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 	Copyright, 1995-2017, all rights reserved, Mark Jarrell (Dept.of 
 Physics and Atronomy, Louisiana State University, LA 70803).  This 
 material may not be reproduced for profit, modified or published in any 
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\title{Chapter 10: Superconductivity}
\author{Bardeen, Cooper, \& Schrieffer}
\newfont{\Sc}{eusm10}
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\tableofcontents

\pagebreak
\Large

\section{Introduction}

     From what we have learned about transport, we know that there is no such 
thing as an ideal ($\rho \ = 0$) conventional conductor.  All materials have 
defects and phonons (and to a lessor degree of importance, electron-electron 
interactions).  As a result, from our basic understanding of metallic 
conduction $\rho$ must be finite, even at $T = 0$.  Nevertheless many 
superconductors, for which $\rho \ = 0$, exist.  The 
first one Hg was discovered by Onnes in 1911. \ It becomes superconducting 
for $T < 4.2^{\circ }K$.  Clearly this superconducting state must be 
fundamentally different than the "normal" metallic state.  I.e., the 
superconducting state must be a different phase, separated by a phase 
transition, from the normal state. 

\subsection{Evidence of a Phase Transition}
 
     Evidence of the phase transition can be seen in the specific heat (See Fig.~\ref{fig:specific}). 
\begin{figure}[htp]
\centerline{\includegraphics[height=2.75in,keepaspectratio,clip=true]{specific.pdf}}
\caption[]{\em{The specific heat of a superconductor $C_S$ and
and normal metal $C_n$.  Below the transition, the superconductor
specific heat shows activated behavior, as if there is a minimum
energy for thermal excitations.}}
\label{fig:specific}
\end{figure}
The jump in the superconducting specific heat $C_s$ indicates that there is a 
phase transition without a latent heat (i.e. the transition is continuous or 
second order).  Furthermore, the activated nature of C for $T < T_c$ 
\beq
	C_s \sim e^{-\beta \Delta}
\eeq
gives us a clue to the nature of the superconducting state.  It is as if 
excitations require a minimum energy $~ \Delta$. 
 
\subsection{Meissner Effect}

     There is another, much more fundamental characteristic which 
distinguishes the superconductor from a normal, but ideal, conductor.  
The superconductor expels magnetic flux, ie., $\B = 0$ within the 
bulk of a superconductor.  This is fundamentally different than an ideal 
conductor, for which $\dot{\B } = 0$ since for any closed path 
\begin{figure}[htb]
\centerline{\includegraphics[height=1.8in,keepaspectratio,clip=true]{closed.pdf}}
\caption[]{\em{A closed path and the surface it contains within
a superconductor.}}
\end{figure}
\beq
0 = IR = V = \oint \CE\cdot d\l = \int_S \grad \times \CE \cdot d\S 
= -\frac 1c\int_S \frac{\partial \B}{\partial t} \cdot d\S\,,
\eeq
or, since S and C are arbitrary 
\beq
	0 = -\frac 1c\dot{\B } \cdot S \Rightarrow \ \dot{\B } = 0
\eeq
Thus, for an ideal conductor, it matters if it is field cooled or 
zero field cooled. 
\begin{figure}[htb]
\centerline{\includegraphics[height=4.5in,keepaspectratio,clip=true]{ideal.pdf}}
\caption[]{\em{For an ideal conductor, flux penetration in the ground
state depends on whether the sample was cooled in a field through
the transition.}}
\end{figure}
Where as for a superconductor, regardless of the external field and its 
history, if $T < T_c$, then $B = 0$ inside the bulk.  This effect, which 
uniquely distinguishes an ideal conductor from a superconductor, is called the 
{\it Meissner effect}. 

     For this reason a superconductor is an ideal diamagnet. \ I.e.

\beq
	\B = \mu H = 0 \Rightarrow \ \mu = 0 \qquad \M = \chi H = \frac{\mu - 	
		1}{4\pi }H
\eeq
\beq
	\chi _{SC} = -\frac 1{4\pi }
\eeq
Ie., the measured $\chi $, Fig.~\ref{fig:SCchi}, in a superconducting metal is 
very large and negative (diamagnetic).
\begin{figure}[htb]
\centerline{\includegraphics[height=2.7in,keepaspectratio,clip=true]{Scchi.pdf}}
\caption[]{\em{LEFT: A sketch of the magnetic susceptibility versus temperature 
of a superconductor.  RIGHT: Surface currents on a superconductor are induced
to expel the external flux.  The diamagnetic response of a superconductor
is orders of magnitude larger than the Pauli paramagnetic response of
the normal metal at $T>T_C$}}
\label{fig:SCchi}
\end{figure}
This can also be interpreted as the presence of persistent 
surface currents which maintain a magnetization of 
\beq
	\M = -\frac 1{4\pi }H_{\mbox{ext}}
\eeq
in the interior of the superconductor in a direction opposite to the applied field.  The energy associated with this currents increases with  $H_{\mbox{ext}}$.  At some point it is then more favorable  (ie., a lower 
free energy is obtained)  if the system returns to a normal metallic state
and these screening currents abate.  Thus there exists an upper critical 
field $H_c$
\begin{figure}[htb]
\centerline{\includegraphics[height=2.4in,keepaspectratio,clip=true]{SCboundary.pdf}}
	\caption[]{\em{Superconductivity is destroyed by either raising
the temperature or by applying a magnetic field.}}
\end{figure} 
 
\section{ \ The London Equations }
\label{sec:london}
     London and London derived a phenomenological theory of superconductivity 
which correctly describes the Meissner effect.  They assumed 
that the electrons move in a frictionless state, so that 

\beq
	m\vd = -e\CE
\eeq
or, since $\pde jt = -en_s\vd$,

\beq
	\pde{\j_s}{t} = \frac{e^2n_s}{m}\CE \qquad \mbox{(First London Eqn.)}
\eeq
Then, using the Maxwell equation 
\beq
	\grad \times \CE = -\frac 1c \pde{\B }t \Rightarrow \frac 		
		m{n_se^2}\grad \times \pde{\j_s}{t} + \frac 1c \pde{\B }t = 0
\eeq
or 
\beq
	\pde{ }t\left( \frac m{n_se^2}\grad \times \j_s + \frac 1c\B\right) = 
		0
\eeq
This described the behavior of an ideal conductor (for which $\rho = 0$), but 
not the Meissner effect.  To describe this, the constant of integration must 
be chosen to be zero.  Then  
 \beq
	\grad \times \j_s = -\frac{n_se^2}{mc}\B \qquad \mbox{(Second London 
		Eqn.)}
\eeq
or defining $\lambda _L = \frac m{n_se^2}$, the London Equations become
\beq
	\frac{\B }c = -\la _L \grad \times \j _s \qquad \CE = 
	\la_L\pde{\j_s}{t}
\eeq
If we now apply the Maxwell equation 
$\grad \times \H = \frac{4\pi }c\j \Rightarrow \ \grad 
\times \B = \frac{4\pi }c\mu \j $ then we get 
\beq
\grad \times (\grad \times \B) = \frac{4\pi }c\mu \grad \times \j = 
		-\frac{4\pi \mu }{c^2\la _L}\B 
\eeq
and 
\beq
	\grad \times (\grad \times \j ) = -\frac 1{\la _Lc}\grad \times \B = 
-\frac{4\pi \mu }{c^2\la _L}\j
\eeq
or since $\grad \cdot \B =0$, $\grad\cdot\j=\frac{1}{c}\pde{\rho}t = 0$ 
and $\grad \times (\grad \times \a 
) = \grad (\grad \cdot \a ) - \grad ^2\a $ we get 
 
\beq
	\grad ^2\B - \frac{4\pi \mu }{c^2\la _L}\B = 0 \qquad \grad ^2\j - 
			\frac{4\pi \mu }{c^2\la _L}\j = 0
\eeq

\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{externalfld.eps}}
	\caption[]{\em{A superconducting slab in an external field.
	The field penetrates into the slab a distance $\La _L =
	\sqrt{\frac{mc^2}{4\pi ne^2\mu}}$.}}
\label{fig:externalfld}
\end{figure} 
Now consider a the superconductor in an external field shown in 
Fig.~\ref{fig:externalfld}.  The field is only in the x-direction, and
can vary in space only in the z-direction, then since $\grad 
\times \B = \frac{4\pi }c\mu \j $, the current is in the y-direction, so
\beq
	\frac{\partial ^2\B _x}{\partial z^2} - \frac{4\pi \mu }{c^2\la _L}\B 
		_x = 0 \qquad \frac{\partial ^2\j _{sy}}{\partial z^2} - 	
		\frac{4\pi \mu }{c^2\la _L}\j _{sy} = 0
\eeq
with the solutions 
 \beq
	\B _x = \B _x^0e^{-\frac{z}{\La _L}} \qquad \j _{sy} = \j 
				_{sy}e^{-\frac{z}{\La _L}}
\eeq
$\La _L = \sqrt{\frac{c^2\la _L}{4\pi \mu }} = \sqrt{\frac{mc^2}{4\pi ne^2\mu 
}}$ is the penetration depth. 

