The BCS Theory
Each quantum system has its ground state, which is pccupied at $T=0$, and a set of excited levels. The idea of quasiparticles if that the lowest energy level of a (uniform) macroscopic system are reasonable approximated by plane waves $$\begin{align}\psi_{\mathbf{k}\alpha}(\mathbf{r})\propto \exp (i\mathbf{k}\mathbf{r}-i\left( \epsilon_{\mathbf{k}\alpha}/\hbar)t\right)\end{align}$$ with quasiparticle wave vectir $\mathbf{k}$ and other quantum number $\alpha$.
If $\psi_{\mathbf{k}\alpha}$ is not exactly an eigenfunction of the Hamiltonian, then the plane wave will transform to other waves with time (quasiparticle decay) or if one starts with combination of quasiparticles then they will transform to other combinations (quasiparticle scattering). For the quasiparticle picture to be useful such processes should be sufficiently slow, so that the transformation time, or lifetime of a quasiparticle $\tau_{life}\gg \hbar/\epsilon_{\mathbf{k}\alpha}$.
1. Landau Fermi liquid
The ground state of a system of non-interacting fermions corresponds to the filled states with energies $E$ below the Fermi energy $E_F$ , determined by the number of fermions.
The ground state of the Fermi gas corresponds to the energy $E_0$. Excitations in the Fermi gas that increase its energy as compared to $E_0$ are created by moving a particle from a state below the Fermi surface to a state above it. This can be understand as two steps.
Here we put $E=p^2/2m$ as appropriate for free particles. Since $E_F = p^2_F /2m$ we have $\epsilon_p = 0$ at $p = p_F$ . As we can create quasiparticles with an arbitrary small energy, in thermal equilibrium they obey Fermi distribution with zero chemical potential $$\begin{align}f(\epsilon_\mathbf{p})=\frac{1}{e^{\epsilon_\mathbf{p}/k_BT}+1}\end{align}$$
Now let us switch on the interaction between particles. Landau supposed that in such Fermi liquid excitations have the same type of spectrum as in the gas. $m$ becomes a material parameter, which is called the effective mass $m^∗$.
We can approximate the spectrun (2) as $$\begin{align}\epsilon_\mathbf{p}=\frac{|p^2-p_F^2|}{2m^*}=\frac{|p-p_F|(p+p_F)}{2m^*}\approx v_F|p-p_F|\end{align}$$ where $v_F=p_F/m^*$ is a material parameter called Fermi velocity.
sum of momenta?
The quantity $N(\epsilon)$ is called the density of states: $N(\epsilon)d\epsilon$ is the number of states (per single spin direction) in the energy interval from $\epsilon$ to $\epsilon+d\epsilon$. In Landau Fermi-liquid, $N (\epsilon_\mathbf{p})$ can be replaced with its value at $\epsilon_\mathbf{p} = 0$ (i.e. $p = p_F$ ) $$\begin{align}N(0)=\frac{m^* p_F}{2\pi^2 \hbar^3}\end{align}$$ and taken out of the integral: $$\begin{align}\sum_\mathbf{p} a(\mathbf{p})=N(0)\int a(\epsilon_\mathbf{p})d\epsilon_\mathbf{p},\quad \sum_{\mathbf{p}\sigma} a(\mathbf{p})=2N(0)\int a(\epsilon_\mathbf{p})d\epsilon_\mathbf{p},\end{align}$$ The constant density of states is one of the most important features of Landau Fermi-liquid, which determines many of its physical properties.
Note that the care should be taken when the summation covers both particle and hole excitations: $$\begin{align}\sum_{\mathbf{p},\epsilon_\mathbf{p}<E_c}a(\mathbf{p})=N(0)\int_{\epsilon_\mathbf{p}<E_c}a(\epsilon_\mathbf{p})d\epsilon_\mathbf{p}=2N(0)\int^{E_c}_0a(\epsilon(\mathbf{p})d\epsilon_\mathbf{p}.\end{align}$$ we assumed that $a(\epsilon_\mathbf{p})$ is the same for the particle and hole excitations.
