# What is the Quantum Transition Time for Photon Emission?

When an electron in an atom changes energy states to emit a photon, how long does the process take? Is this question even meaningful?

• Maybe related to this question: physics.stackexchange.com/q/141685 – an offer can't refuse Oct 21 '14 at 7:18
• It is just as meaningful in classical physics. If we want to measure the frequency of an oscillator or a wave with perfect precision, we need to have an infinite aperture. Quantum mechanics didn't improve on this limitation of classical measurement. – CuriousOne Oct 21 '14 at 8:07

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.

• Hmmm... the coupled QED Hamiltonian is still time independent, it's just a system that, after it is initially prepared will decay once and never go back to its excited state. To include the time dependent field of the photon sounds like some perturbative approximation to me. Can we explain it to the OP using a non-relativistic time independent Hamiltonian? Maybe using a potential that has a local minimum but that also has an energetically lower flat area that extends to infinity? – CuriousOne Oct 21 '14 at 9:32
• @CuriousOne The standard description of this process in quantum optics goes via the Jaynes-Cummings model, which is time-independent. Talking about the time-dependent field only makes sense when the light field is in a semi-classical (e.g. coherent) state. In the full quantum treatment the role of the oscillating electric field is replaced (formally speaking) by the oscillatory time dependence $e^{-iEt/\hbar}$ of the photon states. Nevetheless, the Hamiltonian is time-independent; the point is that the interaction mixes the $n$-photon and atomic eigenstates so transitions occur between them. – Mark Mitchison Oct 21 '14 at 13:34
• @CuriousOne BTW, the purpose of my comment was simply to point out that the Jaynes-Cummings Hamiltonian is easy to understand but also gives extremely good agreement with experiments at optical frequencies, so no need for a contrived model Hamiltonian. I think this answer gives a nice exposition of the relevant details. – Mark Mitchison Oct 21 '14 at 13:39
• @MarkMitchison: I understand the system, I am just not sure that the answer goes to the heart of the OP's question. The time-Energy uncertainty exists already in classical mechanics and it has nothing to do with time dependence of the Hamiltonian that shows up in the semi-classical treatment. – CuriousOne Oct 21 '14 at 18:29
• @CuriousOne To be clear, I was agreeing with you; time dependence is not the explanation. I was just pointing out that the Jaynes-Cummings model is a good place to start along the lines you were suggesting. And indeed that the answer I linked already contains a lot of the necessary elements. – Mark Mitchison Oct 21 '14 at 20:51

If we want to study the decay of an excited atomic state, the atom must be coupled to the radiation field. This can be done (and usually is) non-relativistically. The overall Hamiltonian then consists of the sum of a free atomic part, a free field part and their interaction. Now the atomic excited states are no longer bound states but turn into "resonances". The latter can be characterised by complex energies where their imaginary parts (their width) determine the decay rate of the excited atom. They can be approximately calculated using golden rule type expressions.

Perhaps it is now only a curiosity, but there is a classical answer to this question that can give order of magnitude estimates that are close to the answers provided by quantum mechanics in many cases.

If an electron is in a circular orbit (i.e. thinking of the now obsolete Bohr theory), then if we look at it from a direction that is in the orbital plane, the electron appears to execute a sinusoidal oscillation, with amplitude equal to the electron radius in the atom, and frequency given by the circular motion angular frequency. $$m_{e}\, r\, \omega^{2} = \frac{Z\, e^{2}}{4\pi \epsilon_{0} r^{2}}\, .$$ If the orbital plane is the x-y plane, and we observe from the y-direction, then we see a sinusoidal oscillation in the x-direction of the form $x= r \sin \omega t$. Differentiating twice with respect to time, we can show that the mean acceleration squared is $0.5\, r^{2}\, \omega^{4}$, and this value can be put into Larmor's formula for the amount of power radiated by an accelerated charge. $$P = \frac{e^2 < \ddot{x}^2 >}{6\pi \epsilon_0 c^3}$$ (in SI units).

The kinetic energy of the electron is given by $m_{e} r^{2} \omega^{2}/2$, so if we take the ratio of KE to power radiated, we have the lifetime of the state. $$\tau = \frac{m_{e}\, r^{2}\, \omega^{2}}{e^{2}\, r^{2}\, \omega^{4}}\, 6\pi\epsilon_{0}\, c^{3} = \frac{6\pi\, \epsilon_{0}\,m_{e}\, c^{3}}{e^{2}\, \omega^{2}}\, ,$$ (again, all in SI units) which is the reciprocal of the classical damping constant and sometimes known as the classical radiative lifetime.

Putting in some numbers we can see that, for instance the lifetime of the ground state of hydrogen is going to be very short (like $10^{-8}$ s), so something is clearly wrong and this is one of the things that motivated the search for a better atomic theory.

However, this formula does give the right sort of order of magnitude for the radiative lifetimes of transitions that are allowed by electric dipole transitions, where $\omega$ would be given by $\hbar \omega = E_2 -E_1$.

A proper quantum mechanical treatment will involve Fermi's golden rule and a more correct calculation of the transition probabilities. These result in "oscillator strengths" that modify the classical radiative lifetime.

In many cases though we have transitions that are "forbidden" by electric dipole transitions. That is, the change in quantum numbers results in zero transition probability. In which case radiative transitions may proceed via magnetic dipole or electric quadrupole routes with much lower probabilities and the classical calculation utterly fails.

Thus the timescales for atomic transitions vary enormously, both because of the frequency dependence of the classical lifetime and the quantum mechanical selection rules. e.g. The radiative lifetime for spontaneous emission in hydrogen atoms can vary from as little as $2\times10^{-9}$s for Lyman alpha, through to 10 million years for the forbidden magnetic dipole hyperfine transition responsible for 21 cm radiation.