\section{Cooper Pairing}
 
     The superconducting state is fundamentally different than 
any possible normal metallic state (ie a perfect metal at $T = 0$). \ 
Thus, the transition from the normal metal state to the superconducting state 
must be a phase transition. \ A phase transition is accompanied by 
an instability of the normal state. \ Cooper first quantified this 
instability as due to a small attractive(!?) interaction between two 
electrons above the Fermi surface. 

\subsection{The Retarded Pairing Potential}

     The attraction comes from the exchange of phonons. \  
\begin{figure}[htb]
\centerline{\includegraphics[width=6.2in,keepaspectratio,clip=true]{phonpot.pdf}}
\caption[]{\em{Origin of the retarded attractive potential.
Electrons at the Fermi surface travel with a high velocity $v_F$.
As they pass through the lattice (left), the positive ions respond slowly.
By the time they have reached their maximum excursion, the first electron
is far away, leaving behind a region of positive charge which attracts
a second electron.}}
\label{fig:phonpot}
\end{figure}
The lattice deforms slowly in the time scale of the electron. \ It 
reaches its maximum deformation at a time $\tau \sim \ \frac{2\pi }{\omega 
_D} \sim \ 10^{-13}\mbox{s}$ after the electron has passed.  In this time the 
first electron has traveled $\sim \ v_F\tau \sim 
\ 10^8\frac{\mbox{cm}}{\mbox{s}} \cdot 10^{-13}\mbox{s} \sim \ 1000 \AA $.  
The positive charge of the lattice deformation can then attract another 
electron without feeling the Coulomb repulsion of the first electron.  
Due to retardation, the electron-electron Coulomb repulsion may be neglected! 
 
     The net effect of the phonons is then to create an attractive interaction 
which tends to pair time-reversed quasiparticle states. \ 
\begin{figure}[htb]
\centerline{\includegraphics[height=2.0in,keepaspectratio,clip=true]{pair.pdf}}
\caption[]{\em{To take full advantage of the attractive potential
illustrated in Fig.~\ref{fig:phonpot}, the spatial part of the electronic
pair wave function is symmetric and hence nodeless.  To obey the
Pauli principle, the spin part must then be antisymmetric or a singlet.}}
\end{figure}
They form an antisymmetric spin singlet so that the spatial part 
of the wave function can be symmetric and nodeless and so take 
advantage of the attractive interaction. \ Furthermore they tend 
to pair in a zero center of mass (cm) state so that the two 
electrons can chase each other around the lattice.
 
\subsection{Scattering of Cooper Pairs}

     This latter point may be quantified a bit better by considering two 
electrons above a filled Fermi sphere. \ These two electrons are attracted by 
the exchange of phonons. \ However, the maximum energy which may be exchanged 
in this way is $\sim \ \hbar \omega _D$. \ Thus the scattering in phase space 
is restricted to a narrow shell of energy width $\hbar \omega _D$. \ 
\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{scattering.pdf}}
	\caption[]{\em{Pair states scattered by the exchange of phonons
	are restricted to a narrow scattering shell of width $\hbar\omega_D$
	around the Fermi surface.}}
\end{figure}
Furthermore, the momentum in this scattering process is also conserved 
\beq
	\k _1 + \k _2 = \k _1^{\prime } + \k _2^{\prime } = \K
\eeq
Thus the scattering of $\k_1 $ and $\k_2$ into $\k_1^{\prime }$ and 
$\k_2^{\prime }$ is restricted to the overlap of the two scattering shells, 
Clearly this is negligible unless $\K \approx \ 0$. \ Thus the interaction is 
strongest (most likely) if $\k_1 = -\k_2$ and $\sigma_1 = -\sigma_2$; ie., 
pairing is primarily between time-reversed eigenstates. 
\begin{figure}[htb]
\centerline{\includegraphics[height=2.2in,keepaspectratio,clip=true]{scattering2.pdf}}
	\caption[]{\em{If the pair has a finite center of mass momentum,
	so that $\k _1 + \k _2 = \K$, then there are few states which
	it can scatter into through the exchange of a phonon.}}
\end{figure}
 
\subsection{The Cooper Instability of the Fermi Sea}

Now consider these two electrons above the Fermi surface.
They will obey the Schroedinger equation. 
 \beq
	-\frac{\hbar ^2}{2m}(\grad _1^2 + \grad _2^2)\psi (\r_1\r_2) + 
	V(\r_1\r_2)\psi (\r_1\r_2) = (\ep \ + 2E_F)\psi (\r_1\r_2)
\eeq
If $V = 0$, then $\ep = 0$, and 
\beq
	\psi _{V=0} = \frac 1{L^{3/2}}e^{i\k_1 \cdot \r_1} \frac 		
	1{L^{3/2}}e^{i\k_2 \cdot \r_2} = \frac 1{L^3}e^{i\k (\r_1 - \r_2)},
\eeq
where we assume that $\k _1 = -\k _2 = \k $. \ For small V, we will perturb 
around the $V = 0$ state, so that 
\beq
	\psi ( \r_1\r_2) = \frac 1{L^3}\sum _{\k } g(\k )e^{i\k \cdot (\r_1 
			- \r_2)}
\eeq
The sum must be restricted so that 
\beq
	E_F < \frac{\hbar ^2\k ^2}{2m} < E_F + \hbar \omega _D
\eeq
this may be imposed by $g(\k )$, since $|g(\k )|^2$ is the probability of 
finding an electron in a state $\k $ and the other in $-\k $. \ Thus we 
take 
\beq
	g(\k ) = 0 \quad \mbox{for} \quad \left\{ \begin{array}{ll}
			\k \ < \k _F \\
			\k \ > \frac{\sqrt{2m(E_F + \hbar \omega _D)}}{\hbar }
		\end{array}
		\right.
\eeq

The Schroedinger equations may be converted to a k-space equation by 
multiplying it by 
\beq
	\frac 1{L^3} \int d^3\r \ e^{-i\k ^{\prime } \cdot \ \r } \Rightarrow \ 
		\mbox{S.E.}
\eeq
so that
\beq	
 	\frac{\hbar ^2k^2}mg(\k ) + \frac 1{L^3}\sum _{\k ^{\prime }} g(\k 	
		^{\prime })V_{\k \k ^{\prime }} = (\ep \ + 2E_F)g(\k )
\eeq 
where 
\beq
	V_{\k \k ^{\prime }} = \int V(\r )e^{-i(\k \ - \k ^{\prime }) \cdot \r 
			} d^3\r
\eeq
now describes the scattering from $(\k , -\k )$ to $(\k ^{\prime }, -\k 
^{\prime })$. \ It is usually approximated as a constant for all $\k $ and 
$\k^{\prime}$ which obey the Pauli-principle and scattering shell restrictions 
\beq
	V_{\k \k ^{\prime }} = \left\{ \begin{array}{ll} 
			-V_0 \qquad & E_F < \frac{\hbar ^2\k ^2}{2m}, 
			\frac{\hbar ^2\k ^{\prime ^2}}{2m} < E_F + 		
			\hbar \omega _D \\
			0 \qquad & \mbox{otherwise} 
		\end{array} 
		\right.\,.
\eeq
so 
\beq
	\left( -\frac{\hbar ^2\k ^2}{m} + \ep \ + 2E_F\right) g(\k ) = 		
		-\frac{V_0}{L^3}\sum _{\k ^{\prime }} g(\k ^{\prime }) \equiv \ 
		-A
\eeq
or 
\beq
	g(\k ) = \frac{-A}{-\frac{\hbar ^2\k ^2}m + \ep \ + 2E_F} \qquad 	
		\mbox{(i.e. for $E_F < \frac{\hbar ^2\k ^2}{2m} < E_F + \hbar 	
		\omega _D$)}
\eeq
Summing over $\k $
\beq
	\frac{V_0}{L^3} \sum _{\k } \frac A{\frac{\hbar ^2\k^2}m - \ep \ - 
			2E_F} = +A
\eeq 
or 
\beq
	1 = \frac{V_0}{L^3} \sum _{\k } \frac 1{\frac{\hbar ^2\k^2}m - \ep \ - 
			2E_F}
\eeq
This may be converted to a density of states integral on 
$E = \frac{\hbar ^2\k ^2}{2m}$

