5.3 (1) Two-particle Nuclei

The description of two-particle and two-hole states builds upon the formalism established for one-particle and one-hole states. However, a crucial distinction lies in the necessity of angular momentum coupling. Since two-particle (two-hole) states involve the angular momenta of two distinct particles, their coupling should be considered.

5.3.1 Examples of Two-Particle Nuclei 

The wave function for two like nucleons outside of the core is given as:

 

$$ | a\,b; J\,M\rangle = \mathcal{N}_{ab} (J) [ c_a^\dagger c_b^\dagger ]_{JM} |{\rm CORE} \rangle 
 = \mathcal{N}_{ab} (J) \sum_{m_\alpha m_\beta} (j_a m_\alpha j_b m_\beta |J M ) c_\alpha^\dagger c_\beta^\dagger  |{\rm CORE} \rangle,  $$

where the normalization factor $\mathcal{N}_{ab} (J) $ is given as

$$ \mathcal{N}_{ab}(J) = \frac{\sqrt{1+ \delta_{ab}(-1)^J}}{1+\delta_{ab}}.$$

This implies a normalization factor of $\mathcal{N}(J) = 1$ for $a \neq b$. On the other hand, for identical particles, only even $J$ is allowed, in which $\mathcal{N}(J)=1/\sqrt{2}$.

 

Derivation of the normalization factor

 

First, we can evaluate the value of $\langle a \, b;\, J\, M| a'\, b';\, J'\,M'\rangle $ as follows:

 

$$ \langle a \, b;\, J\, M| a'\, b';\, J'\,M'\rangle = \mathcal{N}_{ab} (J) \mathcal{N}_{a'b'} (J')  \sum_{m_\alpha m_\beta m_{\alpha'} m_{\beta'} } (j_a m_\alpha j_b m_\beta |J M ) (j_{a'} m_{\alpha'} j_{b'} m_{\beta'} |J' M' ) \langle {\rm CORE} | c_\beta c_\alpha  c_{\alpha'}^\dagger c_{\beta'}^\dagger  |{\rm CORE} \rangle.  $$

 

We note that $|\alpha \rangle = |n_a l_a j_a, m_\alpha \rangle = | a, m_\alpha \rangle$. The last term can be evaluated by Wick's theorem:

 

$$ \langle {\rm CORE} | c_\beta c_\alpha c_{\alpha'}^\dagger c_{\beta'}^\dagger  |{\rm CORE} \rangle =   \delta_{\beta \beta'} \delta_{\alpha \alpha'} - \delta_{\beta \alpha'} \delta_{\alpha \beta'}. $$

 

Then,

 

$$ \begin{eqnarray} \langle a \, b;\, J\, M| a'\, b';\, J'\,M'\rangle &=&  \mathcal{N}_{ab} (J) \mathcal{N}_{a'b'} (J')  \sum_{m_\alpha m_\beta m_{\alpha'} m_{\beta'} } (j_a m_\alpha j_b m_\beta |J M ) (j_{a'} m_{\alpha'} j_{b'} m_{\beta'} |J' M' ) \left( \delta_{\beta \beta'} \delta_{\alpha \alpha'} -  \delta_{\beta \alpha'} \delta_{\alpha \beta'} \right)  \\[12pt] \nonumber &=&  \mathcal{N}_{ab} (J) \mathcal{N}_{a'b'} (J')  \sum_{m_\alpha m_\beta } \left[ (j_a m_\alpha j_b m_\beta |J M ) (j_{a} m_{\alpha} j_{b} m_{\beta} |J' M' ) \delta_{aa'} \delta_{bb'}  \right. \\[12pt] \nonumber &-& \left. (j_a m_\alpha j_b m_\beta |J M ) (j_{b} m_{\beta} j_{a} m_{\alpha} |J' M' )\delta_{ab'} \delta_{ba'} \right] \\[12pt] \nonumber &=&  \mathcal{N}_{ab} (J) \mathcal{N}_{a'b'} (J')  \left[ \delta_{JJ'} \delta_{MM'} \delta_{aa'} \delta_{bb'} - (-1)^{j_a + j_b - J'} \delta_{JJ'} \delta_{MM'} \delta_{ab} \delta_{ba'} \right],  \end{eqnarray} $$

