To understand the origin and consequences of pairing correlation, consider a pair of electrons (2\(e^-\)) interacting. The other electrons (\(N-2\) of them) are noninteracting, and form the Fermi sea. We assume that we have a translationally invariant system and spin-independent interactions. The goal of this section is to derive a rewriting of the Schrödinger equation in momentum space, which will allow us to show that, to form a bound state, an attractive interaction is required in at least one angular momentum channel.

To start, we consider the Hamiltonian for these two electrons and a two electron wave function:

\[\begin{equation} \Hhat\ket{\psi} = (\Hhat_0 + \Vhat)\ket{\psi} \end{equation}\]

The orbital wave function and spin wave function are both included in \(\ket{\psi}\). The Schrödinger equation for this situation is given by

\[\begin{equation} -\frac{\hbar^2}{2m}(\nabla_1^2 + \nabla_2^2)\Psi(\x_1,\x_2,\sigma_1,\sigma_2) + (V(\x_1-\x_2) - E)\Psi(\x_1,\x_2,\sigma_1,\sigma_2) = 0 \end{equation}\]

where \(\Psi(\x_1,\x_2,\sigma_1,\sigma_2) = -\Psi(\x_2,\x_1,\sigma_2,\sigma_1)\), as is required by fermion statistics. Because the potential only depends on the distance between the two electrons, we introduce the relative and centre of mass coordinates

\[\begin{equation} \x = \x_1 - \x_2,\quad \mathbf{X} = \frac{1}{2}(\x_1+\x_2) \end{equation}\]

which defines the new masses \(M = 2m\) (total mass) and \(\mu = m/2\) (effective mass). Then, we have the new Schrödinger equation

\[\begin{equation} -\frac{\hbar^2}{2}\left(\frac{\nabla_{\mathbf{X}}^2}{M} + \frac{\nabla_{\x}^2}{\mu}\right)\Psi(\mathbf{X},\x,\sigma_1,\sigma_2) + (V(\r) - E)\Psi(\boldsymbol{R},\r,\sigma_1,\sigma_2) = 0 \end{equation}\]

The new condition on \(\Psi\), in terms of these new variables, is \(\Psi(\mathbf{X},\x,\sigma_1,\sigma_2) = -\Psi(\mathbf{X},-\x,\sigma_2,\sigma_1)\). We see that the potential is independent of \(\mathbf{X}\), which means that the \(\hat{\mathbf{P}}\) operator commutes with the Hamiltonian. Therefore, we can separate the wave function into the centre of mass motion part, the relative part, and the spin part:

\[\begin{equation} \psi(\mathbf{X},\x,\sigma_1,\sigma_2) = e^{i\K\cdot\mathbf{X}}\psi(\x)\chi_{\sigma_1\sigma_2}. \end{equation}\]

The next step is to substitute this simplification into the Schrödinger equation, which yields

\[\begin{equation} -\frac{\hbar^2}{2}\left(-\frac{\K^2}{M} + \frac{\nabla_{\r}^2}{\mu}\right)\psi(\x)\chi_{\sigma_1\sigma_2} + (V(\x) - E)\psi(\x)\chi_{\sigma_1\sigma_2} = 0. \end{equation}\]

We define \(\tilde{E} = E - \frac{\hbar^2\K^2}{2M}\) to be the energy shifted by the centre of mass kinetic energy. We see that \(\mathbf{K} = 0\) minimizes the energy, so we take it to be zero. Then, we get

\[\begin{equation} -\frac{\hbar^2}{2\mu}\nabla_{\x}^2\phi(\x)\chi_{\sigma_1\sigma_2} + (V(\x) - \tilde{E})\phi(\r)\chi_{\sigma_1\sigma_2} = 0 \end{equation}\]

Because the potential is independent of spin, we can factor it out. Thus, we get the spatial Schrödinger equation for only the relative wave function:

\[\begin{equation} -\frac{\hbar^2}{2\mu}\nabla_{\x}^2\phi(\x) + (V(\x) - \tilde{E})\phi(\x) = 0. \end{equation}\]

Since we are working in a large box with periodic boundary conditions, the Fourier transform of the wave function is given by

\[\begin{equation} \phi(\x) = \frac{1}{\sqrt{\V}}\sum_{\k} e^{i\k\cdot\x} \phi_{\k} \end{equation}\]

which gives us

\[\begin{equation} \frac{1}{\sqrt{\V}}\sum_{\k}e^{i\k\cdot\x}\left[\frac{\hbar^2\k^2}{2\mu}\phi_{\k} + (V(\x) - \tilde{E})\phi_{\k}\right] = 0 \end{equation}\]

Then, multiplying by \(e^{-i\k’\cdot\x}\) on both sides and integrating, we get

\[\begin{align} \int \sum_{\k}e^{i(\k-\k')\cdot\x}\left[\frac{\hbar^2\k^2}{2\mu}\phi_{\k} + (V(\x) - \tilde{E})\phi_{\k}\right] \d^3\x = 0 \\ \sum_{\k}\left[2\epsilon_{\k}\phi_{\k}\V\delta_{\k\k'} + (V(\k'-\k) - \tilde{E}\V\delta_{\k\k'})\phi_{\k}\right] = 0 \end{align}\]

where \(\V\) is the volume of the system, and we have defined

\[\begin{equation} V(\k'-\k) = \int e^{i(\k-\k')\cdot\x} V(\x) \d^3\x \end{equation}\]

