1. Consider the two-particle Hamiltonian: $$\begin{equation} \Hhat = -\frac{\hbar^2}{2m_e}\nabla_{\r_1}^2 - \frac{\hbar^2}{2m_e}\nabla_{\r_2}^2 - \left[2\mathbf{S}_1 \cdot \mathbf{S}_2 + \frac{1}{2}\right] V(\r_1 - \r_2). \end{equation}$$
    1. Show that $$\begin{equation} (2\epsilon_{\k} - \tilde{E})\phi_{\ell m}(k) = -\frac{(-1)^\ell}{(2\pi)^3}\int V_\ell(k,k') \phi_{\ell m}(k') 4\pi k^{\prime 2} \d k' \end{equation}$$ where \(V_\ell(k,k')\) is defined as it was in the previous section. Remember what the total spin should be for an even parity (singlet) vs. odd parity (triplet) wave function. What condition on \(V_\ell(k_F,k_F)\) is required such that a bound state exists? (a bound state will have energy \(\tilde{E} < 2\epsilon_F\) because the energy must be lower than the kinetic energy of the 2 electrons comprising the pair)
    2. Next, show that, for a nontrivial solution for \(\phi_{\ell m}\) to exist, we require $$\begin{equation} 1 = -(-1)^\ell\int \frac{V_\ell(k,k')}{2\epsilon_{\k'} - \tilde{E}} \frac{4\pi k^{\prime 2} \d k'}{(2\pi)^3} \end{equation}$$ Recall that \(\phi_{\ell m}(k) = 0\) for \(k < k_F\), due to the Fermi surface.
    3. Defining \(\Delta = 2\epsilon_F - \tilde{E}\), and assuming a high-energy cutoff \(D\), show that there is a bound state with binding energy $$\begin{equation} \Delta \approx 2D e^{-2/\lambda_\ell \rho} \end{equation}$$ where \(\rho\) is the density of states at the Fermi energy, and \(\lambda_\ell = -(-1)^\ell V_\ell(k,k')\) is the approximate value of the potential in the \(\ell\) angular momentum channel on the Fermi surface (\(k,k'=k_F\)).
    4. Suppose the potential is given by \(V(r) = \alpha\frac{j_1(2k_Fr)}{k_Fr^2}\), where \(\alpha\) is a positive constant, and \(j_1(x)\) is the spherical Bessel function of order 1: $$\begin{equation} j_1(x) = \frac{\sin x}{x^2} - \frac{\cos x}{x} \end{equation}$$ If this potential seems very weird (e.g. how can we have an oscillating potential?), please recall that an oscillating potential is possible due to electron-electron overscreening (this was called Friedel oscillations). For \(\ell=0,1,2,3\) plot \((-1)^\ell V_\ell(k,k')\) for \(0 < k,k' < 2k_F\). You should write a short program to do this (these integrals are possible to calculate analytically, but this is a waste of your time). For \(k,k'\) on the Fermi surface, what can you say about the sign of \((-1)^\ell V_\ell(k,k')\) as a function of \(\ell\)? Which value of \(\ell\) gives the leading attractive interaction? Does this yield a spin-singlet or spin-triplet solution?
  2. Show that we still need a threshold if we have no Fermi surface
    1. If we instead had a lower bound of 0 instead of \(\epsilon_F\), then if \(\rho(\epsilon) = c\sqrt{\epsilon}\), the integral would be $$\begin{align} 1 ={}& N_sV_\ell c \int_{\epsilon_L}^{\epsilon_F + D} \frac{\sqrt{\epsilon}}{2\epsilon - \tilde{E}} \d\epsilon \\ ={}& N_s V_\ell c \left(\sqrt{\epsilon} + \frac{\sqrt{\tilde{E}}}{2\sqrt{2}}\log\left|\frac{\sqrt{2\epsilon} - \sqrt{\tilde{E}}}{\sqrt{2\epsilon} + \sqrt{\tilde{E}}}\right| \right)\bigg|_{\epsilon_L}^{\epsilon_F+D} \\ ={}& N_s V_\ell c \left(\sqrt{\epsilon_F+D} + \frac{\sqrt{\tilde{E}}}{2\sqrt{2}}\log\left|\frac{\sqrt{2\epsilon_F+2D} - \sqrt{\tilde{E}}}{\sqrt{2\epsilon_F+2D} + \sqrt{\tilde{E}}}\right| - \sqrt{\epsilon_L} - \frac{\sqrt{\tilde{E}}}{2\sqrt{2}}\log\left|\frac{\sqrt{2\epsilon_L} - \sqrt{\tilde{E}}}{\sqrt{2\epsilon_L} + \sqrt{\tilde{E}}}\right|\right) \end{align}$$ If we look now for a solution \(\tilde{E} < 2\epsilon_F\), one can see by plotting software (define \(z = \tilde{E}/2\epsilon_F\) and \(a = \sqrt{1 + D/\epsilon_F} > 1\)) that we need to solve the equation $$\begin{align} 1 ={}& N_s V_\ell c \sqrt{\epsilon_F}\left(a + \frac{\sqrt{z}}{2}\log\left|\frac{a - \sqrt{z}}{a + \sqrt{z}}\right| \right) \end{align}$$ If we look for a solution for \(z < 1\), (which is equivalent to saying \(\tilde{E} < 2\epsilon_F\))