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BdG理论

发表于 27k 25 分钟

BdG理论

哈密顿量通常可以分为两部分，正常态哈密顿量和相互作用项：

H=H0+H1H=H_0+H_1

H0=∑σ∫drΨσ†(r)h0(r)Ψσ(r)H_0 = \sum_{\sigma} \int dr \Psi_{\sigma}^{\dagger}(r) h_0(r) \Psi_{\sigma}(r)

H1=12∑σ,σ′∫dr∫dr′Ψσ†(r)Ψσ′†(r′)gσσ′(r−r′)Ψσ′(r′)Ψσ(r)H_1 = \frac{1}{2} \sum_{\sigma, \sigma'} \int dr \int dr' \Psi_{\sigma}^{\dagger}(r) \Psi_{\sigma'}^{\dagger}(r') g_{\sigma \sigma'}(r-r') \Psi_{\sigma'}(r') \Psi_{\sigma}(r)

σ\sigma是自旋指标，Ψσ(r)\Psi_{\sigma}(r)是场算符，表示在位置 rr处产生一个自旋为σ\sigma的电子， h0(r)h_0(r)是单电子哈密顿量， gσσ′(r−r′)g_{\sigma \sigma'}(r-r')是电子间的相互作用势。
对于自由电子系统，可写成

h0=−ℏ22m(∇−ieℏcA)+V(r)−μh_0=-\frac{\hbar^2}{2m}\left(\nabla-i\frac{e}{\hbar c}\boldsymbol{A}\right)+V(\boldsymbol{r})-\mu

在 BCS 理论中，有效相互作用可以表示为：

HBCS=∑k⃗ϵk⃗ck⃗†ck⃗−V2∑k⃗,k⃗′ck⃗↑†c−k⃗↓†c−k⃗′↓ck⃗′↑H_{BCS} = \sum_{\vec{k}} \epsilon_{\vec{k}} c_{\vec{k}}^{\dagger} c_{\vec{k}} - \frac{V}{2} \sum_{\vec{k},\vec{k}'} c_{\vec{k}\uparrow}^{\dagger} c_{-\vec{k}\downarrow}^{\dagger} c_{-\vec{k}'\downarrow} c_{\vec{k}'\uparrow}

平均场近似
为了处理相互作用项H1H_1，引入平均场近似，平均场表示的是诸如⟨Ψσ(r)Ψσ′(r′)⟩,⟨Ψσ†(r)Ψσ′†(r′)⟩\langle\Psi_\sigma(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{\overset{\prime}{r}})\rangle,\quad\langle\Psi_\sigma^\dagger(\boldsymbol{r})\Psi_{\sigma^{\prime}}^\dagger(\boldsymbol{\overset{\prime}{r}})\rangle这类库珀对的期望值。
将算符 Ψσ†(r)Ψσ′†(r′)\Psi_\sigma^\dagger(r)\Psi_{\sigma^{\prime}}^\dagger(r^{\prime})和Ψσ′(r′)Ψσ(r)\Psi_\sigma^{\prime}(r^{\prime})\Psi_\sigma(r)表示为

Ψσ†(r)Ψσ′†(r′)=⟨Ψσ†(r)Ψσ′†(r′)⟩+[Ψσ†(r)Ψσ′†(r′)−⟨Ψσ†(r)Ψσ′†(r′)⟩]Psiσ(r)Ψσ′(r′)=⟨Ψσ(r)Ψσ′(r′)⟩+[Ψσ(r)Ψσ′(r′)−⟨Ψσ(r)Ψσ′(r′)⟩]\Psi_{\sigma}^{\dagger}(r)\Psi_{\sigma^{\prime}}^{\dagger}(r^{\prime})=\langle\Psi_{\sigma}^{\dagger}(r)\Psi_{\sigma^{\prime}}^{\dagger}(r^{\prime})\rangle+\left[\Psi_{\sigma}^{\dagger}(r)\Psi_{\sigma^{\prime}}^{\dagger}(r^{\prime})-\langle\Psi_{\sigma}^{\dagger}(r)\Psi_{\sigma^{\prime}}^{\dagger}(r^{\prime})\rangle\right]\\Psi_{\sigma}(r)\Psi_{\sigma^{\prime}}(r^{\prime})=\langle\Psi_{\sigma}(r)\Psi_{\sigma^{\prime}}(r^{\prime})\rangle+[\Psi_{\sigma}(r)\Psi_{\sigma^{\prime}}(r^{\prime})-\langle\Psi_{\sigma}(r)\Psi_{\sigma^{\prime}}(r^{\prime})\rangle]

