BdG Equations in Tight-Binding Model
BdG Equation
The single-particle part of the Hamilonian is of the form: $$\begin{align}H_0=\int\int d\mathbf{r}\mathbf{r}'\psi^\dagger_\alpha(\mathbf{r})h_{\alpha\beta}(\mathbf{r},\mathbf{r}')\psi_\beta(\mathbf{r}')\end{align}$$ where $h_{\alpha\beta}(\mathbf{r},\mathbf{r}')$ is general enough to include the non-local and spin-flip effects.
The field operaotrs are now expressed in the localized-state basis as $$\begin{align}\psi_\alpha(\mathbf{r})=\sum_i w(\mathbf{r}-\mathbf{R}_i)c_{i\alpha},\\ \psi^\dagger_\alpha(\mathbf{r})=\sum_i w^*(\mathbf{r}-\mathbf{R}_i)c^\dagger_{i\alpha}\nonumber\end{align}$$ where $c^\dagger_{i\alpha}$ creates an electron of spin $\alpha$ at site $i$, and $w(\mathbf{r}-\mathbf{R}_i)$ is a localized orbital around the atomic site $\mathbf{R}_i$.
Substitution of these expressions into Eq. (1) gives $$\begin{align}H_0=\sum_{ij,\sigma\sigma'}c^\dagger_{i\sigma}h_{i\sigma,j\sigma'}c_{j\sigma}=-\sum_{i\ne j,\sigma\sigma'}t_{i\sigma,j\sigma'}c^\dagger_{i\sigma}c_{j\sigma'}+\sum_{i\sigma}\epsilon_ic^\dagger_{i\sigma}c_{i\sigma}+\sum_{i,\sigma\sigma'}\Omega_{i,\sigma\sigma'}c^\dagger_{i\sigma}c_{i\sigma'}\end{align}$$
Then we write down an effective model Hamiltonian for superconductivity: $$\begin{align}\mathcal{H}=H_0-\mu\sum_{i\sigma}c^\dagger_{i\sigma}c_{i\sigma}+U\sum_i\left( n_{i\uparrow}-\frac{1}{2}\right)\left(n_{i\downarrow}-\frac{1}{2}\right)-\frac{V}{2}\sum_{i\ne j}n_in_j\\ \nonumber =\sum_{ij,\sigma\sigma'}c^\dagger_{i\sigma]}\left[ h_{i\sigma,j\sigma'}-\left(\mu+\frac{U}{2}\right)\delta_{ij}\delta_{\sigma\sigma'}\right] c-{j\sigma'}+U\sum_i n_{i\uparrow}n_{i\downarrow}-\frac{V}{2}\sum_{i\ne j}n_in_j\end{align}$$
Here $n_i=\sum_\sigma n_{i\sigma}$ with $n_{i\sigma}=c^\dagger_{i\sigma}c_{i\sigma}$ is the particle number operator on site $i$. Note that the single particle energy is measured with respect to the chemical potential $\mu$.
The effective mean-field Hamiltonian with a singlet pairing symmetry is now given by: $$\begin{align} \mathcal{H}_{eff}=\sum_{ij,\sigma\sigma'} c^\dagger_{i\sigma}\tilde{h}_{i\sigma,j\sigma'}c_{j\sigma'}+\sum_{ij}\left[ \Delta_{ij}c^\dagger_{i\uparrow}c^\dagger_{j\downarrow}+\Delta^*_{ij}c_{j\downarrow}c_{i\uparrow}\right] +E_{const}\end{align}$$
Here the effective single-particle Hamiltonian is given by $$\begin{align}\tilde{h}_{i\sigma,j\sigma'}=h_{i\sigma,j\sigma'}-\left(\mu+\frac{U}{2}-U\langle n_{i\bar{\sigma}}\rangle\right)\delta_{ij}\delta_{\sigma\sigma'}\end{align}$$
the singlet-pairing potential $$\begin{align}\Delta_{ij}=\frac{V}{2}\left(\langle c_{i\uparrow} c_{j\downarrow}\rangle -\langle c_{i\downarrow} c_{j\uparrow}\rangle \right)\\ \Delta^*_{ij}=\frac{V}{2}\left(\langle c^\dagger_{j\downarrow}c^\dagger_{i\uparrow}\rangle - \langle c^\dagger_{j\uparrow}c^\dagger_{i\downarrow}\rangle\right)\end{align}$$
and the constant energy term is given by $$\begin{align}E_{const}=-U\langle n_{i\uparrow}\rangle\langle n_{i\downarrow}\rangle + \sum_{ij}\frac{|\Delta_{ij}|^2}{V}\end{align}$$
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.
