# How do you relate the transition probability per unit time to the eigenstates involved in electronic transitions?

Jul 17, 2017

This question is best answered via Quantum Mechanics.

An electron after staying in an excited state for a very short time will jump back down to the ground state and emit its excess energy in the form of a photon. It can make a single jump to come to the ground state and emit a single photon or it can do so by making several jumps through intermediate energy levels. At each jump, the energy of the emitted photon is equal to the difference between the energies of the initial and final energy levels.

Electronic excited states have a non-zero lifetime. The excited states of atoms have mean lifetimes of a few nanoseconds , though the mean lifetime of some forbidden excited states can be as long as 10 million years .

Probability per unit of time of transition ${\Gamma}_{i \to f}$ from eigenstate | i ⟩ to a final eigenstate state | f ⟩ in the continuum is essentially constant and can be found with the help of Fermi's Golden rule as below

Gamma_(i->f)=(2pi)/ (ħ)| ⟨ f | H ′ | i ⟩ |^2rho_f
where ⟨ f | H ′ | i ⟩ is the matrix element of interaction between the two eigenstates | i >> and | f >>, $H '$ is the first-order perturbation Hamiltonian, and ${\rho}_{f}$ is the density of final states of continuum.

All this says is that if two states $i$ and $f$ are interacting, their magnitude is given by the $\mathmr{if}$th element of the $H '$ matrix describing the interactions of state $i$ and state $f$ (the number of each of which gives each dimension of the $N \times M$ matrix), controlled by the closeness of all of the involved energy levels to each other.

So, the farther the energy levels are from each other, the slower the transition, and the smaller the magnitude of ⟨ f | H ′ | i ⟩, the slower (and more forbidden) the transition.

This transition probability is also called "decay probability" and is proportional to the inverse mean lifetime ${\tau}^{- 1}$.