\input phyzzx
\def\a{\hat a^{\vphantom\dagger}}
\def\ad{\hat a^\dagger}
\def\b{\hat b^{\vphantom\dagger}}
\def\bd{\hat b^\dagger}
\def\bx{{\bf x}}
\def\bp{{\bf p}}
\def\bmp{{\bf-p}}
\def\bx{{\bf x}}
\def\by{{\bf y}}
%\def\hilb{{\cal H}^B}
\def\eqdef{\buildrel\rm def\over=}
\hsize=7in \hoffset=-0.35in \vsize=9in \voffset=-0.25in
\quad\noindent PHY--396~K.\qquad
Problem set \#2.\qquad
Due September 14, 2004.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip
\pointbegin
Consider a {\sl massive} relativistic vector field $A^\mu(x)$ with the
Lagrangian density
$$
{\cal L}\ =\ -\coeff14\,F_{\mu\nu}F^{\mu\nu}\
+\ \half m^2\,A_\mu A^\mu\ -\ A^\mu J_\mu
\eqn\Lagran
$$
where $c=\hbar=1$, $F_{\mu\nu}\eqdef \partial_\mu A_\nu-\partial_\nu A_\mu$,
and the current $J^\mu(x)$ is a fixed source for the $A^\mu(x)$ field.
Note that because of the mass term, the Lagrangian~\Lagran\ is {\sl not}
gauge invariant.
\spointbegin
Derive the Euler--Lagrange field equations for the massive vector field
$A^\mu(x)$.
\spoint %b
Show that this field equation {\sl does not require} current conservation;
however, if the current happens to satisfy $\partial_\mu J^\mu=0$, then the
field $A^\mu(x)$ satisfies
$$
\partial_\mu A^\mu\ =\ 0\qquad {\rm and}\qquad
(\partial^2+m^2)A^\mu=J^\mu.
\eqn\TwoEqs
$$
\pointcon
Now, let us derive the Hamiltonian formalism for the massive vector field.
As a first step, we need to identify the canonically
conjugate ``momentum'' fields.
\spoint %c
Show that $\partial{\cal L}/\partial{\bf\dot A}=-{\bf E}$
but $\partial{\cal L}/\partial\dot A_0\equiv0$.
\pointcon
Thus, the canonically conjugate field to ${\bf A}(\bx)$
is $-{\bf E}(\bx)$ but the $A_0(\bx)$ does not have a canonical
conjugate!
Consequently,
$$
H\ =\ -\int\!\!d^3\bx\,{\bf\dot A}(\bx)\cdot{\bf E}(\bx)\ -\ L.
\eqn\HamDef
$$
\spoint %d
Show that in terms of the $\bf A$, $\bf E$ and $A_0$ fields and their
{\sl space} derivatives,
$$
H\ =\int\!\!d^3\bx\left\{ \half{\bf E}^2\,
+\,A_0\left(J_0-\nabla\cdot{\bf E}\right)\,-\,\half m^2 A_0^2\
+\ \half\left(\nabla\times{\bf A}\right)^2\,
+\,\half m^2{\bf A}^2\,-\,{\bf J}\cdot{\bf A} \right\} .
\eqn\HamiltAAE
$$
\pointcon
Because the $A_0$ field does not have a canonical conjugate, the Hamiltonian
formalism does not produce an equation for the time-dependence of this field.
Instead, it gives us a time-independent equation relating the $A_0(\bx,t)$
to the values of other fields {\sl at the same time} $t$.
Specifically, we have
$$
{\delta H\over\delta A_0(\bx)}\ \equiv\
\left.\partder{{\cal H}}{A_0}\right|_\bx\
-\ \nabla\cdot\left.\partder{{\cal H}}{\nabla A_0}\right|_\bx\
=\ 0.\eqn\Constraint
$$
At the same time, the vector fields $\bf A$ and $\bf E$ satisfy the Hamiltonian
equations of motion,
$$
\partder{}{t}{\bf A}(\bx,t)\
=\ -\left.{\delta H\over \delta {\bf E}(\bx)}\right|_t\,,\qquad
\partder{}{t}{\bf E}(\bx,t)\
=\ +\left.{\delta H\over \delta {\bf A}(\bx)}\right|_t\,.
\eqn\HamiltEqs
$$
\spoint %e
Write down the explicit form of all these equations.
