EXAMPLE PROBLEM

Posted by student team #5 (Anthony Kuchera, Jeff Klatsky, Chelsey Morien)

QUESTION:

Consider a transition from $i\rightarrow f$ between two states of a 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. Hint: This is just the ratio between $B(T_{\lambda };i\rightarrow f)$ and $B(T_{\lambda };f\rightarrow i)$

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 $M_{f}$

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

Using the orthogonality condition: $\sum _{m_{1}m_{2}}\left({\begin{array}{lll}j_{1}&j_{2}&j_{3}\\m_{1}&m_{2}&m_{3}\end{array}}\right)\left({\begin{array}{lll}j_{1}&j_{2}&j'_{3}\\m_{1}&m_{2}&m'_{3}\end{array}}\right)={\dfrac {\delta _{j_{3}j'_{3}}\delta _{m_{3}m'_{3}}}{2j_{3}+1}}$

Which leads us to our final result: $B(T_{\lambda };i\rightarrow f)={\dfrac {(f||T_{\lambda }||i)^{2}}{2J_{i}+1}}$

It is obvious that this result does not depend on $M_{i}$

b) All that is missing to find the detailed balance relation is $B(T_{\lambda };f\rightarrow i)$. This is done in the same way as part a).

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

Note, the only difference is the sum over $M_{i}$

Thus, we have $B(T_{\lambda };i\rightarrow f)={\dfrac {(f||T_{\lambda }||i)^{2}}{2J_{f}+1}}$

And the detailed balance relation is: ${\frac {B(T_{\lambda };i\rightarrow f)}{B(T_{\lambda };f\rightarrow i)}}={\frac {2J_{f}+1}{2J_{i}+1}}$