General Seniority 学习笔记(1): BCS

我学习了参考文献 [1,2]，把里面的核心公式推导核对整理了，因为觉得有点意思。做完笔记我顺手写了个代码，还没来得及核对。

下一步可以考虑投影 broken-pair 的优化和投影，即 seniority 取次极小的内秉态做投影。

参考文献：

[1] 贾力源，"Application of the variational principle to a coherent-pair condensate: The BCS case", PRC99, 014302 (2019).

[2] 俞一信 et al., "Nucleon pair truncation of the shell model for medium-heavy nuclei", to be submitted.

1. 时间反演轨道上的配对凝聚

所有轨道和自己的时间反演轨道，进行两两配对 \((\alpha, \bar{\alpha})\)，然后定义时间反演轨道上的配对：

\[\hat{P}^\dagger_\alpha = \hat{c}^\dagger_\alpha \hat{c}^\dagger_\bar{\alpha}. \]

1.1 配对与配对凝聚

配对定义为

\[\hat{P}^\dagger = \sum_{\alpha \in \Theta} \nu_\alpha \hat{P}^\dagger_\alpha, \]

其中 \(\Theta\) 表示非集体对序号集合，序号个数是单粒子轨道数的一半。单粒子轨道可以是形变基，也可以是球形基，只要两两配对即可。General Seniority 的全部变化，就是这样的配对的凝聚、对破缺。

一般假定基态对应着配对凝聚，

\[|\phi_N \rangle = \frac{1}{ \sqrt{\chi_N} } (\hat{P}^\dagger)^N | 0 \rangle, \]

其中 \(\chi_N\) 是归一化因子，

\[\chi_N = \langle 0 | \hat{P}^N (\hat{P}^\dagger)^N | 0 \rangle. \]

Block 掉非集体对 \(\alpha\) 以后，态矢变为

\[|\phi^{[\alpha]}_N \rangle = \frac{1}{\sqrt{\chi^{[\alpha]}_N}}( \hat{P}^\dagger - \nu_\alpha \hat{P}^\dagger_\alpha)^N | 0 \rangle, \]

其中，归一化因子

\[\chi^{[\alpha]}_N = \langle 0 |\hat{P}^N \hat{P}_\alpha \hat{P}^\dagger_\alpha(\hat{P}^\dagger)^N | 0 \rangle, \]

中间加上 \(\hat{P}_\alpha \hat{P}^\dagger_\alpha\) 起的作用就是 block 掉非集体对 \(\alpha\)。

1.2 归一化因子的计算

如果已经 block 掉 \(\gamma_1, \gamma_2, \cdots, \gamma_r\)，则有

\[\chi^{[\gamma_1 \cdots \gamma_r]}_N = N \sum_\alpha \nu^2_\alpha \chi^{[\alpha\gamma_1 \cdots \gamma_r]}_{N-1},　\\ \chi^{[\gamma_1 \cdots \gamma_r]}_N - \chi^{[\alpha\gamma_1 \cdots \gamma_r]}_{N} = N^2 \nu^2_\alpha \chi^{[\alpha\gamma_1 \cdots \gamma_r]}_{N-1}. \]

另外，因为 \(\chi^{[\alpha\gamma_1 \cdots \gamma_r]}_{0} = 0\)，所以可以构造迭代：

\[\left\{\chi^{[\alpha\gamma_1 \cdots \gamma_r]}_{0} \right\} \rightarrow \chi^{[\gamma_1 \cdots \gamma_r]}_{1} \rightarrow \left\{\chi^{[\alpha\gamma_1 \cdots \gamma_r]}_{1} \right\} \rightarrow \cdots \rightarrow\left\{\chi^{[\alpha\gamma_1 \cdots \gamma_r]}_{N-1} \right\} \rightarrow \chi^{[\gamma_1 \cdots \gamma_r]}_{N} \rightarrow \left\{\chi^{[\alpha\gamma_1 \cdots \gamma_r]}_{N} \right\}. \]

也就是说，给定 \(\gamma_1, \cdots, \gamma_r\)，任意粒子 \(N\) 的归一化因子都可以如此迭代计算。因为迭代式非常简单，所以预计是秒出。可以定义矢量

\[\vec{\chi}^{[\gamma_1, \cdots, \gamma_r]}_N \equiv \left( \chi^{[\gamma_1 \cdots \gamma_r]}_{N},\left\{\chi^{[\alpha\gamma_1 \cdots \gamma_r]}_{N} \right\} \right) \]

