Decay of excited electron from Schrödinger equation?

I'm currently studying for a quantum mechanics exam. I found some old exams which has the following question I'm trying to solve. Please tell me if my reasoning is wrong.

Can you see from the Schrödinger equation why an excited electron decays? If yes, explain how. If no, explain why not.

I find this question a bit annoying because there are multiple interpretations possible. The answer could be as simple as saying "Yes because the time-dependent Schrödinger equation allows for the wave function of the electron to change in time and the ground state is energetically much more favorable than an excited state".

But it could also be "no" since the derivation we saw for spontaneous emission (Quantum Mechanics by Bransden & Joachain, section 11.3, p. 527) is based on a thermodynamical argument by Einstein. But that derivation deals more with the how and the how fast rather than why.

I think the answer is "yes" but feel like there's more to it than the simple explanation that I gave. Moreover, the derivation of the decay rates of stimulated emission (which implies an excited electron falling back to the ground state) in the textbook is based on perturbation theory to find the decay and absorption rates of electrons in the presence of an external electromagnetic field. Maybe the actual meaning of the question was "show me how the Schrödinger equation is used to obtain time-dependent perturbation theory"?

Or maybe it's something entirely different.

(Also, don't assume I'm smart! Thanks.)

• I guess they expect the answer "no", because the Schrödinger equation for an atom which is not coupled to a quantized electromagnetic field (or to a time dependent periodic external field) will not show decay processes (as the energy eigenstates are by definition, up to the phase factor, time independent states). – Sebastian Riese Aug 28 '15 at 23:04
• @colouredmirrorball Sebastian is correct provided the "excited electron" is an electron in an excited eigenstate (of atom/molecule/system). Just to cover all cases, you might want to make sure the question does not refer to an excited electron in some sort of arbitrary superposition state. In this case you may want to consider the time-dependent amplitude to transition to the ground state. – udrv Aug 28 '15 at 23:26
• I'm pretty sure you have to add a external Hamiltonian contribution by hand or assume the electron starts in a mixture of time-independent eigenstates. If you assume the electron starts in a pure time-independent eigenstate and there is no perturbing component in the Hamiltonian, then you get time-independent behavior. – dmckee Aug 29 '15 at 0:30
• Since in the lectures we have only dealt with time-independent energy eigenstates (certainly no QFT) of the one-electron atom we can limit ourselves to that case. The Hamiltonian is $$-\dfrac{\hbar^2}{2\mu}\nabla^2 - \dfrac{Ze^2}{4\pi \epsilon_0 r} - \dfrac{i \hbar e}{\mu} \vec{A} \cdot \vec{\nabla} + \dfrac{e^2}{2 \mu } \vec{A}^2$$ with $\mu$ the reduced mass and \vec{A} the vector potential of the electromagnetic interaction in the Coulomb gauge: $\vec{E} = - \dfrac{\partial}{\partial t} \vec{A}$ and $\vec{B} = \vec{\nabla} \times \vec{A}$. The last two terms are taken as the perturbation. – colouredmirrorball Aug 29 '15 at 8:08
• After reading this Wikipedia article: en.wikipedia.org/wiki/Stationary_state#Spontaneous_decay I think the answer might be "no" (unless you indeed take QFT or external radiation into account) given the information we have seen in the lectures. It gives essentially the same answer Sebastian Riese commented. – colouredmirrorball Aug 30 '15 at 16:38