How are the deformation parameters (quadrupole, hexadecapole etc) of a nucleus mathematically related to the reduced transition probabilities $B(El)$ values obtained experimentally?
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The short answer is that the deformation parameters are related to the diagonal matrix elements of the multipole operators, while the reduced transition probabilities $B(El)$ are related to the off-diagonal matrix elements.
As a specific example, the quadrupole moment (directly related to the deformation parameter $\beta$) of a nucleus in its ground state is given by $$Q_{gs} = \langle \Psi_{gs} | \mathcal{O}(E2) | \Psi_{gs} \rangle $$ with the electric quadrupole operator given by $$ \mathcal{O}(E2) = \sum_{i} e_{i} r_{i}^2 Y^{2}_{m}(\theta_{i},\phi_i). $$ The reduced transition probability for an $E2$ transition from an initial state $\Psi_i$ to a final state $\Psi_f$ is given by $$ B(E2;i\rightarrow f) = \frac{\left| \langle \Psi_{f} \| \mathcal{O}(E2) \| \Psi_{i} \rangle \right|^2}{2J_i+1}. $$ An analogous relationship hold for other multipoles.
In the case where the nucleus is modelled as a rigid rotor with an intrinsic quadrupole moment $Q_0$, then (see Bohr and Mottelson Volume 2 for a derivation, or else this) the static (i.e. lab-frame) quadrupole moment is given by $$Q = \langle J K,2 0 | J K \rangle \langle J J, 2 0 | J J \rangle Q_0 = \frac{3K^2-J(J+1)}{(J+1)(2J+3)}Q_0$$ where the brakets are Clebsch-Gordan coefficients and $K$ is the projection of $J$ on the symmetry axis of the intrisic system. The reduced $E2$ transition probability between two rotational states with the same intrinsic deformation is given by $$ B(E2; KJ_i \rightarrow K J_f) = \frac{5}{16\pi}e^2Q_0^2 |\langle J_i K, 2 0 | J_f K \rangle|^2.$$ Finally, relating $Q_0$ to the quadrupole deformation parameter $\beta$, we have $$Q_0 \approx \frac{4}{3}\beta\langle \sum_i e_ir_i^2 \rangle$$ (again, see the previous references for more detail).
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