H1=12∑σ,σ′∫dr∫dr′(⟨Ψσ†(r)Ψσ′†(r′)⟩+[Ψσ†(r)Ψσ′†(r′)−⟨Ψσ†(r)Ψσ′†(r′)⟩])&×gσ,σ′(r−r′)(−⟨Ψσ(r)Ψσ′(r′)⟩−[Ψσ(r)Ψσ′(r′)−⟨Ψσ(r)Ψσ′(r′)⟩])&=12∑σ,σ′∫dr∫dr∫dr′(−gσ,σ′(r−r′)⟨Ψσ′†(r)Ψσ′†(r′)⟩Ψσ′(r′)Ψσ′(r′)−gσ,σ′(r−r′)⟨Ψσ′(r)Ψσ′(r′)⟩Ψσ′†(r)Ψσ′†(r′))&−gσ,σ′(r−r′)⟨Ψσ†(r)Ψσ′†(r′)⟩⟨Ψσ(r)Ψσ′(r′)⟩&−gσ,σ′(r−r′)[Ψσ(r)Ψσ′(r′)−⟨Ψσ(r)Ψσ′(r′)⟩][Ψσ(r)Ψσ′(r′)−⟨Ψσ(r)Ψσ′(r′)⟩]\begin{aligned}H_{1}&=\frac{1}{2}\sum_{\sigma,\sigma^{\prime}}\int dr\int dr^{\prime}\left(\langle\Psi_{\sigma}^{\dagger}(r)\Psi_{\sigma^{\prime}}^{\dagger}(r^{\prime})\rangle+\left[\Psi_{\sigma}^{\dagger}(r)\Psi_{\sigma^{\prime}}^{\dagger}(r^{\prime})-\langle\Psi_{\sigma}^{\dagger}(r)\Psi_{\sigma^{\prime}}^{\dagger}(r^{\prime})\rangle\right]\right)\&\times g_{\sigma,\sigma^{\prime}}(\boldsymbol{r}-r^{\prime})\left(-\langle\Psi_{\sigma}(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})\rangle-[\Psi_{\sigma}(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})-\langle\Psi_{\sigma}(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})\rangle]\right)\&=\frac{1}{2}\sum_{\sigma,\sigma^{\prime}}\int dr\int dr\int dr^{\prime}\left(-g_{\sigma,\sigma^{\prime}}(\boldsymbol{r}-\boldsymbol{r}^{\prime})\langle\Psi_{\sigma^{\prime}}^{\dagger}(\boldsymbol{r})\Psi_{\sigma^{\prime}}^{\dagger}(\boldsymbol{r}^{\prime})\rangle\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})-g_{\sigma,\sigma^{\prime}}(\boldsymbol{r}-\boldsymbol{r}^{\prime})\langle\Psi_{\sigma^{\prime}}(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})\rangle\Psi_{\sigma^{\prime}}^{\dagger}(\boldsymbol{r})\Psi_{\sigma^{\prime}}^{\dagger}(\boldsymbol{r}^{\prime})\right)\&-g_{\sigma,\sigma^{\prime}}(r-\boldsymbol{r}^{\prime})\langle\Psi_{\sigma}^{\dagger}(\boldsymbol{r})\Psi_{\sigma^{\prime}}^{\dagger}(\boldsymbol{r}^{\prime})\rangle\langle\Psi_{\sigma}(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})\rangle\&-g_{\sigma,\sigma^{\prime}}(r-\boldsymbol{r}^{\prime})\left[\Psi_{\sigma}(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})-\langle\Psi_{\sigma}(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})\rangle\right]\left[\Psi_{\sigma}(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})-\langle\Psi_{\sigma}(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})\rangle\right]\end{aligned}

忽略作为常数项的第三项和作为偏离平均场的二次项的第四项整理

H1∼12∑σ,σ′∫dr∫dr′(Ψσ(r)(−gσ,σ′(r−r′)⟨Ψσ†(r)Ψσ′†(r′)⟩)Ψσ′(r′)&+Ψσ†(r)(−gσ,σ′(r−r′)⟨Ψσ(r)Ψσ′(r′)⟩)Ψσ′†(r′))\begin{aligned}H_{1}&\sim\frac{1}{2}\sum_{\sigma,\sigma^{\prime}}\int d\boldsymbol{r}\int d\boldsymbol{r}^{\prime}\Big(\Psi_{\sigma}(\boldsymbol{r})(-g_{\sigma,\sigma^{\prime}}(\boldsymbol{r}-\boldsymbol{r}^{\prime})\langle\Psi_{\sigma}^{\dagger}(\boldsymbol{r})\Psi_{\sigma^{\prime}}^{\dagger}(\boldsymbol{r}^{\prime})\rangle)\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})\&+\Psi_{\sigma}^{\dagger}(\boldsymbol{r})(-g_{\sigma,\sigma^{\prime}}(\boldsymbol{r}-\boldsymbol{r}^{\prime})\langle\Psi_{\sigma}(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})\rangle)\Psi_{\sigma^{\prime}}^{\dagger}(\boldsymbol{r}^{\prime})\Big)\end{aligned}

引入矢量

Ψ⃗(r)≡(Ψ↑(r)Psi↓(r))\left.\vec{\Psi}(r)\equiv\left(\begin{array}{c}\Psi_\uparrow(r)\\Psi_\downarrow(r)\end{array}\right.\right)

第二项进一步改写

12∫dr∫dr′(Ψ↑†(r)Ψ↓†(r))(−g↑,↑(r−r′)⟨Ψ↑(r)Ψ↑(r′)⟩−g↑↓↓(r−r′)⟨Ψ↑(r)Ψ↓(r′)⟩−g↓,↑(r−r′)⟨Ψ↓(r)Ψ↑(r′)⟩−g↓↓↓(r−r′)⟨Ψ↓(r)Ψ↓(r′)⟩)(Ψ↑†(r′)Psi↓†(r′))\frac{1}{2}\int d\boldsymbol{r}\int d\boldsymbol{r}^{\prime}\begin{pmatrix}\Psi_\uparrow^\dagger(\boldsymbol{r})&\Psi_\downarrow^\dagger(\boldsymbol{r})\end{pmatrix}\begin{pmatrix}-g_{\uparrow,\uparrow}(\boldsymbol{r}-\boldsymbol{r}^{\prime})\langle\Psi_\uparrow(\boldsymbol{r})\Psi_\uparrow(\boldsymbol{r}^{\prime})\rangle&-g_{\uparrow\downarrow\downarrow}(\boldsymbol{r}-\boldsymbol{r}^{\prime})\langle\Psi_\uparrow(\boldsymbol{r})\Psi_\downarrow(\boldsymbol{r}^{\prime})\rangle\\-g_{\downarrow,\uparrow}(\boldsymbol{r}-\boldsymbol{r}^{\prime})\langle\Psi_\downarrow(\boldsymbol{r})\Psi_\uparrow(\boldsymbol{r}^{\prime})\rangle&-g_{\downarrow\downarrow\downarrow}(\boldsymbol{r}-\boldsymbol{r}^{\prime})\langle\Psi_\downarrow(\boldsymbol{r})\Psi_\downarrow(\boldsymbol{r}^{\prime})\rangle\end{pmatrix}\begin{pmatrix}\Psi_\uparrow^\dagger(\boldsymbol{r}^{\prime})\\Psi_\downarrow^\dagger(\boldsymbol{r}^{\prime})\end{pmatrix}