The quasiparticle operators satisfy the anti-commutation relations: $$\begin{align} \{\gamma_n,\gamma_m^\dagger\}=\delta_{nm},\\ \{\gamma_n,\gamma_m\}=\{\gamma^\dagger_n,\gamma^\dagger_m\}=0\end{align}$$
With the above canonical transformation, the Hamiltonian is diagonalized in the following form: $$\begin{align}H_{eff}=\sum_n E_n\gamma^\dagger_n\gamma_n+E'_{const},\end{align}$$ where the index $n$ also includes the psedu-spin state index.
We get commutation relations, $$\begin{align}[\gamma^\dagger_n,H_{eff}]&=& -E_n\gamma_n^\dagger,\\ [\gamma_n,H_{eff}] &=& E_n\gamma_n,\end{align}$$ 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 $$\begin{align}\hat{\phi}_i=\begin{pmatrix}u_{i\uparrow}\\ u_{i\downarrow}\\ v_{i\uparrow}\\ v_{i\downarrow}\end{pmatrix}\end{align}$$
This set of BdG equations is subjected to the self-consistency conditions: $$\begin{align}n_{i\uparrow}=\sum^{'}_n [|u^n_{i\uparrow}|^2f(E_n)+|v^n_{i\uparrow}|^2f(-E_n)],\\ n_{i\downarrow}=\sum^{'}_n [|u^n_{i\downarrow}|^2f(E_n)+|v^n_{i\downarrow}|^2f(-E_n)]\\ \Delta_{ij}=\frac{V}{4}\sum^{'}_n [(u^n_{i\uparrow}v^{n*}_{j\downarrow}+u^n_{j\downarrow}v^{n*}_{i\uparrow})+(u^n_{i\downarrow}v^{n*}_{j\uparrow}+u^n_{j\uparrow}v^*_{i\downarrow})]\tanh \left( \frac{E_n}{2k_BT}\right)\end{align}$$
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 $-E_n$. Using this symmetry property, we can also rewrite $$\begin{align}n_{i\uparrow}=\sum_n |u^n_{i\uparrow}|^2f(E_n),\\ n_{i\downarrow}=\sum_n |v^n_{i\downarrow}|^2f(-E_n)\\ \Delta_{ij}=\frac{V}{4}\sum_n [u^n_{i\uparrow}v^{n*}_{j\downarrow}+u^n_{j\uparrow}v^*_{i\downarrow}]\tanh \left(\frac{E_n}{2k_BT}\right)\end{align}$$
No spin-orbit coupling and spin-flip scattering
It means $\tilde{h}_{i\uparrow,j\downarrow}=\tilde{h}_{i\downarrow,j\uparrow}=0$, then BdG equations becomes block-diagonalized into two sets of equations: $$\begin{align}\begin{cases}E_{\tilde{n}1}u^{\tilde{n}1}_{i\uparrow}=\sum_j\tilde{h}_{i\uparrow,j\uparrow}u^{\tilde{n}1}_{j\uparrow}+\sum_j\Delta_{ij}v^{\tilde{n}1}_{j\downarrow},\\ E_{\tilde{n}1}v^{\tilde{n}1}_{i\downarrow}=-\sum_j\tilde{h}^*_{i\downarrow,j\downarrow}v^{\tilde{n}1}_{j\downarrow}+\sum_j\Delta^*_{ji}u^{\tilde{n}1}_{j\uparrow}\end{cases}\end{align}$$ and $$\begin{align}\begin{cases}E_{\tilde{n}2}u^{\tilde{n}2}_{i\downarrow}=\sum_j\tilde{h}_{i\downarrow,j\downarrow}u^{\tilde{n}1}_{j\downarrow}+\sum_j\Delta_{ji}v^{\tilde{n}1}_{j\uparrow},\\ E_{\tilde{n}2}v^{\tilde{n}2}_{i\uparrow}=-\sum_j\tilde{h}^*_{i\uparrow,j\uparrow}v^{\tilde{n}2}_{j\downarrow}+\sum_j\Delta^*_{ij}u^{\tilde{n}2}_{j\uparrow}\end{cases}\end{align}$$
This block-diagonalization structure leads to a distince nature of the canonical transformation: $$\begin{align}c_{i\uparrow}=\sum^{'}_{\tilde{n}}(u^{\tilde{n}1}_{i\uparrow}\gamma_{\tilde{n}1}-v^{\tilde{n}2*}_{i\uparrow}\gamma^\dagger_{\tilde{n}2}),\quad c^\dagger_{i\uparrow}=\sum^{'}_{\tilde{n}}(u^{\tilde{n}1*}_{i\uparrow}\gamma^\dagger_{\tilde{n}1}-v^{\tilde{n}2}_{i\uparrow}\gamma_{\tilde{n}2}),\\ c_{i\downarrow}=\sum^{'}_{\tilde{n}}(u^{\tilde{n}2}_{i\downarrow}\gamma_{\tilde{n}2}+v^{\tilde{n}1*}_{i\downarrow}\gamma^\dagger_{\tilde{n}1}),\quad c^\dagger_{i\downarrow}=\sum^{'}_{\tilde{n}}(u^{\tilde{n}2*}_{i\downarrow}\gamma^\dagger_{\tilde{n}2}+v^{\tilde{n}1}_{i\downarrow}\gamma_{\tilde{n}1})\end{align}$$
The block diagonalization and the symmtery property suggest that we merely need to solve Eq. (30) subject to the self-consistency condition: $$\begin{align}n_{i\uparrow}=\sum_{\tilde{n}}|u^{\tilde{n}1}_{i\uparrow}|^2f(E_{\tilde{n}1}),\\ n_{i\downarrow}=\sum_{\tilde{n}}|v^{\tilde{n}1}_{i\downarrow}|^2f(-E_{\tilde{n}1})\end{align}$$ and $$\begin{align}\Delta_{ij}=\frac{V}{4}\sum_{\tilde{n}}[u^{\tilde{n}1}_{i\uparrow}v^{\tilde{n}1*}_{j\downarrow}+u^{\tilde{n}1}_{j\uparrow}v^{\tilde{n}1*}_{i\downarrow}]\tanh \left(\frac{E_{\tilde{n}1}}{2k_BT}\right)\end{align}$$
Therefore, by reducing the diagonalization of a $4N$ by $4N$ matrix down to that of a $2N$ by $2N$ matrix, the computational efficiency is improved significantly.
