\input phyzzx

\def\pd{{\vphantom\dagger}}
\def\a{\hat a^\pd}
\def\ad{\hat a^\dagger}
\def\kl{{{\bf k},\lambda}}
\def\bx{{\bf x}}
\def\by{{\bf y}}
\def\eqdef{\buildrel\rm def\over=}

\hsize=7in	\hoffset=-0.35in 	\vsize=8.5in	%\voffset=-0.25in
\quad\noindent PHY--396~K.\qquad
Problem set \#4.\qquad
Due October 9, 2000.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
In homework\#2 we developed Hamiltonian formalism for a massive
vector field $A^\mu(x)$.
Upon quantization, the 3--vector field ${\bf A}(x)$ and its canonical
conjugate $-{\bf E}(x)$ become quantum fields subject to equal-time
commutation relations
$$
[\hat A^i(\bx),\hat A^j(\by)]\ =\ 0,\quad
[\hat E^i(\bx),\hat E^j(\by)]\ =\ 0,\quad
[\hat A^i(\bx),\hat E^j(\by)]\ =\ -i\delta^{ij}\delta^{(3)}(\bx-\by)
\eqn\CCR
$$
($\hbar=1,c=1$ units)
governed by the free Hamiltonian
$$
\hat H\ = \int\!\!d^3\bx\left(
	\half{\bf\hat E}^2\ +\ {(\nabla\cdot {\bf\hat E})^2\over 2m^2}\
	+\ \half(\nabla\times{\bf\hat A})^2\ +\ \half m^2{\bf\hat A}^2
	\right)
\eqn\VectorHam
$$
(we assume $J^\mu=0$).
For the non-dynamical $A^0$ field,
its time-independent equation of motion becomes an operatorial
identity
$$
\hat A^0(x)\ =\ -{\nabla\cdot{\bf\hat E}(x)\over m^2}\,.
\eqn\Azero
$$

\pointcon
The purpose of the present exercise is to expand fields in terms
of creation and annihilation operators $\ad_\kl$ and $\a_\kl$
where $\lambda$ labels three different polarization states of
a vector particle ($\rm spin=1$).
Generally, bases for polarization states correspond to $\bf k$--dependent
complex bases ${\bf e}_\lambda({\bf k})$ for ordinary 3--vectors,
$$
{\bf e}^{}_\lambda({\bf k}) \cdot {\bf e}^*_{\lambda'}({\bf k})\
=\ \delta_{\lambda,\lambda'}
\eqn\ComplexBasis
$$
Of particular convenience is the helicity basis of eigenvectors
of the vector product $i{\bf k}\times{}$, namely
$$
i{\bf k}\times {\bf e}^{}_\lambda({\bf k})\
=\ \lambda|{\bf k}|{\bf e}^{}_\lambda({\bf k}),\qquad
\lambda=-1,0,+1.
\eqn\Helicity
$$
By convention, the overall phases of the helicity eigenvectors
are chosen such that
$$
{\bf e}^{}_0({\bf k})\ =\ {{\bf k}\over |{\bf k}|}\,\quad
{\bf e}^*_\lambda({\bf k})\ =\ (-1)^\lambda {\bf e}^{}_{-\lambda}({\bf k}),\quad
{\bf e}^{}_\lambda(-{\bf k})\ =\ -{\bf e}^*_\lambda(+{\bf k}).
\eqn\Phases
$$

\pointcon
Combining Fourier transform with a basis decomposition, we have
$$
{\bf\hat A}(\bx)\ = \int\!\!{d^3{\bf k}\over(2\pi)^3}\sum_\lambda
e^{i\bf kx}{\bf e}^{}_\lambda({\bf k})\,\hat A_\kl\,,\qquad
\hat A_\kl\ = \int\!\!d^3\bx\,e^{-i\bf kx}
	 {\bf e}^*_\lambda({\bf k})\cdot{\bf\hat A}(\bx)
\eqn\Fourier
$$
and ditto for the ${\bf\hat E}(\bx)$ fields and its $\hat E_\kl$ modes.

\spointbegin
Show that $\hat A_\kl^\dagger=-\hat A_{-\kl}$,
$\hat E_\kl^\dagger=-\hat E_{-\kl}$, and derive the equal-time
commutation relations for the $\hat A_\kl$ and $\hat E_\kl$ operators.