\beq
	1 = V_0 \int_{E_F}^{E_F + \hbar \omega _D} Z(E_F) \frac{dE}{2E - \ep \ 
			- 2E_F}
\eeq
\beq
	1 = \frac 12V_0Z(E_F)\ln \left( \frac{\ep \ - 2\hbar \omega _D}{\ep 
}\right) 
\eeq
\beq
	\ep \ = \frac{2\hbar \omega _D}{1 - e^{2/(V_0Z(E_F))}} \simeq \ -2\hbar 
		\omega _D e^{-2/(V_0Z(E_F))} < 0, \qquad \mbox{as} 		
		\frac{V_0}{E_F} \rightarrow \ 0
\eeq

\section{The BCS Ground State }
 
In the preceding section, we saw that the weak phonon-mediated attractive
interaction was sufficient to destabilize the Fermi sea, and promote the 
formation of a Cooper pair $(\k \uparrow ,-\k \downarrow )$. \ The scattering 
 
\beq
	(\k \uparrow ,-\k \downarrow ) \rightarrow \ (\k ^{\prime }\uparrow 
			,-\k ^{\prime }\downarrow )
\eeq
yields an energy $V_0$ if $\k $ and $\k ^{\prime }$ are in the scattering 
shell $E_F < E_{\k }, E_{\k ^{\prime}} < E_F + \hod$. \ Many electrons can 
participate in this process and many Cooper pairs are formed, yielding a new 
state (phase) of the system. \ The energy of this new state is {\it not} just 
$\frac N2\ep $ less than that of the old state, since the Fermi surface is 
renormalized by the formation of each Cooper pair.

\subsection{The Energy of the BCS Ground State}  
     Of course, to study the thermodynamics of this new phase, it is 
necessary to determine its energy.  It will have both kinetic and potential 
contributions.  Since pairing only occurs for electrons {\it above} the Fermi 
surface, the kinetic energy actually increases:  if $w_k$ is the probability 
that a pair state $(\k \uparrow ,-\k \downarrow )$ is occupied then 
\beq
	E_{\mbox{kin}} = 2\sum _{\k } w_k\xi _k, \qquad \xi _k = \frac{\hbar 
			^2\k ^2}{2m} - E_F 
\eeq
The potential energy requires a bit more thought.  It may be 
written in terms of annihilation and creation operators for the 
pair states labeled by $\k$
\beqa
	& \left| 1\right> _k \qquad & (\k \uparrow ,-\k \downarrow ) 		
		\mbox{occupied} \\
	& \left| 0\right> _k \qquad & (\k \uparrow ,-\k \downarrow ) 		
		\mbox{unoccupied}
\eeqa
 or 
\beq
	\left| \psi_k\right> = u_k \left| 0\right>_k + v_k \left| 1\right>_k
\eeq
where $v_k^2 = w_k$ and $u_k^2 = 1 -w_k$.  Then the BCS state, which is a 
collection of these pairs, may be written as 
\beq
	\left| \phi _{BCS}\right> \ \simeq \ \prod _k \left\{ u_k\left| 
			0\right> _k + v_k\left| 1\right> _k\right\}\,.
\eeq
We will assume that $u_k, v_k \in 
\Re $.  Physically this amounts to taking the phase of the order parameter to 
be zero (or $\pi $), so that it is real.  However the validity of this 
assumption can only be verified for a more microscopically based theory.  



     By the Pauli principle, the state $(\k \uparrow ,-\k \downarrow )$ can be, 
at most, singly occupied, thus a ($s=\frac 12$) Pauli representation is 
possible 
\beq
	\left| 1\right> _k = \begin{pmatrix} 1 \cr 0 \cr \end{pmatrix}_k \qquad \left| 0\right> _k 
		= \begin{pmatrix} 0 \cr 1 \cr \end{pmatrix}_k
\eeq
Where $\sigma _k^+$ and $\sigma _k^-$, describe the creation and anhialation of 
the state $(\k \uparrow,-\k \downarrow )$
\beqa
	\sigma _k^+ & = \frac 12 (\sigma _k^1 + i\sigma _k^2) &
	= \begin{pmatrix} 0 & 1 \cr 0 & 0 \cr \end{pmatrix} \\
	\sigma _k^- & = \frac 12 (\sigma _k^1 - i\sigma _k^2) & 
	= \begin{pmatrix} 0 & 0 \cr 1 & 0 \cr \end{pmatrix}
\eeqa
Of course $\sigma _k^+ \begin{pmatrix} 0 \cr 1 \cr\end{pmatrix}_k 
        = \begin{pmatrix} 1 \cr 0 \cr \end{pmatrix} $
\beqa
	\sigma _k^+\left| 1\right> _k = & 0 \qquad \sigma _k^+\left| 		
		0\right> _k = & \left| 1\right> _k \\
	\sigma _k^- & \left| 1\right> _k = \left| 0\right> _k \qquad \sigma 	
		_k^+\left| 0\right> _k = & 0
\eeqa
 
The process $(\k \uparrow,-\k \downarrow ) \rightarrow \ (\k ^{\prime}\uparrow 
,-\k ^{\prime }\downarrow )$, if allowed, is associated with an energy 
reduction $V_0$. \ In our Pauli matrix representation this process 
is represented by operators $\sigma _{\k ^{\prime }}^+\sigma _\k^-$, so 
\beq
	V = -\frac{V_0}{L^3}\sum _{\k \k ^{\prime }} \sigma _{\k ^{\prime 
		}}^+\sigma _\k^- \qquad \mbox{(Note that this {\it is} 				
		Hermitian)}
\eeq
Thus the reduction of the potential energy is given by $\left< \phi 
_{BCS}\left| V\right|\phi _{BCS}\right> $
\beqa
\left< \phi_{BCS}\left| V\right|\phi_{BCS}\right>  =&&	
		-\frac{V_0}{L^3}\left\{ \prod_p \left( u_p \left< 0\right| 	
		 + v_p \left< 1\right| \right) \sum_{\k \k ^{\prime }} 
		\sigma_\k^+\sigma_{\k^{\prime }}^- \right.\nonumber \\
		 &&\left.\prod_{p^{\prime }} 	
		\left( u_{p^{\prime }}\left| 0 \right>_{p^{\prime }}+
		v_{p'}\left| 	
		1\right>_{p^{\prime }}\right) \right\}
\eeqa
Then as $_k\left< 1\arrowvert 1\right> _{k^{\prime }} = \delta _{kk^{\prime }}, 
{}_k\left< 0\arrowvert 0\right> _{k^{\prime }} = \delta _{kk^{\prime }}$ and 
${}_k\left< 0\arrowvert 1\right> _{k^{\prime }} = 0$
\beq
	\left< \phi _{BCS}\left| V\right|\phi _{BCS}\right> \ = 	
		-\frac{V_0}{L^3}\sum _{kk^{\prime }} v_ku_{k^{\prime 		
		}}u_kv_{k^{\prime }}
\eeq
Thus, the total energy (kinetic plus potential) of the system of 
Cooper pairs is 
 \beq	
	W_{BCS} = 2\sum _k v_k^2\xi _k - \frac{V_0}{L^3}\sum 
				_{kk^{\prime }} v_ku_{k^{\prime 
}}u_kv_{k^{\prime }}
\eeq