2. Landau criterion
? 생략이요
3. Phonon-mediated electron attraction
Let consider a charge $Q$ at $\mathbf{r}=0$ in the cloud of electrons. The charge produces electrostatic potential $\varphi$ which changes electron density by $\delta n$. Poisson equation for the potential is $$\nabla \varphi=-4\pi [e\delta n + Q\delta (\mathbf{r})].$$ At low temperatures the change of occupation occurs only in the region of energies $e\varphi$ at the Fermi energy and $\delta n=-2N(0)e\varphi$. The equation for the potential becomes $$\nabla\varphi -8\pi e^2 N(0)\varphi=-4\pi Q\delta(\mathbf{r}).$$
The easiest way to solve this equation is Fourier trnsform. The answer is $$\begin{align}\varphi_\mathbf{k}=\frac{4\pi Q}{k^2+k_s^2},\quad \varphi(\mathbf{r})=\frac{Q}{r}\exp (-k_sr),\quad k_s=\sqrt{8\pi e^2 N(0)}.\end{align}$$ Thus Coulomb interaction becomes short-ranged in metals. We can esimate the range of the interaction as $$k^{-1}_s\sim \left(e^2\frac{n}{E_F}\right)^{-1/2}=\left(\frac{e^2n^{1/3}}{E_F}\right)^{-1/2}n^{-1/3}\sim n^{-1/3}=a_0.$$ Thus the interaction range is limited to the inter-electron (i.e. interatomic) distances.
전자의 움직임 -> 이온의 조화 진동($\omega_D$) -> 다른 전자와 상호작용
On the quantum-mechanical language one describes this process as a scattering of two electrons via exchange of a phonon. The momenta of electrons before interaction are $\mathbf{p}_1$ and $\mathbf{p}_2$ and after the interaction are $\mathbf{p}'_1$ and $\mathbf{p}'_2$, respectively. All initial and final states should be very close to the Fermi surface.
Case 1
Thus in the case of $\mathbf{p}_1\approx \mathbf{p}_2$ only limited number of final states with $\mathbf{p}'_1\approx \mathbf{p}_1$ and $\mathbf{p}'_2\approx \mathbf{p}_2$ are available.
Case 2
On the other hand if $\mathbf{p}_1\approx -\mathbf{p}_2$ we have $\mathbf{p}'_1\approx -\mathbf{p}'_2$. Thus essentially all Fermi surface is available for final states, the probability of scattering is drastically larger and only such processes are important.
Initial momenta $\mathbf{p}_1$ and $\mathbf{p}_2$ to go to the states with momenta $\mathbf{p}'_1=\mathbf{p}_1+\mathbf{q}$ and $\mathbf{p}'_2=\mathbf{p}_2-\mathbf{q}$ via exchange of a phonon. Owing to the energy conservation the energy of the initial state (I) is equal to that of the final state (II): $E_I=\epsilon_{\mathbf{p}_1}+\epsilon_{\mathbf{p}_2}=\epsilon_{\mathbf{p}'_1}+\epsilon_{\mathbf{p}'_2}=E_{II}$.
In intermediate states, phonon have momentum $\pm \mathbf{q}$ and energy $\hbar \omega_\mathbf{q}$. Since for phonons $\omega_{-\mathbf{q}}=\omega_\mathbf{q}$, the energies of the intermediate states are: $$E_1=\epsilon_{\mathbf{p}'_1}+\epsilon_{\mathbf{p}_2}+\hbar \omega_\mathbf{q},\quad E_2=\epsilon_{\mathbf{p}_1}+\epsilon_{\mathbf{p}'_2}+\hbar \omega_\mathbf{q}$$
In the second-order perturbation theory the matrix element connecting initial and final states is $$\begin{align}\langle II|H_{e-ph-e}|I\rangle\\ \nonumber = \frac{1}{2}\sum_{i=1,2}\langle II|H_{e-ph-e}|i\rangle \left(\frac{1}{E_{II}-E_i}+\frac{1}{E_I-E_i}\right)\langle Ii|H_{e-ph-e}|I\rangle\\ \nonumber =W^*_\mathbf{q}\left(\frac{1}{\epsilon_{\mathbf{p}'_1}-\epsilon_{\mathbf{p}_1}-\hbar\omega_\mathbf{q}}+\frac{1}{\epsilon_{\mathbf{p}'_2}-\epsilon_{\mathbf{p}_2}-\hbar\omega_\mathbf{q}}\right)W_\mathbf{q} \\ \nonumber =\frac{|W_\mathbf{q}|^2}{\hbar}\left(\frac{1}{\omega-\omega_\mathbf{q}}-\frac{1}{\omega+\omega_\mathbf{q}}\right)=\frac{2|W_\mathbf{q}|^2}{\hbar}\frac{\omega_\mathbf{q}}{\omega^2-\omega^2_\mathbf{q}}\end{align}$$ Here $W_\mathbf{q}$ is the matrix element for emission of the photon with momentum $\mathbf{q}$, while $\hbar\omega=\epsilon_{\mathbf{p}'_1}-\epsilon_{\mathbf{p}_1}=-(\epsilon_{\mathbf{p}'_2}-\epsilon_{\mathbf{p}_2})$.