\spoint %b
Show that
$$
\hat H\ =\int\!\!{d^3{\bf k}\over(2\pi)^3}\sum_\lambda\left(
	{C_\kl\over 2}\,\hat E_\kl^\dagger\hat E^\pd_\kl\
	+\ {\omega_{\bf k}^2\over 2C_\kl}\hat A_\kl^\dagger\hat A^\pd_\kl
	\right)
\eqn\FourierHam
$$
where $\omega_{\bf k}=\sqrt{{\bf k}^2+m^2}$ and 
$C_\kl=1+\delta_{\lambda,0}({\bf k}^2/m^2)$.

\spoint %c
Define creation and annihilation operators according to
$$
\a_\kl\ =\ \sqrt{\omega_{\bf k}\over 2C_\kl}\,\hat A^\pd_\kl\
	-\ i\sqrt{C_\kl\over2\omega_{\bf k}}\,\hat E^\pd_\kl\,,\quad
\ad_\kl\ =\ \sqrt{\omega_{\bf k}\over 2C_\kl}\,\hat A^\dagger_\kl\
	+\ i\sqrt{C_\kl\over2\omega_{\bf k}}\,\hat E^\dagger_\kl
\eqn\BCR
$$
and verify that they satisfy bosonic commutation relations
(at equal times).

\spoint %d
Show that
$$
\hat H\ = \int\!\!{d^3{\bf k}\over(2\pi)^3}\sum_\lambda
	\omega_{\bf k}\,\ad_\kl\a_\kl\
	+\ \rm const.
\eqn\Quanta
$$

\spoint %e
Next, consider the time dependence of the free vector field
and show that
$$
{\bf\hat A}(\bx,t)\
=\int\!\!{d^3{\bf k}\over(2\pi)^3}\,{1\over\sqrt{2\omega_{\bf k}}}
\sum_\lambda\sqrt{C_\kl}\left(
	e^{-ikx}{\bf e}^{}_\lambda({\bf k})\,\a_\kl(0)\
	+\ e^{+ikx}{\bf e}^*_\lambda({\bf k})\,\ad_\kl(0)
	\right)_{k^0=+\omega_{\bf k}}\, .
\eqn\TimeDep
$$

\spoint %f
Write down a similar formula for the $\hat A^0(\bx,t)$
(use eq.~\Azero).
Together with the previous result, you should get
$$
\hat A_\mu(x)\ 
=\int\!\!{d^3{\bf k}\over(2\pi)^3}\,{1\over\sqrt{2\omega_{\bf k}}}
\sum_\lambda\left(
	e^{-ikx}f^{}_\mu(\kl)\,\a_\kl(0)\
	+\ e^{+ikx}f^*_\mu(\kl)\,\ad_\kl(0)
		\right)_{k^0=+\omega_{\bf k}}
\eqn\Relavector
$$
where
$$
f^\mu(\kl)\ =\cases{
\bigl(0,{\bf e}^{}_\lambda({\bf k})\bigr) &
for $\lambda=\pm1$,\cr
\left({|{\bf k}|\over m},{\omega_{\bf k}\over m}\,{{\bf k}\over|{\bf k}|}\right)&
for $\lambda=0$.\cr }
\eqn\Fdef
$$
Please note that the 4--vectors $f^\mu(\kl)$ are nothing but
purely-spatial vectors ${\bf e}_\lambda({\bf k})$ Lorentz-boosted
into the moving particle's frame.
In particular, for all $(\kl)$, $f^\mu f^*_\mu=-1$ and $f^\mu k_\mu=0$.

\spoint %g
Finally, verify that the vector field~\Relavector\ satisfies the free
equations of motion $\partial_\mu \hat A^\mu(x)=0$ and
$(\partial^2+m^2)\hat A^\mu(x)=0$.

