\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=}
\def\pit{\setbox0=\hbox{$\widetilde{\phantom\pi}$} \dp0=-\ht0
        \vtop{\hbox{$\pi$}\hrule height 0pt\kern 2pt\box0}}

\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 \#4.\qquad
Due September 30, 2003.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
When an {\it exact} symmetry of a quantum field theory is spontaneously
broken down, it gives rise to exactly massless Goldstone bosons.
But when the spontaneously broken symmetry was only approximate to begin with,
the would-be Goldstone bosons are no longer exactly massless but only relatively
light.
The best-known examples of such pseudo-Goldstone bosons are the
pi-mesons $\pi^\pm$ and $\pi^0$, which are indeed much lighter
then other hadrons.
The Quantum ChromoDynamics theory (QCD) of strong interactions has
an approximate chiral isospin symmetry $SU(2)_L\times SU(2)\cong
{\rm Spin}(4)$ which would be exact if the two lightest quark flavors
$u$ and $d$ were exactly massless;
in reality, the current quark masses $m_u$ and $m_d$ do not exactly vanish
but are small enough to be treated as a perturbation.
Exact or approximate, the chiral isospin symmetry
is spontaneously broken down to the ordinary isospin symmetry
$SU(2)\cong {\rm Spin}(3)$, and the 3 generators of the broken
$\rm Spin(4)/Spin(3)$ give rise to 3 (pseudo) Goldstone bosons
$\pi^\pm$ and $\pi^0$.
\pointcon
QCD is a rather complicated theory, so it is often convenient to
describe the physics of the spontaneously broken chiral symmetry
in terms of a simpler effective theory such as the linear sigma model.
This model has 4 real scalar fields; in terms of the unbroken isospin
symmetry, we have an isosinglet $\sigma(x)$ and an isotriplet
$\pit(x)$ comprising $\pi^1(x)$, $\pi^2(x)$ and $\pi^3(x)$
(or equivalently, $\pi^0(x)\equiv\pi^3(x)$ and $\pi^\pm(x)\equiv\bigl(
\pi^1(x)\pm i\pi^2(x)\bigr)/\sqrt{2}$).
The Lagrangian
$$
{\cal L}\ =\ \half(\partial_\mu\sigma)^2\ +\ \half(\partial_\mu\pit)^2\
-\ {\lambda\over8}\left( \sigma^2+\pit^2-f^2\right)^2\ +\ \beta\sigma
\eqn\LSM
$$
is invariant under the $SO(4)$ rotations of the four fields, except
for the last term which we take to be very small.
(In QCD $\beta\sim{m_u+m_d\over 2f}\vev{\overline\Psi\Psi}$ which is indeed
very small because the $u$ and $d$ quarks are very light.)

\pointcon
In class, we discussed this theory for $\beta=0$ and showed that it has
$SO(4)$ spontaneously broken to $SO(3)$ and hence 3 massless Goldstone bosons.
In this exercise, we let $\beta>0$ but $\beta\ll\lambda f^3$ to show
how this leads to massive but light pions.

\spointbegin
Show that the scalar potential of the linear sigma model with $\beta>0$
has a unique minimum at
$$
\vev{\pit}\ =\ 0\quad{\rm and}\quad
\vev{\sigma}\ =\ f\ +\ {\beta\over\lambda f^2}\ +\ O(\beta^2) .
\eqn\LSMVEV
$$
\spoint %b
Expand the fields around this minimum and show that the pions are light
while the $\sigma$ particle is much heavier.
Specifically, $M_\pi^2\approx(\beta/f)$ while $M_\sigma^2\approx\lambda f^2$.

\bigskip
\point %2
Our second exercise is about the Bogolyubov transform.
Let $\a_\bp$ and $\ad_\bp$ be annihilation and creation operators
satisfying the bosonic commutation relations, and let
$$
\b_\bp\ =\ \cosh(t_\bp)\a_\bp\ +\ \sinh(t_\bp)\ad_\bmp\,,\quad
\bd_\bp\ =\ \cosh(t_\bp)\ad_\bp\ +\ \sinh(t_\bp)\a_\bmp
\eqn\Bogolyubov
$$
for some arbitrary real parameters $t_\bp=t_\bmp$.
\spointbegin
Show that the $\b_\bp$ and the $\bd_\bp$ satisfy the same bosonic commutation
relations as the $\a_\bp$ and the $\ad_\bp$.