     As yet $v_k$ and $u_k$ are unknown. \ They may be treated as variational 
parameters. \ Since $w_k = v_k^2$ and $1 - w_k = u_k^2$, we may impose this
constraint by choosing 
\beq
	v_k = \cos \theta _k, \qquad u_k = \sin \theta _k
\eeq
At $T = 0$, we require $W_{BCS}$ to be a minimum.
\beqa 
 	W_{BCS} & = \sum _k 2\xi _k \cos ^2\theta _k - \frac{V_0}{L^3} 	
		\sum _{kk^{\prime }} \cos \thk \sin \thkp \cos \thkp \sin \thk 	
		\nonumber \\
 	& = \sum _k 2\xi _k \cos ^2\theta _k - \frac{V_0}{L^3} \sum 		
		_{kk^{\prime }} \frac 14\sin 2\thk \sin 2\thkp 
\eeqa
\beq
	\pde{W_{BCS}}{\thk} = 0 = -4\xi _k\cos \thk \sin \thk - 		
	\frac{V_0}{L^3}\sum _{\kpr} \cos 2\thk \sin 2\thkp
\eeq
\beq
	\xi _k\tan 2\thk = -\frac 12\frac{V_0}{L^3}\sum _{\kpr} \sin 2\thkp
\eeq
Conventionally, one introduces the parameters $E_k = \sqrt{\xi _k^2 + \Delta 
^2}, \Delta = \frac{V_0}{L^3}\sum _k u_kv_k = \frac{V_0}{L^3}\sum _k \cos \thk 
\sin \thk $. \ Then we get 
\beqa
	\xi _k\tan 2\thk \ = -\Delta \ \Rightarrow \ 2u_kv_k = \sin 2\thk \ = 	
		\frac{\Delta }{E_k} \\
	\cos 2\thk = \frac{-\xi _k}{E_k} = \cos ^2\thk \ - \sin ^2\thk \ = 	
		v_k^2 - u_k^2 = 2v_k^2 - 1
\eeqa
\beq
	w_k = v_k^2 = \frac 12\left( 1 - \frac{-\xi _k}{E_k}\right) \ = \frac 	
		12\left( 1 - \frac{\xi _k}{\sqrt{\xi _k^2 + \Delta ^2}}\right)
\eeq	
\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{energy.pdf}}
\caption[]{\em{Sketch of the ground state pair distribution function.}}
\end{figure}
If we now make these substitutions $\left(2u_kv_k = \frac{\Delta }{E_k}, v_k^2 = 
\frac 12\left( 1 - \frac{\xi _k}{E_k}\right)\right) $ into $W_{BCS}$, then we 
get 
\beq
	W_{BCS} = \sum _k \xi _k\left( 1 - \frac{\xi _k}{E_k}\right) - 		
		\frac{L^3}{V_0}\Delta ^2.
\eeq
 
Compare this to the normal state energy, again measured relative 
to $E_F$ 
 \beq
	W_n = \sum _{k<k_F} 2\xi _k
\eeq
or 
\beqa
	\frac{W_{BCS} - W_n}{L^3} & = & -\frac 1{L^3}\sum _k \xi _k\left( 1 + 	
		\frac{\xi _k}{E_k}\right) \ - \frac{\Delta ^2}{V_0} \\
	& \approx & -\frac 12Z(E_F)\Delta ^2 < 0.
\eeqa	 
So the formation of superconductivity reduces the ground state energy.  
This can also be interpreted as $\Delta Z(E_F)$ electrons pairs per and volume 
condensed into a state $\Delta $ below $E_F$.  The average energy gain per 
electron is $\frac{\Delta}2$. 
 
\subsection{The BCS Gap}
The gap parameter $\Delta $ is fundamental to the BCS theory.  It tells us 
both the energy gain of the BCS state, and about its excitations.  Thus 
$\Delta $ is usually what is measured by experiments.  To see this consider 
\beqa 
	W_{BCS} = \sum _k 2\xi _k\frac 12\left( 1 - \frac{\xi _k}{E_k}\right) - 
		\frac{L^3\Delta ^2}{V_0} \\
	\downarrow \qquad \mbox{Lots of algebra (See I\&L)} \nonumber \\
	W_{BCS} = -\sum 2E_kv_k^4
\eeqa
Now recall that the probability that the Cooper state $(\k \uparrow , \k 
\downarrow )$ was occupied, is given by $w_k = v_k^2$.  Thus the first pair 
breaking excitation 
\begin{figure}[htb]
\centerline{\includegraphics[height=2.4in,keepaspectratio,clip=true]{breaking.pdf}}
\caption[]{\em{Breaking a pair requires an energy 
$2\sqrt{\xi_{\k}^2 + \Delta ^2}\geq 2\Delta$}}
\end{figure} 
takes $v_{\kpr }^2 = 1$ to $v_{\kpr }^2 = 0$, for a change in energy 
\beq
	\Delta E = -\sum _{k\neq \kpr } 2v_k^4E_k + \sum _k 2v_k^4E_k = 	
		2E_{\kpr } = 2\sqrt{\xi _{\kpr }^2 + \Delta ^2}
\eeq
Then since $\xi _{\kpr } = \frac{\hbar ^2k^{\prime 2}}{2m} - E_F$, the smallest 
such excitation is just 
\beq
	\Delta E_{min} = 2\Delta
\eeq
This is the minimum energy required to break a pair, or create an excitation in 
the BCS ground state.  It is what is measured by the specific heat 
$C \sim e^{-\beta 2\Delta }$ for $T < T_c$.
 
Now consider some experiment which adds a {\it single} electron, or perhaps a 
few unpaired electrons, to a superconductor (ie tunneling).
\begin{figure}[htb]
\centerline{\includegraphics[height=2.0in,keepaspectratio,clip=true]{exper.pdf}}
 	\caption[]{}
\end{figure}
This additional electron cannot find a partner for pairing. \ Thus 
it must enter one of the excited states discussed above. \ Since 
it is a single electron, its energy will be 
\beq
	E_k = \sqrt{\xi _k^2 + \Delta ^2}
\eeq
For $\xi _k^2 \gg \ \Delta , E_k = \xi _k = \frac{\hb ^2k^{\prime 2}}{2m} - 
E_F$, which is just the energy of a normal metal state.  Thus for energies 
well above the gap, the normal metal continuum is recovered for unpaired 
electrons. 
 
     To calculate the density of unpaired electron states, recall that the 
density of states was determined by counting k-states.  These are unaffected by 
any phase transition.  Thus it must be that the number of states in $d^3k$ is 
equal.
\begin{figure}[htb]
\centerline{\includegraphics[height=2.0in,keepaspectratio,clip=true]{kspace.pdf}}
\caption[]{\em{The number of k-states within a volume $d^3k$ 
of k-space is unaffected by any phase transition.}}
\end{figure}
\beq
	D_s(E_k) dE_k = D_n(\xi_k) d\xi_k  
\eeq
In the vicinity of $\Delta \sim \ \xi _k, D_n(\xi _k) \approx \ D_n(E_F)$ since 
$|\De | \ll \ E_F$   (we shall see that $\Delta \leq 2w_D$). \ Thus for $\xi _k 
\sim \ \De $
\beq
	\frac{D_s(E_k)}{D_n(E_F)} = \frac{d\xi_x}{dE_k} = \frac 		
		d{dE_k}\sqrt{E_k^2 - \De^2} = \frac{E_k}{\sqrt{E_k^2 - \De 	
		^2}} \qquad E_k > \De 
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[height=2.25in,keepaspectratio,clip=true]{plot.pdf}}
	\caption[]{}
\end{figure}