\bigskip
\point %2
Now consider the Feynman propagator for the massive vector field.
\spointbegin
First, a lemma:  Show that
$$
\sum_\lambda f^\mu(\kl)f^{\nu *}(\kl)\
=\ -g^{\mu\nu}\ +\ {k^\mu k^\nu\over m^2}\,.
\eqn\Lemma
$$

\spoint %b
Next, show that
$$
\eqalign{
\bra{0}\hat A^\mu(x)\hat A^\nu(y) \ket{0}\ &
=\int\!\!{d^3{\bf k}\over(2\pi)^3}\,{1\over{2\omega_{\bf k}}}
    \left[\left(-g^{\mu\nu}+{k^\mu k^\nu\over m^2}\right)e^{-ik(x-y)}\right]
	_{k^0=+\omega_{\bf k}}\cr
&=\ \left(-g^{\mu\nu}-{\partial^\mu \partial^\nu\over m^2}\right) D(x-y) .\cr
}\eqno\eq
$$

\spoint %c
Finally, the Feynman propagator: Show that
$$
\eqalign{
G_F^{\mu\nu}\ \equiv\ \bra{0}{\bf T}\hat A^\mu(x)\hat A^\nu(y) \ket{0}\ &
= \int\!\!{d^4{\bf k}\over(2\pi)^4}\,
	\left(-g^{\mu\nu}+{k^\mu k^\nu\over m^2}\right)\,
	{ie^{-ik(x-y)}\over k^2-m^2+i0}\cr
&=\ \left(-g^{\mu\nu}-{\partial^\mu \partial^\nu\over m^2}\right) D_F(x-y).\cr
}\eqno\eq
$$


\bigskip\goodbreak
\point %3
The last exercise is about superfluid Helium.
For $\mu=\lambda n>0$, we expect a ground state with a non-zero expectation
values of the creation/annihilations fields,
$\vev{\smash{\hat\Psi}\vphantom{\Psi}}=\sqrt{n}$,
up to an arbitrary phase.
Let us therefore define shifted fields $\widetilde\Psi(x)=\hat\Psi(x)-\sqrt{n}$
and $\widetilde\Psi^\dagger(x)=\hat\Psi^\dagger(x)-\sqrt{n}$ and expand the
Hamiltonian (or rather $\hat H-\mu\hat N$) into powers of shifted fields.

\spointbegin
Show that
$$
\hat H-\mu\hat N\ =\ {\rm const}\ +\ \hat H_2\ +\ \hat H_{\rm int}
$$
where (in $\hbar=1$ units)
$$
\hat H_2\ = \int\!\!d^3\bx\left\{
	{1\over 2M}\,\nabla\widetilde\Psi^\dagger\cdot\nabla\widetilde\Psi\
	+\ {\lambda n\over2}\left(
		 \widetilde\Psi^\dagger\widetilde\Psi^\dagger
		+2\widetilde\Psi^\dagger\widetilde\Psi
		+\widetilde\Psi\widetilde\Psi
		\right)\right\}
\eqn\QuadraHam
$$
while $\hat H_{\rm int}$ comprises cubic and quartic terms with respect
to the shifted fields.

\spoint %b
Fourier-transform the shifted fields into shifted creation/annihilation
operators
$\tilde a^\dagger_{\bf k}=\ad_{\bf k}-(2\pi)^3\sqrt{n}\,\delta^{(3)}({\bf k})$
and $\tilde a^\pd_{\bf k}=\a_{\bf k}-(2\pi)^3\sqrt{n}\,\delta^{(3)}({\bf k})$,
then perform a canonical transform
$$
\hat b^\pd_{\bf k}\ =\ \cosh(t_{\bf k})\,\tilde a^\pd_{\bf k}\
	+\ \sinh(t_{\bf k})\,\tilde a^\dagger_{-\bf k}\,,\quad
\hat b^\dagger_{\bf k}\ =\ \cosh(t_{\bf k})\,\tilde a^\dagger_{\bf k}\
	+\ \sinh(t_{\bf k})\,\tilde a^\pd_{-\bf k}\,.
\eqn\Canon
$$
Show that for any $t_{\bf k}=t_{-\bf k}$, the operators $\hat b^\pd_{\bf k}$
and $\hat b^\dagger_{\bf k}$ satisfy bosonic commutation relations.
(At equal times, of course).