\spoint %b
Consider a Hamiltonian of the form
$$
\hat H\ =\,\sum_\bp A_\bp\ad_\bp\a_\bp\
+\ \half\sum_\bp B_\bp\left(\a_\bp\a_\bmp\,+\,\ad_\bp\ad_\bmp\right)
\eqn\MixedH
$$
where $A_\bp=A_\bmp$ and $B_\bp=B_\bmp$.
Show that as long as $|B_\bp|<A_\bp\ \forall\bp$, this Hamiltonian
can be ``diagonalized'' by means of a Bogolyubov transform~\Bogolyubov.
That is, for a suitable choice of the $t_\bp$ parameters,
$$
\hat H\ =\,\sum_\bp\omega_\bp\bd_\bp\b_\bp\ +\ {\rm const}\qquad
{\rm where}\ \omega_\bp\ =\ \sqrt{A_\bp^2-B^2_\bp\,}.
\eqn\Hdiag
$$
In particular, for the liquid helium,
$$
A_\bp\ =\ {\bp^2\over 2M}\ +\ n_0\lambda(\bp)\quad {\rm and}\quad
B_\bp\ =\ n_0\lambda(\bp)\quad\Longrightarrow\quad
\omega_\bp\ =\ \sqrt{{\bp^2\over 2M}
	\left({\bp^2\over 2M}+n_0\lambda(\bp)\right)}.
\eqno\eq
$$

\bigskip
\point %3
The rest of this homework is about 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
$$
{\cal L}\ =\ \partial^\mu\Phi^*\,\partial_\mu\Phi\ -\ m^2\,\Phi^*\Phi .
\eqn\Lagrangian
$$
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
The plane-wave modes
$$
\hat\Phi_\bp\ =\int\!\!d^3\bx\,e^{-i\bp\bx}\,\hat\Phi(\bx),\qquad
\hat\Pi_\bp\ =\int\!\!d^3\bx\,e^{-i\bp\bx}\,\hat\Pi(\bx),\qquad
\eqn\Modes
$$
of non-hermitian fields are completely independent
of each other, thus $\hat\Phi_\bp^\dagger\neq\hat\Phi_\bmp$ and
$\hat\Pi_\bp^\dagger\neq\hat\Pi_\bmp$.
Consequently, we have two independent species of creation and annihilation
operators; in the relativistic normalization
$$
\vtop{\normalbaselines\openup 1\jot
\ialign{&\quad\hfil$\displaystyle{#}$\ &
        $\displaystyle{{}\eqdef\ {E_\bp #}\,}$,\quad\hfil\cr
\a_\bp & \hat\Phi_\bp\,+\,i\hat\Pi_\bp &
\ad_\bp & \hat\Phi^\dagger_\bp\,-\,i\hat\Pi^\dagger_\bp\cr
\b_\bp & \hat\Phi^\dagger_\bmp\,+\,i\hat\Pi^\dagger_\bmp &
\bd_\bp & \hat\Phi_\bmp\,-\,i\hat\Pi_\bmp\cr
}}\eqn\Operators
$$
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 $\bd_\bp$.
%Note the relativistic normalisation of the operators~\Operators.

\spoint %c
Show that  the Hamiltonian of the free charged fields is
$$
\hat H\ = \int\!{d^3\bp\over(2\pi)^3\,2E_\bp}\
\left( E^{}_\bp\ad_\bp\a_\bp\ +\ E^{}_\bp\bd_\bp\b_\bp\right)\
+\ \rm const.
\eqn\FreeHam
$$

