5.5.3 Isospin Representation of Two-Particle and Two-Hole Nuclei

 

Based on the Isospin Representation, we can express the Two-Particle and Two-Hole Nuclei:

$$ |a\, b; J \, M; T \, M_T \rangle = \mathcal{N}_{ab}(JT) [c_a^\dagger c_b^\dagger]_{JM}^{T M_T} | {\rm CORE} \rangle, $$

The normalization factor is given by 

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

This implies that there are two possible non-zero values: $\mathcal{N}_{ab} = 1$ for $a \neq b$, and $\mathcal{N}_{aa} = 1/ \sqrt{2}$ for $J + T = {\rm odd}$. For $a=b$, $T=0$ with odd $J$ and $T=1$ with even $J$ are possible. 

 

We can check the overlap of two wave functions, and derive the normalization factor (Exercise 5.16 & 5.17): 

2024.03.05 - [Nuclear Physics/From Nucleons to Nucleus] - Exercise 5.16 & 5.17.

Exercise 5.16 & 5.17.

 

Two-hole wave functions:

$$ | a^{-1} \, b^{-1} ; J \, M ; T \, M_T \rangle = \mathcal{N}_{ab}(JT) [h_b^\dagger h_b^\dagger ]_{JM}^{T M_T} | {\rm HF} \rangle.$$

Since the hole creation operator should correspond to the annihilation operator $\hat{c}_{a m_{t\alpha}}$, 

$$ h^\dagger_{a m_\alpha m_{t \alpha}} = \hat{c}_{a m_\alpha m_{t\alpha}} = (-1)^{\frac{1}{2}  + m_{t \alpha}} \tilde{c}_{a m_\alpha, -m_{t\alpha}} = (-1)^{j_a + m_\alpha + \frac{1}{2} + m_{t\alpha}} c_{a, -m_\alpha, -m_{t \alpha}}.$$

In the last term, abbreviation of $-\alpha \equiv (a, -m_\alpha, -m_{t\alpha})$ can be used. Considering $m_{t\alpha} = \pm \frac{1}{2}$, 

$$ h^\dagger_{\alpha m_\alpha, \pm \frac{1}{2}}  = \mp \tilde{c}_{a m_\alpha, \mp \frac{1}{2}}. $$

 

One-hole state:

By the relation between $h^\dagger$ and $\tilde{c}$, one-hole state becomes

$$ | a^{-1}; J \, m_\alpha; \frac{1}{2} \pm \frac{1}{2} \rangle = h^\dagger_{a m_\alpha, \pm \frac{1}{2}} | {\rm HF} \rangle = \mp \tilde{c}_{a m_\alpha, \mp \frac{1}{2}} |{\rm HF} \rangle.$$

Since a proton has $m_{t\alpha} = -\frac{1}{2}$, and a neutron $m_{t_\alpha} = +\frac{1}{2}$, 

$$ |a^{-1} ; J \, m_\alpha; \frac{1}{2} | + \frac{1}{2} \rangle = - | p^{-1} ; J \, m_\alpha \rangle,  \ \  | a^{-1}; J \, m_\alpha; \frac{1}{2} -\frac{1}{2} \rangle = | n^{-1}; J \, m_\alpha \rangle.  $$

 

Symmetric property:

For two-particle state,

$$ \begin{eqnarray} | b\, a ; J \, M ; T \, M_T \rangle &=& \mathcal{N}_{ba}(JT) \sum_{ m_{\alpha}, m_{\beta}, m_{t\alpha}, m_{t\beta} } (j_b \, m_{\beta} \, j_a \, m_{\alpha} | J \, M  ) ( \frac{1}{2} \, m_{t\beta} \, \frac{1}{2} \, m_{t \alpha} | T \, M_T ) |\beta \rangle | \alpha \rangle \\[12pt]     &=&\mathcal{N}_{ba}(JT) \sum_{ m_{\alpha}, m_{\beta}, m_{t\alpha}, m_{t\beta} } (-1)^{j_a + j_b - J} (j_a \, m_{\alpha} \, j_b \, m_{\beta} | J \, M  ) \\[12pt] \nonumber &\times& (-1)^{\frac{1}{2} + \frac{1}{2} - T} ( \frac{1}{2} \, m_{t\alpha} \, \frac{1}{2} \, m_{t \beta} | T \, M_T ) (-1)^1 | \alpha \rangle | \beta \rangle \\[12pt] \nonumber &=& (-1)^{j_a + j_ b - J - T} | a\, b ; J \, M ; T \, M_T \rangle = (-1)^{j_a + j_b + J + T} | a\, b; J \, M ; T \, M_T \rangle. \end{eqnarray}$$

