\input phyzzx

\def\bx{{\bf x}}
\def\eqdef{\buildrel\rm def\over=}
\def\a{\hat a^{\vphantom\dagger}}
\def\ad{\hat a^\dagger}
\def\b{\hat b^{\vphantom\dagger}}
\def\bd{\hat b^\dagger}
\def\bp{{\bf p}}
%\def\bpp{{\bf p'}}
\def\bmp{{\bf-p}}


\hsize=6.8in	\hoffset=-0.2in	%\vsize=8.5in	\overfullrule=0pt
\noindent PHY--396~K.\quad
Problem set \#3.\quad
Due September 24, 1998.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
Consider a classical Lagrangian
$$
{\cal L}\ =\ D^\mu\Phi^*\,D_\mu\Phi\
-\ m^2\,\Phi^*\Phi\ -\ \coeff14\lambda\bigl(\Phi^*\Phi\bigr)^2
\eqn\Lagrangian
$$
for a charged relativistic complex field $\Phi(x)$.
Canonical quantization of this system yields non-hermitian quantum fields
$\hat\Phi(x)\neq\hat\Phi^\dagger(x)$ and $\hat\Pi(x)\neq\hat\Pi^\dagger(x)$
and the Hamiltonian
$$
\!\hat H\ = \int\!\!d^3\bx\,\left(
\hat\Pi^\dagger\hat\Pi\,
+\, \coeff{i}{2}q A_0\left(\bigl\{\hat\Pi^\dagger,\hat\Phi^\dagger\bigr\}
	 -\bigl\{\hat\Pi,\hat\Phi\bigr\}\right)\,
+\, {\bf D}\hat\Phi^\dagger\cdot{\bf D}\hat\Phi
+\, m^2\,\hat\Phi^\dagger\hat\Phi\,
+\, \coeff14\lambda\,\hat\Phi^\dagger\hat\Phi^\dagger\hat\Phi\hat\Phi
\right) .
\eqn\Hamiltonian
$$

\spointbegin
Derive the Hamiltonian \Hamiltonian\ and write down the equal-time
commutation relations between the quantum fields
$\hat\Phi(\bx)$, $\hat\Phi^\dagger(\bx)$, $\hat\Pi(\bx)$ and $\hat\Pi^\dagger(\bx)$.

\pointcon
Because of the non-hermiticity of the quantum fields
$\hat\Phi(x)\neq\hat\Phi^\dagger(x)$ and $\hat\Pi(x)\neq\hat\Pi^\dagger(x)$,
their respective plane-wave modes $\hat\Phi_\bp$, $\hat\Phi^\dagger_\bp$, 
$\hat\Pi_\bp$ and $\hat\Pi^\dagger_\bp$ are completely independent
of each other \ie, $\hat\Phi_\bp^\dagger\neq\hat\Phi_\bmp$
and $\hat\Pi_\bp^\dagger\neq\hat\Pi_\bmp$.
Let us therefore define:
$$
\vtop{\normalbaselines\openup 1\jot
\ialign{&\quad\hfil$\displaystyle{#}$\ &
	$\displaystyle{{}\eqdef\ {E_\bp #\over\sqrt{2E_\bp}}\,}$,\quad\hfil\cr
\a_\bp & \hat\Phi_\bp\,+\,i\hat\Pi^\dagger_\bmp &
\ad_\bp & \hat\Phi^\dagger_\bp\,-\,i\hat\Pi_\bmp\cr
\b_\bp & \hat\Phi^\dagger_\bmp\,+\,i\hat\Pi_\bp &
\bd_\bp & \hat\Phi_\bmp\,-\,i\hat\Pi^\dagger_\bp\cr
}}\eqn\Operators
$$
where
$$
E_\bp\ \eqdef\ \sqrt{\bp^2+m^2\,}.
\eqn\Energy
$$

\spoint %b
Verify the bosonic commutation relations (at equal times)
between the annihilation operators $\a_\bp$ and $\b_\bp$ and the
corresponding creation operators $\ad_\bp$ and $\bd_\bp$.