那么上面的迭代即

\[\vec{\chi}^{[\gamma_1, \cdots, \gamma_r]}_0 = (1,1,\cdots, 1), \\ \vec{\chi}^{[\gamma_1, \cdots, \gamma_r]}_{N-1} \rightarrow \vec{\chi}^{[\gamma_1, \cdots, \gamma_r]}_N. \]

test case: \(\nu_i = 1, i \in \Theta\)

此时 \(\chi_N\) 很容易解析地得到：

\[\chi_N = C^N_\Omega (N!)^2 = \frac{\Omega!N!}{(\Omega-N)!}. \]

2. 矩阵元的计算

通过 many-pair density matrix，可以计算由对算符构造的任意算符矩阵元。

体系哈密顿量为

\[\hat{H} = \sum_{\alpha \beta} \epsilon_{\alpha \beta} \hat{c}^\dagger_\alpha \hat{c}_\beta + \frac{1}{4} \sum_{\alpha\beta\gamma\delta} V_{\alpha\beta\gamma\delta} \hat{c}^\dagger_\alpha \hat{c}^\dagger_\beta \hat{c}_\gamma \hat{c}_\delta, \]

其中 \(V_{\alpha \beta\gamma\delta} = - \langle \alpha \beta | \hat{V} | \gamma \delta \rangle\)，有对称性 \(\epsilon_{\alpha\beta} = \epsilon_{\beta \alpha}, V_{\alpha\beta\gamma\delta} = - V_{\beta\alpha\gamma\delta} = V_{\gamma\delta\alpha\beta}\)。贾师兄在这里假定了 \(\epsilon_{\alpha\beta} = \epsilon_{\bar{\beta}\bar{\alpha}}, V_{\alpha\beta\gamma\delta} = V(\bar{\delta}\bar{\gamma}\bar{\beta}\bar{\alpha})\)，据说这是假定 time-even 哈密顿量。他还假定 \(\epsilon_{\alpha\beta}\) 和 \(V_{\alpha\beta\gamma\delta}\) 是实数。

2.1 many-pair density matrix

若给定 \(\gamma_1, \cdots, \gamma_r\)，要求 block 掉 \(\{ \hat{P}^\dagger_{\gamma_i}\}\) ，然后在这个截断空间里计算 \((\hat{P}^\dagger)^{M-p} \hat{P}^\dagger_{\alpha_1} \cdots \hat{P}^\dagger_{\alpha_p} |0\rangle\) 与 \((\hat{P}^\dagger)^{M-q} \hat{P}^\dagger_{\beta_1} \cdots \hat{P}^\dagger_{\beta_q} |0\rangle\) 的 overlap，

\[t^{[\gamma_1 \gamma_2 \cdots \gamma_r], M}_{\alpha_1, \cdots, \alpha_p; \beta_1, \cdots, \beta_q} \equiv \langle 0 | \hat{P}^{M-p} \hat{P}_{\gamma_1} \cdots \hat{P}_{\gamma_r} \hat{P}_{\alpha_1}\cdots \hat{P}_{\alpha_p} \hat{P}^\dagger_{\beta_1} \cdots \hat{P}^\dagger_{\beta_q} (\hat{P}^\dagger)^{M-q} |0\rangle, \]

这里我们约定 \(\{\alpha_i\}, \{\beta_j\}\)没有共同元素，否则可以移到 \(\{\gamma_i\}\)里面去。这样的话，这么写也可以

\[t^{[\gamma_1 \gamma_2 \cdots \gamma_r], M}_{\alpha_1, \cdots, \alpha_p; \beta_1, \cdots, \beta_q} = \langle 0 | \hat{P}^{M-p} \hat{P}_{\gamma_1} \cdots \hat{P}_{\gamma_r} \hat{P}^\dagger_{\beta_1} \cdots \hat{P}^\dagger_{\beta_q} \hat{P}_{\alpha_1}\cdots \hat{P}_{\alpha_p} (\hat{P}^\dagger)^{M-q} | 0 \rangle, \]