Given the experimental and theoretical importance of $\De $, it should be 
calculated. 
\beq
	\De \ = \frac{V_0}{L^3} \sum _k \sin \thk \cos \thk \ = \frac{V_0}{L^3} 
		\sum _k u_kv_k = \frac{V_0}{L^3} \sum _k \frac{\De }{2E_k}
\eeq
\beq
	\De \ = \frac 12 \frac{V_0}{L^3} \sum _k \frac{\De }{\sqrt{\xi _k^2 + 	
		\De ^2}}
\eeq
Convert this to sum over energy states (at $T = 0$ all states with $\xi<0$ 
are occupied since $\xi_k = \frac{\hb ^2\k^2}{2m} - E_F$).
\beq
	\De \ = \frac{V_0}2\De \ \int _{-\hod}^{\hod} \frac{Z(E_F + \xi ) d\xi 	
		}{\sqrt{\xi ^2 + \De ^2}}
\eeq
\beq
	\frac 1{V_0Z(E_F)} = \int _0^{\hod} \frac{d\xi }{\sqrt{\xi ^2 		
		+ \De ^2}}
	\label{eq:delta}
\eeq
\beq
	\frac 1{V_0Z(E_F)} = \sinh ^{-1}\left( \frac{\hb \omega _D}{\De 	
		}\right)
\eeq
For small $\De $,
\beq
	\frac{\hb \omega _D}{\De } \sim \ e^{\frac 1{V_0Z(E_F)}}
\eeq
\beq
	\De \ \simeq \ \hod e^{-\frac 1{V_0Z(E_F)}} 
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[height=2.25in,keepaspectratio,clip=true]{plot2.pdf}}
	\caption[]{}
\end{figure}

\section{ \ Consequences of BCS and Experiment} 
 
\subsection{Specific Heat}
As mentioned before, the gap $\De $ is fundamental to experiment. \ The 
simplest excitation which can be induced in a superconductor has energy 
$2\De $. 
Thus 
\beq
	\De E \sim \ 2\De e^{-\beta 2\De } \qquad T \ll \ T_c
\eeq
\beq
	C \sim \ \pde{\De E}{\beta } \pde{\beta }{T} \sim \ \frac{\De 
^2}{T^2}e^{-\beta 2\De }
\eeq

\subsection{Microwave Absorption and Reflection}
 
     Another direct measurement of the gap is reflectivity/absorption.  
A phonon impacting a superconductor can either be reflected or absorbed. 
Unless $\hb \omega > 2\De $, the phonon cannot create an excitation and 
is reflected.  Only if $\hb \omega > 2\De $ is there absorption. 
Consider a small cavity within a superconductor.  The cavity has a small 
hole which allows microwave radiation to enter the cavity.  
\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{absorb.pdf}}
\caption[]{If $B > B_c$ or $\hbo > 2\De $, then absorption reduces the
intensity to the normal-state value $I = I_n$.  For $B=0$ the microwave
intensity within the cavity is large so long as $\hbo < 2\De $}
\end{figure}
If $\hbo \ < 2\De $ and if $B < B_c$, then the microwave intensity is high $I 
= I_s$. \ On the other hand, if $\hbo \ > 2\De $ ,or $B > B_c$, then the 
intensity falls in the cavity $I = I_n$ due to absorbs ion by the walls. 

Note that this also allows us to measure $\De $ as a function of $T$. \ At $T =
T_c, \De = 0$, since thermal excitations reduce the number of Cooper pairs and 
increase the number of unpaired electrons, which obey Fermi-statistics. 
\begin{figure}[htb]
\centerline{\includegraphics[height=1.9in,keepaspectratio,clip=true]{breaking2.pdf}}
	\caption[]{}
	\label{fig:breaking2}
\end{figure}
The size of (Eqn.~\ref{eq:delta}) is only effected by the presence of a Cooper pair 
. \ The probability that an electron is unpaired is $f\left( \sqrt{\xi ^2 +\De 
^2} + E_F, T\right) = \frac 1{\exp{\beta \sqrt{\xi ^2 + \De ^2}} + 1}$ so, the 
probability that a Cooper pair exists is $1 - 2f\left( \sqrt{\xi ^2 +\De ^2} + 
E_F, T\right) $.  Thus for $T\neq 0$ 
\beq
	\frac 1{V_0Z(E_F)} = \int _0^{\hod } \frac {d\xi }{\sqrt{\xi ^2 + \De 
^2}} \left\{ 1 - 2f\left( \sqrt{\xi ^2 +\De ^2} + E_F, T\right) \right\}
\eeq
Note that as $\sqrt{\xi ^2 +\De ^2} \geq 0$, when $\beta \ \rightarrow \ \infty 
$ we recover the $T = 0$ result. 
 
This equation may be solved for $\De (T)$
\begin{figure}[htb]
\centerline{\includegraphics[height=1.75in,keepaspectratio,clip=true]{temp.pdf}}
\caption[]{\em{The evolution of the gap (as measured by reflectivity)
as a function of temperature.  The BCS approximation is in reasonably
good agreement with experiment.}}
\end{figure} 
and for $T_c$. To find $T_c$ consider this equation as 
$\frac T{T_c} \rightarrow \ 1$, the first solution to the gap 
equation, with $\De = 0^+$, occurs at $T = T_c$.  Here 
\beq
	\frac 1{V_0Z(E_F)} = \int _0^{\hod } \frac {d\xi }{\xi } \tanh \left( 	
		\frac{\xi }{2k_BT_c}\right)
\eeq
which may be solved numerically to yield 
 \beq
	1 = V_0Z(E_F) \ln \frac{1.14 \hod }{k_BT_c}
\eeq
\beq
	k_BT_c = 1.14\hod e^{-1/\left\{ V_0Z(E_F)\right\} }
\eeq
but recall that $\De \ = 2\hod e^{-1/\left\{ V_0Z(E_F)\right\} }$, so 
 \beq
	\frac{\De (0)}{k_BT_c} = \frac 2{1.14} = 1.764
\eeq
\begin{table}[htb]
	\begin{center}
	\begin{tabular}{|c|c|c|c|} \hline 
		metal 	& $T_c {}^{\circ }K$ & $Z(E_F)V_0$ & $\De (0)/k_BT_c$ 	
			\\ \hline
		Zn	& 0.9 	& 0.18	& 1.6 \\
		Al	& 1.2 	& 0.18	& 1.7 \\
		Pb	& 7.22	& 0.39	& 2.15 \\ \hline
	\end{tabular}
\caption[]{{\em Note that the value 2.15 for $\De (0)/k_BT_c$ for Pb is 	
higher than BCS predicts.  Such systems are labeled strong coupling 
superconductors and are better described by the Eliashberg-Migdal theory.}}
	\end{center}
\end{table}

\subsection{The Isotope Effect}

Finally, one should discuss the isotope effect. \ We know that $V_{k\kpr }$, 
results from phonon exchange. \ If we change the mass of one of the vibrating 
members but not its charge, then $V_0 N(E_F)$ etc are unchanged but 
\beq
	\omega _D \sim \ \sqrt{\frac kM} \sim \ M^{-\frac 12}.
\eeq
Thus $T_c \sim \ M^{-\frac 12}$. \ This has been confirmed for most normal 
superconductors, and is considered a "smoking gun" for phonon mediated 
superconductivity. 
 
\section{ \ BCS $\Rightarrow $ Superconducting Phenomenology}

	Using Maxwell's equations, we may establish a relation between
the critical current and the critical field necessary to destroy the
superconducting state.  Consider a long thick wire (with radius 
$r_0 \gg \Lambda _L$) 
\begin{figure}[htb]
\centerline{\includegraphics[height=2.75in,keepaspectratio,clip=true]{wire.pdf}}
\caption[]{\em{Integration contour within a long thick superconducting
wire perpendicular to a circulating magnetic field.  The field only 
penetrates into the wire a distance $\Lambda_L$.}}
\label{fig:longSCwire}
\end{figure}
and integrate the equation
\beq
	\grad \times \H = \frac{4\pi }c {\bf j}
\eeq
along the contour shown in Fig.~\ref{fig:longSCwire}.
\beq
	\int \ \grad \times \H dS = \int \H \cdot d\l = \frac{4\pi }c \int {\bf 
		j} \cdot d\s
\eeq
\beq
	2\pi r_0 H = \frac{4\pi }c 2\pi r_0\Lambda _L{j}_0
\eeq
If ${\bf j}_0 = {\bf j}_c$ (${\bf j}_c$ is the critical current), then 

\beq
	H _c = \frac{4\pi }c \Lambda _L{\j}_c 
\eeq
Since both $H _c$ and ${ j}_c \propto \ \De $, they will share the 
temperature-dependence of $\De $.

