\input phyzzx

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

\hsize=6.8in	\hoffset=-0.2in	\vsize=8.8in	\voffset=-0.25in
\noindent PHY--396~K.\quad
Problem set \#5.\quad
Due October 16, 2000.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
According to Noeter theorem, a system of several classical fields $\phi_a$
has stress-energy tensor
$$
T^{\mu\nu}_{\rm Noeter}\
= \sum_a \partder{{\cal L}}{(\partial_\mu\phi_a)}\,\partial^\nu\phi^a\
-\ g^{\mu\nu}\,{\cal L} .
\eqn\GeneralT
$$
Actually, to assure the symmetry of the stress-energy tensor,
$T^{\mu\nu}=T^{\nu\mu}$
(which is necessary for the angular momentum conservation), one sometimes
has to add a total divergence,
$$
T^{\mu\nu}\ =\ T^{\mu\nu}_{\rm Noeter}\
+\ \partial_\lambda{\cal K}^{[\lambda\mu]\nu} ,
\eqn\Tcorr
$$
where ${\cal K}^{[\lambda\mu]\nu}$ is some 3--index Lorentz tensor antisymmetric
in its first two indices.

\spointbegin
Show that regardless of the specific form of
${\cal K}^{[\lambda\mu]\nu}(\phi,\partial\phi)$,
$$
\eqalign{
\partial_\mu T^{\mu\nu}\ &
=\ \partial_\mu T^{\mu\nu}_{\rm Noether}\ = ({\rm hopefully})=\ 0\cr
P^\mu_{\rm net}\ &
\equiv\int\!\!d^3\bx\,T^{0\mu}\ =\int\!\!d^3\bx\,T^{0\mu}_{\rm Noether}\,.\cr
}\eqno\eq
$$

\pointcon
For the scalar fields, real or complex, $T^{\mu\nu}_{\rm Noeter}$ is
properly symmetric and one simply has $T^{\mu\nu}=T^{\mu\nu}_{\rm Noeter}$.
Unfortunately, the situation is more complicated for the vector, tensor or spinor
fields.
To illustrate the problem, consider the free electromagnetic fields
described by the Lagrangian
$$
{\cal L}(A_\mu,\partial_\nu A_\mu)\ =\ -\coeff14\,F_{\mu\nu} F^{\mu\nu}
\eqn\LEM
$$
where $A_\mu$ is a real vector field and
$F_{\mu\nu}\equiv\partial_\mu A_\nu-\partial_\nu A_\mu$.

\spoint %b
Write down $T^{\mu\nu}_{\rm Noeter}$ for the free electromagnetic fields and
show that it is neither symmetric nor gauge invariant.

\spoint %c
The properly symmetric --- and also gauge invariant ---
 stress-energy tensor for the free electromagnetism is
$$
T_{\rm EM}^{\mu\nu}\ =\ -F^{\mu\lambda} F^\nu_{\,\,\lambda}\
+\ \coeff14\,g^{\mu\nu}\,F_{\kappa\lambda} F^{\kappa\lambda} .
\eqn\TEM
$$
Show that this expression indeed has form \Tcorr\ for some
${\cal K}^{[\lambda\mu]\nu}$.

\spoint %d
Write down the components of the stress-energy tensor~\TEM\ in
non-relativistic notations and make sure you have the familiar
electromagnetic energy density, momentum density and pressure.

\bigskip
\point %2
Now consider electromagnetic fields coupled to some charged ``matter'' fields
which carry an EM current $J^\mu$.

\spointbegin
Use Maxwell equations to show that
$$
\partial_\mu T_{\rm EM}^{\mu\nu}\ =\ - F^{\nu\lambda} J_\lambda\,.
\eqn\Tdiv
$$

\pointcon
Eq.~\Tdiv\ suggests that any system of charged ``matter'' fields should have its
stress-energy tensor $T_{\rm mat}^{\mu\nu}$ obeying
$$
\partial_\mu T_{\rm mat}^{\mu\nu}\ =\ + F^\nu_{\,\,\lambda} J^\lambda_{\rm EM}\,.
\eqn\Tmatdiv
$$
Consequently, the combined stress-energy tensor
$T_{\rm EM}^{\mu\nu}+T_{\rm mat}^{\mu\nu}$ should be divergence-less and thus
lead to a conserved total energy and momentum.

\pointcon
Generally, testing eq.~\Tmatdiv\ for any particular system
of charged ``matter'' fields
makes use of fields' equations of `motion' and also of the fact that
{\it the covariant derivatives $D_\mu$ do not commute with each other}.
Instead, when acting upon a field $\Phi_q$ of charge $q$, one has
$$
(D_\mu D_\nu-D_\nu D_\mu)\Phi_q\ =\ iq\,F_{\mu\nu}\,\Phi_q
\eqn\NonCommute
$$
(in $c=\hbar=1$ units).

\spoint %b
Verify eq.~\NonCommute.

