BCS 발표 2

 

1. free & isotropic system에서 고체의 band 구조

$$\mathcal{H}=\sum_p \epsilon_p c^\dagger_pc_p,\quad \epsilon_p=\frac{p^2}{2m}$$

FD statistics -> filled inside fermi surface.

In this situation, change system is make hole inside make particle ouside.

$$\epsilon_p =\begin{cases}\frac{p^2}{2m}-E_F,& p>p_F\\ E_F-\frac{p^2}{2m},& p<p_F\end{cases}$$

Define hole $$\hat{h}_{p,\sigma} = - \hat{c}^\dagger_{-p,-\sigma}$$

$$\mathcal{H}=\sum_{p>p_F} \epsilon_p c^\dagger_p c_p+\sum_{p<p_F}\epsilon_p h^\dagger_p h_p$$  

hole의 spin은 어떻게 되는가?

2. Quantization of Phonon

For quantized electron, phonon field and electron field interact $$H=\sum_{k,k',q,\nu}g(q,\nu)c^\dagger_{k+q}c_k(b^\dagger_{q,\nu}+b_{-q,\nu})$$

(momentum conservation)

g의 스케일은 무엇이 결정할까?

저 term이 나오는 물리학이 뭘까?

3. Phonon-mediated electron attraction

$$\hat{H}=\hat{H}_0+\hat{H}_I$$

$\hat{H}_0$ makes energy eigenstates, $\hat{H}_I$ makes transition energy -> transition happen!

$$T_{fi}=\langle f |\hat{H}_I|i\rangle + \sum_n \langle f|\hat{H}_I|n\rangle\langle n|\hat{H}_I|i\rangle \left(\frac{1}{E_i-E_n+i\epsilon}+\frac{1}{E_f-E_n-i\epsilon}\right)\cdots$$

Electron-electron scattering is in opposite momentum:

(왜? 대칭적 운동량: 질량중심 고정 -> 에너지/운동량 교환해주는 포논 에너지가 낮음)

$$E_I=\epsilon_{p_1}+\epsilon_{p_2}$$

$$E_{II}=\epsilon_{p'_1}+\epsilon_{p'_2}$$

$$E_1=\epsilon_{p'_1}+\epsilon_{p_2}+\hbar \omega_q$$

$$\langle II|H_{e-ph-e}|I\rangle\\ \nonumber = \frac{1}{2}\sum_{i=1,2}\langle II|H_{e-ph-e}|i\rangle \left(\frac{1}{E_{II}-E_i}+\frac{1}{E_I-E_i}\right)\langle i|H_{e-ph-e}|I\rangle\\ \nonumber =W^*_\mathbf{q}\left(\frac{1}{\epsilon_{\mathbf{p}'_1}-\epsilon_{\mathbf{p}_1}-\hbar\omega_\mathbf{q}}+\frac{1}{\epsilon_{\mathbf{p}'_2}-\epsilon_{\mathbf{p}_2}-\hbar\omega_\mathbf{q}}\right)W_\mathbf{q} \\ \nonumber =\frac{|W_\mathbf{q}|^2}{\hbar}\left(\frac{1}{\omega-\omega_\mathbf{q}}-\frac{1}{\omega+\omega_\mathbf{q}}\right)=\frac{2|W_\mathbf{q}|^2}{\hbar}\frac{\omega_\mathbf{q}}{\omega^2-\omega^2_\mathbf{q}}$$

$W_q$ is the matrix element for emission of the photon with momentum $q$

$\hbar\omega=\epsilon_{p'_1}-\epsilon_{p_1}=-(\epsilon_{p'_2}-\epsilon_{p_2})$

When $|\omega|<\omega_q$, reults negative, attractive.

운동량 보존은 항상 성립(interaction term 형태). 

에너지 보존은 (중간에만) 위배 가능.

$\omega_q$는 뭐가 결정? 음수가 되면 끌어당기는 물리학?

방향에 상관없는 끌어당김 -> $L=0$, orbital wave function symmetric

-> spin wave function antisymmetric -> up/down 끼리 끌림.