定义

Δσσ′(r,r′)≡−gσ,σ′(r−r′)⟨Ψσ(r)Ψσ′(r′)⟩hatΔ(r,r′)≡(Δ↑↑(r,r′)Δ↑↓(r,r′)Delta↓↑(r,r′)Δ↓↓(r,r′))\begin{gathered}\Delta_{\sigma\sigma^{\prime}}(r,r^{\prime})\equiv-g_{\sigma,\sigma^{\prime}}(\boldsymbol{r}-\boldsymbol{r}^{\prime})\langle\Psi_\sigma(\boldsymbol{r})\Psi_{\sigma^{\prime}}(\boldsymbol{r}^{\prime})\rangle\\hat{\Delta}(r,r^{\prime})\left.\equiv\left(\begin{array}{cc}\Delta_{\uparrow\uparrow}(\boldsymbol{r},\boldsymbol{r}^{\prime})&\Delta_{\uparrow\downarrow}(\boldsymbol{r},\boldsymbol{r}^{\prime})\\Delta_{\downarrow\uparrow}(\boldsymbol{r},\boldsymbol{r}^{\prime})&\Delta_{\downarrow\downarrow}(\boldsymbol{r},\boldsymbol{r}^{\prime})\end{array}\right.\right)\end{gathered}

得到

12∫dr∫dr′Ψ⃗†(r)Δ^(r,r′)Ψ⃗†T(r′)\frac{1}{2}\int d\boldsymbol{r}\int d\boldsymbol{r}^{\prime}\vec{\Psi}^\dagger(\boldsymbol{r})\hat{\Delta}(\boldsymbol{r},\boldsymbol{\overset{\prime}{r}})\vec{\Psi}^{\dagger T}(\boldsymbol{\overset{\prime}{r}})

如果哈密顿量HH是厄米矩阵，那么 H1H_1的第一项应该是第二项的厄米共轭。由于哈密顿量可以看作是以坐标r,r′r,r'和自旋指标为下标的无限维矩阵，因此为了取厄米共轭，只需交换坐标rr和 r′r'，并对自旋矩阵进行转置和复共轭。
对H1H_1的第二项取厄米共轭，得到

12∫dr′∫drΨ⃗T(r)Δ^†(r′,r)Ψ⃗(r′)\frac{1}{2}\int d\boldsymbol{r}^{\prime}\int d\boldsymbol{r}\vec{\Psi}^T(\boldsymbol{r})\hat{\Delta}^\dagger(\boldsymbol{r}^{\prime},\boldsymbol{r})\vec{\Psi}(\boldsymbol{r}^{\prime})

Δ^†(r′,r)=(−g↑,↑(r′−r)⟨Ψ↑†(r)Ψ↑†(r′)⟩−g↓,↑(r′−r)⟨Ψ↑†(r)Ψ↓†(r′)⟩−g↑,↓(r′−r)⟨Ψ↓†(r)Ψ↑†(r′)⟩−g↓,↓(r′−r)⟨Ψ↓†(r)Ψ↓†(r′)⟩)\left.\hat{\Delta}^\dagger(\boldsymbol{r}^\prime,\boldsymbol{r})=\left(\begin{array}{cc}-g_{\uparrow,\uparrow}(\boldsymbol{r}^\prime-\boldsymbol{r})\langle\Psi_{\uparrow}^\dagger(\boldsymbol{r})\Psi_{\uparrow}^\dagger(\boldsymbol{r}^\prime)\rangle&-g_{\downarrow,\uparrow}(\boldsymbol{r}^\prime-\boldsymbol{r})\langle\Psi_{\uparrow}^\dagger(\boldsymbol{r})\Psi_{\downarrow}^\dagger(\boldsymbol{r}^\prime)\rangle\\-g_{\uparrow,\downarrow}(\boldsymbol{r}^\prime-\boldsymbol{r})\langle\Psi_{\downarrow}^\dagger(\boldsymbol{r})\Psi_{\uparrow}^\dagger(\boldsymbol{r}^\prime)\rangle&-g_{\downarrow,\downarrow}(\boldsymbol{r}^\prime-\boldsymbol{r})\langle\Psi_{\downarrow}^\dagger(\boldsymbol{r})\Psi_{\downarrow}^\dagger(\boldsymbol{r}^\prime)\rangle\end{array}\right.\right)

利用gσ,σ′(r−r′)=gσ′,σ(r′−r)g_{\sigma,\sigma^{\prime}}(\boldsymbol{r}-\boldsymbol{r}^{\prime})=g_{\sigma^{\prime},\sigma}(\boldsymbol{r}^{\prime}-\boldsymbol{r})可以确认，上式与H1H_1的第一项相等。因此， H1H_1最终可表示为

H1=12∫dr′∫drΨ⃗T(r)Δ^†(r′,r)Ψ⃗(r′)+12∫dr∫dr′Ψ⃗†(r)Δ^(r,r′)Ψ⃗†T(r′)H_{1}=\frac{1}{2}\int d\boldsymbol{r}^{\prime}\int d\boldsymbol{r}\vec{\Psi}^{T}(\boldsymbol{r})\hat{\Delta}^{\dagger}(\boldsymbol{r}^{\prime},\boldsymbol{r})\vec{\Psi}(\boldsymbol{r}^{\prime})+\frac{1}{2}\int d\boldsymbol{r}\int d\boldsymbol{r}^{\prime}\vec{\Psi}^{\dagger}(\boldsymbol{r})\hat{\Delta}(\boldsymbol{r},\boldsymbol{r}^{\prime})\vec{\Psi}^{\dagger T}(\boldsymbol{r}^{\prime})