\pointcon
Now consider a specific example of EM coupled to a charged scalar field,
with a combined Lagrangian density
$$
{\cal L}\ =\ D^\mu\Phi^* D_\mu\Phi\ -\ m^2\Phi^*\Phi\
-\ \coeff14 F^{\mu\nu} F_{\mu\nu}\,.
\eqn\LTotal
$$
\spoint %c
Calculate the Noether stress-energy tensor for this field system and show
that
$$
T^{\mu\nu}_{\rm net}\
\equiv\ T^{\mu\nu}_{\rm Noeter}\ +\ \partial_\lambda{\cal K}^{[\lambda\mu]\nu}\
=\ T^{\mu\nu}_{\rm EM}\ +\ T^{\mu\nu}_{\rm mat}
\eqn\Tnet
$$
where ${\cal K}^{[\lambda\mu]\nu}$ is the same function of EM fields
as in the free EM case~(c), $T^{\mu\nu}_{\rm EM}$ is exactly as in eq.~\TEM\
and
$$
T^{\mu\nu}_{\rm mat}\ =\ D^\mu\Phi^*\,D^\nu\Phi\ +\ D^\nu\Phi^*\,D^\mu\Phi\
-\ g^{\mu\nu}\bigl( D^\lambda\Phi^*\,D^\lambda\Phi\,-\,m^2\Phi^*\Phi \bigr)
\eqn\Tmat
$$
Hint: In the presence of an electric current $J^\mu$, the
$\partial_\lambda{\cal K}^{[\lambda\mu]\nu}$ correction to the electromagnetic
stress-energy tensor contains an extra $J^\mu A^\nu$ term.
This term is important for obtaining a gauge-invariant stress-energy tensor~\Tmat\
for the scalar field.

\spoint %d
Use the scalar field's equations of motion and eq.~\NonCommute\
to verify eq.~\Tmatdiv.


\bigskip
\point %3
Finally, let us quantize the (free) electromagnetic fields.
Unlike the massive vector fields studied in previous homeworks, the Hamiltonian
formalism for the massless EM fields suffers from redundancies associated with
the gauge transformations.
To ameliorate this problem, we would like to separate the gauge-invarian
transverse polarizations of the vector field from the gauge-dependent
longitudinal and scalar polarizations.
The separation is best done in terms of the Fourier-transformed fields
using the helicity basis defined in the previous homework.

\spointbegin
Show that in terms of the $A_\kl(t)$ and $A^0_\bk$ modes of the classical fields,
 the  free EM Lagrangian becomes
$$
\!\!L\ \equiv\int\!\!d^3\bx{\cal L}\ =\ L^\perp\ +\ L^\parallel
\eqn\separated
$$
where
$$
L^\perp\ =\int\!\!{d^3\bk\over (2\pi)^3}\sum_{\lambda=\pm1}
\left( {1\over2}\,{\dot A}_\kl^* {\dot A}_\kl^{}\
    	-\ {\bk^2\over2}\,A_\kl^* A_\kl^{} \right).
\eqn\Ltransverse
$$
and
$$
L^\parallel\ =\int\!\!{d^3\bk\over (2\pi)^3}\,
\half\left| {\dot A}^{}_{\bk,0} + ik A^0_\bk\right|^2 .
\eqn\Llongit
$$

\spoint %b
Show that the transverse ($\lambda=\pm1$) modes $A_\kl$ are gauge invariant
while the longitudinal and scalar modes transform according to
$$
A_{\bk,0}(t)\ \to\ A_{\bk,0}(t)\,+\,ik\Lambda_\bk(t),\qquad
A^0_\bk(t)\ \to\ A^0_\bk(t)\ -\ \dot\Lambda_\bk(t)
\eqn\GT
$$
for arbitrary, independent $\Lambda_\bk(t)$.

\spoint %c
Write down the Hamiltonian $H^\perp$ for the transverse modes and their
canonical conjugates.
Quantize the transverse modes and write down the commutation relations for
the transverse $\hat A_\kl$ and $\hat E_\kl$.

\spoint %d
Construct the creation and annihilation operators for the transverse modes
and show that
$$
\hat H^\perp\
=\int\!\!{d^3\bk\over (2\pi)^3}\sum_{\lambda=\pm1}\,
\omega_k\ad_\kl\a_\kl\ +\ \rm const
\eqn\Photons
$$
with $\omega_k=k\equiv|\bk|$.
The massless particles created by the $\ad_\kl$ operators and annihilated by the
$\a_\kl$ are called {\it photons}.

\pointcon
As to the longitudinal and scalar modes $A_{\bk,0}^{}$ and $A^0_\bk$,
they make for rather redundant dynamical variables in light of gauge
transforms~\GT.
The Hamiltonian formalism abhors such redundancy, so we need to fix a gauge.
%
Let us therefore impose the so-called {\sl transverse} gauge condition
$A_{\bk,0}\equiv0$ --- or equivalently $\nabla\cdot{\bf A}(\bx,t)=0$.

\spoint %e
Show that in the transverse gauge, the scalar modes satisfy time-independent
equations of motion $\bk^2 A^0_\bk=0$.\brk
Also show that for EM fields coupled to an electric current $J^\mu$,
the scalar modes satisfy $\bk^2 A^0_\bk(t)=-J^0_\bk(t)$, or in the field
language, the scalar potential $A^0(\bx,t)$ is the instantaneous Coulomb
potential for the charge density $J^0(\bx,t)$.\brk
For that reason, the transverse gauge is also known as the Coulomb gauge.

\pointcon
Utimately, for the free EM fields, the longitudinal sector contains no dynamics
and thus cab be simply thrown out of the quantum theory.
Consequently, the free quantum EM theory is the theory of photons with
transverse polarizations only, $\lambda=\pm1$.

\spoint %f
Finally, assemble the time-dependend free quantum EM fields $\hat A^\mu(x)$
as
$$
\!\hat A^\mu(x)\ 
=\int\!\!{d^3{\bf k}\over(2\pi)^3}\,{1\over\sqrt{2|\bk|}}
\sum_{\lambda=\pm1}\left( e^{-ikx}f^\mu(\kl)\,\a_\kl(0)\
	+\ e^{+ikx}f^{*\mu}(\kl)\,\ad_\kl(0) \right)_{k^0=+|\bk|}
\eqn\QEMfield
$$
where {\sl in the transverse gauge} $f^0(\kl)\equiv0$ and
${\bf f}(\kl)={\bf e}_\lambda(\bk)$.


\bye
