
=\ \left(-g^{\mu\nu}\,
	-\,{1\over m^2}\partder{}{x_\mu} \partder{}{x_\nu}\right)G_F(x-y).
\eqno\eq
$$
This proves the first line of eq.~(16);
the second line follows from the momentum-space form of the scalar propagator
$$
\eqalign{
\left(-g^{\mu\nu}\,-\,{1\over m^2}\partder{}{x_\mu} \partder{}{x_\nu}\right) &
\left[ G_F(x-y)\ = \int\!\!{d^4{\bf k}\over(2\pi)^4}\,
	{ie^{-ik(x-y)}\over k^2-m^2+i0}\right]\cr
&=\int\!\!{d^4{\bf k}\over(2\pi)^4}\,
	\left(-g^{\mu\nu}+{k^\mu k^\nu\over m^2}\right)\,
	{ie^{-ik(x-y)}\over k^2-m^2+i0}\,.\cr
}\eqno\eq
$$
\QED

\bye