所以可以叫做 many-pair density matrix。容易看出，它可以表示成 \(\chi\) 的表达式：

\[t^{[\gamma_1, \cdots, \gamma_r],M}_{\alpha_1, \cdots, \alpha_p; \beta_1, \cdots, \beta_q} = \nu_{\alpha_1} \cdots \nu_{\alpha_p} \nu_{\beta_1} \cdots \nu_{\beta_q} \frac{(M-p)!(M-q)!}{(M-p-q)!(M-p-q)!} \chi^{[\alpha_1 \cdots \alpha_p \beta_1 \cdots \beta_q \gamma_1 \cdots \gamma_r]}_{M-p-q}. \]

2.2 单体算符矩阵元

这个比较简单，

\[\langle 0 | \hat{P}^N \hat{c}^\dagger_\alpha \hat{c}_\beta (\hat{P}^\dagger)^N | 0 \rangle = \delta_{\alpha\beta} \chi_N \langle \phi_N | \hat{n}_\alpha | \phi_N \rangle = (N\nu_\alpha)^2 \chi^{[\alpha]}_{N-1}. \]

2.3 两体算符矩阵元

只有以下 3 种情况，两体矩阵元不为零：

\[\langle 0 | \hat{P}^N \hat{c}^\dagger_\alpha \hat{c}^\dagger_\bar{\alpha} \hat{c}_\bar{\alpha} \hat{c}_\alpha (\hat{P}^\dagger)^N |0\rangle = (N\nu_\alpha)^2 \chi^{[\alpha]}_{N-1}, \\ \langle 0 | \hat{P}^N \hat{c}^\dagger_\alpha \hat{c}^\dagger_\bar{\alpha} \hat{c}_\bar{\beta} \hat{c}_\beta (\hat{P}^\dagger)^N | 0 \rangle = N^2 \nu_\alpha \nu_\beta \chi^{[\alpha\beta]}_{N-1}, \\ \langle 0 | \hat{P}^N \hat{c}^\dagger_\alpha \hat{c}^\dagger_\beta \hat{c}_\beta \hat{c}_\alpha (\hat{P}^\dagger)^N | 0 \rangle = N^2(N-1)^2 \nu^2_\alpha \nu^2_\beta \chi^{[\alpha\beta]}_{N-2}. \]

2.4 同类核子哈密顿量期望值

哈密顿量定义为

\[\hat{H} = \sum_{\alpha\beta} \epsilon_{\alpha\beta} \hat{c}^\dagger_\alpha \hat{c}_\beta + \frac{1}{4} \sum_{\alpha\beta\gamma\delta} V_{\alpha\beta\gamma\delta} \hat{c}^\dagger_\alpha \hat{c}^\dagger_\beta \hat{c}_\delta \hat{c}_\gamma. \]

\[\langle \phi_N | \hat{H} | \phi_N \rangle = \sum_{\alpha \in \Theta} 2 \epsilon_{\alpha \alpha} \langle \phi_N | \hat{c}^\dagger_\alpha \hat{c}_\alpha | \phi_N \rangle + \sum_{\alpha \in \Theta } V_{\alpha \bar{\alpha} \alpha \bar{\alpha} } \langle \phi_N | \hat{c}^\dagger_\alpha \hat{c}^\dagger_\bar{\alpha} \hat{c}_\bar{\alpha} \hat{c}_\alpha | \phi_N \rangle \\ + \sum^{\alpha \neq \beta}_{\alpha, \beta \in \Theta} V_{\alpha \bar{\alpha} \beta \bar{\beta} } \langle \phi_N | \hat{c}^\dagger_\alpha \hat{c}^\dagger_\bar{\alpha} \hat{c}_\bar{\beta} \hat{c}_\beta | \phi_N \rangle + \sum^{\alpha \neq \beta}_{\alpha, \beta \in \Theta} (2V_{\alpha \beta \alpha \beta } + 2V_{\alpha \bar{\beta} \alpha \bar{\beta} } ) \langle \phi_N |\hat{c}^\dagger_\alpha \hat{c}^\dagger_\beta \hat{c}_\beta \hat{c}_\alpha | \phi_N \rangle. \]