 
At $T = 0$, we could also get an expression for $H_c$ by noting that, 
since the superconducting state excludes all flux, 
\beq
	\frac 1{L^3}\left( W_n - W_{BCS}\right) = \frac 1{8\pi }\H _c^2
\eeq
However, since we have earlier 
\beq
	\frac 1{L^3}\left( W_n - W_{BCS}\right) = \frac 12N(0)\De ^2,
\eeq
we get 
\beq
	H_c = 2\De \sqrt{\pi N(0)}
\eeq

We can use this, and the relation derived above 
${j_c} = \frac c{4\pi \Lambda_L}H_c$, to get a (properly 
derived) relationship for ${j}_c$. 
\beq
	{j_c} = \frac c{4\pi \Lambda _L}2\De \sqrt{\pi N(0)}
\eeq
However, for most metals
\beq
	N(0) \simeq \ \frac n{E_F} 
\eeq
\beq
	\La _L = \sqrt{\frac{mc^2}{4\pi ne^2\mu }}
\eeq 
taking $\mu \ = 1$
\beq
	{j}_c = \frac c{4\pi }\sqrt{\frac{4\pi ne^2}{mc^2}} 2\De 
				\sqrt{\frac{\pi n2m}{\hbar ^2k_F^2}} = 
\sqrt{2}\De 				\frac{ne}{\hbar k_F}
\eeq
 
This gives a similar result to what Ibach and L\"uth get, but for a 
completely different reason. Their argument is similar to 
one originally proposed by Landau.  Imagine that you have a fluid 
which must flow around an obstacle of mass $M$. \ From the 
perspective of the fluid, this is the same as an obstacle moving 
in it. \ 
\begin{figure}[htb]
\centerline{\includegraphics[height=0.6in,keepaspectratio,clip=true]{fluid.pdf}}
\caption[]{\em{A superconducting fluid which must flow around an obstacle 
of mass $M$.  From the perspective of the fluid, this is the same as an 
obstacle, with a velocity equal and opposite the fluids, moving in it.}}
\end{figure}
Suppose the obstacle makes an excitation of energy $\ep $ and momentum $\p $
in the fluid, then 
\beq
	E^{\prime } = E - \ep \qquad \P ^{\prime } = \P - \p
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[height=1.0in,keepaspectratio,clip=true]{excite.pdf}}
\caption[]{\em{A large mass $M$ moving with momentum $\P$ in a superfluid
(a), creates an excitation (b) of the fluid of energy $\ep$ and momentum
$\p$}}
\end{figure}
or from squaring the second equation and dividing by $2M$
\beq
	\frac{P^{\prime 2}}{2M} - \frac{P^2}{2M} = - \frac{\P \cdot \p }M + 	
		\frac{\p ^2}{2M} = E^{\prime } - E = \ep
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[height=1.2in,keepaspectratio,clip=true]{momentum.pdf}}
	\caption[]{}
\end{figure}
\beq
	\ep \ = \frac{p P\cos \th }M - \frac{p ^2}{2M}
\eeq
\beq
	\ep \ = p v \cos \th \ - \frac{p ^2}{2M}
\eeq
If $M \rightarrow \ \infty $ (a defect in the tube which carries the fluid 
could have essentially an infinite mass) then 
\beq
	\frac{\ep }{p } = v \cos \th
\eeq
Then since $\cos\th \ \le \ 1$
\beq
	v\ \ge \ \frac{\ep }{p }
\eeq
Thus, if there is some minimum $\ep $,then there is also a minimum 
velocity below which such excitations of the fluid cannot happen. 
For the superconductor 
\beq	
	v _c = \frac{\ep _{min}}{p } = \frac{2\De }{2\hbar k_F}
\eeq
Or 
\beq
	{j}_c = env _c = \De \frac{ne}{\hbar k _F}
\eeq
 
     This is the same relation as we obtained with the previous thermodynamic 
argument (within a factor $\sqrt{2}$).  However, the former argument is 
more proper, since it would apply even for {\it gapless} superconductors, and 
it takes into account the fact that the S.C. state is a collective phenomena 
ie., a minuet, not a waltz of electric pairs. 
 
 
\section{ \ Coherence of the Superconductor $\Rightarrow $ Meisner effects }
 
Superconductivity {\em{is}} the Meissner effect, but thus far, we have
not yet shown that the BCS theory leads to the second London equation
which describes flux exclusion.  In this subsection, we will see that
this requires an additional assumption: the rigidity of the BCS wave
function.

In the BCS approximation, the superconducting wave function is taken to be 
composed of products of Cooper pairs.  One can estimate the size 
of the pairs from the uncertainty principle 
\beq
	2\De \ = \de \left( \frac{\p ^2}{2m}\right) \ \sim \ \frac{\p _F}m\de 	
		\p \Rightarrow \ \de \p \ \sim \ 2m\frac{\De }{\p _F}
\eeq
\beq
	\xi_{cp} \sim \ \de x \sim \ \frac{\hbar }{\de \p } \sim 		
		\frac{\hbar \p _F}{2m\De } = \frac{\hbar ^2\k _F}{2m\De } = 	
		\frac{E_F}{\k _F\De }
\eeq
\beq
	\xi_{cp} \sim 10^3 - 10^4 \AA \sim \ \mbox{size of Cooper pair wave 	
		function}
\eeq
Thus in the radius of the Cooper pair, about 

\beq
	\frac {4\pi n}{3}\left(\frac{\xi_{cp}}2\right) ^3 \sim \ 10^8
\eeq
other pairs have their center of mass.
\begin{figure}[htb]
\centerline{\includegraphics[height=1.75in,keepaspectratio,clip=true]{CM.pdf}}
	\caption[]{\em{Many electron pairs fall within the volume of a
Cooper wavefunction.  This leads to a degree of correlation between
the pairs and to rigidity of the pair wavefunction.}}
\end{figure}
 
The pairs are thus {\it not} independent of each other (regardless of the BCS 
wave function approximation). \ In fact they are specifically anchored to each 
other; ie., they maintain coherence over a length scale of at least $\xi 
_{cp}$.
\begin{figure}[htb]
\centerline{\includegraphics[height=1.5in,keepaspectratio,clip=true]{coherence.pdf}}
	\caption[]{}
\end{figure} 
 
In light of this coherence, lets reconsider the supercurrent 

\beq
	{\bf j} = -\frac{2e}{4m}\left\{ \psi \p ^*\psi ^* + \psi ^*\p \psi 	
		\right\}
\eeq
where pair mass $= 2m$ and pair charge $= -2e$.
\beq
	\p = -i\hbar \grad \ - \frac{2e}cA
\eeq
A current, or a CM momentum $\K $, modifies the single pair state 
\beq
	\psi (\r _1,\r _2) = \frac 1{L^3}\sum _k g(\k )e^{i\K \cdot \ (\r _1 + 	
		\r _2)/2}e^{i\k \cdot \ (\r _1 - \r _2)}
\eeq
\beq
	\psi (\K, \r_1,\r_2) = \psi (\K=0, \r_1,\r_2) e^{i\K \cdot \R}
\eeq
where $\R =\frac{\r _1 + \r _2}2$ is the cm coordinate and $\hbar \K $ is 
the cm momentum.  Thus 
\beq
	\Phi _{BCS} \simeq \ e^{i\phi } \Phi _{BCS}(\K =0) = e^{i\phi }\Phi (0)
\eeq
\beq
	\phi \ = \K \cdot \ (\R _1 + \R _2 + \cdots )
\eeq 
(In principle, we should also antisymmetrize this wave function; 
however, we will see soon that this effect is negligible).  Due 
to the rigidity of the BCS state it is valid to approximate 
\beq
	\grad \ =  \grad _R + \grad _r \approx\grad _R
\eeq
Thus 
\beqa
	{\bf j}_s \approx \ \frac{2e}{4m} \sum _{\nu } \left\{ \Phi _{BCS}^* 	
		\left( -i\hbar \grad _{R_{\nu }} + \frac{2eA}c\right) \Phi 	
		_{BCS} \right. \nonumber \\
	\left. + \Phi _{BCS} \left( i\hbar \grad _{R_{\nu }} + \frac		
		{2eA}c\right) ^*\Phi _{BCS}^*\right\}
\eeqa
or 
\beq
	{\bf j}_s = -\frac {2e}{2m}\left\{ \left| \Phi (0)\right| ^2\frac{4eA}c + 	
		2\hbar \left| \Phi (0)\right| ^2 \sum _{\nu }\grad _{R_{\nu 	
		}}\phi \right\}
\label{eq:js}
\eeq
Then since for any $\psi , \grad \ \times \ \grad \psi \ = 0$
\beq
	\grad  \times {\bf j}_s = -\frac{2e^2}{mc}\left| \Phi (0)\right| 	
		^2\grad \times A
\eeq
or since $\left| \Phi (0)\right| ^2 = \frac{n_s}2$
\beq
	\grad \times {\bf j} = -\frac{ne^2}{mc}\B
\eeq
which is the second London equation which as we saw in Sec.~\ref{sec:london} 
leads to the Meissner effect.  Thus the second London equation can only 
be derived from the BCS theory by assuming that the BCS state is spatially 
homogeneous. 
 
