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research:memos:reduces_transition_probability [2020/04/01 22:25] – [Definition] kobayashresearch:memos:reduces_transition_probability [2020/04/01 23:25] (現在) – [Reduced matrix element of the electric operator for wave functions in a spherically symmetric potential] kobayash
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 $ is a unit vector, and $e$ is elementary charge. The $e_q$ are the electric charges for the proton ($q=p$) and neutron ($q=n$) in units of $e$. For example, for free nucleon, $e_p = 1$ for proton and $e_n=0$ for neutron (Brown's lecture note). $ is a unit vector, and $e$ is elementary charge. The $e_q$ are the electric charges for the proton ($q=p$) and neutron ($q=n$) in units of $e$. For example, for free nucleon, $e_p = 1$ for proton and $e_n=0$ for neutron (Brown's lecture note).
  
-==== Reduced matrix element of the electric operator ====+==== Reduced matrix element of the electric operator for wave functions in a spherically symmetric potential  ====
 From Bohr & Mottelson Vol. I, Eq. (3C-34), From Bohr & Mottelson Vol. I, Eq. (3C-34),
  
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 \end{align} \end{align}
  
-For wave functions +For the radial part of wave function $R_{nl}(r)$ in a spherically symmetric potential 
 + 
 +\begin{align} 
 +R_{nl}(r)&=N_{nl}r^{l}e^{-\frac{1}{2}{2\nu}r^2}L_{n-1}^{l+\frac{1}{2}}({2\nu}r^2)\\ 
 +N_{nl} &= \sqrt{\frac{2(2\nu)^{l+3/2}(n-1)!}{\Gamma(n+l+1/2)}}\\ 
 +\nu &= \frac{m\omega}{2\hbar}\\ 
 +N &= 2(n-1)+l\\ 
 +E_{nl}&=\left(2(n-1)+l+\frac{3}{2}\right)\hbar\omega=\left(N+\frac{3}{2}\right)\hbar\omega\\ 
 +n &\in\mathbb{N}, 
 +\end{align} 
 + 
 +$\langle j_2 \left| r^\lambda \right| j_1 \rangle$ is 
 + 
 +\begin{align} 
 +\langle j_2 \left| r^\lambda \right| j_1 \rangle 
 +=&\ (-1)^{n_1+n_2}\left(\frac{\hbar}{m\omega}\right)^{\frac{\lambda}{2}}\sqrt{\frac{(n_1-1)!(n_2-1)!}{\Gamma(n_1+l_1+1/2)\Gamma(n_2+l_2+1/2)}}\\ 
 +&\times\Gamma\left(\frac{l_1+l_2+\lambda+3}{2}\right) 
 +\sum_{k=0}^{\min(n_1-1,n_2-1)}\binom{\frac{-l_1+l_2+\lambda}{2}}{n_1-k-1}\binom{\frac{l_1-l_2+\lambda}{2}}{n_2-k-1}\binom{k+\frac{l_1+l_2+\lambda+1}{2}}{k}. 
 +\end{align} 
 + 
 +Therefore, 
 + 
 +\begin{align} 
 +B(E\lambda; j_1 \rightarrow j_2) 
 +=&\ e^2 \frac{2\lambda+1}{4\pi}\left\langle j_1 \frac{1}{2} \lambda 0 \left| j_2 \frac{1}{2}\right.\right\rangle^2 
 +\left[(-1)^{n_1+n_2}\left(\frac{\hbar}{m\omega}\right)^{\frac{\lambda}{2}}\sqrt{\frac{(n_1-1)!(n_2-1)!}{\Gamma(n_1+l_1+1/2)\Gamma(n_2+l_2+1/2)}}\right.\\ 
