5.5 Isospin Representation of Few-Nucleon System - (2) Tensor Operators in Isospin Representation

By introducing the isospin formalism, we can describe the single particle state as follows:

$$ \alpha = (a, m_\alpha, m_{t\alpha}). $$

With the new quantum number,$m_t$, the Kronecker delta is generalized to

$$ \begin{eqnarray} &&\delta_{ab} = \delta_{n_a n_b} \delta_{l_a l_b} \delta_{j_aj_b}, \\[12pt] && \delta_{\alpha \beta} = \delta_{ab} \delta_{m_\alpha m_\beta} \delta_{m_{t\alpha} m_{t\beta}}. \end{eqnarray} $$ 

Two isospins can also be coupled, which can be expressed by using the Clebsch-Gordan coefficients:

$$ |a\, b; J\ M; T\, M_T \rangle = \mathcal{N} \sum_{\substack{m_\alpha m_\beta \\ m_{t\alpha} m_{t \beta}}} (j_a\, m_\alpha \, j_b \, m_\beta | J \, M ) (\frac{1}{2} \, m_{t \alpha} \, \frac{1}{2} \, m_{t \beta} | T \, M_T ) | \alpha \rangle | \beta \rangle.   $$

Note that the $t$ of a single particle is only 1/2. 

 

Considering the coupling of the isospin, the tensor operators can be rewritten as follows: 

$$ \begin{eqnarray} O_{\lambda \mu}^{T M_T} &=& \sum_{\alpha \beta} \langle \alpha | O_{\lambda \mu}^{T M_T} | \beta \rangle c_\alpha^\dagger c_\beta^{} = \sum_{\substack{a b \\ m_\alpha m_\beta \\ m_{t\alpha} m_{t\beta}} } \langle a \, m_\alpha \, m_{t \alpha} | O_{\lambda \mu}^{T  M_T} | a\, m_\alpha \, m_{t\alpha} \rangle c^\dagger_{a m_\alpha m_{t \alpha}} c_{b m_\beta m_{t \beta}}^{} \\[12pt] \nonumber
&=& \hat{\lambda}^{-1} \sum_{\substack{a b \\ m_{t \alpha} m_{t\beta} }} (a \, m_{t \alpha} || {\pmb O}_\lambda^{T M_T} || b \, m_{t\beta} ) [c^\dagger_{a m_{t\alpha}}  \tilde{c}_{b m_{t \beta} } ]_{\lambda \mu}, 
\end{eqnarray} $$ 

where $\hat{\lambda} \equiv \sqrt{2 \lambda + 1}.$ In the second line, the Wigner-Eckart theorem is applied. Furthermore, the double-barred matrix element can be reduced in isospin space as follows:

$$ \begin{eqnarray} (a\,m_{t\alpha} || {\pmb O}_\lambda^{T M_T} || b\,m_{t \beta} ) &=& (-1)^{\frac{1}{2} - m_{t\alpha }}
\begin{pmatrix} \frac{1}{2} & T & \frac{1}{2} \\ -m_{t \alpha} & M_T & m_{t \beta} 
\end{pmatrix} ( a ||| {\pmb O}_\lambda^{T} ||| b) \\[12pt] \nonumber
&=& (-1)^{\frac{1}{2} - m_{t\alpha}} (-1)^{\frac{1}{2} - T - m_{t \beta}} (\sqrt{2})^{-1} ( \frac{1}{2}  -m_{t\alpha} \, T \, M_T |\frac{1}{2} \, -m_{t \beta} ) ( a ||| {\pmb O}_\lambda^{T} ||| b) \\[12pt] \nonumber 
&=& (-1)^{\frac{1}{2} - m_{t\alpha}} (-1)^{\frac{1}{2} - T - m_{t \beta}} (\sqrt{2})^{-1}  (-1)^{\frac{1}{2} + m_{t\alpha}} \frac{\sqrt{2}}{\hat{T}} ( \frac{1}{2} \, -m_{t\beta}  \, \frac{1}{2} \, m_{t \alpha} | T \, M_T ) ( a ||| {\pmb O}_\lambda^{T} ||| b)\\[12pt] \nonumber
&=& (-1)^{\frac{1}{2} - m_{t\alpha}} (-1)^{\frac{1}{2} - T - m_{t \beta}} (\sqrt{2})^{-1}  (-1)^{\frac{1}{2} + m_{t\alpha}} \frac{\sqrt{2}}{\hat{T}} (-1)^{\frac{1}{2} + \frac{1}{2} - T} ( \frac{1}{2} \, m_{t\alpha}  \, \frac{1}{2} \, -m_{t \beta} | T \, M_T ) ( a ||| {\pmb O}_\lambda^{T} ||| b) \\[12pt] \nonumber 
&=& (-1)^{2 + \frac{1}{2}} (-1)^{-2T} (-1)^{- m_{t \beta}} \hat{T}^{-1} ( \frac{1}{2} \, m_{t\alpha}  \, \frac{1}{2} \, -m_{t \beta} | T \, M_T ) ( a ||| {\pmb O}_\lambda^{T} ||| b) \\[12pt] \nonumber 
&=& \hat{T}^{-1} (-1)^{\frac{1}{2} - m_{t \beta}}  ( \frac{1}{2} \, m_{t\alpha}  \, \frac{1}{2} \, -m_{t \beta} | T \, M_T ) ( a ||| {\pmb O}_\lambda^{T} ||| b).
\end{eqnarray}$$

