For atomic systems most observed transitions are electric dipoles with only a few transitions allowed.
💼 Case: consider a system coupled to an oscillating electric field $E_0\vec \epsilon\cos(\omega t)$ where $\vec \epsilon$ is a unit vector
$$ \hat V(t)=\left \{ \begin{matrix} 0 & \text{for } & t\le 0 \\ eE_0 \cos(\omega t) \vec \epsilon \cdot \hat r =\frac{eE_0}{2} (e^{i\omega t}+e^{-i\omega t})\vec \epsilon \cdot \hat r & \text{for} & t>0 \end{matrix} \right . $$
Fermi’s golden rule gives
$$ \lim_{t\to\infin}[R_{i\to f}] \propto |\bra{f} \vec \epsilon \cdot \hat r\ket{i} |^2 $$
🗒️ Note: since we are in first order approximation we cannot conclude a transition cannot happen, however these will be slow and difficult to observe
⚽ Goal: find the transitions allowed, we need $\bra f \vec \epsilon \cdot \hat r \ket{i}$ to be non zero
💼🤯 Caseception: Lets consider a Hydrogen-like state in the basis $\ket{n,l,s,j,m_j}$ wher $n,l,s,j$ are the principal, orbital angular momentum, spin and total angular momentum quantum numbers, and $m_j$ is the projection of $j$ on the $z$-axis
We need to evaluate the matrix element
$$ \bra{n',l',s',j',m_j'}\vec \epsilon \cdot \hat r \ket{n,l,s,j,m_j} $$
For Hydrogen we have $s'=s=\frac 12$, $j=l\pm\frac 12$
Hydrogen atom wavefunctions are in spherical harmonic so lets start writing in this basis
$$ \hat r = \sqrt{\frac{4\pi}{3}}r(Y^1_1e_-+Y^0_1 e_0 + Y^{-1}1e+) $$
where $e_\pm=\pm\sqrt{1}{2}(e_x\pm ie_y)$, $e_0=e_z$ and $Y^{0,\pm 1}_{1}$ are the $l=1$ spherical harmonics
📜 Allowed transitions:
- Transitions differ by at most one unit $\Delta j=j-j'\in \{ 0,\pm 1\}$ and $\Delta m_j \in \{0,\pm 1\}$
- The angular momentum needs to be positive $j'j\ge 0$
- The case of $j=0\to j'=0$ is not allowed
- $\hat r$ is odd parity which makes its integral is 0 when integrating overall space. To avoid this, we require the parity $(-1)^l$ to change at transitions meaning we only have $\Delta l=\pm 1$
💼 Case: multiple electron atoms, a light atom.
The dipole operator becomes $\vec \epsilon\cdot \hat R$ where $\hat R-\sum_i \hat r_i$
In the light atom approximation $\hat S=\sum_i \hat S_i$ and $\hat L=\sum_i \hat L_i$
The total angular momentum is $\hat J=\sum_i (\hat L_i + \hat S_i)$
The parity of $(n_1,l_1)(n_2,l_2)\ldots(n_z,l_z)$ is given by $(-1)^{\sum^z_i l_i}$
In this case we get the following transitions
$$ \Delta J =0,\pm 1 \;(\text{not}\; J=0\to J'=0)\quad \Delta M_j=0,\pm 1 \quad \Delta L=0,\pm 1 \quad \Delta S=0 $$
where $\Delta S=0$ is because the dipole operator is spin-independent, and we need the parity to flip each time (💃 Eg: $L=0\to L'=0$ is not allowed as they are both even parity)
🧠 Remember: we found that as $t\to\infin$ we get a linear increase in probability between $\ket{i}$ and $\ket{f}$
We can thus write the probability for the state $\ket{i}$ to remain intact in the same limit as
$$ \lim_{t\to \infin} [P_i(t)]=1-t\sum_{f\ne i}R_{i\to f} $$
where the sum is over all accessible states (energy conservation + selection rules)
🗒️ Note: this probability is weird, we are assuming large time but there probability is negative
Through some math which is not covered in this course the actual term is
$$ P_i(t)=e^{-Rt} \quad \text{where}\quad R=\sum_{f\ne i} R_{i\to f} $$
Thus the state lifetime is
$$ \tau = \frac{1}{R} $$
which can be found using first order perturbation theory