\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\at{\tilde a^{\vphantom\dagger}}
\def\atd{\tilde a^\dagger}
\def\bx{{\bf x}}
\def\bp{{\bf k}}
\def\bmp{{\bf-k}}
\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 28, 2004.
\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.
\par\goodbreak
\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\goodbreak
\point %2
The rest of this homework is about the Bogolyubov transform
and the superfluid helium.
Let us start with some kind of annihilation and creation operators
$\a_\bp$ and $\ad_\bp$ be satisfying the bosonic commutation relations
$$
[\a_\bp,\a_{\bp'}]\ =\ [\ad_\bp,\ad_{\bp'}]\ =\ 0,\qquad
[\a_\bp,\ad_{\bp'}]\ =\ \delta_{\bp,\bp'}\,.
\eqn\Bosonic
$$
Let us define new operators $\b_\bp$ and $\bd_\bp$ according to
$$
\b_\bp\ =\ \cosh(t_\bp)\a_\bp\ +\ \sinh(t_\bp)\ad_\bmp\,,\quad
\bd_\bp\ =\ \cosh(t_\bp)\ad_\bp\ +\ \sinh(t_\bp)\a_\bmp
\eqn\Bogolyubov
$$
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 $\ad_\bp$.
\pointcon
The Bogolyubov transform --- replacing the `original' creation and
annihilation operators $\ad_\bp$ and $\a_\bp$ with the `transformed'
operators $\bd_\bp$ and $\b_\bp$ --- is useful for diagonalizing
quadratic Hamiltonians 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 for all momenta $\bp$, $A_\bp=A_\bmp$, $B_\bp=B_\bmp$,
and $A_\bp>|B_\bp|$.

\spoint %b
Show that for a suitable choice of the $t_\bp$ parameters,
$$
\hat H\ =\,\sum_\bp\omega_\bp\bd_\bp\b_\bp\ +\ {\rm const}\qquad
{\rm where}\ \omega_\bp\ =\ \sqrt{A_\bp^2-B^2_\bp\,}.
\eqn\Hdiag
$$

\goodbreak
\pointcon
Now consider the quantum field theory of superfluid helium.
As discussed in class, we start with a semi-classical ground state
with a symmetry-breaking expectation value $\vev{\hat\Psi(\bx)}\equiv
\sqrt{n}=\sqrt{\mu/\lambda}$ and shift the quantum fields according to
$$
\hat\Psi(\bx)\ =\ \sqrt{n}\,+\,\hat\varphi(\bx),\qquad
\hat\Psi^\dagger(\bx)\ =\ \sqrt{n}\,+\,\hat\varphi^\dagger(\bx).
\eqn\Shifted
$$
In terms of the shifted fields the free energy operator becomes
$$
\hat H-\mu\hat N\ =\ {\rm const}\ +\ \hat H_{\rm free}\ +\ \hat H_{\rm int}
$$
where
$$
\hat H_{\rm free}\ = \int\!\!d^3\bx\left\{
	{1\over 2M}\,\nabla\hat\varphi^\dagger\cdot\nabla\hat\varphi\
	+\ {\lambda n\over2}\left(
		 \hat\varphi^\dagger\hat\varphi^\dagger
		+2\hat\varphi^\dagger\hat\varphi
		+\hat\varphi\hat\varphi
		\right)\right\}
\eqn\QuadraHam
$$
is quadratic with respect to the shifted fields while
$\hat H_{\rm int}$ comprises the cubic and the quartic terms.

\spoint %c
The momentum modes of the shifted fields are shifted creation and annihilation
operators $\atd_\bp=\ad_\bp-\sqrt{N}\delta_{\bp,{\bf0}}$
and $\at_\bp=\a_\bp-\sqrt{N}\delta_{\bp,{\bf0}}$.
Apply Bogolyubov transform to the shifted operators and re-write the
free Hamiltonian~\QuadraHam\ as
$$
\hat H_{\rm free}\ =\,\sum_\bp\omega_\bp\bd_\bp\b_\bp\ +\ {\rm const}
\eqn\BogoHam
$$
where
$$
\omega_\bp\ =\ |\bp|\times\sqrt{{\lambda n\over M}\,+\,{\bp^2\over 4M^2}}.
\eqn\Dispersion
$$