将上一小节的结果代入上式，若记 \(G_{\alpha\beta} = V_{\alpha\bar{\alpha}\beta\bar{\beta}}, \Lambda_{\alpha\beta} = V_{\alpha\beta\alpha\beta} + V_{\alpha \bar{\beta} \alpha \bar{\beta} }\)，得到

\[\bar{E} \equiv \langle \phi_N | \hat{H} | \phi_N \rangle = \frac{N^2}{\chi_N} \left( \sum_{\alpha \in \Theta} (2\epsilon_{\alpha\alpha} + G_{\alpha\alpha}) \nu^2_\alpha \chi^{[\alpha]}_{N-1} + \sum^{\alpha < \beta}_{\alpha, \beta \in \Theta} 2G_{\alpha \beta} \nu_\alpha \nu_\beta \chi^{[\alpha\beta]}_{N-1} + (N-1)^2 \sum^{\alpha < \beta}_{\alpha, \beta \in \Theta} 2 \Lambda_{\alpha\beta} \nu^2_\alpha \nu^2_\beta \chi^{[\alpha\beta]}_{N-2} \right). \]

这个与贾力源的公式一致，只是我的后两个求和符号上有 \(\alpha < \beta\)，导致相应的有个 2 倍。最后一项有点 tricky，本就只需考虑 \(\sum^{\alpha<\beta}_{\alpha,\beta \in \Theta}\) 的情况，\(V_{\alpha\beta\beta\alpha}, V_{\alpha \bar{\beta} \alpha \bar{\beta} }, V_{\bar{\alpha} \bar{\beta} \bar{\alpha} \bar{\beta} }, V_{\bar{\alpha} \beta \bar{\alpha} \beta }\)，这四项+时间反演对称，得到 \(2\Lambda_{\alpha\beta}\)。

感觉可以设置 \(\nu_i = 1\)，来做简单测试。

2.5 同类核子哈密顿量期望值的偏导数

先算 overlap：

\[\delta \chi_N = 2N \langle 0 | \hat{P}^N \hat{P}^\dagger_\alpha (\hat{P}^\dagger)^{N-1} |0\rangle \delta \nu_\alpha = 2 N^2 \nu_{\alpha}\chi^{[\alpha]}_{N-1} \delta \nu_\alpha = \frac{ 2\chi_N}{ \nu_\alpha} \langle \phi_N | \hat{n}_\alpha |\phi_N\rangle\delta \nu_\alpha. \]

这可以推广： $ \alpha \notin { \gamma_1, \cdots , \gamma_r }$ 时，

\[\delta \chi^{[\gamma_1 \cdots \gamma_r]}_N = 2 N^2 \nu_{\alpha}\chi^{[\alpha \gamma_1 \cdots \gamma_r]}_{N-1} \delta \nu_\alpha. \]

Anyway, \(\chi_N\) 的偏导数有了

\[\frac{ \partial }{\partial \nu_\alpha} \chi_N = 2 N^2 \nu_{\alpha}\chi^{[\alpha]}_{N-1}. \]

考虑哈密顿量矩阵元的偏导数，

\[\delta \langle \hat{P}^N | \hat{H} |(\hat{P}^\dagger)^N \rangle = 2N \langle \hat{P}^{N-1} \hat{P}_\alpha | \hat{H} | (\hat{P}^\dagger)^N \rangle \delta \nu_\alpha， \]

下面计算 \(\langle \hat{P}^{N-1} \hat{P}_\alpha | \hat{H} |(\hat{P}^\dagger)^N \rangle\) ，可以如下从容计算。

• 先考虑 \(\hat{H}\) 中不含 \(\alpha\) 的项的贡献，得到

\[N \nu_\alpha \langle (\hat{P} - \nu_\alpha \hat{P}_\alpha)^{N-1} | \hat{H} | (\hat{P}^\dagger - \nu_\alpha \hat{P}^\dagger_\alpha)^{N-1} \rangle = N \nu_\alpha \langle \phi^{[\alpha]}_{N-1} | \hat{H} | \phi^{[\alpha]}_{N-1} \rangle \chi^{[\alpha]}_{N-1}. \]

• 再分别考虑 \(\hat{H}\) 中包含 \(\alpha\) 的各项贡献。

  • 源于 \(\hat{c}^\dagger_\alpha \hat{c}_\alpha, \hat{c}^\dagger_\bar{\alpha} \hat{c}_\bar{\alpha}, \hat{P}^\dagger_\alpha \hat{P}_\alpha\)形式的相互作用，与\(2\epsilon_{\alpha \alpha} + G_{\alpha \alpha}\) 相关的贡献（对单粒子能）为

    \[(2\epsilon_{\alpha \alpha} + G_{\alpha\alpha}) N \nu_\alpha \chi^{[\alpha]}_{N-1}; \]