Δ^†(r′,r))σσ′=−(Δ^(r,r′))σσ′\hat{\Delta}^\dagger(\boldsymbol{r}^{\prime},\boldsymbol{r})){\sigma\sigma^{\prime}}=-(\hat{\Delta}(\boldsymbol{r},\boldsymbol{r}^{\prime})){\sigma\sigma^{\prime}}始终成立。
考虑对角化无电子间相互作用的哈密顿量H0H_0的二次量子化表示：

H0=∑∫drΨσ†(r)h0Ψσ(r)H_0=\sum\int d\boldsymbol{r}\Psi_\sigma^\dagger(\boldsymbol{r})h_0\Psi_\sigma(\boldsymbol{r})

通过求解薛定谔方程：

h0fν,σ(r)=Eνfν,σ(r)∫drfν,σ∗(r)fν′,σ′(r)=δν,ν′δσ,σ′∑νfν,σ(r)fν,σ′∗(r)=δ(r−r′)δσ,σ′\begin{aligned}h_0f_{\nu,\sigma}(\boldsymbol{r})&=E_\nu f_{\nu,\sigma}(\boldsymbol{r})\\\int d\boldsymbol{r}f_{\nu,\sigma}^*(r)f_{\nu^{\prime},\sigma^{\prime}}(r)&=\delta_{\nu,\nu^\prime}\delta_{\sigma,\sigma^\prime}\\\sum_{\nu}f_{\nu,\sigma}(r)f_{\nu,\sigma^{\prime}}^{*}(r)&=\delta(r-r^{\prime})\delta_{\sigma,\sigma^{\prime}}\end{aligned}

得到本征值和归一化的波函数EνE_{\nu}和 fν,σ(r)f_{\nu,\sigma}(r)
对算符Ψσ(r)\Psi_{\sigma}(r)进行幺正变换：

Ψσ(r)=∑νcν,σfν,σ(r)\Psi_\sigma(r)=\sum_\nu c_{\nu,\sigma}f_{\nu,\sigma}(r)

cν,σc_{\nu,\sigma}是湮灭自旋为σ\sigma、本征值为EiE_{i}的电子的算符。
其结果是，H0H_0可对角化为：

H0=∑σ∫dr∑ν,ν′cν,σ†cν′,σfν,σ∗(r)h0fν′,σ(r)=∑σ∫dr∑ν,ν′cν,σ†cν′,σfν,σ∗(r)Eν′fν′,σ(r)=∑σ∑νcν,σ†cν,σEν\begin{aligned}H_{0}&=\sum_{\sigma}\int d\boldsymbol{r}\sum_{\nu,\nu^{\prime}}c_{\nu,\sigma}^{\dagger}c_{\nu^{\prime},\sigma}f_{\nu,\sigma}^{*}(\boldsymbol{r})h_{0}f_{\nu^{\prime},\sigma}(\boldsymbol{r})\\&=\sum_{\sigma}\int d\boldsymbol{r}\sum_{\nu,\nu^{\prime}}c_{\nu,\sigma}^{\dagger}c_{\nu^{\prime},\sigma}f_{\nu,\sigma}^{*}(\boldsymbol{r})E_{\nu^{\prime}}f_{\nu^{\prime},\sigma}(\boldsymbol{r})\\&=\sum_\sigma\sum_\nu c_{\nu,\sigma}^\dagger c_{\nu,\sigma}E_\nu\end{aligned}

变形可得

∑σ∑νcν,σ†cν′,σEν=∑σ∫dr∑ν,ν′cν,σ†cν′,σEνfν,σ∗(r)fν′,σ(r)=∑σ∫dr∑ν,ν′cν,σ†cν′,σ(Eνfν,σ(r))∗fν′,σ(r)=−∑σ∫dr∑ν,ν′cν′,σcν,σ†(h0fν,σ(r))∗fν′,σ(r)=−∑σ∫drΨσ(r)(h0∗Ψσ†(r))=−∑σ∫drΨσ(r)h0∗Ψσ†(r)\begin{aligned}\sum_{\sigma}\sum_{\nu}c_{\nu,\sigma}^{\dagger}c_{\nu^{\prime},\sigma}E_{\nu}&=\sum_{\sigma}\int d\boldsymbol{r}\sum_{\nu,\nu^{\prime}}c_{\nu,\sigma}^{\dagger}c_{\nu^{\prime},\sigma}E_{\nu}f_{\nu,\sigma}^{*}(\boldsymbol{r})f_{\nu^{\prime},\sigma}(\boldsymbol{r})\\&=\sum_{\sigma}\int d\boldsymbol{r}\sum_{\nu,\nu^{\prime}}c_{\nu,\sigma}^{\dagger}c_{\nu^{\prime},\sigma}(E_{\nu}f_{\nu,\sigma}(\boldsymbol{r}))^{*}f_{\nu^{\prime},\sigma}(\boldsymbol{r})\\&=-\sum_\sigma\int d\boldsymbol{r}\sum_{\nu,\nu^{\prime}}c_{\nu^{\prime},\sigma}c_{\nu,\sigma}^\dagger(h_0f_{\nu,\sigma}(\boldsymbol{r}))^*f_{\nu^{\prime},\sigma}(\boldsymbol{r})\\&=-\sum_\sigma\int d\boldsymbol{r}\Psi_\sigma(\boldsymbol{r})\left(h_0^*\Psi_\sigma^\dagger(\boldsymbol{r})\right)\\&=-\sum_\sigma\int d\boldsymbol{r}\Psi_\sigma(\boldsymbol{r})h_0^*\Psi_\sigma^\dagger(\boldsymbol{r})\end{aligned}

