Electron Transitions: Superposition of States

It'd be great if you could help me clarify a few things in my head.

Firstly I've got written in my notes "quantum mechanics forbids spontaneous transitions from one energy level to another because energy eigenfunctions are time independent".

However this seems a bit of a circular argument. I was under the impression that all electron wavefunctions are in fact decaying exponentially with time, but instead the probability density is time independent. Is this true?

Secondly, I'm willing to accept a spontaneous emission can be induced by vacuum fluctutions, but I'm not too sure what this means.

Now the main problem I've got is: during a transition, the atom is in a superposition of 2 states (lets say 2p0 and 1s0 where these represent the quantum number n, l and ml). In my notes I'm told to consider a "50:50" superposition and calculate the distribution at various times.

From this I can see that the charge distribution "sloshes" backwards and forwards with time which I appreciate will emit radiation. However, how do you get out of these seemingly eternal oscillations and finally settle down into the ground state? Surely the 50:50 ratio will have to change, so after a certain time it becomes a 0:100 ratio? Am I thinking about this in the correct way?

Any help would be appreciated.

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SpectraCat
It'd be great if you could help me clarify a few things in my head.

Firstly I've got written in my notes "quantum mechanics forbids spontaneous transitions from one energy level to another because energy eigenfunctions are time independent".
Add the words, "in the absence of an external perturbation", and you've got it.

However this seems a bit of a circular argument. I was under the impression that all electron wavefunctions are in fact decaying exponentially with time, but instead the probability density is time independent. Is this true?
Eigenstates have a time-oscillating phase (the complex exponential $$e^\frac{-iEt}{\hbar}$$ is an oscillating function, not a decaying one), which is canceled out in the probability density expression $$\psi^* \psi d\tau$$, yielding a time-independent probability.

Secondly, I'm willing to accept a spontaneous emission can be induced by vacuum fluctutions, but I'm not too sure what this means.
First of all, in order to properly describe this, you need to use quantum field theory to account for the quantum nature of the electromagnetic fields representing both the Coulomb potential between the electron and nucleus, as well as the emitted photon. I do not have a dettailed understanding of QFT, but the phenomenological version of the process is as follows: as you said above, eigenstates are stable in the absence of an external perturbation. Therefore, although conservation of energy says that they can decay to a lower state by emission, this process requires some external agent to perturb the state so that it can occur. "Vacuum fluctuations" is a catch-all term used to describe the spontaneous creation and destruction of virtual particle pairs, a process that is predicted by QFT to be occurring all the time. These vacuum fluctuations induce a coupling (i.e. a perturbation) between the ground and excited states, allowing the atom to emit radiation "spontaneously". [One of the great early successes of quantum field theory was the Dirac's derivation from first principles of the known phenomenological Einstein A-coefficient describing wavelength-dependent spontaneous emission processes.]

Now the main problem I've got is: during a transition, the atom is in a superposition of 2 states (lets say 2p0 and 1s0 where these represent the quantum number n, l and ml). In my notes I'm told to consider a "50:50" superposition and calculate the distribution at various times.

From this I can see that the charge distribution "sloshes" backwards and forwards with time which I appreciate will emit radiation. However, how do you get out of these seemingly eternal oscillations and finally settle down into the ground state? Surely the 50:50 ratio will have to change, so after a certain time it becomes a 0:100 ratio? Am I thinking about this in the correct way?
The "sloshing" of the charge density that you describe is indeed expected to emit radiation from a *classical* point of view, but the situation in QM is a bit different. In the absence of a perturbation, the expansion coefficients will not vary with time, and your supposition they will oscillate forever is the correct one (for non-relativistic QM in the semi-classical approximation of quantized matter and classical EM radiation). However, as you have noticed, the probability density is oscillating between the ground state (which cannot decay), and the excited state (which can decay spontaneously).

Thanks very much for that reply, it cleared a lot of things up. I'm just still a little confused about this time dependence

In this video, it shows a transition from 2p to 1s.

Let u2 = the wavefunction for the excited state and u1 = the ground state. At time 0, the total probability density appears to be u$$^{2}$$ = u2$$^{2}$$. Then as time goes on, we seem to have u$$^{2}$$ = [a(t)u1 + b(t)u2]$$^{2}$$ and then at the end we have u$$^{2}$$ = u1$$^{2}$$.

Am I correct in saying this transition has been caused by "switching on" a perturbation? I know b(t$$_{o}$$)$$^{2}$$ must equal 1, and at the end of the transition a(t$$_{end}$$)$$^{2}$$ must equal 1, but how do I calculate these coefficients at other times? I have a feeling they are dependent on the perturbation...

Thanks

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SpectraCat
Thanks very much for that reply, it cleared a lot of things up. I'm just still a little confused about this time dependence

In this video, it shows a transition from 2p to 1s.

Let u2 = the wavefunction for the excited state and u1 = the ground state. At time 0, the total probability density appears to be u$$^{2}$$ = u2$$^{2}$$. Then as time goes on, we seem to have u$$^{2}$$ = [a(t)u1 + b(t)u2]$$^{2}$$ and then at the end we have u$$^{2}$$ = u1$$^{2}$$.

Am I correct in saying this transition has been caused by "switching on" a perturbation? I know b(t$$_{o}$$)$$^{2}$$ must equal 1, and at the end of the transition a(t$$_{end}$$)$$^{2}$$ must equal 1, but how do I calculate these coefficients at other times? I have a feeling they are dependent on the perturbation...

Thanks
I have to say that I have no idea what is intended to be shown by that video, and unfortunately the author didn't post any details about how the simulation was constructed. It could be that it is a purely classical simulation, where radiation is continuously emitted from the moving charge distribution, in which case it has no bearing on our discussion. Or perhaps it is just one half-cycle of the oscillations of the probability density of the 50-50 superposition of a 2p and 1s orbital. In any case, it does *not* describe a quantum mechanical transition as I understand it, because that would involve an effectively instantaneous transition between the states, with the concomitant emission (or absorption) of a photon with an energy matching the level spacing.

Anyway, if you are not considering any perturbations (including spontaneous emission), then the 50-50 superposition will stay stable forever in the semi-classical description. Furthermore, there is no time-dependence of the expansion coefficients .. in other words, if it starts out 50-50, it stays 50-50. However, the probability density will be time-dependent and will show periodic oscillations akin to what was shown in that video. If you write down the complete time-dependent equations for both the wavefunction and the probability density of the superposition state, you should be able to understand why this is so.

Now, if you then want to consider what happens to the superposition when an external perturbation is suddenly switched on, then I suggest you read about time-dependent perturbation theory. A good low-level description of this theory for a two-state systems in the semi-classical limit is given in chapter 6 of the textbook, "Molecular Quantum Mechanics", by Peter Atkins and Ronald Friedman.

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