  • 源于 \(\hat{P}^\dagger_\alpha \hat{P}_\beta\) 形式的相互作用，与 \(G_{\alpha \beta}\) 有关的贡献（对散射？）为

    \[\sum_{\beta \neq \alpha} N \nu_\beta G_{\alpha\beta} \chi^{[\alpha\beta]}_{N-1}; \]

  • 源于 \(\hat{c}^\dagger_\alpha \hat{c}_\alpha \hat{c}^\dagger_\beta \hat{c}_\beta\) 形式的相互作用，与 \(\Lambda_{\alpha\beta}\)有关的贡献为

    \[\sum_{\beta \neq \alpha} 2 N(N-1)^2 \nu_\alpha \nu^2_\beta \chi^{[\alpha\beta]}_{N-2} \Lambda_{\alpha\beta}, \]

  ​ 这里有个 2 倍，因为相互作用可以是 \(\hat{c}^\dagger_\alpha \hat{c}_\alpha \hat{c}^\dagger_\beta \hat{c}_\beta\)，也可以是 \(\hat{c}^\dagger_\beta \hat{c}_\beta \hat{c}^\dagger_\alpha \hat{c}_\alpha\)。

所以最终得到，

\[\langle \hat{P}^{N-1} \hat{P}_\alpha | \hat{H} |(\hat{P}^\dagger)^N \rangle = N \nu_\alpha \langle \phi^{[\alpha]}_{N-1} | \hat{H} | \phi^{[\alpha]}_{N-1} \rangle \chi^{[\alpha]}_{N-1} + (2\epsilon_{\alpha \alpha} + G_{\alpha\alpha}) N \nu_\alpha \chi^{[\alpha]}_{N-1} \\ + \sum_{\beta \neq \alpha} N \nu_\beta G_{\alpha\beta} \chi^{[\alpha\beta]}_{N-1} + \sum_{\beta \neq \alpha} 2 N(N-1)^2 \nu_\alpha \nu^2_\beta \chi^{[\alpha\beta]}_{N-2} \Lambda_{\alpha\beta}, \]

所以得到了能量偏导数

\[ \frac{\partial}{\partial \nu_\alpha} \bar{E} = - \frac{1}{\chi^2_N} 2N^2 \nu_\alpha \chi^{[\alpha]}_{N-1} \langle \hat{P}^N| \hat{H} |(\hat{P}^\dagger)^N \rangle \\ + \frac{2N^2}{\chi_N} \left\{ \nu_\alpha \langle \phi^{[\alpha]}_{N-1}| \hat{H} | \phi^{[\alpha]}_{N-1} \rangle \chi^{[\alpha]}_{N-1} + (2\epsilon_{\alpha \alpha} + G_{\alpha\alpha}) \nu_\alpha \chi^{[\alpha]}_{N-1} + \sum_{\beta \neq \alpha} \nu_\beta G_{\alpha\beta} \chi^{[\alpha\beta]}_{N-1} + \sum_{\beta \neq \alpha} 2 (N-1)^2 \nu_\alpha\nu^2_\beta \chi^{[\alpha\beta]}_{N-2} \Lambda_{\alpha\beta} \right\} \\ = \frac{2N^2}{\chi_N} \left\{ - \nu_\alpha \chi^{[\alpha]}_{N-1} \bar{E} + \nu_\alpha \langle \phi^{[\alpha]}_{N-1}| \hat{H} | \phi^{[\alpha]}_{N-1}\rangle \chi^{[\alpha]}_{N-1} + (2\epsilon_{\alpha \alpha} + G_{\alpha\alpha}) \nu_\alpha \chi^{[\alpha]}_{N-1} + \sum_{\beta \neq \alpha} \nu_\beta G_{\alpha\beta} \chi^{[\alpha\beta]}_{N-1} + \sum_{\beta \neq \alpha} 2 (N-1)^2 \nu_\alpha \nu^2_\beta \chi^{[\alpha\beta]}_{N-2} \Lambda_{\alpha\beta} \right\} \\ = \frac{2N^2}{\chi_N} \left\{ \nu_\alpha \chi^{[\alpha]}_{N-1} [ - \bar{E} + \langle \phi^{[\alpha]}_{N-1} | \hat{H} | \phi^{[\alpha]}_{N-1} \rangle + 2\epsilon_{\alpha \alpha} + G_{\alpha\alpha} ] + \sum_{\beta \neq \alpha} \nu_\beta G_{\alpha\beta} \chi^{[\alpha\beta]}_{N-1} + \sum_{\beta \neq \alpha} 2 (N-1)^2 \nu_\alpha \nu^2_\beta \chi^{[\alpha\beta]}_{N-2} \Lambda_{\alpha\beta} \right\} \]

