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\noindent PHY--396~K.\quad
Problem set \#5.\quad
Due March 4, 1999.
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\pointbegin
Once again, consider the scalar QED.
At the one-loop level, the 1PI two-photon amplitude is
$$
\epsfxsize=4in
\epsfbox{h05-sqed.eps}
$$
where the last diagram results from the counterterm $-\coeff14\delta_3F_{\mu\nu}^2$
in the renormalized Lagrangian.

\pointcon
Your task is to evaluate the two one-loop digrams, to determine the counterterms
coefficient $\delta_3$ and to obtain the finite 1PI amplitude
$\Sigma^{1\,\rm loop}_{\rm incl}(k^2)$ --- and hence the effective coupling
$\alpha_{\rm eff}(k^2)$.

\pointcon
Hint: It is convenient to bring the two one-loop diagrams to a common
denominator, \eg
$$
\int\!{d^4p\over(2\pi)^4}\,{\rm\sum numerators\over common\ denominator}
\eqno\eq
$$
{\sl before} introducing the Feynman parameter and shifting the integration
variable $p\mapsto\ell$.
After that, use dimensional regularization.


\medskip
\point
Now consider the renormalized $\lambda\Phi^4$ theory.
%Consider the $\half\delta_Z(\partial_\mu\Phi)^2$ counterterm
%in the $\lambda\Phi^4$ theory.
As discussed in class, the $\half\delta_Z(\partial_\mu\Phi)^2$ counterterm
does not show up at the one-loop level of the renormalized perturbation theory
because the one-loop diagram yields a momentum-independent $\Sigma_{1\,\rm loop}$
amplitude.
At the two-loop level however, there is a momentum-dependent 1PI diagram, namely
$$
\epsfxsize=2in
\epsfbox{h05-2loop.eps}
$$
and hence the second-order $\delta_Z^{(2)}\neq0$.

\pointcon
Your task is to evaluate this two-loop diagram and to calculate the counterterm
$\delta_Z^{(2)}$.

\pointcon
This is a difficult calculation, so proceed with care.
I suggest you first combine the three propagators using Feynman parameters and
shift the two independent loop momenta to rewrite the loop integral as
$$
\int\!\!\!\int\!\!\!\int\!\!dxdydz\,\delta(x+y+z-1)
\int\!\!\!\int\!{d^4\ell_1\over(2\pi)^4}\,{d^4\ell_2\over(2\pi)^4}\,
{1\over\left[\alpha\ell_1^2+\beta\ell_2^2+\gamma p^2-m^2+i0\right]^3}
\eqno\eq
$$
for some $(x,y,z)$ dependent coefficients $\alpha,\beta,\gamma$.
Second, Wick-rotate $\ell_1$ and $\ell_2$ into the Euclidean space,
dimensionally regularize to $d<4$ and evaluate the momentum integrals.
Third, for the purpose of calculating the $\delta_Z$ counterterm,
take the derivative $d\Sigma/dp^2$ {\sl before} integrating over Feynman parameters
or taking the $d\to4$ limit.
Fourth, do take the $d\to4$ limit and write the counterterm as
$$
\delta_Z^{(2)}\ =\int\!\!\!\int\!\!\!\int\!\!dxdydz\,\delta(x+y+z-1)\,F(x,y,z)\times
\left\{{1\over\epsilon}\,+\,{\rm const}\,+\,\log G(x,y,z)\right\}
\eqn\Mess
$$
for some rational functions $F$ and $G$ of the Feynman parameters.
\subpar
Finally, the simplest way of evaluating the integral~\Mess\ over the Feynman
parameters is to use Mathematica.
To help it along, replace the $(x,y,z)$ variables to $(w,\xi)$
according to $x=\xi w$, $y=(1-\xi)w$, $z=1-w$, then integrate
over the $w$ variable first and over the $\xi$ second.
Here is a couple of integrals I did this way you might find useful:
$$
\eqalign{
\int\!\!\!\int\!\!\!\int\!\!dxdydz\,\delta(x+y+z-1)\,
{xyz\over(xy+xz+yz)^3}\ &
=\ {1\over2}\,,\crr
\int\!\!\!\int\!\!\!\int\!\!dxdydz\,\delta(x+y+z-1)\,
{xyz\over(xy+xz+yz)^3}\,\log{(xy+xz+yz)^3\over (xy+xz+yz-xyz)^2}\ &
=\ -{3\over4}\,.\cr
}\eqno\eq
$$



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