The equation can then be rewritten by swapping \(\k,\k’\) to get

\[\begin{align} 2\epsilon_{\k}\phi_{\k} + \frac{1}{\V}\sum_{\k'}V(\k-\k')\phi_{\k'} - \tilde{E}\phi_{\k} ={}& 0 \\ (2\epsilon_{\k} - \tilde{E})\phi_{\k} ={}& -\frac{1}{\V}\sum_{\k'} V(\k-\k') \phi_{\k'} \end{align}\]

Based on the definition of \(V(\k-\k’)\), we observe that it can only depend on the lengths of \(\k,\k’\), as well as the angle between them. We can expand it in partial waves by recalling the formula

\[\begin{equation} e^{ikr\cos\theta} = \sum_{\ell=0}^\infty i^\ell (2\ell+1)j_\ell(kr) P_\ell(\cos\theta) \end{equation}\]

where \(j_\ell\) is the spherical Bessel function of order \(\ell\) and \(P_\ell\) is the Legendre polynomial of order \(\ell\). Then, if we orient the \(\x\) coordinate system so that its \(z\) axis lies along \(\k\), then we have that \(\k = k\zhat\). Thus, the above identity applies identically. For the \(e^{i\k’\cdot\x}\), this is a bit more difficult. We of course have \(\x = r(\cos\varphi\sin\theta,\sin\varphi\sin\theta,\cos\theta)\), and we also write \(\k’\) in spherical coordinates as \(\k’ = k’(\cos\varphi’\sin\theta’,\sin\varphi’\sin\theta’,\cos\theta’)\). Then, we have that

\[\begin{equation} \x\cdot\k' = k'r[\cos(\varphi-\varphi') \sin\theta\sin\theta' + \cos\theta\cos\theta'] \end{equation}\]

This means that

\[\begin{align} e^{-i\k'\cdot\x} ={}& e^{-ik'r(\cos(\varphi-\varphi') \sin\theta\sin\theta' + \cos\theta\cos\theta')} \\ ={}& \sum_{\ell'=0}^\infty (-i)^{\ell'}(2\ell'+1) j_{\ell'}(k'r) P_{\ell'}(\cos(\varphi-\varphi') \sin\theta\sin\theta' + \cos\theta\cos\theta') \end{align}\]

Substituting this in to the potential integral, we get

\[\begin{align} V(\k' - \k) ={}& \sum_{\ell,\ell'=0}^\infty \int_0^{2\pi} \int_0^\pi \int_0^\infty i^\ell (2\ell+1)j_\ell(kr) P_\ell(\cos\theta) \\ &\times (-i)^{\ell'}(2\ell'+1) j_{\ell'}(k'r) P_{\ell'}(\cos(\varphi-\varphi') \sin\theta\sin\theta' + \cos\theta\cos\theta') V(r) r^2\sin\theta \d r \d\theta \d\varphi \end{align}\]

We can then use the identity

\[\begin{equation} P_{\ell}(\cos(\varphi-\varphi') \sin\theta\sin\theta' + \cos\theta\cos\theta') = \sum_{m=-\ell}^\ell (-1)^m P^{-m}_\ell(\cos\theta) P^m_n(\cos\theta') \cos(m(\varphi-\varphi')) \end{equation}\]

and then, since

\[\begin{equation} \int_0^{2\pi} \cos(m\varphi) \d\varphi = 2\pi \delta_{m,0} \end{equation}\]

we get that

\[\begin{align} \int_0^{2\pi} P_{\ell}(\cos(\varphi-\varphi') \sin\theta\sin\theta' + \cos\theta\cos\theta') \d\varphi ={}& \sum_{m=-\ell}^\ell (-1)^m P^{-m}_\ell(\cos\theta) P^m_n(\cos\theta') \int_0^{2\pi} \cos(m(\varphi-\varphi')) \d\varphi \\ ={}& 2\pi\sum_{m=-\ell}^\ell (-1)^m P^{-m}_\ell(\cos\theta_1) P^m_n(\cos\theta_2) \delta_{m,0} \\ ={}& 2\pi P^0_\ell(\cos\theta)P^0_\ell(\cos\theta') \\ ={}& 2\pi P_\ell(\cos\theta)P_\ell(\cos\theta') \end{align}\]

Thus, we have a vast simplification, and find that

\[\begin{align} V(\k' - \k) ={}& 2\pi \sum_{\ell,\ell'=0}^\infty \int_0^\pi \int_0^\infty i^\ell (2\ell+1)j_\ell(kr) P_\ell(\cos\theta) (-i)^{\ell'}(2\ell'+1) j_{\ell'}(k'r) V(r) P_{\ell'}(\cos\theta)P_{\ell'}(\cos\theta') r^2\sin\theta \d r \d\theta \end{align}\]