It is straightforward for us to obtain the BdG equations for an s-wave superconductor in the lattice model. There we first replace $U$ by $-U$ $(U > 0)$ so that the Hamiltonian becomes $$\begin{align}\mathcal{H}=\sum_{ij,\sigma\sigma'}c^\dagger_{i\sigma}\left[ h_{i\sigma,j\sigma'}-\mu\delta_{ij}\delta_{\sigma\sigma'}\right] c_{j\sigma'}-U\sum_i n_{i\uparrow}n_{i\downarrow},\end{align}$$ which lead to the BdG equations: $$\begin{align}E_nu^n_{i\uparrow}=\sum_{j\sigma'}\tilde{h}_{i\uparrow,j\sigma'}u_{j\sigma'}+\Delta_{ii}v^n_{i\downarrow},\\ E_nu^n_{i\downarrow}=\sum_{j\sigma'}\tilde{h}_{i\downarrow,j\sigma'}u_{j\sigma'}+\Delta_{ii}v^n_{i\uparrow},\\ E_nv^n_{i\uparrow}=-\sum_{j\sigma'}\sigma'\tilde{h}^*_{i\uparrow,j\sigma'}v_{j\sigma'}+\Delta_{ii}^*u^n_{i\downarrow},\\ E_nv^n_{i\downarrow}=\sum_{j\sigma'}\sigma'\tilde{h}^*_{i\downarrow,j\sigma'}v_{j\sigma'}+\Delta_{ii}^*u^n_{i\uparrow} \end{align}$$ Here $$\begin{align}\tilde{h}_{i\sigma,j\sigma'}=h_{i\sigma,j\sigma'}-\mu\delta_{ij}\delta_{\sigma\sigma'},\end{align}$$ and the self-consistency condition $$\begin{align}\Delta_{ii}=\frac{U}{2}\sum^{'}_n[u^n_{i\uparrow}v^{n*}_{i\downarrow}+u^n_{i\downarrow}v^{n*}_{i\uparrow}]\tanh \left(\frac{E_n}{2k_BT}\right)\end{align}$$
In the absence of spin-orbit coupling and spin-flip scattering, the BdG equations become $$\begin{align}\begin{cases}E_{\tilde{n}1}u^{\tilde{n}1}_{i\uparrow}=\sum_j\tilde{h}_{i\uparrow,j\uparrow}u^{\tilde{n}1}_{j\uparrow}+\Delta_{ii}v^{\tilde{n}1}_{i\downarrow},\\ E_{\tilde{n}1}v^{\tilde{n}1}_{i\downarrow}=-\sum_j\tilde{h}^*_{i\downarrow,j\downarrow}v^{\tilde{n}1}_{j\downarrow}+\Delta_{ii}^*u^{\tilde{n}1}_{i\uparrow} \end{cases}\end{align}$$ and $$\begin{align}\begin{cases}E_{\tilde{n}2}u^{\tilde{n}2}_{i\downarrow}=\sum_j\tilde{h}_{i\downarrow,j\downarrow}u^{\tilde{n}2}_{j\downarrow}+\Delta_{ii}v^{\tilde{n}2}_{i\uparrow},\\ E_{\tilde{n}2}v^{\tilde{n}2}_{i\uparrow}=-\sum_j\tilde{h}^*_{i\uparrow,j\uparrow}v^{\tilde{n}2}_{j\uparrow}+\Delta_{ii}^*u^{\tilde{n}2}_{i\downarrow} \end{cases}\end{align}$$ with $$\begin{align}\Delta_{ii}=\frac{U}{2}\sum_{\tilde{n}}u^{\tilde{n}1}_{i\uparrow}v^{\tilde{n}1*}_{j\downarrow}\tanh \left( \frac{E_{\tilde{n}1}}{2k_BT}\right)\end{align}$$
$N$개의 상태가 있는데, 왜 $2N$가 되고, $4N$ 이 되는가?
결과적으로 자유도는 $N$이 되어야 할 것.
positive sum의 이유. negative energy는 무슨 의미?
근데 지금은 mean field 근사로 풀어놨기 때문에 방정식을 만족하는 wavefunction $u$, $v$ 만 찾으면 물리량을 다 구한다.
mean field가 안되면 particle-hole symmetry는 있을까?
그렇지 않다. 따라서 QMC를 여기다 적용했을 때 생기는 negative-sign problem이 골치가 아픈 것이다.