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\noindent PHY--396~L.\quad
Problem set \#21.\quad
Due March 31, 2005.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
In QED, all physical on-shell amplitudes are gauge-invariant,
but the off-shell amplitudes may depend on the gauge-fixing condition
used to derive the photon's propagator.
Also, the individual loop diagrams are often gauge-dependent, and to
compensate for that, the counterterms are also gauge-dependent.
\pointcon
In class, we have used the Feynman gauge for all calculations.
In this problem, we use more general Lorentz-invariant gauges
$$
\pspicture[shift=-0.42](-2,-0.5)(0,+0.5)
\rput[l](-2.125,0){\psCoil{90}{1530}}
\endpspicture\
=\ {-i\over k^2+i0}\left(g^{\mu\nu}\,
 +\,(\xi-1){k^\mu k^\nu\over k^2+i0}\right)
\eqn\Gauge
$$
and study $\xi$--dependence of various quantities.


\spointbegin
Consider the one-loop vertex correction $ie\Gamma^\mu_{\rm 1\,loop}(p',p)$.
Show that for general $\xi$,
$$
\Gamma^\mu_\xi(p',p)\ =\ \Gamma^\mu_F(p',p)\
+\ e^2(\xi-1)\gamma^\mu\times\int_{\rm reg}\!\!{d^4k\over(2\pi)^4}\,
{-i\over (k^2+i0)^2}
\eqn\Vertex
$$
where $\Gamma^\mu_F(p',p)$ obtains in the Feynman gauge $\xi=1$.
Note that the second term on the right hand side needs both UV and
IR regulators, but you don't need the specifics of such regulators
at this point.
Hint: Use
$$
{1\over\D k+\D p-m}\times\D k\
=\ 1\ -\ {1\over\D k+\D p-m}\times(\D p-m)
\eqno\eq
$$
and ditto for the $\D p'$.

\spoint %b
Use eq.~\Vertex\ to show that the physical form factors $F_1(q^2)$
and $F_2(q^2)$ are gauge-invariant (at least at the one-loop level)
but the $\delta_1$ counterterm is gauge dependent:
At the one-loop level
$$
\delta_1(\xi)\ =\ \delta_1({\rm Feynman})\
-\ e^2(\xi-1)\times\int_{\rm reg}\!\!{d^4k\over(2\pi)^4}\,
{-i\over (k^2+i0)^2}\,.
\eqn\Done
$$

\spoint %c
Now consider the one-loop correction to the electron's propagator
and show that
$$
\Sigma_\xi(\D p)\ =\ \Sigma_F(\D p)\
-\ (\D p-m)\times e^2(\xi-1)\int_{\rm reg}\!\!{d^4k\over(2\pi)^4}\,
 {-i\over (k^2+i0)^2}\
+\ O\bigl((\D p-m)^2\bigr)
\eqn\Prop
$$
where $\Sigma_F(\D p)$ obtains in the Feynman gauge $\xi=1$.

\spoint %d
Finally, use eq.~\Prop\ to show that at the one-loop level
$$
\delta_2(\xi)\ =\ \delta_2({\rm Feynman})\
-\ e^2(\xi-1)\times\int_{\rm reg}\!\!{d^4k\over(2\pi)^4}\,
{-i\over (k^2+i0)^2}\,.
\eqn\Dtwo
$$
and hence $\delta_1=\delta_2$ for any gauge parameter~$\xi$.


\medskip
\point %2
In the minimal Standard Model, the anomalous magnetic moment
$a_\mu=\half(g-2)$ of the muon has been calculated to an extremely
high accuracy of $\Delta_{\rm th}a_\mu\sim 10^{-11}$ and experimentally
measured to an almost as high accuracy of $\Delta_{\rm exp}a_\mu\approx
80\cdot 10^{-11}$.
At present, the theory and the experiment agree with each other,
but future refinement may lead to a discrepancy indicating some new physics.

\spointbegin
Suppose a non-minimal version of the Standard Model contains
a heavy neutral scalar of mass $M_S^{}\simeq 200$~GeV and a Yukawa coupling
to the muon spinor, $g\Phi\overline\Psi\Psi$.
\subpar
Calculate the contribution of this field to the muon's magnetic moment
at the one-loop level of the perturbation theory.
Then use your result to derive an upper limit on the Yukawa coupling~$g$.

\spoint %b
A different non-minimal Standard model contains an {\it axion}, a
pseudoscalar field which couples to
leptons according to
$$
{\partial_\mu\phi \over f^{}_a}\,\overline\Psi\gamma^5\gamma^\mu\Psi\
\approx\ {2im_{\rm lepton}\over f_a^{}}\,\phi\overline\Psi\gamma^5\Psi\
+\ \rm a\ total\ derivative.
\eqno\eq
$$
The axion is a pseudo-Goldstone boson resulting from spontaneous breakdown of
an axial symmetry at a very high energy scale $f_a^{}\gg 100$~GeV; the
symmetry is inexact but very good, and hence the axion is not exactly massless
but very light, $M_A\lsim 1$~MeV.
\subpar
Calculate the axion's contribution to the muon's magnetic moment
at the one-loop level.
Then use your result to derive a lower limit on the axion scale~$f_a^{}$.

\bye
