核结构单体跃迁算符

这都是教科书上都有的内容,我只是整理整理,把一些约定也统一一下,方便以后写代码的时候参照。

1. 约定

1.1 约化矩阵约定

\[\langle J'M' | s \sigma | J M \rangle = \langle J' || s || J \rangle (JM,s \sigma| J' M'). \]

1.2 时间反演算符约定

\[\tilde{b}_\beta = (-1)^{ b + \beta } b_{-\beta}. \]

2. 单体算符的二次量子化

\[Q_{s\sigma} = \sum^N_{i=1} q(\vec{r}_i) = \sum_{\alpha, \beta} \langle \alpha | q | \beta \rangle \alpha^\dagger \beta. = \sum_{a, b} \langle a || q || b \rangle \frac{ [ j_a ] }{ [t] } ( a^\dagger \otimes \tilde{b} )_{s \sigma} \equiv \sum_{a,b} q_{ab} ( a^\dagger \otimes \tilde{b} )_{s \sigma}. \]

证明在下面。为了叙述方便,我用了爱因斯坦求和约定,

\[Q_{s\sigma} = \langle a \alpha | q_{s \sigma} | b \beta \rangle a^\dagger_\alpha b_\beta = \langle a || q || b \rangle ( b \beta; s \sigma | a \alpha ) a^\dagger_\alpha b_\beta \\ = \langle a || q || b \rangle (-1)^{ b - \beta} [a] [s]^{-1} (b \beta; a -\alpha | s -\sigma) a^\dagger_\alpha b_\beta = \langle a || q || b \rangle [a] [s]^{-1} ( a^\dagger \otimes \tilde{b} )_{s \sigma}. \]

另外可以推出约化矩阵元的关系:

\[(-1)^{b+\sigma-a}\langle a || q || b \rangle [a] = \langle b || q^\dagger || a \rangle ^* [b]. \]

如果带角动量的厄米算符有(纯属猜测!)\(q^\dagger_{s\sigma} = (-1)^\sigma q_{s -\sigma}\),以及实数的约化矩阵元,则有 \(q_{ab} = (-1)^{1+a+b} q_{ba}\)。相应地会有 \(Q^\dagger_{s, \sigma } = (-1)^\sigma Q_{s, -\sigma}\)。电多极跃迁算符是 \(q_{\lambda \mu} \equiv r^\lambda Y^\mu_\lambda\) 是厄米的,且有\(q^\dagger_{\lambda \mu} = (-1)^\mu q_{\lambda -\mu}\)(见下面的球谐函数定义),所以有 \(q_{ab} = (-1)^{j_a - j_b} q_{ba}\)

\[Y^m_n(\theta,\phi) \equiv \epsilon \sqrt{ \frac{2n-1}{4\pi} \frac{(n-|m|)!}{(n+|m|)!} } P^{|m|}_n(\cos \theta) e^{im\phi}, \epsilon = \left\{ \begin{aligned} &(-1)^m, &m \geq 0, \\ &1, &else \end{aligned} \right. \]

如果带角动量的反厄米算符(纯属猜测!)有 \(q^\dagger_{s\sigma} = - (-1)^\sigma q_{s -\sigma}\),则有 \(q_{ab} = (-1)^{a+b} q_{ba}\)。相应地会有 \(Q^\dagger_{s, \sigma } = - (-1)^\sigma Q_{s, -\sigma}\)

3. E2 跃迁的约化矩阵元

3.1 \(\langle \alpha || Y_\lambda || \beta \rangle\)

根据 Lawson 的书 P435(电子版445页),(A2.23)式,

\[\langle \alpha || Y_\lambda || \beta \rangle = (-1)^{l_\beta + l_\alpha + j_\beta - j_\alpha} \sqrt{ \frac{ (2\lambda+1)(2j_\beta+1) }{ 4\pi (2j_\alpha+1) } } ( j_\beta \frac{1}{2}; \lambda 0 | j_\alpha \frac{1}{2} ) \frac{ 1 + (-1)^{l_\alpha + l_\beta + \lambda} }{2} \int R_\alpha R_\beta r^2 dr. \]

这个推导过程有点巧妙,利用了\(P_n(1)=1, Y^m_l(\theta = 0, \phi ) = \delta_{m0} \sqrt{\frac{ 2l+1 }{4} }\)

3.2 \(\int R_{n' l'}(\alpha r) R_{nl }(\alpha r) r^{\lambda+2} dr\)