4. Cooper problem

Two electron

$$\psi(r_1,r_2)$$

$$\left[ \hat{H}_e(r_1)+\hat{H}_e(r_2)+W(r_1-r_2)\right] \Psi(r_1,r_2)=E\Psi(r_1,r_2)$$

이 때 $\hat{H}=\frac{\hat{p}^2}{2m}=-\frac{\hbar^2}{2m}\nabla^2$이고, 정확히 하 $\hat{H}_e(\nabla_1)\psi_{p\uparrow}(r_1)=\epsilon_p\psi_{p\uparrow}(r_1)$임. 

$$\psi_{p\uparrow}(r_1)\propto e^{ipr_1/\hbar},\quad \psi_{-p\downarrow}(r_2)\propto e^{-ipr_2/\hbar}$$

$$\Psi(r_1,r_2)=\sum_p c_p\psi_{p\uparrow}(r_1)\psi_{-p\downarrow}(r_2)=\sum_p a_pe^{ipr/\hbar}=\Psi(r)$$

Fourier transform

$$2\epsilon_p a_p+\sum_{p'}W_{p,p'}a_{p'}=Ea_p$$

Surface 근처에서 $-W$인 $W_{p,p'}$ 대입, $\epsilon_p$는 0부터 $E_c=\hbar \omega_D$까지. $E_c\ll E_F$ (정확히 안따라감)

상호작용하는 $p$의 범위, 왜 포논의 속도가 아닌 $v_F$가 들어가는지.

Ground state energy

$$|E_b|=\frac{2E_c}{e^{1/N(0)W}-1}$$

Strong coupling $N(0)W\gg 1$, $$|E_b|=2N(0)WE_c$$

두개 묶인 상태는 spin 0이네? 보존이네? 온도 0이면 BEC하네?

결합 에너지가 커지면 그만큼 Pair가 많이 생길 것.

이걸 깨면 exited state가 되는 것.

5. Mean Field

$$H=\sum_{k\sigma}\xi_kc^\dagger_{k\sigma}c_{k\sigma}+\sum_{kk'}W_{kk'}c^\dagger_{k\uparrow}c^\dagger_{-k\downarrow}c_{-k'\downarrow}c_{k'\uparrow}$$

superconductor order parameter( Cooper pair 개수)가 평균장이 된다고 가정.

fluctuation 0이라는 가정?

quadratic Hamiltonian -> 다시 Band (H를 diag)를 보겠다!

self-consistent 하다: 적어도 틀리지는 않음. (모르겠다. 직접 해봐야 감이 잡힐듯)

$$E_k=\sqrt{\xi^2_k+|\Delta_k|^2}$$ ($\xi$는 interaction 없을 때 energy.)

즉 gap 이상의 에너지가 있어야 ground state인 Cooper pair를 깨고 Bogoliubov quasiparticle을 만들 수 있음. Cooper Pair가 많이 있으면 그만큼 많이 깨야함.

6. Bogoliubov Quasiparticle

Cooper Pair는 hole끼리, particle끼리 pair이룸.

Bogoliubov quasiparticle은 hole과 particle을 함께 생각함.

$$\gamma_{k\uparrow}=u^*_k c_{k\uparrow}+v^*_kh_{k\uparrow}=u^*_k c_{k\uparrow}+v^*_kc^\dagger_{-k\downarrow}\\ \gamma^\dagger_{-k\downarrow}=u_kc^\dagger_{-k\downarrow}-v_k h^\dagger_{-k\downarrow}=u_kc^\dagger_{-k\downarrow}-v_kc_{k\uparrow}$$

$c$입장에서 보면,

$$c_{k\uparrow}=u_k \gamma_{k\uparrow}-v^*_k\gamma^\dagger_{-k\downarrow}$$

$$c^\dagger_{-k\downarrow}=u^*_k\gamma^\dagger_{-k\downarrow}+v_k\gamma_{k\uparrow}$$

따라서 $k$를 +랑 - 다 세지 말고 한쪽으로만 세야함.

Constraint: $$\{\gamma_{k\uparrow},\gamma^\dagger_{k\uparrow}\}=|u_k|^2+|v_k|^2=1$$

This make transformation unitary.

$|u_k|$는 $\Delta$ 없을 때 surface 밖에만 있어서 particle fraction

$|v_k|$는 $\Delta$ 없을 때 surface 안에만 있어서 hole fraction

$\Delta$ 생기면 주황 부분 값이 약간 달라짐.

7. BdG Equation

We obtain the following commutation relations: $$\begin{align}[c_{i\uparrow},\mathcal{H}_{eff}] &=& \sum_{j\sigma'}\tilde{h}_{i\uparrow,j\sigma'}c_{j\sigma'}+\sum_j \Delta_{ij}c^\dagger_{j\downarrow},\\ \nonumber [c_{i\uparrow}^\dagger,\mathcal{H}_{eff}] &=& -\sum_{j\sigma'}\tilde{h}_{j\sigma',i\uparrow}c_{j\sigma'}^\dagger-\sum_j \Delta_{ij}^*c^\dagger_{j\downarrow},\\ \nonumber[c_{i\downarrow},\mathcal{H}_{eff}] &=& \sum_{j\sigma'}\tilde{h}_{i\downarrow,j\sigma'}c_{j\sigma'}-\sum_j \Delta_{ji}c^\dagger_{j\uparrow},\\ \nonumber[c_{i\downarrow}^\dagger,\mathcal{H}_{eff}] &=& -\sum_{j\sigma'}\tilde{h}_{j\sigma',i\downarrow}c_{j\sigma'}^\dagger+\sum_j \Delta_{ji}^*c^\dagger_{j\uparrow}\end{align}$$