 

Similarly, for a two-hole state,

$$ | b^{-1} \, a^{-1} ; J \, M; T \, M_T \rangle = (-1)^{j_a + j_b + J +T} | a^{-1} \, b^{-1} ; J \, M ; T \, M_T \rangle.$$

 

Example 5.1: $a=b = 0 f_{7/2}$

Possible $J$ states are given from $ |7/2- 7/2| \le J \le  7/2 + 7/2 $. Since $T=0$ with odd $J$ and $T=1$ with even $J$ are possible for $a=b$:

$$ T = 0,  \ J = 1,3,5,7 \ \ \ \ \ \ \ T=1, J = 0, 2,4,6.$$

Since $M_T = 0, \pm 1$ for $T=1$, the total number of states is $4\ (T=0, J=1,3,5,7) + 3\ (M_T=0, \pm 1) \times 4\ (T=1, J=0,2,4,6) = 16.$  

 

This is the same result in the proton-neutron representation: $2 (pp, nn) \times 4 (J=0,2,4,6) + 8 (np, J=0,1,2,3,4,5,6,7) = 16. $ 

 

Relation between Isospin and proton-neutron representations: (1) Two particle states

For $T=1$ with $M_T = \pm 1$, we can find the relation between isospin and proton-neutron representations:

$$ |a_1 \, a_2 ; J \, M ; 1 \, \pm 1 \rangle = \mathcal{N}_{a_1 a_2}(J1) [c_{a_1}^\dagger c_{a_2}^\dagger]^{1, \pm 1}_{JM} | {\rm CORE} \rangle.  $$ 

Using the result of the Clebsch-Gordan coefficient, 

$$ |a_1 \, a_2 ; J \, M ;1 \pm 1 \rangle = \mathcal{N}_{a_1 a_2} (J1) [c_{a1, \pm\frac{1}{2}}^\dagger, c_{a_2, \pm \frac{1}{2}}^\dagger ]_{JM} |{\rm CORE} \rangle.$$

This means that

\begin{eqnarray} && |a_1 \, a_2 ; J \, M ; 1 \, +1 \rangle = | n_1 \ n_2 ; J \ M \rangle, \\[12pt] && | a_1 \, a_2 ; J \, M ; 1\, -1 \rangle = |p_1 \, p_2; J \, M \rangle. \end{eqnarray}

 

For $M_T = 0$, since there are two possible states of $m_{t \alpha} = \pm \frac{1}{2} $ and $m_{t \beta} = \mp 1/2$, 

\begin{eqnarray} |a_1 \, a_2 ; J \, M ; T \, 0 \rangle = \frac{\mathcal{N}_{a_1 a_2} (JT)}{\sqrt{2}} \left( \left[ c_{a_1, \frac{1}{2}}^\dagger c_{a_2, - \frac{1}{2}}^\dagger  \right]_{JM} + (-1)^{T+1}\left[ c_{a_1, - \frac{1}{2}}^\dagger  c_{a_2, \frac{1}{2}}^\dagger \right]_{JM} \right) |{\rm CORE} \rangle, \end{eqnarray}

where the coefficient can be derived from the following results:

$$ \begin{eqnarray} && ( \frac{1}{2} \, \frac{1}{2} \, \frac{1}{2} \, -\frac{1}{2} | T \, 0 ) = \sqrt{\frac{1}{2}}, \\[12pt] && ( \frac{1}{2} \, \frac{1}{2} \, \frac{1}{2} \, -\frac{1}{2} | T \, 0 ) = (-1)^{\frac{1}{2} + \frac{1}{2} + T} ( \frac{1}{2} \, -\frac{1}{2} \, \frac{1}{2} \, \frac{1}{2} | T \, 0 ). \end{eqnarray} $$

Therefore,

$$ \begin{eqnarray} &&|a_1 \, a_2 ; J \, J ; 0 \, 0 \rangle = \frac{\mathcal{N}_{a_1 a_2}(J0)}{\sqrt{2}} \left( | n_1 \, p_1; J \, M \rangle - | p_1 \, n_2 ; J \, M \rangle \right) \\[12pt] &&|a_1 \, a_2 ; J \, M; 1 \, 0 \rangle = \frac{\mathcal{N}_{a_1 a_2}(J1)}{\sqrt{2}} \left( | n_1 \, p_2 ; J \, M \rangle + | p_1 \, n_2 ; J \, M \rangle \right). \end{eqnarray} $$

