It can be shown that the probability of absorption or emission between two stationary states m and n is proportional to the square of the integral $\mu_{mn}$ called the dipole transition moment.

\begin{equation} \mu_{mn}=\int{\psi^{\ast}_m\hat{\mu}\psi_n}dq \end{equation}

Being $\hat{\mu}$ the electric dipole moment, defined as: $\hat{\mu}=\sum_{i}Q_ir_i$

  • When $\mu_{mn}=0$, the transition between states m and n has zero probability, transition forbidden.
  • When $\mu_{mn}\neq 0$, the transition between states m and n is allowed.

Example: For the particle in a box, obtain the dipole transition moment and obtain the selection rule

Solution: The wave function of the particle in a box is given by $\psi_n=\sqrt{\frac{2}{a}}sin\frac{n\pi x}{a}$ and the electric dipole moment is $\hat{\mu}=Qx$. Substituting both expressions into $\mu_{mn}=\int{\psi_{m}^{\ast}\hat{\mu}\psi_n dq}$. Solving the integral of the transition moment we obtain:

\begin{equation} \mu_{mn}=\frac{Qa}{\pi^2}\left[\frac{cos[(mn)\pi-1]}{(mn)^2}-\frac{ cos[(m+n)\pi-1]}{(m+n)^2}\right] \end{equation}

Only if mn and m+n are odd $\mu_{mn}\neq 0$. Selection rule $\Delta n=\pm 1, \pm 3, \pm 5 ...$