Christmas Gift: Radiating Atom

Bohr atom model as a planetary system with fixed orbits

Here is a small gift to the quantum mechanics community in search for mathematical models of reality with the basic question: Why is the minimal energy ground eigenstate of an atom stable without radiation, while excited eigenstates with larger energies radiate and give rise to a spectrum? 

The common answer is that the ground state of minimal energy does not radiate because if it did it would not be stable and so not exist over time. But this is a form of circular reasoning expressing that an atom is stable because if was not stable,  then it would not exist. Not so illuminating. 

This answer connects to the collapse of classical atomic models with electrons in swirling motion around the kernel which radiate, because electrons in accelerated motion do generate varying electrical fields and so radiate. In particular, the ground state would be radiating but it does not and so electrons cannot be swirling around the kernel as in the classical Bohr atom model with a planetary type system of electrons. 

A different more precise answer is given by RealQM: The ground state does not radiate because its electron charge density does not vary in time and so does not accelerate and give rise to radiation. On the other hand the charge density of a superposition of the ground state with minimal energy $E_1$ and eigenstate of higher energy $E_j>E_1$ is time dependent with frequency $E_j-E_1$ and so does radiate.

To see some details consider the following 1d Schrödinger equation, which is analysed in detail at Computational Blackbody Radiation:      

• $i\frac{d\Psi}{dt} = H\Psi$    (*)

where the Hamiltonian $H$ is given by

• $Hu = -\frac{d^2u}{dx^2} - \gamma \frac{d^3u}{dt^3}$

acting on functions $u(x,t)$ defined for $0<x<\pi$ and $t>0$ satisfying the boundary condition $u (0,t)=u (\pi,t)=0$ for $t>0$, and $\gamma\ge 0$ is a small constant. The $\gamma$-term with $\gamma >0$ corresponds to outgoing radiation of intensity 

• $\gamma (\frac{d^2u}{dt^2})^2$

including the acceleration $\frac{d^2u}{dt^2}$.

For $\gamma =0$ the eigenfunctions of $H$ are given by $\sin(jx)$ for $j=1,2,3...$ with corresponding eigenvalues $j^2$, and so the solution of (*) in this case can be expressed as 

• $\Psi (x,t) =\sum_{j=1}^\infty c_j\exp(-ij^2t)\sin(jx)$,

where the $c_j$ are real coefficients. The ground state with $c_j=0$ for $j>1$ is given by 

• $\exp( -it)\sin(x)$

with modulus squared as charge density independent of time. On the other hand a superposition 

• $\exp(-it)\sin(x)+\exp(-ij^2t)\sin(jx)$
• $= \exp(-it)(sin(x)+\exp(-ij^2t+ it)\sin(jx))$

does have a time dependent charge density for $j>1$ and so does radiate. Merry Christmas!

PS1 To get perspective, try to Google the following question:

• Why do not atoms in ground state radiate?

See that you do not get a meaningful answer. Ask yourself why?

PS2 Pure eigenstates do not radiate, but it is not clear how to excite a higher eigenstate for an atom. For a vibrating string this is possible suppressing the ground state with a light left finger touch on the middle of the string, thus exciting (mainly) the second harmonic (flageolet). In RealQM a superposition of ground state and higher energy eigenstate involves a mean-value Hamiltonian and so is a true "complex mixed state" which needs exterior forcing to sustain over time. In standard QM with a linear Schrödinger equations exact superposition of eigenstates is formally possible without radiation and forcing which formally would sustain over time. But again it is most unclear how to excite higher eigenstates, which puts quantum computing building on such superpositions in doubt as clarified in upcoming post.