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An oft cited problem with the planetary model of hydrogen is that, if the electron were in fact classically orbiting the proton, then it would radiate away all of its energy and fall into the nucleus.

The quantum mechanical approach fixes this by saying that there exist states for the electron orbiting the proton whose probability density has no time dependence. No time dependence means that there is no oscillating field, which means that no energy is radiated away.

My question is, why do the electrons tend to be in energy eigenstates? In principle, a bound electron could be in a state $|{\psi}\rangle$ given by $$ |\psi \rangle = \sum_{n = 1}^\infty c_{nlms} | nlms \rangle $$ where the $|nlms\rangle$ are the energy eigenstates of the electron. In these states, the probability density would have nontrivial time dependence, which would lead to radiation. It seems to me that if there is no reason for an electron to be in an energy eigenstate, then the quantum mechanical model has the same problem as the planetary model. How does the quantum mechanical model assure us that the electron eventually falls into a particular energy eigenstate? And why does the electron stay there?

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  • $\begingroup$ I dont know much about eigenstates but doesnt the temperature provide energy to the electrons to radiate. Like for the ideal case of black bodies it radiates at all frequencies with the peak depending on the temperature. $\endgroup$ – ChemEng Nov 16 '19 at 3:23
  • $\begingroup$ I also don't know very well what I'm talking about, but I think it's misleading to speak of temperature causing anything at the quantum level. My understanding is that it's more of a description of the distribution of energy eigenstates. $\endgroup$ – Charles Hudgins Nov 16 '19 at 4:18
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In general - yes, if an electron is in superposition of eigenstates it can radiate its energy. In order to describe this, of course, we need also to introduce into our model the electromagnetic field, so the electron will be able to radiate its energy to something. We can do this and calculate transition probabilities and rates etc. And this is of course done - fluorescent light is gas of atoms that their electrons are excited to higher states and then radiate away their energy in the form of light.

However, an electron cannot decay past its lowest energy state. So if we take an electron and just leave it a long time 'at peace', it will decay until it will reach the ground state and then just sit there, in a state which is very very close to an eigenstate.

The case of atoms with more electrons is similar, only the decay can be only to the lowest unoccupied level.

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  • $\begingroup$ So is the idea that we observe stable hydrogen because there is a lowest energy state? Is the quantum mechanical answer to the classical question "why doesn't the electron radiate away all of its energy?" just "because it can't" ? That is, the lowest lying state still has some momentum and energy. $\endgroup$ – Charles Hudgins Oct 21 '19 at 21:17
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    $\begingroup$ Basically, yes. It cannot decay without giving its energy to something else and transitioning into a lower energy state, and there is no lower energy state (at least not on the time-scale of the universe. The proton itself might decay, but don't hold your breath en.wikipedia.org/wiki/Proton_decay). $\endgroup$ – yu-v Oct 21 '19 at 21:26
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    $\begingroup$ Generally speaking, a "long time" for electrons moving to a lower energy state will be usec, sometimes msec. Slower than that is exceptional or requires special circumstances. All of this is of course assuming a low-energy environment.. Good answer and +1. $\endgroup$ – Gloweye Oct 22 '19 at 11:38
  • $\begingroup$ @Charles Hudgins: Not only does the lowest lying state of the electron "going around" the atom still have some energy and momentum quantum-mechanically, so does the lowest state of the vacuum (IE completely empty space). $\endgroup$ – Dast Oct 22 '19 at 15:58
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It's not that they tend to be in energy eigenstates. It is that if the energy of the electron is measured (somehow) then you will measure it to be in an energy eigenstate. This does not mean the electron has to be in an energy eigenstate before measurement though.

In general the state can be expressed as a linear combination of energy eigenstates, as you have given in your question. In order to say more about the time dependence, I think you need to specify a particular example. Certainly the coefficients can have nontrivial time dependence, but they typically don't unless there is something else going on (i.e. you should get some pretty simple time dependence for just the single electron in a unperturbed hydrogen atom).