根据Lawson的书(1.11a),稍作调整(使用\(\Gamma\)函数),得到 \(l+l'+\lambda\) 为偶数时(为奇数时宇称不守恒,暂时不用考虑),

\[\int R_{n_1 l_1 }(\alpha r) R_{n_2 l_2}(\alpha r) r^{\lambda+2} dr = (-1)^{n_1 + n_2}\sqrt{ \frac{ n_1! n_2! }{ \Gamma( n_1 + l_1 + \frac{3}{2} ) \Gamma( n_2 + l_2 + \frac{3}{2} ) } } \Gamma( (l_1 - l_2 + \lambda)/2 + 1 ) \Gamma( (l_2 - l_1 + \lambda)/2 +1 ) \sum_k \frac{ \Gamma( k+(l_1+l_2+\lambda)/2 + \frac{3}{2} ) }{ k! (n_1-k)! (n_2-k)! \Gamma( k+1-n_2 + \frac{l_1-l_2+\lambda}{2} ) \Gamma( k+1-n_1 + \frac{l_2-l_1+\lambda}{2} ) } \frac{1}{\alpha^\lambda}, \]

其中,\(\alpha \equiv \sqrt{ \frac{m\omega}{\hbar}} = \frac{1}{b_0}\) 量纲为长度的负一次幂,\(b_0\)为谐振子长度,取 \(\hbar c = 197 MeV fm\), \(mc^2=938MeV\)

\[b^2_0 = \frac{ (\hbar c)^2 }{ (mc^2)(\hbar \omega) } = \frac{ 41.4 MeV fm^2 }{ \hbar \omega }, \]

另外,对于原子核一般取 \(\hbar \omega = 41 / A^{1/3}\) MeV,所以有

\[b_0 \approx A^{1/6} fm. \]

3.3 电多极跃迁算符

E\(\lambda\) 跃迁的跃迁算符是\(r^\lambda Y_\lambda\)

\[Q_{\lambda\mu} = r^\lambda Y_{\lambda\mu} = \sum_{\alpha \beta } q(\alpha \beta) ( \alpha^\dagger \otimes \tilde{b} )_{\lambda \mu}, \\ q(\alpha \beta) = \frac{[j_\alpha]}{ [\lambda] } (-1)^{l_\beta + l_\alpha + j_\beta - j_\alpha} \sqrt{ \frac{ (2\lambda+1)(2j_\beta+1) }{ 4\pi (2j_\alpha+1) } } ( j_\beta \frac{1}{2}; \lambda 0 | j_\alpha \frac{1}{2} ) \frac{ 1 + (-1)^{l_\alpha + l_\beta + \lambda} }{2} \int R_\alpha R_\beta r^{\lambda+2} dr\\ = (-1)^{l_\beta + l_\alpha + j_\beta - j_\alpha} \sqrt{ \frac{ (2j_\beta+1) }{ 4\pi } } ( j_\beta \frac{1}{2}; \lambda 0 | j_\alpha \frac{1}{2} ) \frac{ 1 + (-1)^{l_\alpha + l_\beta + \lambda} }{2} \int R_\alpha R_\beta r^{\lambda+2} dr. \]

例如,sd 壳 E2 跃迁的 \(q(\alpha \beta)\) 如下,轨道顺序为 \(d_{3/2}, d_{5/2}, s_{1/2}\)

-0.883096  0.578122  0.797885
-0.578122  -1.15624  -0.977205
-0.797885  -0.977205  0

如果使用有效电荷,如 \(e_p = 1.5 e, e_n = 0.5e\),则将上述矩阵元乘上\(1.5, 0.5\)即可。

3.4 约化跃迁概率

约化跃迁概率(reduced transition probability)就是所谓 \(B\) value,按照本文使用的约定,

\[B(E2, i \rightarrow f) = \frac{2J_f+1}{2J_i+1} | \langle f || Q_\lambda || i \rangle |^2. \]

3.4 单位

使用上面的公式,得到的 B value 的单位是 \(e^2 b^{2\lambda}_0 = A^{\lambda/3} e^2 fm^{2\lambda}\)。另一种常用的单位是 Weisskopf 的单位,这个单位是以单粒子跃迁的 B value 为参照:

\[B_W(E\lambda) = \frac{1}{4\pi} [\frac{3}{3+\lambda}]^2 (1.2 A^{1/3})^{2\lambda} e^2 fm^{2\lambda}, \\ B_W(M\lambda) = \frac{10}{\pi} [\frac{3}{3+\lambda}]^2 (1.2 A^{1/3})^{2\lambda-2} \mu^2_N fm^{2\lambda-2}. \]

对于 E1, E2 和 M1,则为

\[B_W(E1) = 0.0645 A^{2/3} e^2 fm^2, \\ B_W(E2) = 0.0594 A^{4/3} e^2 fm^4, \\ B_W(M1) = 1.790 \mu^2_N. \]

所以用上面的公式计算得到的 B(E2) 的单位也是 \(1/(0.0594A^{2/3}) W.u.\)

posted on 2021-06-23 12:26  luyi07  阅读(384)  评论(0编辑  收藏  举报

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