When an electron in an atom changes energy states to emit a photon, how long does the process take? Is this question even meaningful?
Answer
Both the ground state and the excited state are eigenfunctions of the time independant Schrodinger equation. That means they are also time independant so the excited state will never decay to the ground state i.e. the transition time would be infinite.
However the excited state doesn't decay to just the ground state, it decays to the ground state plus a photon, and the oscillating electric field of the photon is not time independant. The electric field of the photon has to be included in the Hamiltonian, and when you do this the Schrodinger equation is no longer time independant and the excited state is no longer an eigenfunction. There is now a non-zero probability for the excited state to decay, so the decay rate will be greater than zero and the (average) transition time will be less than infinite.
Strictly speaking the presence of the photon makes this a relativistic system, and we should calculate the decay rate using quantum field theory. There are some comments about this in the answers to the question Why and how, in QED, can excited atoms emit photons?. However for most purposes we can calculate transition rates using Fermi's golden rule. There is a detailed discussion of the calculation in the question Interpretation of "transition rate" in Fermi's golden rule and the answers to it. If you're interested in seeing worked examples, e.g. for the hydrogen atom, a quick Google will find you lots of articles on the subject.
No comments:
Post a Comment