\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\eqdef{\buildrel\rm def\over=}

\hsize=7in	\hoffset=-0.35in	\vsize=9in	\voffset=-0.25in
\hsize=6.8in	\hoffset=-0.2in 	\vsize=8.5in	%\voffset=-0.25in
\quad\noindent PHY--396~K.\qquad
Problem set \#3.\qquad
Due October 2, 2000.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
Consider coherent states of a harmonic oscillator.
\spointbegin
Show that for any complex number $\alpha$,
$$
\ket\alpha\ \eqdef\ \exp\bigl(\alpha\ad-\alpha^*\a\bigr)\ket0 \
=\ e^{-|\alpha|^2/2}\,e^{\alpha\ad}\ket0 \quad
{\rm and}\quad \a\ket\alpha\ =\ \alpha\ket\alpha .
\eqn\Coherent
$$

\spoint %b
Calculate the uncertainties $\Delta q$ and $\Delta p$ for a coherent state
$\ket\alpha$ and verify their minimality: $\Delta q\Delta p=\half\hbar$.
Also, verify $\delta n=\sqrt{\bar n}$ where $\bar n\eqdef\vev{\hat n}=|\alpha|^2$.
\subpar
Hint: use $\a\ket\alpha=\alpha\ket\alpha$ and $\bra\alpha\ad=\alpha^*\bra\alpha$.

\spoint %c
Show that for any initial coherent state $\ket{\alpha_0}$,
$$
\ket{\psi(t)}\ \equiv\ e^{-i\omega t/2} \ket{\alpha=\alpha_0 e^{-i\omega t}}
\eqn\Dynamics
$$
satisfies the time-dependent Schr\"odinger equation.

\spoint %d
The coherent states are not quite orthogonal to each other.
Calculate their overlap.

\pointcon
Now consider coherent states of
 multi-oscillator systems and hence quantum fields.
In particular, let us focus on creation and annihilation fields
$\hat\Psi^\dagger(\bx)$ and $\hat\Psi(\bx)$
for non-relativistic spinless bosons.
\spoint %e
Generalize (a) and construct coherent states $\ket\Phi$ which satisfy
$$
\hat\Psi(\bx)\ket\Phi\ =\ \Phi(\bx)\ket\Phi
\eqn\CoherentQFT
$$
for any given classical complex field $\Phi(\bx)$.

\spoint %f
Show that for any such coherent state, $\Delta N=\sqrt{\bar N}$ where
$$
\bar N\ \eqdef\ \bra\Phi\hat N\ket\Phi\ = \int\!\!d\bx\,|\Phi(\bx)|^2.
\eqno\eq
$$

\spoint %g
Let
$$
\hat H\ = \int\!\!d\bx\left(
{\hbar^2\over 2M}\nabla\hat\Psi^\dagger\cdot\nabla\hat\Psi\
+\ v(\bx)\hat\Psi^\dagger\hat\Psi\right) 
$$
and show that for any classical field configuration $\Phi(\bx,t)$
that satisfies the classical field equation
$$
i\hbar\partder{}{t}\Phi(\bx,t)\
=\ \left(-{\hbar^2\over 2M}\nabla^2+V(\bx)\right)\Phi(\bx,t) ,
$$
the time-dependent coherent state $\ket{\Phi}$ satisfies
the true Schr\"odinger equation
$$
i\hbar\partder{}{t}\ket{\Phi}\ =\ \hat H \ket{\Phi} .
\eqno\eq
$$

\spoint %h
Finally, show that the quantum overlap $|\vev{\Phi_1|\Phi_2}|^2$
between two different coherent states is exponentially small
for any {\sl macroscopic} difference $\delta\Phi(\bx)=\Phi_1(\bx)-\Phi_2(\bx)$
between the two field configurations.


\bigskip
\point
Consider a  complex relativistic field $\Phi(x)$ with a Lagrangian density
$$
{\cal L}\ =\ \partial^\mu\Phi^*\,\partial_\mu\Phi\
-\ m^2\,\Phi^*\Phi\ -\ \coeff14\lambda\bigl(\Phi^*\Phi\bigr)^2.
\eqn\Lagrangian
$$
This Lagrangian has a symmetry $\Phi(x)\mapsto e^{i\theta} \Phi(x)$.
According to Noether theorem (which we shall study later in class),
this symmetry gives rise to a conserved current
$$
J^\mu\ =\ i\Phi^*\partial^\mu\Phi\ -\ i(\partial^\mu\Phi^*)\Phi .
\eqn\Current
$$
\spointbegin
Write down classical field equations for $\Phi(x)$ and $\Phi^*(x)$
(treat them as independent fields!)
and verify that indeed $\partial_\mu J^\mu=0$.\brk

\pointcon
Canonical quantization of the complex field yields non-hermitian quantum fields\brk
$\hat\Phi(x)\neq\hat\Phi^\dagger(x)$ and $\hat\Pi(x)\neq\hat\Pi^\dagger(x)$
and the Hamiltonian
$$
\!\hat H\ = \int\!\!d^3\bx\,\left(
\hat\Pi^\dagger\hat\Pi\,
+\, \nabla\hat\Phi^\dagger\cdot\nabla\hat\Phi
+\, m^2\,\hat\Phi^\dagger\hat\Phi\,
+\, \coeff14\lambda\,\hat\Phi^\dagger\hat\Phi^\dagger\hat\Phi\hat\Phi
\right) .
\eqn\Hamiltonian
$$

\spoint %b
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 %c
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 %d
Now, let us turn off the interactions (\ie, set $\lambda=0$).
Show that  the Hamiltonian of free charged fields is
$$
\eqalign{
\hat H_{\rm free}\ &
\eqdef \int\!\!d^3\bx\,\left(\hat\Pi^\dagger\hat\Pi\,
    +\,\nabla\hat\Phi^\dagger\cdot\nabla\hat\Phi\,
    +\,m^2\,\hat\Phi^\dagger\hat\Phi\right)\cr
&= \int\!{d^3\bp\over(2\pi)^3}\;E_\bp\left( \ad_\bp\a_\bp\ +\ \bd_\bp\b_\bp\right)\
    +\ \rm const.\cr
}\eqn\FreeHam
$$

\spoint %e
Next, consider the electric charge operator $\hat Q= \int\!d^3\bx\,\hat J_0(\bx)$.
Show that for the system at hand
$$
\hat Q\
= \int\!\!d^3\bx\,\left(\coeff{i}{2}\bigl\{\hat\Pi^\dagger,\hat\Phi^\dagger\bigr\}
         -\coeff{i}{2}\bigl\{\hat\Pi,\hat\Phi\bigr\}\right)\
= \int\!{d^3\bp\over(2\pi)^3}\left( \ad_\bp\a_\bp\ -\ \bd_\bp\b_\bp\right).
\eqn\ElCharge
$$
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 $\hat\Phi(\bx)$
(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 neutrality
of the vacuum state.

\pointcon
Finally, consider the stress-energy tensor for the complex field $\Phi(x)$.
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 %f
Consider the total mechanical momentum operator of the fields
$\hat P^i_{\rm mech}=\int\!d^3\bx\,\hat T^{0i}(\bx)$ and show that in terms
of creation and annihilation operators
$$
{\bf\hat P}_{\rm mech}\
=\int\!{d^3\bp\over(2\pi)^3}\;\bp\left( \ad_\bp\a_\bp\ +\ \bd_\bp\b_\bp\right)
\eqn\Momentum
$$

\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 $\pm 1$.


\bye