\spoint %f
Finally, verify that the equations you have just written down are equivalent
to the Euler--Lagrange equations you derived in question~(a).
\bigskip
\point
Next, consider the quantum electromagnetic fields.
Canonical quantization of the massless vector field $A_\mu(x)$
is rather difficult because of the redundancy associated with
the gauge symmetry, so let me simply state without proof a few
key properties of the quantum tension fields
$\hat{\bf E}(\bx,t)$ and $\hat{\bf B}(\bx,t)$.
In the absence of electric charges and currents, these fields
satisfy time-independent operatorial identities
$$
\nabla\cdot{\bf\hat E}(\bx,t)\ =\nabla\cdot{\bf\hat B}(\bx,t)\ =\ 0
\eqn\MaxwellConstraints
$$
(we assume free EM fields, \ie\ no electric charges or currents),
and have equal-time commutation relations
$$
\eqalign{
\bigl[\hat E_i(\bx,t),\hat E_j(\bx',t'=t)\bigr]\ & =\ 0,\cr
\bigl[\hat B_i(\bx,t),\hat B_j(\bx',t'=t)\bigr]\ & =\ 0,\cr
\bigl[\hat E_i(\bx,t),\hat B_j(\bx',t'=t)\bigr]\ &
=\ -i\hbar c\epsilon_{ijk}\partder{}{x_k}\delta^{(3)}(\bx-\bx').\cr
}\eqn\ETCR
$$
\spointbegin
Verify that the commutation relations \ETCR\ are consistent with
the time-independent Maxwell equations~\MaxwellConstraints.
\pointcon
In the Heisenberg picture, the quantum EM fields also obey the
time-dependent Maxwell equations
$$
\eqalign{
\partder{\bf\hat B}{t}\ &= -\nabla\times{\bf\hat E}\,,\cr
\partder{\bf\hat E}{t}\ &= +\nabla\times{\bf\hat B}\,.\cr
}\eqn\MaxwellDynamics
$$
\spoint %b
Derive eqs.~\MaxwellDynamics\ from the free electromagnetic Hamiltonian
$$
\hat H_{\it EM}\ = \int\!\!d^3\bx\,\left(
\half{\bf\hat E}^2 + \half{\bf\hat B}^2 \right)
\eqn\FreeEMHam
$$
{\sl and the equal-time commutation relations}~\ETCR.
\bigskip
\point %3
Finally, let us quantize a charged relativistic scalar field~$\Phi(x)$.
A conserved charge implies a complex field with a $U(1)$ symmetry
$\Phi(x)\mapsto e^{i\theta} \Phi(x)$ which gives rise to a conserved
Noether current
$$
J^\mu\ =\ i\Phi^*\partial^\mu\Phi\ -\ i(\partial^\mu\Phi^*)\Phi .
\eqn\Current
$$
For simplicity, let the $\Phi$ field be free, thus classically
$$
{\cal L}\ =\ \partial^\mu\Phi^*\,\partial_\mu\Phi\ -\ m^2\,\Phi^*\Phi .
\eqn\Lagrangian
$$
In the Hamiltonian formalism, we trade the time derivatives $\partial_0\Phi(x)$
and $\partial_0\Phi^*(x)$ for the canonically conjugate fields $\Pi^*(x)$ and
$\Pi(x)$.
(Note that for complex fields $\Pi(\bx)$ is canonically conjugate
to the $\Phi^*(\bx)$
while $\Pi^*(\bx)$ is canonically conjugate to the $\Phi(\bx)$.)
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 operator
$$
\!\hat H\ = \int\!\!d^3\bx\,\left(
\hat\Pi^\dagger\hat\Pi\,
+\, \nabla\hat\Phi^\dagger\cdot\nabla\hat\Phi\,
+\, m^2\,\hat\Phi^\dagger\hat\Phi
\right) .
\eqn\Hamiltonian
$$
\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
Next, let us expand the quantum fields into plane-wave modes:
$$
\hat\Phi(\bx)\ =\,\sum_\bp L^{-3/2}e^{i\bx\bp}\hat\Phi_bp,\qquad
\hat\Phi_\bp\ =\int\!\!d^3\bx\,L^{-3/2}e^{-i\bp\bx}\,\hat\Phi(\bx),
\eqn\Modes
$$
and ditto for the $\hat\Phi^\dagger(\bx)$, $\hat\Pi(\bx)$,
and $\hat\Pi^\dagger(\bx)$ fields.