Then, through the orthogonality relation of Legendre polynomials, which is

\[\begin{equation} \int_0^\pi P_\ell(\cos\theta) P_{\ell'}(\cos\theta)\sin\theta \d\theta = \frac{2\delta_{\ell\ell'}}{2\ell+1} \end{equation}\]

we find that

\[\begin{align} V(\k' - \k) ={}& \sum_{\ell=0}^\infty \int_0^\infty (2\ell+1)j_\ell(kr) j_{\ell}(k'r) V(r) P_{\ell}(\cos\theta') r^2 \d r \end{align}\]

Since \(\cos\theta’\) is the angle between \(\k\) and \(\k’\), we write \(\cos\theta’ = \khat\cdot\khat’\), and thus

\[\begin{align} V(\k' - \k) ={}& \sum_{\ell=0}^\infty (2\ell+1) P_{\ell}(\khat\cdot\khat') V_\ell(k,k') \end{align}\]

whereby the partial wave component is given by

\[\begin{equation} V_\ell(k,k') = 4\pi \int_0^\infty j_\ell(kr) j_{\ell}(k'r) V(r) r^2 \d r \end{equation}\]

Now that we have expanded the potential in partial waves, we can do the same thing for the wave function. The wave function can be written as

\[\begin{equation} \phi(\k) = \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell \phi_{\ell m}(k) Y^m_\ell(\khat). \end{equation}\]

Plugging in the expanded form of the wave function as well as the expanded form of the potential, we find

\[\begin{align} (2\epsilon_{\k} - \tilde{E})\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell \phi_{\ell m}(k) Y^m_\ell(\khat) = -\frac{1}{\V}\sum_{\k'}\sum_{\ell=0}^\infty (2\ell+1)V_{\ell}(k,k')P_{\ell}(\cos\theta)\sum_{\ell'=0}^\infty\sum_{m'=-\ell'}^{\ell'} \phi_{\ell' m'}(k') Y^{m'}_{\ell'}(\khat') \end{align}\]

Then, we plug in the addition theorem for spherical harmonics:

\[\begin{equation} P_\ell(\cos(\khat\cdot\khat')) = \frac{4\pi}{2\ell+1}\sum_{m=-\ell}^\ell Y^m_\ell(\khat)[Y^m_\ell]^*(\khat') \end{equation}\]

which gives us

\[\begin{align} (2\epsilon_{\k} - \tilde{E})\sum_{\ell m} \phi_{\ell m}(k) Y^m_\ell(\khat) = -\frac{1}{\V}\sum_{\k'}\sum_{\ell m \ell' m'} 4\pi V_{\ell}(k,k') Y^m_\ell(\khat) [Y^m_\ell]^*(\khat') \phi_{\ell' m'}(k') Y^{m'}_{\ell'}(\khat') \end{align}\]

The next step is to evaluate the sum (integral) over the angular part of \(\k’\), which gives us (using \(\int Y^{m*}_\ell(\khat’) Y^{m’}_{\ell’}(\khat’)\d\Omega’ = \delta_{\ell\ell’}\delta_{mm’}\))

\[\begin{align} (2\epsilon_{\k} - \tilde{E})\sum_{\ell m} \phi_{\ell m}(k) Y^m_\ell(\khat) ={}& -\frac{1}{(2\pi)^3}\int \sum_{\ell m \ell' m'} 4\pi V_{\ell}(k,k') Y^m_\ell(\khat) [Y^m_\ell]^*(\khat') \phi_{\ell' m'}(k') Y^{m'}_{\ell'}(\khat') k^{\prime 2} \d k' \d\Omega' \\ ={}& -\frac{(4\pi)}{(2\pi)^3}\int \sum_{\ell m \ell' m'} \delta_{\ell\ell'}\delta_{mm'} V_{\ell}(k,k') Y^m_\ell(\khat) \phi_{\ell' m'}(k') k^{\prime 2} \d k' \\ ={}& -\frac{(4\pi)}{(2\pi)^3}\int \sum_{\ell m} V_{\ell}(k,k') Y^m_\ell(\khat) \phi_{\ell m}(k') k^{\prime 2} \d k' \end{align}\]

Thus, we find that, taking the components of each side to be equal to one another (equivalent to multiplying both sides by \(Y^{m’}_{\ell’}(\khat)\) and then integrating over \(\d\Omega\)) gives us

\[\begin{equation} (2\epsilon_k - \tilde{E})\phi_{\ell m}(k) = -\frac{1}{(2\pi)^3}\int V_\ell(k,k') \phi_{\ell m}(k') (4\pi) k^{\prime 2} \d k' \end{equation}\]

The point of this, is that if we want a bound state for the system, then the energy eigenvalue \(\tilde{E}\) should satisfy \(\tilde{E} < 2\epsilon_F\). This is because the energy of the bound state must be less than just the kinetic energies of the two electrons comprising it. We see from this that if \(\tilde{E}<2\epsilon_F\), then this implies that \(V_\ell < 0\) when \(\phi_{\ell m} \neq 0\). Thus, a bound state requires an effective attraction.