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--- research:memos:reduces_transition_probability [2020/04/01 22:27] – [Reduced matrix element of the electric operator] kobayash
+++ research: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
@@ 行 28: / 行 28: @@
 $ 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),
@@ 行 36: / 行 36: @@
 \end{align}
-For wave functions, the solutions are
+For the radial part of wave function $R_{nl}(r)$ in a spherically symmetric potential
 \begin{align}
-R(r)&=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)\\
+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 &= E_{nl}=\left(2(n-1)+l+\frac{3}{2}\right)\hbar\omega=\left(N+\frac{3}{2}\right)\hbar\omega\\
+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}.
+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 ====
 The radial part of the Schrödinger equation is