有了 \(\bar{E}\) 和 \(\frac{\partial}{\partial \nu_\alpha} \bar{E}\) 以后，就可以做变分了，改变 \(\nu_\alpha\)，指导 \(\bar{E}\) 取得极小值。

为了使得这个式子看起来紧凑一点，贾师兄[1]约定：

\[d_\alpha = 2 \epsilon_{\alpha\alpha} + G_{\alpha\alpha} + 2(N-1)^2 \sum^{\beta \neq \alpha}_{\beta} \Lambda_{\alpha\beta} \nu^2_\beta \frac{ \chi^{[\alpha\beta]}_{N-2}}{\chi^{[\alpha]}_{N-1}}, \]

那么就有

\[\nu_\alpha = \frac{ - \sum_{\beta \neq \alpha} \nu_\beta G_{\alpha\beta} \chi^{[\alpha\beta]}_{N-1} }{ \chi^{[\alpha]}_{N-1} [ d_\alpha + \langle \phi^{[\alpha]}_{N-1} | \hat{H} | \phi^{[\alpha]}_{N-1} \rangle - \bar{E} ] }. \]

然后可以整个迭代式，即 \(\partial \bar{E} / \partial \nu_\alpha = 0\)，得到

\[\nu_\alpha = \frac{ - \sum_{\beta \neq \alpha} \nu_\beta G_{\alpha\beta} \chi^{[\alpha\beta]}_{N-1} }{ \chi^{[\alpha]}_{N-1} [ d_\alpha + \langle \phi^{[\alpha]}_{N-1} | \hat{H} | \phi^{[\alpha]}_{N-1} \rangle - \bar{E} ] }. \]

2.6 质子-中子相互作用

上面的公式都是关乎同类核子的，如果哈密顿量中存在质子-中子相互作用，即

\[\hat{H}_{\pi\nu} = \sum_{\alpha_\pi \beta_\nu \gamma_\pi \delta_\nu} V_{\alpha_\pi \beta_\nu \gamma_\pi \delta_\nu} \hat{c}^\dagger_{\alpha_\pi} \hat{c}^\dagger_{\beta_\nu} \hat{c}_{\delta_\nu} \hat{c}_{\gamma_\pi}, \]

对于波函数

\[|\Phi\rangle = | \phi_{n_\pi} \rangle |\phi_{n_\nu} \rangle, \]

有

\[\langle \Phi | \hat{H}_{\pi \nu} |\Phi\rangle = \frac{1}{\chi_{n_\pi} \chi_{n_\nu}} \sum_{\alpha_\pi \beta_\nu } (n_\pi \nu_\alpha)^2 (n_\nu \nu_\beta)^2 \chi^{[\alpha]}_{n_\pi -1} \chi^{[\beta]}_{n_\nu -1 } (V_{\alpha_\pi \beta_\nu \alpha_\pi \beta_\nu} + V_{\alpha_\pi \bar{\beta}_\nu \alpha_\pi \bar{\beta}_\nu} + V_{\bar{\alpha}_\pi \beta_\nu \bar{\alpha}_\pi \beta_\nu} + V_{\bar{\alpha}_\pi \bar{\beta}_\nu \bar{\alpha}_\pi \bar{\beta}_\nu} ) \\ = \frac{1}{\chi_{n_\pi} \chi_{n_\nu}} \sum_{\alpha_\pi \beta_\nu } (n_\pi \nu_\alpha)^2 (n_\nu \nu_\beta)^2 \chi^{[\alpha]}_{n_\pi -1} \chi^{[\beta]}_{n_\nu -1 } 2(V_{\alpha_\pi \beta_\nu \alpha_\pi \beta_\nu} + V_{\alpha_\pi \bar{\beta}_\nu \alpha_\pi \bar{\beta}_\nu} ), \]