Since $T = 0\ (M_T = 0)\ {\rm or }\  T=1\ (M_T = 0, \pm 1)$, $(-1)^{-2 T} = 1$. Then, considering the given possibility of $T$, we can evaluate the triple-barred matrix element in terms of the double-barred matrix element.

 

For $(a ||| {\pmb O}_\lambda^0 ||| b)$, 

$$ \begin{eqnarray}
&& (a \frac{1}{2} || {\pmb O}_\lambda^{00} || b \frac{1}{2} ) = (\sqrt{0 + 1})^{-1} ( \frac{1}{2} \, \frac{1}{2}  \, \frac{1}{2} \, -\frac{1}{2} | 0 \, 0 ) ( a ||| {\pmb O}_\lambda^{0} ||| b) = \frac{1}{\sqrt{2}} ( a ||| {\pmb O}_\lambda^0 ||| b) \Rightarrow (a ||| {\pmb O}_\lambda^0 ||| b) = \sqrt{2} (a \frac{1}{2} || {\pmb O}_\lambda^{00} || b \frac{1}{2} ), \\[12pt] \nonumber
&& (a -\frac{1}{2} || {\pmb O}_\lambda^{00} || b -\frac{1}{2} ) = (\sqrt{2 \cdot 1 + 1})^{-1} ( \frac{1}{2} \, -\frac{1}{2}  \, \frac{1}{2} \, +\frac{1}{2} | 1 \, 0 ) ( a ||| {\pmb O}_\lambda^{0} ||| b) = -\frac{1}{\sqrt{2}} ( a ||| {\pmb O}_\lambda^0 ||| b) \Rightarrow (a ||| {\pmb O}_\lambda^0 ||| b) = -\sqrt{2} (a -\frac{1}{2} || {\pmb O}_\lambda^{00} || b -\frac{1}{2} ),
\end{eqnarray}$$

where I used $(\frac{1}{2} \, \frac{1}{2} \, \frac{1}{2} \, -\frac{1}{2} | 0\, 0 ) = \sqrt{\frac{1}{2}}$ and $(\frac{1}{2} \, -\frac{1}{2} \, \frac{1}{2} \, \frac{1}{2} | 0\, 0 ) = -\sqrt{\frac{1}{2}}$.

Therefore, 

$$( a ||| {\pmb O}_\lambda^0 ||| b) = \sqrt{2} (a \frac{1}{2} || {\pmb O}_\lambda^{00} || b \frac{1}{2} ) = -\sqrt{2} ( a -\frac{1}{2} || {\pmb O}_\lambda^{00} || b - \frac{1}{2} ).$$

 

For $(a ||| {\pmb O}_\lambda^1 ||| b)$, 

$$ \begin{eqnarray}
&& (a \frac{1}{2} || {\pmb O}_\lambda^{10} || b \frac{1}{2} ) = (\sqrt{2\cdot1 + 1})^{-1} ( \frac{1}{2} \, \frac{1}{2}  \, \frac{1}{2} \, -\frac{1}{2} | 1 \, 0 ) ( a ||| {\pmb O}_\lambda^{1} ||| b) = \frac{1}{\sqrt{6}} ( a ||| {\pmb O}_\lambda^1 ||| b) \Rightarrow (a ||| {\pmb O}_\lambda^1 ||| b) = \sqrt{6} (a \frac{1}{2} || {\pmb O}_\lambda^{10} || b \frac{1}{2} ), \\[12pt] \nonumber
&& (a -\frac{1}{2} || {\pmb O}_\lambda^{10} || b -\frac{1}{2} ) = (\sqrt{2 \cdot 1 + 1})^{-1} ( \frac{1}{2} \, -\frac{1}{2}  \, \frac{1}{2} \, \frac{1}{2} | 1 \, 0 ) ( a ||| {\pmb O}_\lambda^{1} ||| b) = \frac{1}{\sqrt{6}} ( a ||| {\pmb O}_\lambda^1 ||| b) \Rightarrow (a ||| {\pmb O}_\lambda^1 ||| b) = \sqrt{6} (a -\frac{1}{2} || {\pmb O}_\lambda^{10} || b -\frac{1}{2} ), \\[12pt] \nonumber 
&& (a \frac{1}{2} || {\pmb O}_\lambda^{11} || b -\frac{1}{2} ) = (\sqrt{2 \cdot 1 +1})^{-1} ( \frac{1}{2} \,  \frac{1}{2} \, \frac{1}{2} \, \frac{1}{2} | 1\, 1) ( a ||| {\pmb O}_\lambda^1 ||| b) = \frac{1}{\sqrt{3}} (a ||| {\pmb O}_\lambda^1 ||| b) \Rightarrow ( a||| {\pmb O}_\lambda^1 ||| b) = \sqrt{3} ( a \frac{1}{2} || {\pmb O}_\lambda^{11} || b -\frac{1}{2} ), \\[12pt] \nonumber  
&& (a -\frac{1}{2} || {\pmb O}_\lambda^{1-1} || b \frac{1}{2} ) = ( \sqrt{2 \cdot 1 +1})^{-1} ( \frac{1}{2} \, -\frac{1}{2} \, \frac{1}{2} \, -\frac{1}{2} | 1 \, -1) ( a ||| {\pmb O}_\lambda^1 ||| b) = \frac{1}{\sqrt{3}} ( a ||| {\pmb O}_\lambda^{1} ||| b) \Rightarrow ( a ||| {\pmb O}_\lambda^{1} ||| b) = \sqrt{3} ( a - \frac{1}{2} || {\pmb O}_\lambda^{1 -1} || b \frac{1}{2} ).
\end{eqnarray}$$