Electron field operator can be expressed as a linear combination of electron- and hole-liker quasiparticle excitations, which enables us to perform a Bogoliubov canonical transformation: $$\begin{align} c_{i\sigma}=\sum^{'}_n (u^n_{i\sigma} \gamma_n -\sigma v^{n*}_{i\sigma}\gamma^\dagger_n),\quad c_{i\sigma}^\dagger=\sum^{'}_n (u^{n*}_{i\sigma} \gamma_n^\dagger -\sigma v^{n}_{i\sigma}\gamma_n)\end{align}$$

where $\sigma=\pm 1$ denotes  the up an down spin orientations. The operators  $\gamma_n^\dagger$ create a Bogoliubov quasiparticle at state $n$. The prime sign above the summation in the transformation means only those states with positive energy are counted. 

we compare the coeffecients of the term with $\gamma_n$ and $\gamma_n^\dagger$ and arrive at $$\begin{align}E_n u^n_{i\uparrow} &=& \sum_{j\sigma'}\tilde{h}_{i\uparrow,j\sigma'}u^n_{j\sigma'}+\sum_j \Delta_{ij}v^n_{j\downarrow},\\  E_n u^n_{i\downarrow} &=& \sum_{j\sigma'}\tilde{h}_{i\downarrow,j\sigma'}u^n_{j\sigma'}+\sum_j \Delta_{ji}v^n_{j\uparrow},\\ E_n v^n_{i\uparrow} &=& -\sum_{j\sigma'}\sigma'\tilde{h}^{*}_{i\uparrow,j\sigma'}v^n_{j\sigma'}+\sum_j \Delta_{ij}^{*}u^n_{j\downarrow},\\ E_n v^n_{i\downarrow} &=& -\sum_{j\sigma'}\sigma'\tilde{h}^{*}_{i\downarrow,j\sigma'}v^n_{j\sigma'}+\sum_j \Delta_{ij}^{*}u^n_{j\uparrow}  \end{align}$$

The set of BdG equations can be case into a matrix form: $$\begin{align}\sum_j \hat{M}_{ij}\hat{\phi}_j =E_n\hat{\phi}_i\end{align}$$ where $$\begin{align}  \hat{M}_{ij}=\begin{bmatrix} \tilde{h}_{i\uparrow j\uparrow}& \tilde{h}_{i\uparrow j\downarrow}&0 & \Delta_{ij}\\ \tilde{h}_{i\downarrow j\uparrow}& \tilde{h}_{i\downarrow j\downarrow}&\Delta_{ji} &0 \\ 0& \Delta^*_{ij}& -\tilde{h}^*_{i\uparrow j\uparrow}& \tilde{h}^*_{i\uparrow j\downarrow}\\ \Delta^*_{ji} &0& \tilde{h}^*_{i\downarrow j\uparrow}& -\tilde{h}^*_{i\downarrow j\downarrow} \end{bmatrix} \end{align}$$ and $$\begin{align}\hat{\phi}_i=\begin{pmatrix}u_{i\uparrow}\\ u_{i\downarrow}\\ v_{i\uparrow}\\ v_{i\downarrow}\end{pmatrix}\end{align}$$

Particle-Hole Symmetry

If $(u^n_{i\uparrow},v^n_{i\downarrow},u^n_{i\downarrow},v^n_{i\uparrow})$ is the solution to the BdG equations with eigenvalue $E_n$, then $(-v^{n*}_{i\uparrow},u^{n*}_{i\downarrow},v^{n*}_{i\downarrow},-u^{n*}_{i\uparrow})$ is the solution to the same equations with eigenvalue $-E_n$. 

Hole 만드는거랑 Particle 없애는게 같음.

찐연구시작

Real space에서 BdG equation을 self-consistent하게 풀어봄. -> gap, supercurrent 봄

피보나치는 1000개 

AB는 2000개 (r=30)  t=1, 10-3를 error로 잡고 대충 해보고 코드 맞으면, 그다음에 줄임.

목표: 2D honeycomb transformation into dodecagonal quasicrystals driven by electrostaticforce