第二个等号利用了 \(V\) 的时间反演对称性，即 \(V_{\alpha\beta\gamma\delta} = V_{\bar{\gamma}\bar{\delta} \bar{\beta} \bar{\alpha}}\)。

稍微写紧凑点，记 \(\Lambda_{\alpha_\pi \beta_\nu} = 2(V_{\alpha_\pi \beta_\nu \alpha_\pi \beta_\nu} + V_{\alpha_\pi \bar{\beta}_\nu \alpha_\pi \bar{\beta}_\nu})\)，利用 \((n_\pi \nu_\alpha)^2 \chi^{[\alpha]}_{n_\pi -1} = \chi_{n_\pi} - \chi^{[\alpha]}_{n_\pi}\)，得到[2],

\[\langle \Phi | \hat{H}_{\pi \nu} |\Phi\rangle = \sum_{\alpha_\pi \beta_\nu } 2 \Lambda_{\alpha_\pi \beta_\nu} (1 - \frac{\chi^{[\alpha]}_{n_\pi}}{\chi_{n_\pi}}) (1 - \frac{\chi^{[\alpha]}_{n_\nu}}{\chi_{n_\nu}}). \]

这个形式确实好看。

2.7 \(\langle\Phi | \hat{H}_{\pi \nu} | \Phi \rangle\)的偏导数

\[\langle \hat{P}^{n_\pi-1} \hat{P}_\alpha; \phi_{n_\nu} | \hat{H}_{\pi\nu} | \hat{P}^{n_\pi}; \phi_{n_\nu} \rangle = \frac{1}{ \chi_{n_\nu}} \sum_{ \beta_\nu } n_\pi \nu_{\alpha_\pi} \chi^{[\alpha_\pi]}_{n_\pi -1} (n_\nu \nu_{\beta_\nu})^2 \chi^{[\beta_\nu]}_{n_\nu -1 } 2\Lambda_{\alpha_\pi \beta_\nu} \\ + \frac{1}{ \chi_{n_\nu}} \sum^{\gamma_\pi \neq \alpha_\pi}_{ \gamma_\pi \beta_\nu } n_\pi \nu_{\alpha_\pi}(n_\pi-1)^2 \nu^2_{\gamma_\pi} \chi^{[\alpha_\pi \gamma_\pi]}_{n_\pi -2} (n_\nu \nu_\beta)^2 \chi^{[\beta]}_{n_\nu -1 } 2 \Lambda_{\gamma_\pi \beta_\nu} ). \]

其中 \(\Lambda_{\alpha_\pi \beta_\nu} = V_{\alpha_\pi \beta_\nu \alpha_\pi \beta_\nu} + V_{\alpha_\pi \bar{\beta}_\nu \alpha_\pi \bar{\beta}_\nu}\)，所以有

\[\frac{\partial}{\partial \nu_\alpha} \langle \Phi | \hat{H}_{\pi \nu} | \Phi \rangle = \frac{1}{ \chi_{n_\pi} \chi_{n_\nu}} \sum_{ \beta_\nu } 2 n^2_\pi \nu_{\alpha_\pi} \chi^{[\alpha_\pi]}_{n_\pi -1} (n_\nu \nu_{\beta_\nu})^2 \chi^{[\beta_\nu]}_{n_\nu -1 } 2\Lambda_{\alpha_\pi \beta_\nu} \\ + \frac{1}{ \chi_{n_\pi} \chi_{n_\nu}} \sum^{\gamma_\pi \neq \alpha_\pi}_{ \gamma_\pi \beta_\nu } 2n^2_\pi \nu_{\alpha_\pi}(n_\pi-1)^2 \nu^2_{\gamma_\pi} \chi^{[\alpha_\pi \gamma_\pi]}_{n_\pi -2} (n_\nu \nu_\beta)^2 \chi^{[\beta]}_{n_\nu -1 } 2 \Lambda_{\gamma_\pi \beta_\nu} \\ - \frac{2 n^2_\pi \nu_{\alpha_\pi} \chi^{[\alpha_\pi]}_{n_\pi-1}}{\chi_{n_\pi}} \langle \Phi | \hat{H}_{\pi\nu} | \Phi \rangle. \]