H0=12∑σ∫drΨσ†(r)h0Ψσ(r)+12∑σ∫drΨσ†(r)h0Ψσ(r)=12∑σ∫drΨσ†(r)h0Ψσ(r)−12∑σ∫drΨσ(r)h0∗Ψσ†(r)=12∑σ,σ′∫dr∫dr′δσ,σ′δ(r−r′)Ψσ†(r)h0(r′)Ψσ(r)−12∑σ,σ′∫dr∫dr′δσ,σ′δ(r−r′)Ψσ(r)h0∗(r′)Ψσ†(r)\begin{aligned} H_{0} & =\frac{1}{2} \sum_{\sigma} \int d \boldsymbol{r} \Psi_{\sigma}^{\dagger}(\boldsymbol{r}) h_{0} \Psi_{\sigma}(\boldsymbol{r})+\frac{1}{2} \sum_{\sigma} \int d \boldsymbol{r} \Psi_{\sigma}^{\dagger}(\boldsymbol{r}) h_{0} \Psi_{\sigma}(\boldsymbol{r}) \\ & =\frac{1}{2} \sum_{\sigma} \int d \boldsymbol{r} \Psi_{\sigma}^{\dagger}(\boldsymbol{r}) h_{0} \Psi_{\sigma}(\boldsymbol{r})-\frac{1}{2} \sum_{\sigma} \int d \boldsymbol{r} \Psi_{\sigma}(\boldsymbol{r}) h_{0}^{*} \Psi_{\sigma}^{\dagger}(\boldsymbol{r}) \\ & =\frac{1}{2} \sum_{\sigma, \sigma^{\prime}} \int d \boldsymbol{r} \int d \boldsymbol{r}^{\prime} \delta_{\sigma, \sigma^{\prime}} \delta\left(\boldsymbol{r}-\boldsymbol{r}^{\prime}\right) \Psi_{\sigma}^{\dagger}(\boldsymbol{r}) h_{0}\left(\boldsymbol{r}^{\prime}\right) \Psi_{\sigma}(\boldsymbol{r})-\frac{1}{2} \sum_{\sigma, \sigma^{\prime}} \int d \boldsymbol{r} \int d \boldsymbol{r}^{\prime} \delta_{\sigma, \sigma^{\prime}} \delta\left(\boldsymbol{r}-\boldsymbol{r}^{\prime}\right) \Psi_{\sigma}(\boldsymbol{r}) h_{0}^{*}\left(\boldsymbol{r}^{\prime}\right) \Psi_{\sigma}^{\dagger}(\boldsymbol{r}) \end{aligned}

H0=12∫dr∫dr′Ψ⃗†(r)h0(r′)δ(r−r′)σ^0Ψ⃗(r)−12∫dr∫dr′Ψ⃗T(r)h0∗(r′)σ^0δ(r−r′)Ψ⃗†T(r)H_0=\frac{1}{2}\int d\boldsymbol{r}\int d\boldsymbol{r}^{\prime}\vec{\Psi}^\dagger(\boldsymbol{r})h_0(\boldsymbol{r}^{\prime})\delta(\boldsymbol{r}-\boldsymbol{r}^{\prime})\hat{\sigma}_0\vec{\Psi}(\boldsymbol{r})-\frac{1}{2}\int d\boldsymbol{r}\int d\boldsymbol{r}^{\prime}\vec{\Psi}^T(\boldsymbol{r})h_0^*(\boldsymbol{r}^{\prime})\hat{\sigma}_0\delta(\boldsymbol{r}-\boldsymbol{r}^{\prime})\vec{\Psi}^{\dagger T}(\boldsymbol{r})

引入Nambu 空间

为了把哈密顿量写成更简洁的形式，引入矢量

Φ⃗=(Ψ⃗(r)Ψ⃗†T(r))=(Ψ↑(r)Ψ↓(r)Ψ↑†(r)Ψ↓†(r))\begin{aligned}\vec{\Phi}&=\begin{pmatrix}\vec{\Psi}(\boldsymbol{r})\\\vec{\Psi}^{\dagger T}(\boldsymbol{r})\end{pmatrix}\\&=\begin{pmatrix}\Psi_{\uparrow}(\boldsymbol{r})\\\Psi_{\downarrow}(\boldsymbol{r})\\\Psi_{\uparrow}^{\dagger}(\boldsymbol{r})\\\Psi_{\downarrow}^{\dagger}(\boldsymbol{r})\end{pmatrix}\end{aligned}

得到

H=12∫dr∫dr′Φ⃗(r)†(h0(r′)δ(r−r′)σ^0Δ^(r,r′)Δ^†(r′,r)−h0(r′)δ(r−r′)σ^0)Φ⃗(r′)\left.H=\frac{1}{2}\int d\boldsymbol{r}\int d\boldsymbol{r}^{\prime}\vec{\Phi}(\boldsymbol{r})^\dagger\left(\begin{array}{cc}h_0(\boldsymbol{r}^{\prime})\delta(\boldsymbol{r}-\boldsymbol{r}^{\prime})\hat{\sigma}_0&\hat{\Delta}(\boldsymbol{r},\boldsymbol{r}^{\prime})\\\hat{\Delta}^\dagger(\boldsymbol{r}^{\prime},\boldsymbol{r})&-h_0(\boldsymbol{r}^{\prime})\delta(\boldsymbol{r}-\boldsymbol{r}^{\prime})\hat{\sigma}_0\end{array}\right.\right)\vec{\Phi}(\boldsymbol{r}^{\prime})

