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\begin{document}
\textbf{\LARGE Application of the second quantization method: Phonons}\\

\underline {Key Concepts}:\\

Adiabatic continuity, elementary excitations; Broken symmetry; Goldstone 
nodes; rigidity; wave function. 

He3 is one of the simplest fermi systems that can be studied in the 
laboratory. It has 2 electrons in a closed shell and behaves like a fermion, 
spin $ \raise.5ex\hbox{$ \scriptstyle 1$ }\kern-.1em/ 
\kern-.15em\lower.25ex\hbox{$ \scriptstyle 2$ } $  particle since its nuclei 
contain 2 protons and one neutron coupled to spin 
$ \raise.5ex\hbox{$ \scriptstyle 1$ }\kern-.1em/ 
\kern-.15em\lower.25ex\hbox{$ \scriptstyle 2$ } $ .

The other common isotope of the He4 has 2p and 2s with zero net angular 
momentum, it behaves very differently at low temperature as discussed in a 
later section (see picture).

%\parbox[t]{8cm} {
\begin{figure}[t]
%\centering
\includegraphics[width=2.in,height=2.in]{fig1.eps}
%\epsfig{file=fig1.eps,width=2.in,height=2.in}
\caption{The phase diagram of $ ^{4}$ He. }
\end{figure}
%}

%\hfill
%\parbox[t]{8cm}{
\begin{figure}[t]
%\centering
\includegraphics[width=2.in,height=2.in]{fig2.eps}
\caption{The phase diagram of $ ^{3}$ He. }
\end{figure}
%}

%\textit{(a) The phase diagram of} $ ^{4}$ \textit{He (b) The phase diagram of} $ ^{3}$ \textit{He}

The interaction between two He atoms can be modified by a Lenard Jones 
interaction, describing a hard core repulsive due to the Pauli principle and 
a week Van der Waals interaction at long distances

\begin{equation}
\label{eq1}
V(r) = 4\varepsilon {\left[ {\left( {{\frac{{\sigma}} {{r}}}} \right)^{12} - 
\left( {{\frac{{\sigma}} {{r}}}} \right)^{6}} \right]} \quad ,
\end{equation}

\noindent
where$ \sigma $ is of the order of the atomic radius 2.5 {\AA} and the scale 
of the interaction is of the order 10K. The parameter is such that $ V(r) $ has no 
bound states but has a low-lying response (almost bound state) leading to a 
rapid variation of the phase shifts at low energy. Quantum effects are 
important where the De Brogile wavelength $ \lambda ({\frac{{1}}{{2m\lambda 
^{2}}}} \equiv KT)$  is of the order of $ \lambda = 7$ {\AA} at 10K.

If we ignore the superfluid phase, which occurs at very low temperature, we 
find that at low temperature the system is either at fluid or a solid 
depending on pressure.

The detailed understanding of these two phases is one old and still open 
problem in the theory of strongly correlated systems.

Two scales are important: $ \sigma $ sets the characteristic ``cage size'' in 
which the He3 atom moves, while $ \varepsilon $  sets the scale of the 
binding. $ {\frac{{h^{2}}}{{2m\sigma ^{2}}}}$  is a kinetic energy of the 
particles and the ratio $ {\frac{{h^{2}}}{{2m\sigma ^{2}\varepsilon}} } = 
\Lambda $ determines the importance of quantum effects. 

If $ \Lambda \gg 1$  there is no liquid phase;

If $ \Lambda \ll 1$  there is a solid at zero temperature at zero pressure. 
For $ \Lambda \sim 1$ the system does not solidify at zero temperature. The 
Hamiltonian modeling He3 is easily written down using our second quantized 
formalism:

\begin{equation}
\label{eq2}
H = H_{0} + H_{1} \quad ,
\end{equation}

\begin{equation}
\label{eq3}
H_{0} = {\sum\limits_{p,\sigma}  {}} \varepsilon _{p} a_{p\sigma} ^{\dag}  
a_{p\sigma} ^{} .
\end{equation}

For H4 the spin index is omitted.

\begin{equation}
\label{eq4}
H_{1} = {\frac{{1}}{{2V}}}{\sum\limits_{p_{1} ,p_{2} ,p_{3} ,p_{4}}  {} 
}a_{p_{1} \sigma} ^{\dag}  a_{p_{2} \sigma} ^{\dag}  a_{p_{3} \sigma} ^{} 
a_{p_{4} \sigma} ^{} V(p_{4} - p_{3} )\delta _{p_{1} + p_{2} + p_{3} + p_{4} 
} ,
\end{equation}