\section{ \ Quantization of Magnetic Flux} 
 
     The rigidity of the wave function (superconducting coherence) also 
guarantees that the flux penetrating a superconducting loop is quantized.  
This may be seen by integrating Eq.~\ref{eq:js} along a contour within the 
superconducting bulk (at least a distance $\Lambda_L$ from the surface). 
\beq
	{\bf j}_s = -\frac{e^2n_s}{mc}A - \frac{e\hbar n_s}{2m}\sum _{\nu 	
		}\grad _{R_{\nu }}\phi
\eeq
\beq
	\int\!\!\!\!\!\:\!\circ {\bf j}_s \cdot \ dl = -\frac{e^2n_s}{ms} 	
		\int\!\!\!\!\!\:\!\circ A \cdot \ dl - \frac{e\hbar 		
		n_s}{2m}\sum _{\nu } \int\!\!\!\!\!\:\!\circ \grad _{R_{\nu 	
		}}\phi \ \cdot \ dl
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[height=2.25in,keepaspectratio,clip=true]{flux.pdf}}
\caption[]{\em{Magnetic flux penetrating a superconducting loop
is quantized. This may be seen by integrating Eq.~\ref{eq:js} along a contour
within the superconducting bulk (a distance $\Lambda_L$ from the
surface).}}
\end{figure}
Presumably the phase of the BCS state $\Phi _{BCS} = e^{i\phi }\Phi (0)$ is 
single valued, so 
\beq
	\sum _{\nu } \int \grad _{R_{\nu }}\phi \ \cdot \ dl = 2\pi N \qquad N 	
		\in \CZ
\eeq
Also since the path $l$ may be taken inside the superconductor by a depth 
of more than $\La _L$, where $\j_s = 0$, we have that 
\beq
	\int {\j_s} \cdot \ dl = 0
\eeq
so 
\beq
	-\frac{e^2n_s}{ms} \int A \cdot \ dl = -\frac{e^2n_s}{ms} \int \B \cdot 
		\ ds = 2N\pi \frac{e\hbar n_s}{2m}
\eeq
Ie., the flux in the loop is quantized. 
 
\section{Tunnel Junctions }
 
     Imagine that we have an insulating gap between two metals, and that a 
plane wave (electronic Block State) is propagating towards this barrier from 
the left 
\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{junction.pdf}}
	\caption[]{}
\end{figure} 
\beqa
	\psi _a = A_1e^{i\k x} + B_1e^{-i\k x} \qquad \psi _b = A_2e^{i\k 	
		^{\prime }x} + B_2e^{-i\k^{\prime }x} \nonumber \\
	\psi _c = B_3e^{-i\k x}
\eeqa
These are solutions to the S.E. if 

\beq
	k \ = \frac{\sqrt{2mE}}{\hbar } \qquad \mbox{in a \& c}
\eeq
\beq
	k ^{\prime } = \frac{\sqrt{2m(E - V_0)}}{\hbar } \qquad \mbox{in b}
\eeq
The coefficients are determined by the BC of continuity of $\psi $ and $\psi 
^{\prime }$ at the barriers $x = 0$ and $x = d$. \ If we take $B_3 = 1$ and $E 
< V_0$, so that 

\beq
	k ^{\prime } = i\kappa \ = \frac{\sqrt{2m(E - V_0)}}{\hbar }
\eeq
then, the probability of having a particle tunnel from left to right is 
 
\beq
	P_{l\rightarrow r} \propto \ \frac {|B_3|^2}{|B_1|^2} = 
	\frac {1}{|B_1|^2}=
	\left\{ \frac 12 - 	
		\frac 18\left( \frac k{\kappa } - \frac{\kappa}k\right) ^2 + 	
		\frac 18\left( \frac k{\kappa } + \frac{\kappa}k\right) ^2\cosh 
		2\kappa d\right\} ^{-1}
\eeq
For large $\kappa d$

\beqa
	P_{l\rightarrow r} & \propto & 8\left( \frac k{\kappa } + 		
		\frac{\kappa}k\right)^{-2}e^{-2\kappa d} \\
	& \propto & 8\left( \frac k{\kappa } + \frac{\kappa}k\right)^{-2}\exp 	
		\left\{ -\frac{2d\sqrt{2m(V_0-E)}}{\hbar }\right\}
\eeqa
Ie, the tunneling probability falls exponentially with distance. 
 
Of course, this explains the physics of a single electron tunneling across a 
barrier, assuming that an appropriate state is filled on the left-hand side 
and available on the right-hand side.  This, as can be seen in 
Fig.~\ref{fig:tunneling}, is not always the case, especially in a 
conductor.  Here, we must take into account the densities of 
states and their occupation probabilities $f$.  We will be 
interested in applied voltages $V$ which will shift the chemical 
potential $eV$.  To study the gap we will apply 
\begin{figure}[htb]
\centerline{\includegraphics[height=2.25in,keepaspectratio,clip=true]{tunneling.pdf}}
	\caption[]{\em{Electrons cannot tunnel accross the barrier since
no unoccupied states are available on the left with correspond in
energy to occupied states on the right (and vice-versa).  However, the
application of an appropriate bias voltage will promote the state on the 
right in energy, inducing a current.}}
\label{fig:tunneling}
\end{figure}

\beq
	eV \sim \ \De
\eeq
We know that $\frac{2\De }{k_BT_c} \sim \ 4, \De \sim \ \frac{4k_BT_c}2 \sim 
10^{\circ }K$.  However typical metallic densities of states have features on 
the scale of electron-volts $\sim \ 10^4{}^{\circ }K$. \ Thus, on this 
energy scale we may approximate the metallic density of states as 
featureless.
\beq
	N_r(\ep ) = N_{metal}(\ep ) \approx \ N_{metal}(E_F)
\eeq
The tunneling current is then, roughly, 
\beqa
	I \propto \ P\int d\ep f(\ep - eV)N_r(E_F)N_l(\ep )(1 - f(\ep )) 	
		\nonumber \\
	- P\int d\ep f(\ep )N_l(\ep )N_r(E_F)(1 - f(\ep - eV))
\eeqa
\begin{figure}[htb]
\centerline{\includegraphics[height=2.25in,keepaspectratio,clip=true]{tunneling2.pdf}}
	\caption[]{\em{If eV$=0$, but there is a small overlap of occupied
and unoccupied states on the left and right sides, then there still will
be no current due to a balance of particle hopping.}}
\label{fig:tunneling2}
\end{figure} 
For $eV = 0$, clearly $I = 0$ i.e. a balance is achieved. \ For $eV \neq \ 0$ a 
current may occur.  Let's assume that $eV > 0$ and $k_BT \ll \ \De $. \ Then 
the rightward motion of electrons is suppressed. \ Then 
\beq
	I \sim \ PN_r(E_F) \int d\ep f(\ep - eV)N_l(\ep )
\eeq
and 
\beq
	\frac{dI}{dV} \sim \ PN_r(E_F) \int d\ep \pde{f(\ep - eV)}VN_l(\ep )
\eeq
\beq
	\pde fV \sim \ e\de (\ep - eV - E_F) \qquad (T \ll \ E_F)
\eeq
\beq
	\frac{dI}{dV} \simeq \ PN_r(E_F) N_l(eV + E_F)
\eeq
Thus the low temperature differential conductance $\frac{dI}{dV}$ is a measure 
of the superconducting density of states. 
\begin{figure}[htb]
\centerline{\includegraphics[height=2.0in,keepaspectratio,clip=true]{graphs.pdf}}
\caption[]{\em{At low temperatures, the differential conductance in a normal
metal--superconductor tunnel junction is a measure of the quasiparticle 
density of states.}}
\end{figure}   

