6.2.2 Examples: Transitions in One-Hole Nuclei 15N and 15O

In the previous posting, we derived the reduced transition probability for one-particle and one-hole nuclei:

2024.04.14 - [Nuclear Physics/From Nucleons to Nucleus] - 6.2 Electromagnetic Transitions in One-Particle and One-Hole Nuclei - 6.2.1 Reduced Transition Probabilities

6.2 Electromagnetic Transitions in One-Particle and One-Hole Nuclei - 6.2.1 Reduced Transition Probabilities

In this posting, we will show examples of reduced transition probability for one-hole nuclei: $^{15}$N (one-proton hole) and $^{15}$O (one-neutron hole). 

 

Both nuclei have the following states:

\begin{eqnarray} && | ^{15}{\rm N}; 1/2^{-}_{\rm gs} \rangle =h_{\pi 0 p_{1/2}}^\dagger |{\rm HF} \rangle, \\[12pt] && | ^{15}{\rm N}; 3/2^{-} \rangle = h_{\pi 0 p_{3/2}}^\dagger |{\rm HF} \rangle, \\[12pt] && | ^{15}{\rm O}; 1/2^{-}_{\rm gs} \rangle = h_{\nu 0 p_{1/2}}^\dagger |{\rm HF} \rangle, \\[12pt] && | ^{15}{\rm O}; 3/2^{-} \rangle = h_{\nu 0 p_{3/2}}^\dagger |{\rm HF} \rangle, \end{eqnarray}

where $|{\rm HF}\rangle = |{\rm HF} (0p) \rangle_\pi | {\rm HF} (0p) \rangle_\nu.$

These examples have initial state of $J^\pi_i = 3/2^-$ and $J^\pi_f = 1/2^-$, so $\Delta J = 1$ and $\pi_i \pi_f = +1 $. In addition, the triangular condition allows $\lambda = 1$ or $2$. Combining both the parity conservation and triangular condition, one can know that E2 and M1 transition can be allowed. 

 

 

E2 Transition

For the E2 transition, we can write the reduced transition probability as follows:

\begin{eqnarray} B({\rm E}2; (0p_{3/2})^{-1} \rightarrow (0p_{1/2})^{-1}) = \frac{1}{4} (0 p_{1/2} || {\pmb Q}_2 || 0 p_{3/2})^2 = \frac{1}{4} \left( 1.410 e b^2 \right)^2 = 4.270 e^2 \,{\rm fm}^4 .\end{eqnarray}

Here, we adopt $b=\hbar c / \sqrt{m_Nc^2 (\hbar \omega) } = 1.712\,{\rm fm}$  where $\hbar \omega = (45 A^{-1/3} - 25 A^{-2/3})\,{\rm MeV}.$ Using the Weisskopf units, $B_W({\rm E}2) = 2.197 e^2 {\rm fm}^4$, the result becomes 1.943 W. u.. This form can be available for only proton hole Nucleus $^{15}$N. For the neutron hole, the charge $e$ should be changed to a non-zero effective neutron charge. 

 

 

M1 Transition

We can also write the M1 transition probability as follows:

\begin{eqnarray} B({\rm M}1; (0p_{3/2}^{-1}) \rightarrow (0 p_{1/2})^{-1} ) = \frac{1}{4} (0 p_{1/2} || {\pmb M}_1 || 0p_{3/2} )^2 = \frac{1}{4} \left( -0.564 g_l + 0.564 g_s \right)^2 (\mu_N/c)^2. \end{eqnarray}

For the proton-hole nuclei ($^{15}N$), since $g_l=1$ and $g_p = 5.586$,

\begin{eqnarray} B({\rm M}1; (0p_{3/2}^{-1} \rightarrow (0 p_{1/2})^{-1}) =\frac{1}{4} ) -0.564 \times 1 + 0.564 \times 5.586)^2 (\mu_N/c)^2 = 1.673 (\mu_N/c)^2 = 0.034 \, {\rm W. u.}. \end{eqnarray}

For the neutron-hole nuclei ($^{15}$O), since $g_l = 0$ and $g_p = -3.826$, 

\begin{eqnarray} B({\rm M}1; ^{15}O )= \frac{1}{4} 0.564 \times (-3.826)^2 (\mu_N/c)^2 = 1.164 (\mu_N/c)^2 = 0.650\,{\rm W.u.}. \end{eqnarray}

 

$^{15}$O

Since the E2 transition in only possible for $^{15}$O with bare charge, the transition probability is

\begin{eqnarray} T({\rm M}1; ^{15}{\rm O}) = 1.779 \times 10^{13} \times (6.176^3 \times 1.164 /s = 4.878 \times 10^{15} /s . \end{eqnarray}

where $E = 6.176\,{\rm MeV}$ denotes the energy gap between initial and final states. Then, the decay half-life is

\begin{eqnarray} t_{1/2} ({\rm M}1; ^{15}{\rm O}) = \frac{ln2}{T ({\rm M} 1 )} = 1.421 \times 10^{-16} \,{\rm s} = 0.14\,{\rm fs}. \end{eqnarray} 

 

$^{15}$N

For $^{15}$N, each transition probability is respectively given as

\begin{eqnarray} && T({\rm E}2; ^{15}{\rm N}) = 1.223 \times 10^9 \times 6.324^5 \times 4.270 /s = 5.282 \times 10^{13} /s \\[12pt] && T({\rm M}1; ^{15}{\rm N}) = 1.779 \times 10^{13} \times 6.324^3 \times 1.673 /s = 7.527 \times 10^{15} /s. \end{eqnarray}

As a result, the M1 transition is dominant in this case. Besides, the transition probabilities are additive, so the total transition probability is

\begin{eqnarray} T(^{15}{\rm N}) = T({\rm E2} + {\rm M1}) = T({\rm E}2) + T({\rm M}1) = 7.580 \times 10^{15} / {\rm s} \end{eqnarray}

Therefore, the decay half-life is

\begin{eqnarray}t_{1/2} ({\rm E2 + M1}; ^{15}{\rm N}) = \frac{\ln 2}{T({\rm E2 + M1}) } = 9.145 \times 10^{-17} {\rm s} = 0.09\,{\rm fs}.  \end{eqnarray}

The experimental result is $t^{\rm exp}_{1/2} (^{15}{\rm N}; 3/2^- \rightarrow 1/2^- ) = 0.5\,{\rm fs}.$ 

 

One can evaluate the measurable quantity of the mixing ratio for this case as follows:

\begin{eqnarray} \Delta ({\rm E2}/{\rm M1}) = \frac{1.410 eb^2}{2.587 \mu_N/c} = \frac{4.133 e \,{\rm fm}^2}{2.587 \mu_N /c} = + 0.01597 \frac{e {\rm barn}}{\mu_N/c} \end{eqnarray}

or

\begin{eqnarray} \delta(E2/M1) = 0.835 \times 6.324 \times 0.01597 = + 0.084. \end{eqnarray}