\noindent
or in first quantized notation:

\begin{equation}
\label{eq5}
H_{0} = {\sum\limits_{i} {}} {\frac{{p_{i}^{2}}} {{2m}}},
\end{equation}

\begin{equation}
\label{eq6}
H_{1} = {\frac{{1}}{{2}}}{\sum\limits_{i \ne j} {}} V(r_{i} - r_{j} ).
\end{equation}

The existence of two phases at T = 0 can be understood from simple 
dimensional arguments. The kinetic energy term scales at $ {\frac{{p_{F} 
}}{{2m}}} \sim {\frac{{n^{2 / 3}h^{2}}}{{2m}}}$  while the interaction energy 
scales as $ \varepsilon n\sigma ^{3}$ . Therefore at low densities $ H_{0}$ 
is the dominant interaction and the system is a fluid. At high densities 
$ H_{1} $  is dominant. From its first quantized expression, we have to 
minimize the potential energy without worrying too much about $ H_{0} $ , and 
the resulting structure is a FCC crystal.

Both solid and liquid helium three are in a regime where $ H_{0} \sim H_{1} $  
and serves as a beautiful example of a strongly correlated system, i.e. a 
system in neither itinerant nor the localized aspects of the physics need to 
be taken into account. 

There are two aspects to this problem, a) the accurate evaluation of the 
ground state energy per particle of both phases to determine the phase 
diagram; b) the qualitative understanding of the low energy properties (low 
lying excitations and long distance behavior of the correlation function in 
both phases).

(a) is a very hard problem and is to a large degree independent of (b). This 
is because the short-range correlations, which determine almost the full 
total energy, are independent of He3 and He4 at low temperature in both the 
solid and fluid phase have a similar density correlation function at short 
distances. 

The best variational wave function to date has completely wrong distance 
behaviour.

While these are very important facts seem well established in He3 quantum 
magnetism, the energy difference between RVB and Neel type of wave functions 
is also very small. 

To answer the questions rose in (b) the ideas of adiabatic continuity and 
broken symmetries are essential. 

Understand the solid phase, imagine that one starts from very high densities 
and one can neglect $ H_{0} $ compared to $ V$ . 

The ground states $ V$  is $ {\left| {R_{1} ...R_{N}}  \right\rangle} $ where 
$ R_{i} $ minimize the classical energy$ {\sum\limits_{i \ne j} {V(R_{i}}}   - 
R_{j} )$ .

The minimum is a crystal with $ R_{i} \sim \sigma $ . If we turn on $ H_{0} 
 $  the particles will not stay completely localized but if $ H_{0} \ll 
V$ , the coordinates $ r_{i} $  will not depart two much from $ R_{i} $ . In 
these circumstances we will show that we can neglect the antisymmetry 
requirement of the wave function.

We write $ r_{i} = R_{i} + U_{i} $ ,$ U_{i} \ll 1$  and expand to quadratic 
order.

\begin{equation}
\label{eq7}
H_{} \simeq {\sum\limits_{i = 1}^{N} {}} {\frac{{p_{i}^{2}}} {{2m}}} + 
{\frac{{1}}{{2}}}{\sum\limits_{i \ne j} {}} {\sum\limits_{\alpha \beta}  {} 
}{\frac{{1}}{{2}}}{\frac{{\partial ^{2}V}}{{\partial r_{i}^{\alpha}  
\partial r_{j}^{\beta}} } }(U_{i}^{\alpha}  - U_{j}^{\alpha}  )(U_{i}^{\beta 
} - U_{j}^{\beta}  ) + E_{0} + O(U_{3} ),
\end{equation}

\noindent
defining:

\[
C_{(r_{i} - r_{j} )} \,\,_{\alpha \beta}  \, \equiv {\left\{ 
{\begin{array}{l}
 { - {\frac{{\partial ^{2}V}}{{\partial r_{i}^{\alpha}  \partial 
r_{j}^{\beta}} } }(r_{i} - r_{j} ),\,i \ne j} \\ 
 {{\sum\limits_{j} {}} {\frac{{\partial ^{2}V}}{{\partial r_{i}^{\alpha}  
\partial r_{j}^{\beta}} } }(r_{i} - r_{j} ),\,i = j} \\ 
 \end{array}} \right.},
\]