\section{Unconventional Superconductors}
From different review articles, it is clear that there are different definitions of unconventional
superconductivity.  For example, some define it as beyond BCS, or to only include supererconductors,
such as odd-frequency superconductors, that clearly cannot be described by the BCS equations
(although it may be possible to describe them with the Eliashberg equations).  Another definition,
which I will adopt, is to define unconventional superconductors that are not described by the simple
discussion of the BCS equations that we have discussed so far in this chapter.  These will include
(and will be finined below) triplet, lower symmetry (e.g., d-wave) of often magnetically mediated
supererconductors, to very unusual ideas, that have not yet been observed, such as odd-frequency
superconductors.  In each case, I will assume that the superconducting state is formed via 
Cooper pairing of fermions, so that the order parameter must remain odd under a product of 
symmetry operations, such as parity, time reversal, spin, etc. as summarized in Tab.~\ref{table:CPT}.

\begin{table}[h!]
\centering
\begin{tabular}{||c c c c||} 
 \hline
 Type & spin symmetry & inversion symmetry & time reversal symmetry \\ [0.5ex] 
 \hline\hline
 odd-frequency         & triplet (even)     & even         & odd \\ 
 odd-frequency         & singlet (odd)      & odd (p-wave) & odd \\ 
 d-wave  S=0 L=2       & singlet (odd)      & even         & even \\
 triplet S=1 L=1      & triplet (even)     & odd          & even \\ [1ex] 
 \hline
\end{tabular}
\caption{Types and characteristics of the order parameter of unconventional 
superconductors formed from electron pairs.}
\label{table:CPT}
\end{table}

Below, we will discuss each of these unconventional superconductors, and identify their 
properties and experimental signatures.  

\subsection{D-wave Superconductors}

Cuprate superconductors which are nearly antiferromagnetic, and so highly anisotropic
that they may be viewed as nearly two-dimensional.  
\begin{wrapfigure}{l}{0.4\textwidth}
\includegraphics[width=0.39\textwidth,clip=true]{./UnconventionalSuperconductivity/HubbFermi.pdf}
\protect\caption{\emph{The Fermi surface of the 2D non-interacting half-filled Hubbard 
model with near-neighbot hopping.  The pairing interactions due to the magentic correlations 
are strongest at half filling and on the Fermi survace near the magnetic order vectors 
$(\pm\pi.\pm\pi)$.  Since they are purely repulsive, the order parameter, which is largest 
in between the ording vectors (i.e., $(0,\pi)$) must change sign like a d0-wave orbital 
(see text).\label{fig:HubbFermi}
}}
\end{wrapfigure}
In fact the modeling of Zhang and Rice reduces the cuprates to a 2D t-J or Hubbard model, 
which are related by a Schrieffer-Wolf transformation, and the models are close to half 
filling where the non-interacting fermi surface forms a square as illustrated in 
Fig.~\ref{fig:HubbFermi} and strong antiferromagnetic correlations begin to form.  The 
latter are believed to provide the pairing interaction, despite the fact that the 
pairing interaction from antiferromagnetic correlations is purely repulsive.


Whereas conventional s-wave superconductors form spin singlet pairs with s-wave symmetry
(S=0, L=0), d-wave superconductors form lower symmetry pairs (S=0, L=2).  This pairing 
may be described by the BCS formalism with a k-dependent $\Delta(\k)$.  We assume that 
the pairing interaction is strongly peaked at the antiferromagnetic ordering 
wavenumbers $\k=(\pm\pi,\pm\pi)$, or $V(\k)= -V_0 \delta(\k-(\pm\pi,\pm\pi))$ where 
$V_0 >0$ and is only finite at energies near the fermi surface.  Since the pairing is 
due to magnetic correlations, the width of the scattering shell is now roughly $J$
which is assumed to be $J \ll E_F$.  In this case the gap equation becomes
\beq
\De(\k') \ = \frac{1}{2L^3} 
\sum _k \frac{\De(\k) V(\k-\k')}
{\sqrt{\xi _k^2 + \De(\k)^2}}
\eeq
In order to have a solution, the minus sign in $V$ must be canceled.  The large contributions
to the sum comes when $\k-\k'$ is a magnetic ordering vector where $V$ is large.  Suppose 
$\k-\k'=(\pi,\pi)$, and $\k=(\pi,0)$ and $\k'=(0,-\pi)$, then to cancel the minus sign, we need 
$\De(\k)=-\De(\k')$, so that the order parameter changes sign for every rotation by
$\pi/2$ and presumably is zero along the diagonal, just like a $d_{x^2-y^2}$ orbital.
Hence the name d-wave superconductivity.

Note that the lower symmetry of the order parameter has experimental consequences.  
First the d-wave geometry of the order parameter may be directly confirmed by
creating a tunnel junction with a conventional s-wave superconductor, as illustrated
in Fig.~\ref{fig:corner}.  
\begin{wrapfigure}{r}{0.4\textwidth}
\includegraphics[width=0.39\textwidth,clip=true]{./UnconventionalSuperconductivity/corner_junction.png}
\protect\caption{\emph{Cartoon of a corner junction between a conbentional s-wave
and a d-wave, cuprate, superconductor.\label{fig:corner}
}}
\end{wrapfigure}
Here, the phase difference across one of the s-d junctions
causes a persistant current and a trapped magnetic flux measurable by a SQUID.
d-wave superconductivity is very sensitive to disorder, such as Zn doping for Cu 
where only a few percent of impurities can destroy superconductivity.  The reason 
for this sensitivity is that the elastic scattering from the Zn impurities is nearly 
local, and hence mixes $\Delta(\k)$ with all other $\k$ values on the fermi surface, 
and when averaged over the fermi surface, the order parameter is zero.  Furthermore
since the gap has a range of values extending to zero, so do the excitations across 
the gap.  As a result, the activated T-dependence seen in the specific heat is 
replaced by algebraic or power-law T-dependence seen in the nuclear magnetic resonance 
relaxation rate and specific heat.



\subsection{Triplet Superconductors}
As illustrated in Tab.~\ref{table:CPT}, triplet superconductivity is also possible
which is even in spin or odd in orbital symmetry.  More complex triplet states are
also possbile, but will not be discussed here.  Triplet superconductivity may actually
be an old subject if it include condensation of $^3$He which is spin 1/2 and forms 
a triplet condensate which is not a superconductor since $^3$He carries no charge.
as (S=1, L=1).  The pairing is believed to be mediated by magnetic fluctuations
enhanced by the proximity to a ferromagnetic transition (similar to the case for the 
cuprates where the magnetic fluctuations are enhanced by proximity to half filling).
It is believed that the triplet state is favored by the exchange hole that keeps 
the pair of electrons apart, avoiding the short ranged repulsive interaction between
them.  

In addition to $^3$He triplet superconductivity, with a real supercurrent, is believed
to exist in a number of solids, including Sr$_2$RuO$_4$ which is believed to have
a chiral p-wave state and a complex gap function which breaks time-reversal so
that the pairs have a finite magnetic moment.  

This triplet state should have a number of experimental consequences.  Perhaps the most
obvious is that, like the d-wave superconductors, the pairing should be very sensitive
to diorder, at least disorder with a mean-free path that is shorter than the pairing
length.  I.e., the stronger the pairing, the less sensitive the state is do disorder.

\subsection{Odd-frequency Superconductors}

Another, not yet observed (to the best of my knowledge) type of pairing is odd in
frequency or in time.  In this case both the spin and orbital part of the pairing can 
be even or both odd.  Of course, it is difficult to treat such a state in the BCS 
formalism since the frequency-dependence of the order parameter is suppressed.

\edo
