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--- research:memos:reduces_transition_probability [2020/03/28 18:45] – [Special cases] 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 27: / 行 27: @@
 where $Y^\lambda_\mu(\hat{\boldsymbol{r}})(=Y^\lambda_\mu(\theta, \varphi))$ is a spherical harmonics, $\hat{\boldsymbol{r}}
 $ 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 for wave functions in a spherically symmetric potential  ====
+From Bohr & Mottelson Vol. I, Eq. (3C-34),
+\begin{align}
+B(E\lambda; j_1 \rightarrow j_2) = e^2 \frac{2\lambda+1}{4\pi} \left\langle j_2 \left| r^\lambda \right| j_1 \right\rangle^2
+\left\langle j_1 \frac{1}{2} \lambda 0 \left| j_2 \frac{1}{2}\right.\right\rangle^2.
+\end{align}
+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 ====
@@ 行 356: / 行 433: @@
 Another derivation is the follows.
 \begin{align}
@@ 行 371: / 行 447: @@
 &(\mathfrak{R}(\beta)>-1;\ \mathfrak{R}(\sigma)>0;\ m{\ge}n{\ge}0\ (m,n\in\mathbb{N}_0)).
 \end{align}
+From http://functions.wolfram.com/GammaBetaErf/Pochhammer/introductions/FactorialBinomials/ShowAll.html,
+\begin{align}
+\binom{n}{k}=\frac{(-1)^k(-n)_k}{k!}.
+\end{align}
+([[https://ja.wikipedia.org/wiki/%E3%83%9D%E3%83%83%E3%83%9B%E3%83%8F%E3%83%9E%E3%83%BC%E8%A8%98%E5%8F%B7|Wikipedia - Pochhammer symbol (in Japanese)]] seems to be wrong.)
 Therefore,
+\begin{align}
+&\int_0^{\infty}x^{\beta}e^{-{\sigma}x}L_{m}^{\alpha}(x)L_{n}^{\beta}(x)dx
+=(-1)^{m+n}\frac{\Gamma(\beta+1)}{\sigma^{\beta+1}}\binom{n+\beta}{n}\binom{\alpha-\beta}{m-n}\\
+&(\mathfrak{R}(\beta)>-1;\ \mathfrak{R}(\sigma)>0;\ m{\ge}n{\ge}0\ (m,n\in\mathbb{N}_0)).
+\end{align}
+Using this formula,
 \begin{align}
 \int_0^{\infty}t^{l_2+\frac{1}{2}}e^{-t}L_{n_1-1}^{l_1+\frac{1}{2}}(t)L_{n_2-1}^{l_2+\frac{1}{2}}(t)dt
-=\Gamma(l_2+3/2)\binom{n_2+l_2-1/2}{n_2-1}\frac{(l_1-l_2)_{n_1-n_2}}{(n_1-n_2)!}\\
+&=(-1)^{n1+n2}\Gamma(l_2+3/2)\binom{n_2+l_2-1/2}{n_2-1}\binom{l_2-l_1}{n_1-n_2},
 \end{align}
+and
+\begin{align}
+&\langle{n_2l_2}\left|r^\lambda\right|{n_1l_1}\rangle\\
+&= (-1)^{n1+n2}\left(\frac{\hbar}{m\omega}\right)^{\frac{l_2-l_1}{2}}\sqrt{\frac{(n_1-1)!(n_2-1)!}{\Gamma(n_1+l_1+1/2)\Gamma(n_2+l_2+1/2)}}
+\Gamma\left(l_2+\frac{3}{2}\right)
+\binom{n_2+l_2-\frac{1}{2}}{n_2-1}\binom{l_2-l_1}{n_1-n_2}\\
+&= (-1)^{n1+n2}\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}
+==== $n_1=2,\ n_2=1,\ l_1=0,\ l_2=1,\ \lambda=1$ ====
+From wolfram alpha,
+<code>
+n1 = 2, n2 = 1, l1 = 0, l2=1, lambda = 1, Integrate[x^((l1+l2+lamda+1)/2)*Exp[-x]*LaguerreL[n1-1,l1+1/2,x]*LaguerreL[n2-1,l2+1/2,x],{x,0,Infinity}]
+n1 = 2, n2 = 1, l1 = 0, l2=1, Gamma[l2+3/2]*Binomial[n2+l2-1/2,n2-1]*Binomial[l2-l1,n1-n2]
+</code>
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