\spoint %c
Show that for a suitable choice of $t_{\bf k}$,
$$
\hat H_2\ =\int\!\!{d^3{\bf k}\over(2\pi)^3}\,
\omega_{\bf k}\hat b^\dagger_{\bf k}\hat b^\pd_{\bf k}\
+\ \rm const
\eqn\SFHam
$$
where
$$
\omega_{\bf k}\ =\ |k|\sqrt{{\lambda n\over M}+{k^2\over 4M^2}\,}.
\eqn\Dispersion
$$

\pointcon
Please note that the ground state of the $\hat H_2$ is the state
$\ket{\Omega_2}$ annihilated
by all the $\hat b^\pd_{\bf k}$ operators.
To construct this ground state, we start with the coherent state $\ket{\rm coh}$
--- which is annihilated by all the shifted $\tilde a^\pd_{\bf k}$ ---
and then modify according to
$$
\ket{\Omega_2}\ =\ e^{\hat F}\ket{\rm coh},\qquad
\hat F\ =\int\!\!{d^3{\bf k}\over(2\pi)^3}\,{t_{\bf k}\over2}
\left(\tilde a^\pd_{\bf k}\tilde a^\pd_{-\bf k}
	-\tilde a^\dagger_{\bf k}\tilde a^\dagger_{-\bf k}\right).
\eqn\OpTrans
$$
\sitem{(*)}
Optional exercise:
Show that $\hat b^\pd_{\bf k}=e^{\hat F}\tilde a^\pd_{\bf k}e^{-\hat F}$,
$\hat b^\dagger_{\bf k}=e^{\hat F}\tilde a^\dagger_{\bf k}e^{-\hat F}$,
and hence $\hat b^\pd_{\bf k}\ket{\Omega_2}=0$ as well as automatic
bosonic commutation relations for the $\hat b^\pd_{\bf k}$
and $\hat b^\dagger_{\bf k}$ operators.

\pointcon
The excited states of the $\hat H_2$ Hamiltonian can be constructed
by applying the $\hat b^\dagger_{\bf k}$ operators to the ground state
$\ket{\Omega_2}$.
Thus, one can say that {\sl the $\hat b^\dagger_{\bf k}$ operators
create {\it quasiparticles} and the the $\hat b^\pd_{\bf k}$ operators
annihilate them}; from this point of view, the $\ket{\Omega_2}$
ground state is the {\sl quasiparticle vacuum}.

\spoint %d
Show that the net mechanical momentum of the superfluid Helium is
$$
{\bf\hat P}\ =\int\!\!{d^3{\bf k}\over(2\pi)^3}\,
{\bf k}\,\hat b^\dagger_{\bf k}\hat b^\pd_{\bf k}\,,
\eqn\Momentum
$$
thus quasiparticles do have well-defined momenta $\bf k$.

\pointcon
On the other hand, the quasiparticles 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 quasiparticles interpolate between phonons in the superfluid
(for small $k$) and atoms knocked out of the Bose condensate (for large $k$)
--- note the appropriate limits of the dispersion relation~\Dispersion.
\subpar
Actually, in the real helium with a finite-range interatomic potential
$V_2(\bx-\by)$, 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 $k$ --- but the small--$k$ and the
large--$k$ limits work exactly as in this exercise.
In particular, there is a positive lower bound on quasi-particle phase
velocities, $\forall{\bf k},\ \omega_{\bf k}\ge v_0|{\bf k|}$.
This fact plays a crucial role in superfluidity.

\spoint %e
Now consider the superfluid in a state of uniform motion with velocity
$\bf v$.
Use Galilean invariance to argue that quasiparticles in moving Helium
are governed by the
$$
\hat H'_2\ =\ \hat H_2\ +\ {\bf v}\cdot{\bf\hat P}\
=\int\!\!{d^3{\bf k}\over(2\pi)^3}\,\left(\omega_{\bf k}+{\bf vk}\right)
\hat b^\dagger_{\bf k}\hat b^\pd_{\bf k}\,.
\eqn\Moving
$$
Therefore, as long as $|{\bf v}|<v_0$, 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} $v_0$ is governed by the dispersion
relation for the quasiparticles:
$v_0=\min(\omega_k/k)$.
For the superfluid, $v_0>0$.
In comparison, the ideal gas has $\omega_k=k^2/2m$, thus $v_0=0$
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 $v_0$ obtaining from the microscopic theory.
Specifically, turbulence leads to spontaneous generations of
vortex rings, which move much slower than the quasiparticles 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 $v_0$.

\bye