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    $\begingroup$ This is roughly what I suspected, but this still raises a few questions for me. What constitutes measurement in this scenario? There is a lot of very stable hydrogen in the universe. Has it all been measured? What is the quantum mechanical description of radiation (don't necessarily want a long answer to this, just a brief account)? $\endgroup$ – Charles Hudgins Oct 21 '19 at 19:59
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    $\begingroup$ Hi Charles- I enjoyed your 'Has it all been measured?' I think the answer is precisely that electrons in mixed states will radiate energy (emphasis on will), and in doing so will morph into a stationary state. $\endgroup$ – Marco Ocram Oct 21 '19 at 20:11
  • $\begingroup$ Won't we have time dependence any time there is a superposition of energy eigenstates. Consider the toy example $|\psi \rangle = \frac{1}{\sqrt{2}} (|0\rangle + |1\rangle)$. Then $\psi(x,t) = \frac{1}{\sqrt{2}}( e^{iE_0 t} \psi_0(x) + e^{iE_1 t} \psi_1(x))$, which will have probability density $\rho(x,t) = \frac{1}{2} (\psi_0(x))^2 + \frac{1}{2} (\psi_1(x))^2 + \cos((E_1 - E_0)t) \psi_0(x) \psi_1(x)$ (I have chosen $\psi_0(x)$ and $\psi_1(x)$ to be real). Shouldn't that time dependence correspond to a moving charge, hence radiation?(this is in response to Aaron's edit) $\endgroup$ – Charles Hudgins Oct 21 '19 at 20:14
  • $\begingroup$ @CharlesHudgins Doesn't $\langle 0|1\rangle=\langle 1|0\rangle=0$? I thought you were talking about energy measurements, not position? $\endgroup$ – Aaron Stevens Oct 21 '19 at 20:19
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    $\begingroup$ Fair enough. I probably just need to reread the part in my quantum book about radiation. $\endgroup$ – Charles Hudgins Oct 21 '19 at 21:31
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If an electron is in a superposition of two eigenstates, its wave function is the sum of those two eigenstates. Each eigenstate evolves independently of the other in time. The time dependent wave function has the form

$$\phi(x, t) = \phi(x)\cdot e^{iat}$$

where the $a$ depends on the energy of the eigenstate. Now, what happens when you sum two such wave functions with different $a$ together? Well, they interfere. Wherever both wave functions overlap there will be times when $\frac{\phi_1(x)}{|\phi_1(x)|}\cdot e^{ia_1t} = \frac{\phi_2(x)}{|\phi_2(x)|}\cdot e^{ia_2t}$ (constructive interference), and times when $\frac{\phi_1(x)}{|\phi_1(x)|}\cdot e^{ia_1t} = -\frac{\phi_2(x)}{|\phi_2(x)|}\cdot e^{ia_2t}$ (destructive interference). And that means that the amplitude of the superposition $\phi_1(x)\cdot e^{ia_1t} + \phi_2(x)\cdot e^{ia_2t}$ oscillates with a frequency of $\frac{a_2 - a_1}{2\pi}$.

So, the probability cloud of an electron in a state of superposition is not static. It's oscillating with a fixed frequency that's proportional to the energy difference, and thus actively interacting with the electromagnetic field. The result of this interaction may be that the electron drops into the lower state, or that it gets exited into the upper state. But until it reaches a state without an oscillating probability cloud (usually a pure eigenstate), the electron won't rest until it does.

The preference of the lowest energy eigenstate is only due to our preference for cool environments in experiments: When there's no photon around to be absorbed, the only way out of superposition is to emit a photon. However, there are cases where the electrons prefer a high eigenstate. One such case is lasers: They need to get more electrons into the exited state than there are in the base state (this is called inversion), because that's the prerequisite for the light amplification process. That's quite a bit of science actually, but it happens in every single CD player.

I believe the desire of identifying eigenstates is largely driven by the fact, that it's easy to derive the time dependent wave function once you have your wave function separated into eigenstates: Each eigenstate has its own $e^{iat}$ factor, and that's easy enough to calculate for the entire wave function. And the superposition is also easy enough to calculate. You could simulate the time dependent Schrödinger Equation directly, but that's computationally expensive, error fraught, and imprecise on large timescales. The separation of the wave function into eigenstates allows us to come up with analytical, and thus precise solutions easily.

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