用 (37) 式，以及 \(\frac{ \partial }{\partial \nu_\alpha} \chi_N = 2 N^2 \nu_{\alpha}\chi^{[\alpha]}_{N-1}\), \(\frac{ \partial }{\partial \nu_\alpha} \chi^{[\gamma]}_N = 2 N^2 \nu_{\alpha}\chi^{[\alpha\gamma]}_{N-1}\)，得到

\[\partial E_{\pi\nu} / \partial \nu_{\alpha_\pi} = \frac{2n^2_\pi \nu_\alpha}{\chi^2_{n_\pi}} \sum_{\gamma_\pi \beta_\nu} 2 \Lambda_{\gamma_\pi \beta_\nu} ( \chi^{[\gamma]}_{n_\pi} \chi^{[\alpha]}_{n_\pi -1} - \chi_{n_\pi} \chi^{[\alpha\gamma]}_{n_\pi -1})( 1- \frac{ \chi^{[\beta]}_{n_\nu}}{\chi_{n_\nu}}), \]

\(\partial E_{\pi\nu} / \partial \nu_{\beta_\nu}\)也可以类似地写出。因此，整个体系的能量 \(E = E_\pi + E_\nu + E_{\pi\nu}\) 的偏导数可以写出

\[\frac{\partial E}{\partial \nu_{\alpha_\pi} } = \frac{\partial E_\pi}{\partial \nu_{\alpha_\pi}} + \frac{ \partial E_{\pi\nu}}{\partial \nu_{\alpha_\pi}} \\ = \frac{ 2n^2_\pi }{ \chi_{n_\pi} } \left\{ \nu_{\alpha_\pi} \chi^{[\alpha_\pi]}_{n_\pi-1} [ d_{\alpha_\pi} + \langle \phi^{[\alpha_\pi]}_{n_\pi-1} | \hat{H}_\pi | \phi^{[\alpha_\pi]}_{n_\pi - 1}\rangle - \bar{E}_\pi ] + \sum_{\beta_\pi \neq \alpha_\pi} \nu_{\beta_\pi} G_{\alpha_\pi \beta_\pi} \chi^{[\alpha_\pi \beta_\pi ]}_{n_\pi - 1} + \sum_{\gamma_\pi \beta_\nu} 2 \Lambda_{\gamma_\pi \beta_\nu} \frac{ \chi^{[\gamma_\pi]}_{n_\pi} \chi^{[\alpha_\pi]}_{n_\pi -1} - \chi_{n_\pi} \chi^{[\alpha_\pi \gamma_\pi]}_{n_\pi -1}}{\chi_{n_\pi}}( 1- \frac{ \chi^{[\beta_\nu]}_{n_\nu}}{\chi_{n_\nu}}) \right\} \]

把这个式子做 \(\pi \leftrightarrow \nu\) 替换，就得到 \(\partial E / \partial \nu_{\beta_\nu}\) 的值。

如果整一个迭代式，即 \(\partial E / \partial \nu_{\alpha_\pi} = 0\)，得到

\[\nu_{\alpha_\pi} = \frac{ - \sum_{\beta_\pi \neq \alpha_\pi} \nu_{\beta_\pi} G_{\alpha_\pi \beta_\pi} \chi^{[\alpha_\pi \beta_\pi]}_{n_\pi - 1} }{ \chi^{[\alpha_\pi]}_{n_\pi - 1} [ d_{\alpha_\pi} + \langle \phi^{[\alpha_\pi]}_{n_\pi - 1} | \hat{H}_\pi | \phi^{[\alpha_\pi]}_{n_\pi - 1} \rangle - \bar{E}_\pi ] + \sum_{\gamma_\pi \beta_\nu} 2 \Lambda_{\gamma_\pi \beta_\nu} \frac{ \chi^{[\gamma_\pi]}_{n_\pi} \chi^{[\alpha_\pi]}_{n_\pi -1} - \chi_{n_\pi} \chi^{[\alpha_\pi \gamma_\pi]}_{n_\pi -1}}{\chi_{n_\pi}}( 1- \frac{ \chi^{[\beta_\nu]}_{n_\nu}}{\chi_{n_\nu}}) }. \]

这样就可以写开壳核的梯度下降，来做变分了。

标签：chi,General,beta,alpha,Seniority,pi,hat,nu,BCS
来源： https://www.cnblogs.com/luyi07/p/16383019.html