Bogoliubov变换

需要求解的薛定谔方程

∫dr′(h0(r′)δ(r−r′)σ^0Δ^(r,r′)Δ^†(r′,r)−h0(r′)δ(r−r′)σ^0)(fν(r′)gν(r′))=Eν(fν(r)gν(r))\int d\boldsymbol{r}^{\prime}\begin{pmatrix}h_0(\boldsymbol{r}^{\prime})\delta(\boldsymbol{r}-\boldsymbol{r}^{\prime})\hat{\sigma}_0&\hat{\Delta}(\boldsymbol{r},\boldsymbol{r}^{\prime})\\\hat{\Delta}^\dagger(\boldsymbol{r}^{\prime},\boldsymbol{r})&-h_0(\boldsymbol{r}^{\prime})\delta(\boldsymbol{r}-\boldsymbol{r}^{\prime})\hat{\sigma}_0\end{pmatrix}\begin{pmatrix}\boldsymbol{f}_\nu(\boldsymbol{r}^{\prime})\\\boldsymbol{g}_\nu(\boldsymbol{r}^{\prime})\end{pmatrix}=E_\nu\begin{pmatrix}\boldsymbol{f}_\nu(\boldsymbol{r})\\\boldsymbol{g}_\nu(\boldsymbol{r})\end{pmatrix}

(hˇ0ΔˇΔˇ†−hˇ0(r′))(f⃗νg⃗ν)=Eν(f⃗νg⃗ν)\left.\left(\begin{array}{cc}\check{h}_0&\check{\Delta}\\\check{\Delta}^\dagger&-\check{h}_0(\boldsymbol{r}^{\prime})\end{array}\right.\right)\begin{pmatrix}\vec{f}_\nu\\\vec{g}_\nu\end{pmatrix}=E_\nu\begin{pmatrix}\vec{f}_\nu\\\vec{g}_\nu\end{pmatrix}

H=(hˇ0ΔˇΔˇ†−hˇ0),ϕ⃗ν=(f⃗νg⃗ν)\mathcal{H}=\begin{pmatrix}\check{h}_0&\check{\Delta}\\\check{\Delta}^\dagger&-\check{h}_0\end{pmatrix},\vec{\phi}_{\nu}=\begin{pmatrix}\vec{f}_{\nu}\\\vec{g}_{\nu}\end{pmatrix}

Hϕ⃗ν=Eνϕ⃗ν\mathcal{H}\vec{\phi}_{\nu}=E_{\nu}\vec{\phi}_{\nu}

由于矩阵HH是厄米矩阵，所以本征矢量组构成完全系。此时，各个归一化的本征矢量相互正交，即

ϕ⃗ν⋅ϕ⃗μ=δνμ\vec{\phi}_\nu\cdot\vec{\phi}_\mu=\delta_{\nu\mu}

∫dr′(fν†(r′)gν†(r′))(fμ(r′)gμ(r′))=δνμ∫dr′(fν↑∗(r′)fν↓∗(r′)gν↑∗(r′)gν↓∗(r′))(fμ↑(r′)fμ↓(r′)gμ↑(r′)gμ↓(r′))=δνμ\begin{aligned}\int d\boldsymbol{r}^{\prime}\begin{pmatrix}{f_{\nu}^{\dagger}(\boldsymbol{r}^{\prime})}&{\boldsymbol{g}_{\nu}^{\dagger}(\boldsymbol{r}^{\prime})}\end{pmatrix}\begin{pmatrix}{f_{\mu}(\boldsymbol{r}^{\prime})}\\{\boldsymbol{g}_{\mu}(\boldsymbol{r}^{\prime})}\end{pmatrix}&=\delta_{\nu\mu}\\\left.\int dr^{\prime}\left(\begin{array}{ccc}f_{\nu\uparrow}^{*}(r^{\prime})&f_{\nu\downarrow}^{*}(r^{\prime})&g_{\nu\uparrow}^{*}(r^{\prime})&g_{\nu\downarrow}^{*}(r^{\prime})\end{array}\right.\right)\begin{pmatrix}f_{\mu\uparrow}(r^{\prime})\\f_{\mu\downarrow}(r^{\prime})\\g_{\mu\uparrow}(r^{\prime})\\g_{\mu\downarrow}(r^{\prime})\end{pmatrix}&=\delta_{\nu\mu}\end{aligned}

当ν=μν=μ时，有

∫dr∑σ(∣fνσ(r)∣2+∣gνσ(r)∣2)=1\int d\boldsymbol{r}\sum_{\sigma}\left(|f_{\nu\sigma}(\boldsymbol{r})|^2+|g_{\nu\sigma}(\boldsymbol{r})|^2\right)=1

定义

a(r)=∑ν[cν(fν↑(r)fν↓(r)gν↑(r)gν↓(r))+dν(gν↓∗(r)gν↓∗(r)fν↑∗(r)fν↓∗(r))\left.\boldsymbol{a}(\boldsymbol{r})=\sum_\nu\left[c_\nu\left(\begin{array}{c}f_{\nu\uparrow}(\boldsymbol{r})\\f_{\nu\downarrow}(\boldsymbol{r})\\g_{\nu\uparrow}(\boldsymbol{r})\\g_{\nu\downarrow}(\boldsymbol{r})\end{array}\right.\right.\right)+d_\nu\begin{pmatrix}g_{\nu\downarrow}^*(\boldsymbol{r})\\g_{\nu\downarrow}^*(\boldsymbol{r})\\f_{\nu\uparrow}^*(\boldsymbol{r})\\f_{\nu\downarrow}^*(\boldsymbol{r})\end{pmatrix}