 +& \left.\times\Gamma\left(\frac{l_1+l_2+\lambda+3}{2}\right) 
 +\sum_{k=0}^{\min(n_1-1,n_2-1)}\binom{\frac{-l_1+l_2+\lambda}{2}}{n_1-k-1}\binom{\frac{l_1-l_2+\lambda}{2}}{n_2-k-1}\binom{k+\frac{l_1+l_2+\lambda+1}{2}}{k}\right]^2\\ 
 +=&\ e^2 \frac{2\lambda+1}{4\pi}\left\langle j_1 \frac{1}{2} \lambda 0 \left| j_2 \frac{1}{2}\right.\right\rangle^2 
 +\left(\frac{\hbar}{m\omega}\right)^{\lambda}\frac{(n_1-1)!(n_2-1)!}{\Gamma(n_1+l_1+1/2)\Gamma(n_2+l_2+1/2)}\\ 
 +& \times\left[\Gamma\left(\frac{l_1+l_2+\lambda+3}{2}\right) 
 +\sum_{k=0}^{\min(n_1-1,n_2-1)}\binom{\frac{-l_1+l_2+\lambda}{2}}{n_1-k-1}\binom{\frac{l_1-l_2+\lambda}{2}}{n_2-k-1}\binom{k+\frac{l_1+l_2+\lambda+1}{2}}{k}\right]^2\\ 
 +\end{align} 
 + 
 +If $l_1+\lambda=l_2$, 
 + 
 +\begin{align} 
 +\langle{j_2}\left|r^\lambda\right|{j_1}\rangle 
 += (-1)^{n_1+n_2}\left(\frac{\hbar}{m\omega}\right)^{\frac{l_2-l_1}{2}}\sqrt{\frac{(n_1-1)!\Gamma(n_2+l_2+1/2)}{(n_2-1)!\Gamma(n_1+l_1+1/2)}} 
 +\binom{l_2-l_1}{n_1-n_2}. 
 +\end{align} 
 + 
 +Therefore, 
 + 
 +\begin{align} 
 +B(E\lambda; j_1 \rightarrow j_2) 
 +=&\ e^2 \frac{2\lambda+1}{4\pi}\left\langle j_1 \frac{1}{2} \lambda 0 \left| j_2 \frac{1}{2}\right.\right\rangle^2 
 +\left[(-1)^{n_1+n_2}\left(\frac{\hbar}{m\omega}\right)^{\frac{l_2-l_1}{2}}\sqrt{\frac{(n_1-1)!\Gamma(n_2+l_2+1/2)}{(n_2-1)!\Gamma(n_1+l_1+1/2)}} 
 +\binom{l_2-l_1}{n_1-n_2}\right]^2\\ 
 +=&\ e^2 \frac{2\lambda+1}{4\pi}\left\langle j_1 \frac{1}{2} \lambda 0 \left| j_2 \frac{1}{2}\right.\right\rangle^2 
 +\left(\frac{\hbar}{m\omega}\right)^{\lambda}\frac{(n_1-1)!\Gamma(n_2+l_2+1/2)}{(n_2-1)!\Gamma(n_1+l_1+1/2)} 
 +\binom{\lambda}{n_1-n_2}^2. 
 +\end{align} 
 + 
 + 
 +If $l_1+\lambda=l_2$ and $\lambda=1$, 
 + 
 + 
 +\begin{align} 
 +B(E\lambda; j_1 \rightarrow j_2) 
 +=&\ e^2 \frac{2+1}{4\pi}\left\langle j_1 \frac{1}{2} 1 0 \left| j_2 \frac{1}{2}\right.\right\rangle^2 
 +\frac{\hbar}{m\omega}\frac{(n_1-1)!\Gamma(n_2+l_2+1/2)}{(n_2-1)!\Gamma(n_1+l_1+1/2)} 
 +\binom{1}{n_1-n_2}^2\\ 
 +=&\ e^2 \frac{3}{4\pi}\left\langle j_1 \frac{1}{2} 1 0 \left| j_2 \frac{1}{2}\right.\right\rangle^2 
 +\frac{\hbar}{m\omega}\frac{(n_1-1)!\Gamma(n_2+l_2+1/2)}{(n_2-1)!\Gamma(n_1+l_1+1/2)}. 
 +\end{align} 
 + 
  
 ==== For wave functions in a spherically symmetric potential ==== ==== For wave functions in a spherically symmetric potential ====
research/memos/reduces_transition_probability.1585747551.txt.gz · 最終更新: 2020/04/01 22:25 by kobayash
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