Point 1
We find that when $|\omega|<\omega_\mathbf{q}$ the matrix element in Eq. (12) is negative, i.e. it corresponds to the attraction between electrons.
Point 2
Another important conclusion is that this attraction does not depend on the directions of $\mathbf{p}_1$, $\mathbf{p}_2$ and thus electrons interact in a state with the orbital momentum $L = 0$. This means that the orbital part of their wave function is symmetric with respect to the particle interchange.
singlet 어쩌고 그래서 up과 down이 만나고 어쩌고
4. The Cooper problem
Attractive -> bound state
3d space integral -> 1d energy (near Fermi surf.) integral 때문이라는데?
Interaction of quasiparticles with opposite momenta and spin: $\psi_{\mathbf{p}\uparrow}(\mathbf{r}_1)\propto e^{i\mathbf{p}\mathbf{r}_1/\hbar}$ and $\psi_{-\mathbf{p}\downarrow}(\mathbf{r}_2)\propto e^{-i\mathbf{p}\mathbf{r}_2/\hbar}$. Then pair wave function as a linear combination of single-particle wave functions $$\Psi(\mathbf{r}_1,\mathbf{r}_2)=\sum_\mathbf{p}c_\mathbf{p}\psi_{\mathbf{p}\uparrow}(\mathbf{r}_1)\psi_{-\mathbf{p}\downarrow}(\mathbf{r}_2)=\sum_\mathbf{p}a_\mathbf{p}e^{i\mathbf{p}\mathbf{r}/\hbar}=\Psi(\mathbf{r}).$$
The $a_\mathbf{p}$ are thus coefficients in Fourier transformation of $\Psi(\mathbf{r})$. The inverse transformation is $$a_\mathbf{p}=V^{-1}\int \Psi(\mathbf{r})e^{-i\mathbf{p}\mathbf{r}/\hbar}d^3\mathbf{r},$$ where $V$ is the volume of the system.
We take Fourier transform of both sides of Eq. (13), that is multiply by $e^{-i\mathbf{p}\mathbf{r}/\hbar}$ and integrate over the volume. Since $\hat{H}_e(\mathbf{r}_1)\psi_{\mathbf{p}\uparrow}(\mathbf{r}_1)=\epsilon_\mathbf{p}\psi_{\mathbf{p}\uparrow}(\mathbf{r}_1)$ and $\hat{H}_e(\mathbf{r}_2)\psi_{-\mathbf{p}\downarrow}(\mathbf{r}_2)=\epsilon_{-\mathbf{p}}\psi_{-\mathbf{p}\downarrow}(\mathbf{r}_2)=\epsilon_{\mathbf{p}}\psi_{-\mathbf{p}\downarrow}(\mathbf{r}_2)$ we obtain $$[\hat{H}_e(\mathbf{r}_1)\Psi(\mathbf{r}_1,\mathbf{r}_2)]_\mathbf{p}=[\hat{H}_e(\mathbf{r}_2)\Psi(\mathbf{r}_1,\mathbf{r}_2)]_\mathbf{p}=\epsilon_\mathbf{p}a_\mathbf{p}.$$
For the product $W(\mathbf{r})\Psi(\mathbf{r})$ we use the convolution theorem $$[W(\mathbf{r})\Psi(\mathbf{r})]_\mathbf{p}=\sum_{\mathbf{p}'}W_{\mathbf{p}-\mathbf{p}'}.$$ Here the Fourier transform of the potential is simultaneously the matrix element for the transition of the pair of particles from states $\mathbf{p}$ and $-\mathbf{p}$ to states $\mathbf{p}'$ and $-\mathbf{p}'$ :$$W_{\mathbf{p}-\mathbf{p}'}=V^{-1}\int W(\mathbf{r})e^{-i(\mathbf{p}-\mathbf{p}')\mathbf{r}/\hbar}d^3\mathbf{r}=\langle e^{i\mathbf{p}'\mathbf{r}_1/\hbar}e^{-i\mathbf{p}'\mathbf{r}_2/\hbar}|W(\mathbf{r}_1,\mathbf{r}_2)|e^{i\mathbf{p}\mathbf{r}_1/\hbar}e^{-i\mathbf{p}\mathbf{r}_2/\hbar}\rangle=W_{\mathbf{p},\mathbf{p}'}.$$
