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.

  • 4
    $\begingroup$ The short answer is that the Hamiltonian you're considering is incomplete, because it doesn't include the interaction between the atoms and the electromagnetic field. But because the interaction is small compared to the other terms, the eigenstates of the incomplete Hamiltonian are still useful, both conceptually and when performing calculations. $\endgroup$ Dec 1, 2011 at 21:31

2 Answers 2


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.


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.

  • $\begingroup$ You do not get the right answer any other way, this is completely wrong. The coupling is nonsensical. What you are describing is is the BKS theory, and it doesn't work. If you radiate an electromagnetic field over some atoms, and the electromagnetic field has the energy of one photon, and one of the atoms absorbs it, how does the photon know it can't excite a far away atom? The signal has to go at the speed of light, and this can't prevent a second absorption event. This means that if you don't quantize EM field, you violate energy conservation, and this is why BKS is totally wrong. $\endgroup$
    – Ron Maimon
    Dec 1, 2011 at 20:34
  • $\begingroup$ I am not describing BKS theory. BKS predated the Schroeding equation, and my theory depends on the Schroedinger atom. And in my theory there is no need for an atomic transition to be driven completely. Since there are no such things as photons, a passing light wave can partially excite the p state in any number of hydrogen atoms. $\endgroup$ Dec 1, 2011 at 20:44
  • $\begingroup$ What you're doing here is coupling a quantum system (the atom) to a classical system (the electromagnetic field). That's conceptually a bit awkward, hence Ron's objection. However, it can be formally justified; it's a routine procedure in quantum optics, though ironically in that case the EM field is quantized and the matter is classical! In this case, the result won't be exactly right, because you're neglecting corrections due to the quantization of the EM field, but that doesn't mean it doesn't provide a good mental model of the decay process. $\endgroup$ Dec 1, 2011 at 21:22
  • $\begingroup$ I wouldn't object to Ron so much if he called my method "conceptually a bit awkward", although I might disagree with that characterization. I find it tiresome that he calls my reasoning "nonsensical" and "completely wrong". $\endgroup$ Dec 1, 2011 at 22:16
  • $\begingroup$ @Marty: You are treating the EM field as classical, and the atom as quantum, and claiming that this works to give the transition from excited to ground state. This does not work, it is physically preposterous, it is exactly like BKS (although admittedly BKS was pre-Schrodinger, the Schrodinger equation doesn't change anything regarding the failures of BKS), and it does not conserve energy. The only way to treat radiation is in a quantum way, as photon emission. This doesn't mean that you can't describe aspects of the process as atom+classical radiation, in fact, you can infer... $\endgroup$
    – Ron Maimon
    Dec 2, 2011 at 4:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.