\pointcon
The ground state of the free Hamiltonian~\BogoHam\ is the state
$\ket{\Omega_2}$ annihilated by all the $\b_\bp$ operators.
To construct this state, we start with the semi-classical ground state
--- the coherent state $\ket{\rm coh}$ in which
$\hat\Psi(\bx)\ket{\rm coh}\equiv\sqrt{n}\ket{\rm coh}$
and therefore $\at_\bp\ket{\rm coh}=0$ for all $\bp$ ---
and then act with a unitary operator $e^{\hat F}$ according to
$$
\ket{\Omega_2}\ =\ e^{\hat F}\ket{\rm coh}\quad{\rm where}\quad
\hat F\ =\,\sum_\bp{t_\bp\over2}
\left(\atd_\bp\atd_\bmp\,-\,\at_\bmp\at_\bp\right)\
=\ -\hat F^\dagger
\eqn\OpTrans
$$
\sitem{(*)}
Optional exercise:
Show that $\b_\bp=e^{\hat F}\at_\bp e^{-\hat F}$,
and hence $\b_\bp\ket{\Omega_2}=0$ for all $\bp$ as well as automatic
bosonic commutation relations for the $\b_\bp$ and $\bd_\bp$ operators.

\pointcon
The excited states of the free Hamiltonian~\BogoHam\ can be constructed
by applying the $\bd_\bp$ operators to the ground state $\ket{\Omega_2}$.
Thus, one can say that {\sl the $\bd_\bp$ operators
create {\it quasi-particles} and the the $\b_\bp$ operators annihilate them};
from this point of view, the $\ket{\Omega_2}$
ground state is the {\sl quasi-particle vacuum}.

\spoint %d
Show that the quasi-particle created by the $\bd_\bp$ operator
and annihilated by the $\b_\bp$ operator has a definite mechanical
momentum~$\bp$, and that
$$
\hat{\bf P}_{\rm tot}\ =\,\sum_\bp \bp\,\bd_\bp\b_\bp.
\eqn\Momentum
$$

\pointcon
On the other hand, the quasi-particles do not have well-defined atomic numbers.
This is related to the spontaneous breakdown of the phase symmetry,
which is generated by the atom number operator $\hat N$.
Physically, the quasi-particles interpolate between phonons in the superfluid
(for small $\bp$) and atoms knocked out of the Bose condensate
(for large $\bp$)
--- note the appropriate limits of the dispersion relation~\Dispersion.
\spoint %e
Check that for large momenta $\bd_\bp\approx\ad_\bp$ and therefore
the quasi-particle is approximately an atom, while for small momenta
$\bd_\bp\approx({\rm coeff})\times(\ad_\bp+\a_\bp)$ and therefore
the quasiparticle is approximately an atom.
\pointcon
Actually, in the real helium with a finite-range interatomic potential
$V_2(\bx-{\bf y})$, the dispersion relation is a bit more complicated than
eq.~\Dispersion\ --- \eg, there is a so-called `roton dip' at intermediate
values of the quasiparticle momenta $\bp$ --- but the small--$\bp$ and the
large--$\bp$ limits work exactly as in this exercise.
In particular, there is a positive lower bound on quasi-particle phase
velocities: $\forall\bp,\ \omega_\bp\ge v_c|{\bp|}$.
This fact plays a crucial role in superfluidity.

\spoint %f
Consider the superfluid in a state of uniform motion with velocity
$\bf v$.
Use Galilean invariance to argue that quasi-particles in moving Helium
are governed by the
$$
\hat H'_{\rm free}\ =\ \hat H_{\rm free}\ +\ {\bf v}\cdot{\bf\hat P}\
=\,\sum_\bp\left(\omega_{\bf k}+{\bf vk}\right)\bd_\bp\b_\bp\,.
\eqn\Moving
$$
Therefore, as long as $|{\bf v}|<v_c$, all excitations have positive
energies, hence there is no spontaneous decay of the flowing ``ground''
state and no energy dissipation!
This is the physical origin of superfluidity.
\subpar
On the other hand, when the Helium flows too fast, $|{\bf v}|>v_0$,
some quasiparticle modes acquire negative energies, which leads
to spontaneous quasiparticle production, hence energy dissipation
and loss of superfluidity.
\subpar
The {\sl critical velocity} $v_c$ is governed by the dispersion
relation for the quasi-particles:
$v_c=\min(\omega_\bp/|\bp|)$.
For the superfluid, $v_c>0$.
In comparison, the ideal gas has $\omega_\bp=\bp^2/2m$, thus $v_c=0$
and no superfluidity at any velocity.

\pointcon
Actually, under most experimental conditions, there is an additional
mechanism for losing superfluidity beyond a much smaller critical
velocity than the $v_c$ obtaining from the microscopic theory.
Specifically, turbulence leads to spontaneous generations of
vortex rings, which move much slower than the quasi-particles and hence
quench superfluidity at much slower speeds.
In very thin capillaries however, the vortex rings do not form because
of size limitations and the superfluidity persists until the microscopic
critical velocity $v_c$.

\bye