\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
$$
\hat Q\
= \int\!\!d^3\bx\,\left(
	\coeff{i}{2}\bigl\{\hat\Pi^\dagger,\hat\Phi\bigr\}
        -\coeff{i}{2}\bigl\{\hat\Pi,\hat\Phi^\dagger\bigr\}
	\right)\
= \int\!{d^3\bp\over(2\pi)^3\,2E_\bp}
	\left(\ad_\bp\a_\bp\ -\ \bd_\bp\b_\bp\right).
\eqn\ElCharge
$$
\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
Now 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
$$
{\bf\hat P}_{\rm mech}\
\eqdef\int\!\!d^3\bx\,\hat T^{0,\bf i}\
=\int\!{d^3\bp\over(2\pi)^3\,2E_\bp}\
\bp\left( \ad_\bp\a_\bp\ +\ \bd_\bp\b_\bp\right)
\eqn\Momentum
$$

\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~$\pm 1$.

\bigskip
\point %4
Finally, consider the time-dependence of the free charged fields
$\hat\Phi(\bx,t)$ and $\hat\Phi^\dagger(\bx,t)$.
\spointbegin
Compare the Schr\"odinger and the Heisenberg pictures of the creation and
annihilation operators~\Operators\ and show that for the free
Hamiltonian~\FreeHam,
$$
\hat a_\bp^{H}(t)=e^{-itE_\bp}\hat a_\bp^S\,,\quad
\hat b_\bp^{H}(t)=e^{-itE_\bp}\hat b_\bp^S\,,\quad
\hat a_\bp^{\dagger H}(t)=e^{+itE_\bp}\hat a_\bp^{\dagger S}\,,\quad
\hat b_\bp^{\dagger H}(t)=e^{+itE_\bp}\hat b_\bp^{\dagger S}\,.
\eqn\Pictures
$$

\spoint %b
Now assemble the Heisenberg-picture quantum fields in terms of the
operators~\Pictures\ and show that in relativistic notations
$$
\eqalign{
\hat\Phi^H(x)\ &
=\int\!{d^3\bp\over(2\pi)^3\,2E_\bp}\,\left(
	e^{-ipx}\,\hat a_\bp^S\ +\ e^{+ipx}\,\hat b_\bp^{\dagger S}
	\right)_{p^0=E_\bp}\,,\cr
\hat\Phi^{\dagger H}(x)\ &
=\int\!{d^3\bp\over(2\pi)^3\,2E_\bp}\,\left(
	e^{-ipx}\,\hat b_\bp^S\ +\ e^{+ipx}\,\hat a_\bp^{\dagger S}
	\right)_{p^0=E_\bp}\,,\cr
}\eqn\QFields
$$
where $px\eqdef p^\mu x_\mu=E_\bp t-\bp\bx$.

\pointcon
Eqs.~\QFields\ allow us to derive the commutation relations between
the quantum fields at un-equal times.
\spoint %c
Show that $[\hat\Phi^H(x),\hat\Phi^H(x')]=0$ and
$[\hat\Phi^{\dagger H}(x),\hat\Phi^{\dagger H}(x')]=0$ even
for un-equal times $x_0\neq x'_0$.

\spoint %d
Show that for un-equal times $x_0\neq x'_0$,
$$
\left[\hat\Phi^H(x),\hat\Phi^{\dagger H}(x')\right]\
=\ D(x-x')\ -\ D(x'-x)
\eqno\eq
$$
where
$$
D(x)\ =\int\!{d^3\bp\over(2\pi)^3\,2E_\bp}\,
\left.e^{-ipx}\right|_{p^0=E_\bp}\,.
\eqno\eq
$$

\pointcon
In class, we shall see that $D(x-x')$ is invariant under orthochronous
Lorentz transformations.
Consequently, for space-like $(x-x')$, $D(x-x')=D(x'-x)$ and therefore
$[\hat\Phi^H(x),\hat\Phi^{\dagger H}(x')]=0$.
\pointcon
In general, for any consistent quantum field theory, for any measurable
local operators $\hat O_1(x_1)$ and $\hat O_2(x_2)$ constructed from
the quantum fields and their space or time derivatives
at respective points $x_1$ and $x_2$, the commutator
$[\hat O_1(x_1),\hat O_2(x_2)]$ \undertext{must} vanish for any space-like
interval $(x_1-x_2)^2<0$.



\bye