(\ref{eq7}) can be diagalized by a Fourier transformation:

\[
\begin{array}{l}
 U_{i}^{} = {\frac{{1}}{{\sqrt {N}}} }{\sum\limits_{k} {}} e^{ikR_{i}} U_{k} 
\\ 
 p_{i}^{} = {\frac{{1}}{{\sqrt {N}}} }{\sum\limits_{k} {}} e^{ikR_{i}} p_{k} 
\\ 
 \end{array},
\]

\noindent
which reduces the Hamiltonian to

\[
H_{} = {\sum\limits_{i = 1}^{N} {}} {\frac{{p_{i}^{2}}} {{2m}}} + 
{\frac{{1}}{{2}}}{\sum\limits_{k} {}} u_{k}^{\alpha}  c_{\alpha \beta}  u_{ 
- k}^{\beta}  + E_{0} .
\]

Where we defined

\[
C_{\alpha \beta}  (k) = {\sum\limits_{R \ne 0} {}} {\frac{{\partial 
^{2}V}}{{\partial R_{i} \partial R_{j}}} }(1 - e^{ikR}).
\]

 $ C_{\alpha \beta}  (k)$  has several important properties $ C_{\alpha \beta}  
(k)^{\dag}  = C_{\alpha \beta}  ( - k)$ . If the interaction $ \nu 
$  is short ranged $ C_{\alpha \beta}  (k) \sim K^{2} $  as $ K \to 0$ . For 
each $ k$  it as the eigenvalues and eigenvectors;

\[
{\sum\limits_{\beta}  {}} C_{\alpha \beta}  (k)\varepsilon _{s}^{\beta}  (k) 
= m\omega _{s}^{2} (k)\varepsilon _{s}^{\alpha}  (k),\,\,\,\,s = 1,2,3
\]

 $ \omega _{s}^{2} $  are the classical frequencies of oscillation of the crystal. 
We will assume that $ V(r) = V( - r)$ do that $ \omega _{k} = \omega _{ - k} 
$ and $ \varepsilon _{s} (k)^{\dag}  = \varepsilon _{s} ( - k)$ \textbf{. 
}Defining normal coordinates $ U_{s} (k)P_{s} (k)$ by

\[
\begin{array}{l}
 u_{k}^{\alpha}  = {\sum\limits_{s} {}} \varepsilon _{s}^{\alpha}  
(k)u_{s}^{} (k) \\ 
 p_{k}^{\alpha}  = {\sum\limits_{s} {}} \varepsilon _{s}^{\alpha}  
(k)p_{s}^{\alpha}  (k) \\ 
 \end{array},
\]

\noindent
we can rewrite the $ H$  as

\begin{equation}
\label{eq8}
H_{0} = {\sum\limits_{k,s}^{} {}} {\frac{{1}}{{2m}}}p_{ks}^{} p_{ - ks}^{} + 
m\omega _{0}^{2} (k)u_{ks}^{} u_{ - ks}^{} .
\end{equation}

By analogy with the harmonic oscillator (problem l A) where

\[
H_{} = {\frac{{p_{}^{2}}} {{2m}}} + m\omega _{}^{2} x^{2},
\]

\noindent
was diagonalized by defining $ x = \sqrt {{\frac{{1}}{{2m\omega}} }} \left( 
{a + a^{\dag}}  \right);\,\,p = \sqrt {{\frac{{m\omega}} {{2}}}} i\left( {a 
- a^{\dag}}  \right)$ we define

\[
\begin{array}{l}
 U_{s} = \sqrt {{\frac{{1}}{{2m\omega _{sk}}} }} \left( {a_{ks} + a_{ - 
ks}^{\dag}}   \right)\,\, \\ 
 p_{s} = \sqrt {{\frac{{m\omega _{ks}}} {{2}}}} i\left( {a_{ks} - a_{ - 
ks}^{\dag}}   \right) \\ 
 \end{array}.
\]

The operator $ a_{ks}^{\dag}  $  meets an excitation (phonon) with momentum 
$ k$; since$ a_{ - ks}^{} $  also add momentum $ k$  by removing a phonon from the 
ground state $ u_{s}^{\dag}  (k)$  adds momentum $ k$ , substitution in (\ref{eq8}) yields:

\[
H_{0} = {\sum\limits_{k,s}^{} {}} \omega _{ks}^{} \left( {a_{ks}^{\dag}  
a_{ks}^{} + {\frac{{1}}{{2}}}} \right)
\]

\end{document}
