Decay from excited state to ground state

Question

People frequently speak about an atomic system decaying from an excited state to the ground state. However, both the ground states and the excited states are defined as eigenstates of the Hamiltonian operator for the system. This implies then that up to a time-dependent complex phase, they are invariant under evolution according to the Hamiltonian for the system. How can it be then that there is a decay from an excited state to a ground state?

I have tried to give this an interpretation in terms of unstable equilibrium (that is, if we have a excited state, it is actually an eigenstate, but if we modify it a little bit, it becomes something will evolve to the ground state). However, I don't think this works, since the evolution under the Hamiltonian of the system will leave invariant (up to a time-dependent complex phase) the amplitudes of the state when expanded in a basis of eigenstates.

My current guess is that it is necessary to consider some kind of noise to explain this, but I don't have any idea about how would the particular details work.

Answer 1

The true eigenstates, when they exist, don't decay. They sit and spin around in phase forever. But atomic eigenstates are not true eigenstates. The reason atomic states decay is because they are coupled to photon states, and the combined photon-atom Hamiltonian doesn't have excited atom eigenstates.

When you have an atom in a box of mirrors, there are true eigenstates of the combined photon-atom system inside the box. These are states where a quantum of energy is absorbed by the atom, remitted into the box, in a steady way, so that it is sometimes in the atom, sometimes in the photons of the box. But when you make the box big, the energy will be in the photons nearly all the time, and the atom will be in its ground state, just because there are infinitely many more photon states than atomic states. In the limit of no box, the excited atomic states are never true eigenstates, they always decay into photons irreversibly. This process was described by Fermi, and the rate of irreversible decay is given by Fermi's golden rule.

For atoms and radiation, the coupling is mostly by a term in the Hamiltonian equal to $p\cdot A(x)$, where p is the momentum of the electron and A is the vector potential at the position of the electron, plus a direct two-photon term $A(x)^2$ which you can usually ignore. You evaluate the transition by expanding A in plane waves, the coefficients of which are photon creation operators, and approximating the exponential of the X operator by the first two terms of a Taylor expansion. This is called the dipole approximation.

The resulting Hamiltonian describes transitions between the pure atom stationary states into states of the atom plus a photon, and for long times, the transitions conserve energy, so that the outgoing photon carries the energy difference that is lost by the decay. The dipole approximation is essentially exact for transitions which are dipole-allowed because the atomic motion is nonrelativistic, so that the wavelength of the light is enormous compared to the atom. The result is that there are small matrix elements for transitions between the states, accompanied by creating one photon, and these give the dipole atomic transitions. This is worked out in Sakurai's book, among others.

Answer 2

Ron Maimon is correct that the Hamiltonian as usually written is not really the full Hamiltonian, because it does not include terms for the coupling of the atomic system to the radiation field. Maimon shows one way of doing this formally, but there is a more intuitive and understandable explanation, which also gives the same transition rate as Ron's formal method. I have posted this on other occasions and whenever I have done so, Ron has ridiculed my analysis. Nevertheless, at the risk of further ridicule, I will post it again.

The simplest case to consider is the 2p=>1s transition of the hydrogen atom. If the 2p state is just the tiniest bit perturbed, so that there is just a little bit of ground state mixed in, the combination is unstable. It is unstable because you can easily verify that the superposition of s and p states has an oscillating dipole moment, and this must radiate according to Maxwell's equations. No more and no less than Maxwell's equations are needed to give the correct transition rates between states, and there is no need to quantize the radiation field.

How does this relate to the apparent fact that in the solution of the differential equation, the coefficients of the various eigenstates are time-invariant? This is again because the equation is only an approximation, neglecting as it does the coupling of the atom to the radiation field. Since the total energy of the radiation field is readily calculated from antenna theory, and the sum of the squares of the coefficients of the eigenstates must be unity, it is easy to solve for the value of those coefficients as a function of time. In effect, the excited atom smoothly decays to the ground state: as it does, it oscillates like a tiny antenna, emitting electromagnetic waves. There is no need to talk about photons and quantization.

Oddly enough, you can also analyze the system in terms of pure eigenstates and quantum leaps, the way Ron does, and you actually get correct answers that way. But just because you get the right answer doesn't mean your physical picture has any connection with reality.