ass5: Is this 17?

superconductivity_assignment5_kvkempen.tex

@@ -58,12 +58,71 @@ with $V$ Cooper's approximate potential and $g(\epsilon_F)$ the density of state
 A thorough discussion can be found in Annett's book \cite[chapter 6]{annett} and in the slides of week 6 of this course.
 The binding energy of the Cooper pairs (i.e. the energy gain of forming these pairs) is
 \[
-	-E = 2\hbar\omega_De^{-1/\lambda}.
+	 -E = 2\hbar\omega_De^{-1/\lambda} =: \Delta_0,
 \]
+which is also called the gap parameter $\Delta_0$ at zero temperature for BCS.
 In the weak coupling limit of the BCS theory, the case we have considered so far, it is assumed that $\lambda << 1$.
+It should be noted that this weak limit also means that the gap is smaller than the thermal energy of the highest excited energy phonon, which corresponds to the Debye temperature
+\[
+	\Delta < k_B\Theta_D.
+\]
 It is when this assumption breaks down, BCS does not work and we find an upper limit to the critical temperature $T_c$.
-We will look at a way to express the critical temperature in terms we can derive, and than look at the values that maximize this critical temperature whilst still following BCS theory.
+We will look at a way to express the critical temperature in terms we can derive, and then look at the values that maximize this critical temperature whilst still following BCS theory.
 
+From the derivation of the BCS coherent state, this gap parameter at finite temperature is found.
+There is a temperature dependence $\Delta(T)$ as in figure \ref{fig:gap-T}.
+\begin{figure}
+	\centering
+	\includegraphics[width=.4\textwidth]{Lecture-7-slides-for-printing-slide-13-gap-parameter.pdf}
+	\caption{By taking the gap parameter to zero, we find the critical temperature. Figure from the slides of lecture 7.}
+	\label{fig:gap-T}
+\end{figure}
+For larger temperatures, thermal energy is increased, and less energy is required to break up Cooper pairs, thus degrading the superconductivity.
+This puts a limit $T_c$.
+
+We will mostly follow the derivation by Waldram \cite[paragraph 7.9, mostly p.128--130]{waldram}.
+The superconducting state breaks down at high temperature, at which also $\Delta$ vanishes so that the gap parameter is a good order parameter for the state.
+
+Let's consider the gap parameter
+\[
+	\Delta_{\vec{k}} = -\sum_{\vec{k'}}(1-2f_{\vec{k'}})u_{\vec{k'}}v_{\vec{k}}V_{\vec{k'}\vec{k}},
+\]
+with $u$ and $v$ occupation functions for the BCS state, $f$ the Fermi occupation number, and $V$ the potential between the states.
+Minimizing $\Delta_{\vec{k}}$ and taking that $V_{\vec{k'}\vec{k}} = -V$ is constant gives us a self-consistent relation for the gap parameter.
+We also recognize that the states that we sum over all all those states such that they have smaller energy than the highest excited phonon.
+\[
+	\Delta_{\vec{k}} = V\sum_{\epsilon_{\vec{k'}}}(1-2f_{\vec{k'}})\frac{\Delta_{\vec{k'}}}{2E_{\vec{k'}}}.
+\]
+Now the right-hand side is independent of $\vec{k}$ but does contain $\Delta_{\vec{k'}}$.
+We can thus conclude that the gap parameter should be constant over all states $\vec{k}$!
+That means we can divide both sides by it, giving us
+\[
+	1 = V\sum_{\epsilon_{\vec{k'}}}(1-2f_{\vec{k'}})\frac{1}{2E_{\vec{k'}}}.
+\]
+Converting the equation to an integral, and substituting in $f(E) = [\exp{(E/(k_BT))}+1]^{-1}$ and $E = \sqrt{\epsilon^2 + \Delta(T)^2}$ yields
+\[
+	1 = 2g(\epsilon_F)V\int_0^{k_B\Theta_D}\frac{1-2[\exp{(E/(k_BT))}+1]^{-1}}{2\sqrt{\epsilon^2 + \Delta(T)^2}} \textup{d}\epsilon.
+\]
+
+I believe Waldram that one could find that
+\[
+	T_c = 1.14\Theta_D\exp{(-1/(g(\epsilon_F)V))} = 1.14\Theta_D\exp{(-1/(\lambda)}
+\]
+from this nice equation.
+
+As limiting value, we take $\lambda = 0.3$, as was posed as a reasonable limit for the weak coupling by Alix in lecture 7,
+although Waldram \cite{waldram} thinks it is more like $\lambda \approx 0.4$.
+
+For metals, Waldram thinks $\Theta_D \leq \SI{300}{\kelvin}$ is a good limit.
+This leads to our final maximum
+\[
+	T_c \leq 1.14 \cdot 300 \cdot \exp{(-1/0.4)} \approx \SI{28}{\kelvin}.
+	%https://www.wolframalpha.com/input?i=1.14*300*e%5E%28-1%2F.4%29
+\]
+(Using $\lambda = 0.3$ yields an even lower $T_c \leq \SI{12}{\kelvin}$.)
+
+For larger $T_c$ values, larger binding of Cooper pairs would be needed to overcome the thermal energy.
+This means our assumption of weak coupling breaks down, making most of the derivation invalid without further arguments.
 
 \section{Energy gap $\Delta$ et al.}