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

##### 1 Answer

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->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

So, the farther the energy levels are from each other, the slower the transition, and the smaller the magnitude of

This transition probability is also called "decay probability" and is proportional to the inverse mean lifetime