Φ⃗(r)=(Ψ↑(r)Ψ↓(r)Ψ↑†(r)Ψ↓†(r))=∑ν[αν(fν↑(r)fν↓(r)gν↑(r)gν↓(r))+βν(gν↑∗(r)gν↓∗(r)fν↑∗(r)fν↓∗(r))]\vec{\Phi}(\boldsymbol{r})=\begin{pmatrix}\Psi_\uparrow(\boldsymbol{r})\\\Psi_\downarrow(\boldsymbol{r})\\\Psi_\uparrow^\dagger(\boldsymbol{r})\\\Psi_\downarrow^\dagger(\boldsymbol{r})\end{pmatrix}=\sum_\nu\left[\alpha_\nu\begin{pmatrix}f_{\nu\uparrow}(\boldsymbol{r})\\f_{\nu\downarrow}(\boldsymbol{r})\\g_{\nu\uparrow}(\boldsymbol{r})\\g_{\nu\downarrow}(\boldsymbol{r})\end{pmatrix}+\beta_\nu\begin{pmatrix}g_{\nu\uparrow}^*(\boldsymbol{r})\\g_{\nu\downarrow}^*(\boldsymbol{r})\\f_{\nu\uparrow}^*(\boldsymbol{r})\\f_{\nu\downarrow}^*(\boldsymbol{r})\end{pmatrix}\right]

α\alpha是湮灭 f,gf,g的算符， β\beta是湮灭 f∗,g∗f^*,g^*的算符。
Ψσ(r)\Psi_\sigma(r)有两种形式

Ψσ(r)=∑ν∫[ανfνσ(r)+βνgνσ∗(r)]Ψσ(r)=(Ψσ†(r))†=∑ν[ανgνσ(r)+βνfνσ∗(r)]†=∑[αν†gνσ∗(r)+βν†fνσ(r)]\begin{aligned}\Psi_{\sigma}(r)&=\sum_{\nu}\int[\alpha_{\nu}f_{\nu\sigma}(r)+\beta_{\nu}g_{\nu\sigma}^{*}(r)]\\\Psi_{\sigma}(r)&=\left(\Psi_\sigma^\dagger(r)\right)^\dagger\\&=\sum_{\nu}\left[\alpha_{\nu}g_{\nu\sigma}(r)+\beta_{\nu}f_{\nu\sigma}^{*}(r)\right]^{\dagger}\\&=\sum\left[\alpha_{\nu}^{\dagger}g_{\nu\sigma}^{*}(r)+\beta_{\nu}^{\dagger}f_{\nu\sigma}(r)\right]\end{aligned}

若 βν=αν†\beta_\nu=\alpha_\nu^\dagger

Φ⃗(r)=∑ν[ανa⃗ν+(r)+αν†a⃗ν−(r)]\vec{\Phi}(r)=\sum_{\nu}\left[\alpha_{\nu}\vec{a}_{\nu+}(\boldsymbol{r})+\alpha_{\nu}^{\dagger}\vec{a}_{\nu-}(\boldsymbol{r})\right]

a⃗ν+(r)=(fν↑(r)fν↓(r)gν↑(r)gν↓(r))a⃗ν−(r)=(gν↑∗(r)gν↓∗(r)fν↑∗(r)fν↓∗(r))\begin{aligned}\vec{a}_{\nu+}(r)&\left.=\left(\begin{array}{c}{f_{\nu\uparrow}(r)}\\{f_{\nu\downarrow}(r)}\\{g_{\nu\uparrow}(r)}\\{g_{\nu\downarrow}(r)}\end{array}\right.\right)\\\vec{a}_{\nu-}(r)&\left.=\left(\begin{array}{c}g_{\nu\uparrow}^*(\boldsymbol{r})\\g_{\nu\downarrow}^*(\boldsymbol{r})\\f_{\nu\uparrow}^*(\boldsymbol{r})\\f_{\nu\downarrow}^*(\boldsymbol{r})\end{array}\right.\right)\end{aligned}

Bogoliubov-de Gennes 方程

将 Φ⃗(r)\vec{\Phi}(r)带入哈密顿量

H=12∫dr∫dr′∑ν[αν†a⃗ν+†(r)+ανa⃗ν−†(r)](h0(r′)δ(r−r′)σ^0Δ^(r,r′)Δ^†(r′,r)−h0(r′)δ(r−r′)σ^0)∑ν′[αν′a⃗ν′+(r′)+αν′†a⃗ν′−(r′)]=12∫dr∑ν,ν′∫dr′[αλ†aˉν+†(r)+ανaˉν−†(r)](h0(r′)δ(r−r′)σ^0Δ^(r,r′)Δ^†(r′,r)−h0(r′)δ(r−r′)σ^0)[αν′aˉν′+(r′)+αν′†aˉν′−(r′)]=12∫dr∑ν,ν′[αν†aˉν+†(r)+ανaˉν−†(r)][αν′Eν′aˉν′+(r)−αν′†Eν′aˉν′−(r)]\begin{aligned}H&=\frac{1}{2}\int dr\int d\boldsymbol{r}^{\prime}\sum_{\nu}\left[\alpha_{\nu}^{\dagger}\vec{a}_{\nu+}^{\dagger}(\boldsymbol{r})+\alpha_{\nu}\vec{a}_{\nu-}^{\dagger}(\boldsymbol{r})\right]\begin{pmatrix}h_0(\boldsymbol{r}^{\prime})\delta(\boldsymbol{r}-\boldsymbol{r}^{\prime})\hat{\sigma}_0&\hat{\Delta}(\boldsymbol{r},\boldsymbol{r}^{\prime})\\\hat{\Delta}^{\dagger}(\boldsymbol{r}^{\prime},\boldsymbol{r})&-h_0(\boldsymbol{r}^{\prime})\delta(\boldsymbol{r}-\boldsymbol{r}^{\prime})\hat{\sigma}_0\end{pmatrix}\sum_{\nu^{\prime}}\left[\alpha_{\nu^{\prime}}\vec{a}_{\nu^{\prime}+}(\boldsymbol{r}^{\prime})+\alpha_{\nu^{\prime}}^{\dagger}\vec{a}_{\nu^{\prime}-}(\boldsymbol{r}^{\prime})\right]\\&=\frac{1}{2}\int dr\sum_{\nu,\nu'}\int dr'\left[\alpha_{\lambda}^{\dagger}\bar{a}_{\nu+}^{\dagger}(\boldsymbol{r})+\alpha_{\nu}\bar{a}_{\nu-}^{\dagger}(\boldsymbol{r})\right]\begin{pmatrix}h_{0}(\boldsymbol{r}')\delta(\boldsymbol{r}-\boldsymbol{r}')\hat{\sigma}_{0}&\hat{\Delta}(\boldsymbol{r},\boldsymbol{r}')\\\hat{\Delta}^{\dagger}(\boldsymbol{r}',\boldsymbol{r})&-h_{0}(\boldsymbol{r}')\delta(\boldsymbol{r}-\boldsymbol{r}')\hat{\sigma}_{0}\end{pmatrix}\left[\alpha_{\nu'}\bar{a}_{\nu'+}(\boldsymbol{r}')+\alpha_{\nu'}^{\dagger}\bar{a}_{\nu'-}(\boldsymbol{r}')\right]\\&=\frac{1}{2}\int dr\sum_{\nu,\nu'}\left[\alpha_{\nu}^{\dagger}\bar{a}_{\nu+}^{\dagger}(\boldsymbol{r})+\alpha_{\nu}\bar{a}_{\nu-}^{\dagger}(\boldsymbol{r})\right]\left[\alpha_{\nu'}E_{\nu'}\bar{a}_{\nu'+}(\boldsymbol{r})-\alpha_{\nu'}^{\dagger}E_{\nu'}\bar{a}_{\nu'-}(\boldsymbol{r})\right]\end{aligned}