\spoint %c
Show that in the absence of an electromagnetic background or self interactions
(\ie, for $\lambda=0$ and $A_\mu(x)\equiv0$), the Hamiltonian of free charged
fields becomes
$$
\eqalign{
\hat H_{\rm free}\ &
\eqdef \int\!\!d^3\bx\,\left(\hat\Pi^\dagger\hat\Pi\,
    +\,\nabla\hat\Phi^\dagger\cdot\nabla\hat\Phi\,
    +\,m^2\,\hat\Phi^\dagger\hat\Phi\right)\cr
&= \int\!{d^3\bp\over(2\pi)^3}\;E_\bp\left( \ad_\bp\a_\bp\ +\ \bd_\bp\b_\bp\right)\
    +\ \rm const.\cr
}\eqn\FreeHam
$$

\spoint %d
Next, consider the electric charge operator $\hat Q= \int\!d^3\bx\,\hat J_0(\bx)$.
Show that for the system at hand
$$
\hat Q\
=\ {iq\over2} \int\!\!d^3\bx\,\left(\bigl\{\hat\Pi^\dagger,\hat\Phi^\dagger\bigr\}
	 -\bigl\{\hat\Pi,\hat\Phi\bigr\}\right)\
= \int\!{d^3\bp\over(2\pi)^3}\;q\left( \ad_\bp\a_\bp\ -\ \bd_\bp\b_\bp\right).
\eqn\ElCharge
$$
Actually, the classical formula for the current $J_\mu(x)$ determines eq~\ElCharge\
only up to ordering of the non-commuting operators $\hat\Pi(\bx)$ and $\hat\Phi(\bx)$
(and likewise of the $\hat\Pi^\dagger(\bx)$ and $\hat\Phi^\dagger(\bx)$).
The anti-commutators in eq.~\ElCharge\ provide a solution to this ordering
ambiguity, but any other ordering would be just as legitimate.
\subpar
The net effect of changing operator ordering in $\hat J_0$ amounts to changing
the total charge $\hat Q$ by an infinite constant (prove this!).
The specific ordering in eq.~\ElCharge\ provides for the electric neutrality
of the vacuum state.

\pointcon
Finally, consider the stress-energy tensor for the complex field $\Phi(x)$.
Classically,
$$
T^{\mu\nu}\ =\ D^\mu\Phi^*\,D^\nu\Phi\ +\ D^\mu\Phi\,D^\nu\Phi^*\
-\ g^{\mu\nu}{\cal L} .
\eqn\ClassicT
$$
Quantization of this formula is straightforward (modulo ordering ambiguity);
for example, $\hat{\cal H}\equiv\hat T^{00}$ is precisely the integrand
on the right hand side of eq.~\Hamiltonian.

\spoint %e
Consider $\hat T^{0i}$ and show that in the absence of the electromagnetic
background, the total mechanical momentum operator
$\hat P^i_{\rm mech}=\int\!d^3\bx\,\hat T^{0i}(\bx)$ becomes
$$
{\bf\hat P}_{\rm mech}\
=\int\!{d^3\bp\over(2\pi)^3}\;\bp\left( \ad_\bp\a_\bp\ +\ \bd_\bp\b_\bp\right)
\eqn\Momentum
$$

\pointcon
Physically, eqs.~\Momentum, \FreeHam\ and \ElCharge\ show that a complex field
$\Phi(x)$ describes both a particle and its antiparticle; they have exactly
the same rest mass $m$ but exactly opposite charges $\pm q$.


\bigskip
\point %2
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.

\pointcon
For the scalar fields, real or complex, $T^{\mu\nu}_{\rm Noeter}$ is
properly symmetric and one simply has $T^{\mu\nu}=T^{\mu\nu}_{\rm Noeter}$
(\cf\ eq.~\ClassicT).
Unfortunately, the situation is more complicated in the electromagnetic case.

\pointcon
Free electromagnetic fields are 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
$F_{\mu\nu}\equiv\partial_\mu A_\nu-\partial_\nu A_\mu$.

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

\spoint %b
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 %c
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.

\spoint %d
Next, consider the electromagnetic field coupled to a current $J^\mu(x)$
and show that for such a field
$$
\partial_\mu T_{\rm EM}^{\mu\nu}\ =\ - F^{\nu\lambda} J_\lambda\,.
\eqn\Tdiv
$$

\pointcon
Eq.~\Tdiv\ suggests that for any system of ``matter'' fields that couple
to electromagnetism and therefore carry EM current $J^\mu$, the
stress-energy tensor $T_{\rm mat}^{\mu\nu}$ for such system should satisfy
$$
\partial_\mu T_{\rm mat}^{\mu\nu}\ =\ + F^\nu_{\,\,\lambda} J^\lambda_{\rm mat}\,.
\eqn\Tmatdiv
$$
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 ``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 %e
Verify eq.~\NonCommute.

\spoint %f
Finally,
verify eq.~\Tmatdiv\ for the complex scalar field with the Lagrangian~\Lagrangian\
and the stress-energy tensor~\ClassicT.


\bye