 

where I used the following orthogonality

 

$$ \begin{eqnarray} \sum_{m_\alpha m_\beta} (j_a m_\alpha j_b m_\beta | JM) (j_a m_{\alpha'} j_b m_{\beta} | J' M' ) = \delta_{JJ'} \delta_{MM'} \end{eqnarray} $$

 

and symmetric properties of the Clebsch-Gordon coefficients:

 

$$ \begin{eqnarray} (j_a m_\alpha j_b m_\beta | JM ) = (-1)^{j_a + j_b - J}(j_b m_\beta j_a m_\alpha | J M ) \end{eqnarray}. $$

 

We can rewrite the $\langle a b J M | a' b' J' M'$ as follows:

$$ \langle a b J M | a' b' J' M' \rangle = (\mathcal{N}_{ab} )^2 \delta_{JJ'} \delta_{MM'} \left[ \delta_{aa'} \delta_{bb'} - (-1)^{j_a+j_b+J} \delta_{ab} \delta_{ba'} \right], $$

where I used $(-1)^{-J} = (-1)^{-J + 2J} = (-1)^J$ because $J$ is integer.

 

Since the normalization factor is given from this condition: $\langle a b ; J M | a b ; J M \rangle = 1$,

$$ \langle a b ; J M | a b l J M \rangle = (\mathcal{N}_{ab}(J))^2 [1- (-1)^{j_a + j_b +J} \delta_{ab} \delta_{ba}] = 1 \Rightarrow \mathcal{N}_{ab} (J) = \sqrt{\frac{1}{1 + (-1)^J \delta_{ab}}},$$

 

whose result is equivalent to the given form of $\mathcal{N}_{ab}(J) = \sqrt{1+ (-1)^J \delta_{ab}}/(1+ \delta_{ab}).$

 

Proton-neutron two-particle state

Using the derived form, we can write the proton-neutron two-particle state as follows:

$$ | p n ; J M \rangle = [c_p^\dagger c_n^\dagger]_{JM} |{\rm CORE} \rangle = \sum_{m_\pi m_\nu} (j_p m_\pi j_n m_\nu | J M \rangle c_\pi^\dagger c_\nu^\dagger |{\rm CORE} \rangle, $$

 

where $\pi$ and $\nu$ denote proton and neutron states, respectively, and $| {\rm CORE} \rangle$ is effective vacuum.

 

The above form has the following symmetric property:

$$ |p n; J M \rangle = (-1)^{j_n + j_p + J +1} | n p; J M \rangle.$$

 

The factor of $(-1)^{j_n + j_p + J}$ is derived from the symmetric property of the Clebsch-Gordon coefficient, and $(-1)^1$ from the anti-commutation relation of the particle creation operators.

 

 

Example: A=6 isobars ($^6_2{\rm He}_4, \ ^6_3{\rm Li}_3, \ {\rm and} \ ^6_4{\rm Be}_2$)

For $|{\rm CORE} \rangle = |{\rm CORE(0s)} \rangle_\pi | {\rm CORE}(0s) \rangle_\nu$, we can describe each isobar as follows:

$$ \begin{eqnarray} &&|^6{\rm He}; 0^+, 2^+ \rangle = \frac{1}{\sqrt{2}} [c_{\nu 0 p_{3/2}}^\dagger c_{\nu 0 p_{3/2}}^\dagger]_{0^+, 2^+} |{\rm CORE} \rangle,              \\ &&|^6{\rm Li}; 0^+, 1^+, 2^+, 3^+ \rangle = \frac{1}{\sqrt{2}} [c_{\pi 0 p_{3/2}}^\dagger c_{\nu 0 p_{3/2}}^\dagger]_{0^+, 1^+, 2^+, 3^+} |{\rm CORE} \rangle           \\&& |^6{\rm Be}; 0^+, 2^+ \rangle = \frac{1}{\sqrt{2}} [c_{\pi 0 p_{3/2}}^\dagger c_{\pi 0 p_{3/2}}^\dagger]_{0^+, 2^+} |{\rm CORE} \rangle .         \end{eqnarray}$$

Compared to experimental values demonstrating the split energy levels, these formalisms predict the degenerated energy states for $J$. To address this discrepancy, the careful selection of residual interactions is necessary. (For examples in A=18 and 42, please refer to the textbook.)