The equation (13) becomes $$2\epsilon_\mathbf{p}a_\mathbf{p}+\sum_{\mathbf{p}'}W_{\mathbf{p},\mathbf{p}'}a_{\mathbf{p}'}=Ea_\mathbf{p}.$$ For the interaction model introduced at the end of the previoussection we write ???? $$\begin{align}W_{\mathbf{p},\mathbf{p}'}=\begin{cases}-W,& \epsilon_\mathbf{p} \text{ and } \epsilon_{\mathbf{p}'}<E_c,\text{ i.e. }p_F-E_c/v_F<p(\text{and }p')<p_F+E_c/v_F\\ 0,&\text{otherwise}\end{cases}\end{align}$$ where $E_c\ll E_F$. We thus have $$\begin{align}a_\mathbf{p}=-\frac{W}{E-2\epsilon_\mathbf{p}}\sum_{\mathbf{p}',\epsilon_{\mathbf{p}'}<E_c}a_{\mathbf{p}'}\end{align}$$ Let us denote $$C=\sum_{\mathbf{p}',\epsilon_{\mathbf{p}'}<E_c}a_{\mathbf{p}'}$$ Eq. (15) yields $$a_\mathbf{p}=-\frac{WC}{E-2\epsilon_\mathbf{p}}$$ whence $$C=-WC\sum_{\mathbf{p}',\epsilon_{\mathbf{p}'}<E_c}\frac{1}{E-2\epsilon_\mathbf{p}}$$ This gives $$\begin{align}-\frac{1}{W}=\sum_{\mathbf{p}',\epsilon_{\mathbf{p}'}<E_c}\frac{1}{E-2\epsilon_\mathbf{p}}\equiv \Phi(E)\end{align}$$ (슈뢰딩거 방정식에서 여기까지 온거임)
Equation (16) is illustrated in Fig. 2.5.
This implies that there is a bound state with $E = E_b < 0$ satisfying Eq. (16): $$\frac{1}{W}=\sum_{\mathbf{p},\epsilon_\mathbf{p}<E_c}\frac{1}{2\epsilon_\mathbf{p}-E_b}$$ ($E_b$는 결합 에너지!)
We replace the sum over momentum with the integral over energy (Eq. 8): $$\begin{align}\frac{1}{W}=2N(0)\int^{E_c}_0\frac{d\epsilon_\mathbf{p}}{2\epsilon_\mathbf{p}+|E_b|}=N(0)\ln \left( \frac{|E_b|+2E_c}{|E_b|}\right)\end{align}$$ (최종 결론식)
From this equation we obtain $$\begin{align}|E_b|=\frac{2E_c}{e^{1/N(0)W}-1}\end{align}$$ For weak coupling, $N(0)W\ll 1$, we find $$\begin{align}|E_b|=2E_ce^{-1/N(0)W}\end{align}$$ For strong coupling, $N(0)W\gg 1$, $$|E_b|=2N(0)WE_c$$
In strong coupling (interact near Fermi surface), big coupling energy makes Cooper pair.
Cooper pair is boson (why: singlet) so BEC situation makes all Cooper pair share same quantum state.
$$\Psi(\mathbf{r})=\sum_\mathbf{p}a_\mathbf{p}e^{i\mathbf{p}\mathbf{r}/\hbar}.$$
5. The BCS model
계속 했던거
BdG Equations in Tight-Binding Model (balisada.blogspot.com)
6. The gap equation
잘 풀어서 구함
7. Condensation energy
8. Bogoliubov quasiparticles
superposition of particle & hole.
When Cooper pair is broken (not ground state), Bogoliubov quasiparticle excited.
그래서 어떤 입자가 생기는거임?
Gap의 에너지를 가진 입자가 생김.
9. Heat capacity
10. The Bogoliubov-de Gennes equations
11. Electric current
(아직 안읽음)