Quantum mechanics says that if a system is in an eigenstate of the Hamiltonian, then the state ket representing the system will not evolve with time. So if the electron is in, say, the first excited state then why does it change its state and relax to the ground state (since it was in a Hamiltonian eigenstate it should not change with time)?
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$\begingroup$ Note that depending on where you put the zero energy, the ket does evolve in time, with a phase. This is relevant when you make superpositions of different states. While you are probably aware of this, I think it might be useful for other readers of this discussion. $\endgroup$– AntoineCommented Nov 29, 2016 at 20:34
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3$\begingroup$ You can find the answer here $\endgroup$– valerioCommented Nov 30, 2016 at 9:28
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3 Answers
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The atomic orbitals are eigenstates of the Hamiltonian $$ H_0(\boldsymbol P,\boldsymbol R)=\frac{\boldsymbol P^2}{2m}+\frac{e}{R} $$
On the other hand, the Hamiltonian of Nature is not $H_0$: there is a contribution from the electromagnetic field as well $$ H(\boldsymbol P,\boldsymbol R,\boldsymbol A)=H_0(\boldsymbol P+e\boldsymbol A,\boldsymbol R)+\frac12\int_\mathbb{R^3}\left(\boldsymbol E^2+\boldsymbol B^2\right)\,\mathrm d\boldsymbol x $$ (in Gaussian units, and where $\boldsymbol B\equiv\nabla \times\boldsymbol A$ and $\boldsymbol E\equiv \dot{\boldsymbol A}-\nabla\phi$)
Therefore, atomic orbitals are not stationary: they depend on time and you get transitions from different states.
The problem is that what determines time evolution is the total Hamiltonian of the system, and in Nature, the total Hamiltonian includes all forms of interactions. We usually neglect most interactions to get the overall description of the system, and then add secondary effects using perturbation theory. In this sense, the atom is very accurately described by $H_0$, but it is not the end of the story: there are many more terms that contribute to the real dynamics.
answered Nov 28, 2016 at 14:55
AccidentalFourierTransformAccidentalFourierTransform
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$\begingroup$ That was helpful thanks. Just wondering apart from the electromagnetic field what other things will contribute to the Hamiltonian in this case? $\endgroup$ Commented Nov 28, 2016 at 21:02
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5$\begingroup$ @AbhaySrivastav your welcome, I'm glad I could help :-) in the case of atomic phenomena, other contributions come from: electrons interacting with other electrons; the spin of the nucleus interacting with the spin of the electrons; relativistic corrections (due to $E^2=p^2+m^2$ instead of $E=\frac{p^2}{2m}$). There are more contributions, such as the weak force, but these are truly negligible and nobody considers them to be relevant at all. $\endgroup$ Commented Nov 28, 2016 at 22:06
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6$\begingroup$ Though this answer is correct (+1), I think that when reading it some people will think that you need some kind of source (charge or current) in order to have an interaction with the EM field, when instead there will be interaction also in absence of sources because of quantum fluctuations in the EM field ground state (vacuum state). Maybe it is worth mentioning this. $\endgroup$– valerioCommented Nov 30, 2016 at 9:34
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The hydrogen atom in an excited state is not really in an energy eigenstate.
There are two ways of looking at it. One way is to recognize that the atom is not isolated. It is always coupled to the electromagnetic field. Even if field itself is in the ground state, there are "zero-point" fluctuations in the field amplitude. Thus, the atom is always feeling the influence of an external field. The zero-point fluctuations have components at all frequencies, including the atomic transition frequency. So the spontaneous decay of an excited atom can be thought of as stimulated emission due to the zero-point fluctuations.
The second way of looking at it is to take the system of interest to be the atom and the electromagnetic field. In this case, the state with no excitation of the field, and the atom in an excited state is not an energy eigenstate. The total wave function amplitude will start entirely atomic, but evolve to include field excitation.
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$\begingroup$ That was helpful thanks.But I think after completing my course in qft I will be in a better position to appreciate it $\endgroup$ Commented Nov 28, 2016 at 21:03
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One way to look at this is the Fermi's Golden Rule:
If we add a time dependent perturbation to the time independent Hamiltonian, $$ H=H_{0}+H_{1} $$ plug it in the schrodinger's equation,
$$ (H_{0}+H_{1})|\phi(t)\rangle=i\hbar\partial_{t}|\phi(t)\rangle $$
calculate the transition coefficients, $$ c_{f\to i}(t)=\frac{1}{i\hbar}\int_{0}^{t}\langle f|H_{1}|i\rangle e^{i\omega_{fi}t}dt $$
add an approximation like the dipole approximation for atoms
$$ H_{1} = q\,\vec{r}\cdot\vec{E} $$
and compute for an actual atom, we get that the transition rate between the excited state f and ground state i is non-zero even if these are time-independent states.
As others have said you can't get rid of EM fields in the "real world". EM fields, which are waves of electric & magnetic fields generated by charged particles are omnipresent because electric fields decay by $$E = \frac{\text{constant}}{r^2}$$ disappear only at infinity. So the transition rates would always be non-zero. If transition rates are non-zero, it would de-excite even if there is no external time-dependent disturbance.
Further Reading
http://staff.ustc.edu.cn/~yuanzs/teaching/Fermi-Golden-Rule-No-II.pdf
http://www.chemie.unibas.ch/~tulej/Spectroscopy_related_aspects/Lecture25_Spec_Rel_Asp.pdf
http://farside.ph.utexas.edu/teaching/qmech/Quantum/node117.html
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1$\begingroup$ This is actually the best of the three answers so far. $\endgroup$– my2ctsCommented Jun 21, 2019 at 19:45