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research:memos:reduces_transition_probability [2020/03/28 15:20] – [Special cases] 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
行 27: 行 27:
 where $Y^\lambda_\mu(\hat{\boldsymbol{r}})(=Y^\lambda_\mu(\theta, \varphi))$ is a spherical harmonics, $\hat{\boldsymbol{r}} 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). $ 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 ==== ==== For wave functions in a spherically symmetric potential ====
行 331: 行 408:
 &= (-1)^{n_1+n_2}\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)}} &= (-1)^{n_1+n_2}\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) \Gamma\left(l_2+\frac{3}{2}\right)
-\sum_{k=0}^{n_1-1}\binom{\lambda}{n_1-k-1}\binom{0}{n_2-k-1}\binom{k+l_2+\frac{1}{2}}{k}\\+\sum_{k=0}^{n_1-1}\binom{l_2-l_1}{n_1-k-1}\binom{0}{n_2-k-1}\binom{k+l_2+\frac{1}{2}}{k}\\
 &=0. &=0.
 \end{align} \end{align}
行 341: 行 418:
 &= (-1)^{n_1+n_2}\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)}} &= (-1)^{n_1+n_2}\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) \Gamma\left(l_2+\frac{3}{2}\right)
-\sum_{k=0}^{n_2-1}\binom{\lambda}{n_1-k-1}\binom{0}{n_2-k-1}\binom{k+l_2+\frac{1}{2}}{k}\\+\sum_{k=0}^{n_2-1}\binom{l_2-l_1}{n_1-k-1}\binom{0}{n_2-k-1}\binom{k+l_2+\frac{1}{2}}{k}\\
 &= (-1)^{n_1+n_2}\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)}} &= (-1)^{n_1+n_2}\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) \Gamma\left(l_2+\frac{3}{2}\right)
行 351: 行 428:
 \Gamma\left(l_2+\frac{3}{2}\right) \Gamma\left(l_2+\frac{3}{2}\right)
 \binom{l_2-l_1}{n_1-n_2}\frac{\Gamma(n_2+l_2+1/2)}{\Gamma(n_2)\Gamma(l_2+3/2)}\\ \binom{l_2-l_1}{n_1-n_2}\frac{\Gamma(n_2+l_2+1/2)}{\Gamma(n_2)\Gamma(l_2+3/2)}\\
 +&= (-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} \end{align}
 +
 +Another derivation is the follows.
 +
 +\begin{align}
 +\langle{n_2l_2}\left|r^\lambda\right|{n_1l_1}\rangle
 += \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)}}
 +\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.
 +\end{align}
 +
 +From Eq. (18) of [[https://doi.org/10.1016/S0893-9659(03)90106-6|https://doi.org/10.1016/S0893-9659(03)90106-6]],
 +
 +\begin{align}
 +&\int_0^{\infty}x^{\beta}e^{-{\sigma}x}L_{m}^{\alpha}(x)L_{n}^{\beta}(x)dx
 +=\frac{\Gamma(\beta+1)}{\sigma^{\beta+1}}\binom{n+\beta}{n}\frac{(\alpha-\beta)_{m-n}}{(m-n)!}\\
 +&(\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
 +&=(-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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research/memos/reduces_transition_probability.1585376421.txt.gz · 最終更新: 2020/03/28 15:20 by kobayash
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