利用各个本征函数的正交性

H=12∫dr∑ν,ν′[αν†aˉν+†(r)+ανaˉν−†(r)][αν′Eν′aˉν′+(r)−αν′†Eν′aˉν′−(r)]=12∑ν[Eναν†αν−Eναναν†]\begin{aligned} H &= \frac{1}{2} \int d\boldsymbol{r} \sum_{\nu,\nu'} \left[ \alpha_{\nu}^{\dagger} \bar{a}_{\nu +}^{\dagger}(\boldsymbol{r}) + \alpha_{\nu} \bar{a}_{\nu -}^{\dagger}(\boldsymbol{r}) \right] \left[ \alpha_{\nu'} E_{\nu'} \bar{a}_{\nu' +}(\boldsymbol{r}) - \alpha_{\nu'}^{\dagger} E_{\nu'} \bar{a}_{\nu' -}(\boldsymbol{r}) \right] \\ &= \frac{1}{2} \sum_{\nu} \left[ E_{\nu} \alpha_{\nu}^{\dagger} \alpha_{\nu} - E_{\nu} \alpha_{\nu} \alpha_{\nu}^{\dagger} \right] \end{aligned}

如果 ανα _ν满足费米子的反对易关系

H=∑ναν†ανH=\sum_{\nu}\alpha_{\nu}^{\dagger}\alpha_{\nu}

其中Eν=(ϵk⃗−μ)2+∣Δ∣2E_{\nu}= \sqrt{(\epsilon_{\vec{k}} - \mu)^2 + |\Delta|^2}
是准粒子的能量，ϵkϵ_k是电子的动能，μμ是化学势。
准粒子能量的表达式揭示了超导能隙的存在，能隙大小为 2∣Δ∣2∣Δ∣。

能隙

为了确定能隙函数ΔΔ，我们需要解自洽方程。在平均场近似下，能隙函数满足：

Δ(k)=−1N∑k′Vkk′⟨c−k′↓ck′↑⟩\Delta(\boldsymbol{k})=-\frac{1}{N}\sum_{k^{\prime}}V_{\boldsymbol{k}\boldsymbol{k}^{\prime}}\left\langle c_{-\boldsymbol{k}^{\prime}\downarrow}c_{\boldsymbol{k}^{\prime}\uparrow}\right\rangle

通过 Bogoliubov 变换，可以证明：

⟨c−k⃗↓ck⃗↑⟩=−uk⃗vk⃗∗\langle c_{-\vec{k}\downarrow} c_{\vec{k}\uparrow} \rangle = -u_{\vec{k}} v_{\vec{k}}^*

假设+E(k)+E(\boldsymbol{k})的本征值,其对应的本征矢量为u(1)(k)=(uk,−vk)u^{(1)}(k)=(u_k,-v_k)，另外一个本征态 u(2)(k)=(uk,vk)u^{(2)}(k)=(u_k,v_k)

uk⃗=12(1+ϵk⃗−μEk⃗)u_{\vec{k}} = \sqrt{\frac{1}{2} \left(1 + \frac{\epsilon_{\vec{k}} - \mu}{E_{\vec{k}}}\right)}

vk⃗=12(1−ϵk⃗−μEk⃗)eiθv_{\vec{k}} = \sqrt{\frac{1}{2} \left(1 - \frac{\epsilon_{\vec{k}} - \mu}{E_{\vec{k}}}\right)} e^{i\theta}

能隙方程改写为：

Δ(k)=−1N∑k′Vkk′Δ(k′)2E(k′)tanh⁡(E(k′)2kBT)\Delta(\boldsymbol{k}) = -\frac{1}{N} \sum_{\boldsymbol{k}'} V_{\boldsymbol{k}\boldsymbol{k}'} \frac{\Delta(\boldsymbol{k}')}{2E(\boldsymbol{k}')} \tanh\left( \frac{E(\boldsymbol{k}')}{2k_B T} \right)