 

Finally, for possible states, we can write it as follows:

$$ \begin{eqnarray} |a \, a ; J \ {\rm even}, M; 1 \, 0 \rangle &=&  \frac{\sqrt{2}}{2 \sqrt{2}} \left( |n \, p ; J \, M \rangle + (-1)^{T(=1)+1} | n \, p ; J \, M \rangle \right) = | n \, p; \, J \, M \rangle \\[12pt] &=& (-1)^{j_{a,n} + j_{a,p} + J + 1} | p \, n ; J \, M \rangle = | n \, p ; J\, M \rangle, \end{eqnarray}$$

and

$$ \begin{eqnarray} | a\, a ; J\ {\rm odd}, M ; 0 \, 0 \rangle = | n \, p ; J \, M \rangle = - | p \, n; J\, M \rangle. \end{eqnarray} $$

Inversely, using the coupled equations of $|a_1 \, a_2; J\, M; 0 \, 0 \rangle$ and $|a_1 \, a_2 ; J\, M; 1\, 0 \rangle, we can write

$$ \begin{eqnarray} && |n_1 \, p_2 ; J\, M \rangle = \frac{1}{\sqrt{2}} \left( \sqrt{1 + \delta_{a_1 a_2} (-1)^J } | a_1 \, a_2; J\, M; 1; 0 \rangle + \sqrt{1 - \delta_{a_1 a_2} (-1)^J } | a_1 \, a_2; J\, M; 0; 0 \rangle \right), \\[12pt] && |p_1 \, n_2 ; J\, M \rangle = \frac{1}{\sqrt{2}} \left( \sqrt{1 + \delta_{a_1 a_2} (-1)^J } | a_1 \, a_2; J\, M; 1; 0 \rangle - \sqrt{1 - \delta_{a_1 a_2} (-1)^J } | a_1 \, a_2; J\, M; 0; 0 \rangle \right) .\end{eqnarray}$$

 

Relation between Isospin and proton-neutron representations: (2) Two-hole states

Similarly, we can write the Two-hole states as follows:

$$ \begin{eqnarray}&& | a_1^{-1}  \, a_2^{-1} ; J \, M; 1 \, +1 \rangle = | p_1^{-1} \, p_2^{-1} ; J \, M \rangle, \\[12pt] && | a_1^{-1}  \, a_2^{-1} ; J \, M; 1 \, -1 \rangle = | n_1^{-1} \, n_2^{-1} ; J \, M \rangle, \\[12pt] && | a_1^{-1}  \, a_2^{-1} ; J \, M; 0 \, 0 \rangle = \frac{\mathcal{N}_{a_1 a_2} (J0)}{\sqrt{2}} \left( | n_1^{-1} \, n_2^{-1} ; J \, M \rangle - | p_1^{-1} \, n_2^{-1} ; J \, M \rangle \right), \\[12pt] && | a_1^{-1}  \, a_2^{-1} ; J \, M; 1 \, 0 \rangle = -\frac{\mathcal{N}_{a_1 a_2} (J1)}{\sqrt{2}} \left( | n_1^{-1} \, n_2^{-1} ; J \, M \rangle + | p_1^{-1} \, n_2^{-1} ; J \, M \rangle \right). \\[12pt] \end{eqnarray}$$

Inverse relations are given as

$$ \begin{eqnarray}&& | p_1^{-1}  \, n_2^{-1} ; J \, M \rangle = -\frac{1}{\sqrt{2}} \left( \sqrt{1+ \delta_{a_1 a_2} (-1)^J } | a_1^{-1}\, a_2^{-1}; J \, M ; 1\, 0 \rangle + \sqrt{1 - \delta_{a_1 a_2} (-1)^J } | a_1^{-1}\, a_2^{-1}; J \, M ; 0\, 0 \rangle\right), \\[12pt] && | n_1^{-1}  \, p_2^{-1} ; J \, M \rangle = -\frac{1}{\sqrt{2}} \left( \sqrt{1+ \delta_{a_1 a_2} (-1)^J } | a_1^{-1}\, a_2^{-1}; J \, M ; 1\, 0 \rangle - \sqrt{1 - \delta_{a_1 a_2} (-1)^J } | a_1^{-1}\, a_2^{-1}; J \, M ; 0\, 0 \rangle\right). \end{eqnarray}$$