Therefore,

$$ \begin{eqnarray} ( a ||| {\pmb O}_\lambda^1 ||| b ) = \sqrt{6} (a \frac{1}{2} || {\pmb O}_\lambda^{10} || b \frac{1}{2}) = \sqrt{6} (a -\frac{1}{2} || {\pmb O}_\lambda^{10} || b -\frac{1}{2}) = \sqrt{2} ( a \frac{1}{2} || {\pmb O}_\lambda^{11} || b -\frac{1}{2} ) = \sqrt{2} ( a -\frac{1}{2} || {\pmb O}_\lambda^{1-1} || b \frac{1}{2}). \end{eqnarray}$$

 

 

The given one-body operator can be rewritten by using the triple-barred matrix elements:

$$ \begin{eqnarray}
O_{\lambda \mu}^{T M_T} &=& \hat{\lambda}^{-1} \sum_{\substack{a b \\ m_{t \alpha} m_{t\beta} }} (a \, m_{t \alpha} || {\pmb O}_\lambda^{T M_T} || b \, m_{t\beta} ) [c^\dagger_{a m_{t\alpha}}  \tilde{c}_{b m_{t \beta} } ]_{\lambda \mu} \\[12pt] \nonumber 
&=& \hat{\lambda}^{-1} \hat{T}^{-1} \sum_{ \substack{a b \\ m_{t \alpha} m_{t \beta}} }(-1)^{\frac{1}{2} - m_{t \beta}} (\frac{1}{2} \, m_{t\alpha} \, \frac{1}{2} \, -m_{t \beta} | T \, M_T ) ( a ||| {\pmb O}_\lambda^T ||| b) [c^\dagger_{a m_{t\alpha}}, \tilde{c}_{b m_{t \beta}}]_{\lambda \mu} \\[12pt] \nonumber &=& \hat{\lambda}^{-1} \hat{T}^{-1} \sum_{a b} ( a ||| {\pmb O}_{\lambda}^T ||| b) [c^\dagger_{a}, \hat{c}_b ]_{\lambda \mu}^{T M_T},
\end{eqnarray}$$ 
where $\hat{c}_{a m_{t\alpha}} \equiv (-1)^{\frac{1}{2} + m_{t \alpha} } \tilde{c}_{a, -m_{t \alpha}}$.  

 

Finally, the matrix element for the one-body operator is given as

$$ \begin{eqnarray} (\xi_f \, J_f \, T_f \, m_f \, m_{tf} | O_{\lambda \mu}^{T M_T} | \xi_i \, J_i \, T_i \, m_i\, m_{ti} ) = \hat{\lambda}^{-1} \hat{T}^{-1} \sum_{ab} (a ||| {\pmb O} ||| b) (\xi_f\, J_f\, T_f \, m_f \, m_{tf} | [c_a^\dagger \hat{c}_b]_{\lambda \mu}^{T M_T} | \xi_i \, J_i \, T_i \, m_i \, m_{ti} ),
\end{eqnarray}$$

whose form can be reduced by the Wigner-Eckart theorem:

$$ \begin{eqnarray} (\xi_f  \, J_f \, T_f ||| O_{\lambda \mu}^{T M_T} ||| \xi_i \, J_i \, T_i ) = \hat{\lambda}^{-1} \hat{T}^{-1} \sum_{ab} ( a ||| {\pmb O} ||| b) (\xi_f\, J_f   ||| [c_a^\dagger \hat{c}_b]_{\lambda \mu}^{T M_T} ||| \xi_i \, J_i \, T_i ).
\end{eqnarray} $$

Note that coefficients resulting from the Wigner-Eckart theorem on both sides cancel out.