DetailedBalance

Consider a transition from $i\rightarrow f$ between two states of the nucleus with spins $J_{i}$ and $J_{f}$, respectively. The transition probability is proportional to the squared matrix element $|\langle J_{f}M_{f}|T_{\lambda \mu }|J_{i}M_{i}\rangle |^{2}$ where $T_{\lambda \mu }$ is a hermitian tensor operator of rank $\lambda$ responsible for the process. Define the reduced transition probability

$B(T_{\lambda };i\rightarrow f)=\sum _{\mu M_{f}}|\langle J_{f}M_{f}|T_{\lambda \mu }|J_{i}M_{i}\rangle |^{2}$

as a sum of squared matrix elements over final projections $M_{f}$ and operator projections $\mu$.

a) Express $B(T_{\lambda };i\rightarrow f)$ in terms of the reduced matrix element $(f||T_{\lambda }||i)$ and show that it does not depend on the initial projection $M_{i}$.

b) Establish the detailed balance between the reduced transition probabilities of the direct $i\rightarrow f$, and inverse $f\rightarrow i$ processes.

SOLUTION:

a) According to the Wigner-Eckert theorem, the entire dependence of the matrix element of a tensor operator on the magnetic quantum numbers is concentrated in the vector coupling coefficients,

$\langle J_{f}M_{f}|T_{\lambda \mu }|J_{i}M_{i}\rangle =(-)^{J_{f}-M_{f}}\left({\begin{array}{lll}J_{f}&\lambda &J_{i}\\-M_{f}&\mu &M_{i}\end{array}}\right)(f||T_{\lambda }||i)$

We obtain the rate by squaring this and summing over $\mu$ and $M_{f}$

$B(T_{\lambda };i\rightarrow f)=(f||T_{\lambda }||i)^{2}\sum _{\mu M_{f}}\left({\begin{array}{lll}J_{f}&\lambda &J_{i}\\-M_{f}&\mu &M_{i}\end{array}}\right)\left({\begin{array}{lll}J_{f}&\lambda &J_{i}\\-M_{f}&\mu &M_{i}\end{array}}\right)$