Note that for the {\sl non--hermitian} fields $\hat\Phi_\bp^\dagger\neq\hat\Phi_\bmp$ and
$\hat\Pi_\bp^\dagger\neq\hat\Pi_\bmp$;
instead, all the mode operators $\hat\Phi_\bp^{}$, $\hat\Phi_\bp^\dagger$,
$\hat\Pi_\bp^{}$, and $\hat\Pi_\bp^\dagger$ are completely independent
of each other.
Consequently, we have two independent species of creation and annihilation
operators, \ie\ for each mode $\bp$ we have independent operators
$$
\vcenter{\normalbaselines\openup 1\jot\ialign{
&\quad\hfil$\displaystyle{#}$\ &
$\displaystyle{{}\eqdef\ {E^{}_\bp #\over\sqrt{2E_\bp}}\,}$,\quad\hfil\cr
\a_\bp & \hat\Phi^{}_\bp\,+\,i\hat\Pi^{}_\bp &
\ad_\bp & \hat\Phi^\dagger_\bp\,-\,i\hat\Pi^\dagger_\bp\cr
\omit and\cr
\b_\bp & \hat\Phi^\dagger_\bmp\,+\,i\hat\Pi^\dagger_\bmp &
\bd_\bp & \hat\Phi^{}_\bmp\,-\,i\hat\Pi^{}_\bmp\cr
}}\eqn\Operators
$$
where $E_\bp=\sqrt{\bp^2+m^2\,}$.
\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 the Hamiltonian of the free charged fields is
$$
\hat H\
=\int\!\!d^3\bx\left(
\Pi^\dagger\,\Pi\,
+\,\nabla\Phi^\dagger\cdot\nabla\Phi\,
+\,m^2\Phi^\dagger\,\Phi
\right)\
= \,\sum_\bp\left(
E^{}_\bp\ad_\bp\a_\bp\,
+\, E^{}_\bp\bd_\bp\b_\bp
\right)\
+\ \rm const.
\eqn\FreeHam
$$
\pointcon
Next, consider the charge operator $\hat Q= \int\!d^3\bx\,\hat J_0(\bx)$.
\spoint %d
Show that for the system at hand
$$
\hat Q\
= \int\!\!d^3\bx\,\left(
\coeff{i}{2}\bigl\{\hat\Pi^\dagger,\hat\Phi\bigr\}
-\coeff{i}{2}\bigl\{\hat\Pi,\hat\Phi^\dagger\bigr\}
\right)\
=\,\sum_\bp \left(\ad_\bp\a_\bp\ -\ \bd_\bp\b_\bp\right).
\eqn\ElCharge
$$
\pointcon
Actually, the classical formula~\Current\
for the current $J_\mu(x)$ determines eq.~\ElCharge\
only up to ordering of the non-commuting operators
$\hat\Pi(\bx)$ and $\hat\Phi^\dagger(\bx)$
(and likewise of the $\hat\Pi^\dagger(\bx)$ and $\hat\Phi(\bx)$).
The anti-commutators in eq.~\ElCharge\ provide a solution to this ordering
ambiguity, but any other ordering would be just as legitimate.
%\pointcon
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 neutrality
of the vacuum state.
\pointcon
Finally, consider the stress-energy tensor of the charged field.
Classically, Noether theorem gives
$$
T^{\mu\nu}\
=\ \partial^\mu\Phi^*\,\partial^\nu\Phi\
+\ \partial^\mu\Phi\,\partial^\nu\Phi^*\
-\ g^{\mu\nu}{\cal L} .
\eqn\ClassicT
$$
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
Show that the total mechanical momentum operator of the fields is
$$
{\bf\hat P}_{\rm mech}\
\eqdef\int\!\!d^3\bx\,\hat T^{0,\bf i}\
=\,\sum_\bp\left( \bp\,\ad_\bp\a_\bp\ +\ \bp\,\bd_\bp\b_\bp\right)
\eqn\Momentum
$$
\pointcon
Physically, eqs.~\Momentum, \FreeHam\ and \ElCharge\ show that a complex field
$\Phi(x)$ describes a relativistic particle together with its antiparticle;
they have exactly the same rest mass $m$ but exactly opposite charges~$\pm 1$.
\bye