
#1
Apr2509, 04:53 PM

P: 49

Hello,
Nonrelativistic quantum mechanics doesn't explain why the electron in the hydrogen atom (for example) "decays" from excited states to the ground state. Which theory does explain this phenomenon (from basic principles) ? Thanks. 



#2
Apr2509, 08:04 PM

P: 1

All systems in nature tend to be in a state of minimum energy. For example, if you have a ball at a height of a level and go down her loose, because the lower level has a lower energy than the top. A drop of liquid, for example, in the vacuum acquires a spherical shape and this is because this configuration is a state of minimum energy.
It is a law of nature. 



#3
Apr2509, 10:49 PM

P: 200





#4
Apr2609, 05:57 AM

Sci Advisor
PF Gold
P: 2,197

Excited state to ground state
Yes, QED can explain it as long as you allow for coupling to vacuum states (effectivly the enivornment).
You can find a discussusion about this in e.g. CohenTannoudji's book "atomphoton interactions" 



#5
Apr2609, 04:45 PM

P: 49

I bought this book a few weeks ago, but I still have to read it. Thanks !




#6
Apr2609, 07:54 PM

Sci Advisor
P: 1,867

You'd need to pose that question more specifically.
Is the question: "How can that transition occur, in nonrelativistic QM?" then the answer is that there's no real problem with that, you can calculate the transition probability and everything without knowing one thing about photons. Or is your question: "Why do things tend to a lower energy level?"  that's general thermodynamics Or is your question: "How, _exactly_ are the photons emitted/absorbed?"  that's QED. 



#7
Apr2709, 09:35 PM

P: 4,664

Of particular interest are the transition rates from the n=2 levels. The only lower state is the 1s, so the energy difference and transition energy is about 10.2 eV. The calculated lifetime for the 2p state is 1.6 nanoseconds (the 2p > 1s transition), but the lifetime of the 2s state is infinity!!! The 2s > 1s transition is forbidden. Only after resorting to other transitions can the 2s state eventually get to the 1s. 



#8
Apr2809, 04:23 PM

P: 398

I don't know why people seem reluctant to put forward the obvious explanation for these transitions. The mixture of a ground state with an excited state gives you an oscillating charge distribution, which radiates energy like an ordinary antenna. It's not really mysterious.




#9
Apr2809, 05:28 PM

P: 4,664





#10
Apr2809, 05:53 PM

P: 1,160

Let me explain. In the initial non perturbed state you have only an excites atomic state. If there is no other interaction, of any kind, no transition to or superposition with the ground state can arise. Now, let us remember that charges are coupled to the electromagnetic field. In the nonrelativistic case the interaction energy (perturbation term) is jA_rad. This interaction is always on but it is small in many cases so one can prepare the initial atomic exited state and wait. What happens due to the interaction? The quantized electromagnetic field in our simple case can be represented by a harmonic (resonant) oscillator in its ground state. Then the probability of its exciting (populating its level) starts to "grow" with time and the probability to find the atom in its initial state "decreases" with time. Finally the atom gets in its ground state and the photon oscillator gets excited. I took the words in double commas because the corresponding amplitudes (or probabilities) oscillate in time. You can imagine that as a transition from one standing wave to another. In meantime everything oscillates. The radiation takes many "jumps" to and fro. In this sense the previous answer is right, but we must keep in mind why the ground state population (probability) "grows" (grows on average): there is another system that takes the energy difference. Bob. 



#11
Apr2809, 06:17 PM

P: 398





#12
Apr2809, 06:28 PM

P: 1,160

Bob. 



#13
Apr2809, 08:46 PM

P: 4,664

(posted by Bob S)
But why doesn't the hydrogen 2s state decay to the 1s state by the same method all the other states decay? 



#14
Apr2809, 10:28 PM

P: 398





#15
Apr2909, 06:14 PM

P: 4,664

(posted by Bob S)
But why doesn't the hydrogen 2s state decay to the 1s state by the same method all the other states decay? (Posted by Conway) Because the oscillations are spherically symmetric, like a pulsating balloon. There is no net radiation from such an antenna. (Posted by Bob S. ) But then why does the 3s initial state have a lifetime of only 160 nanoseconds, when the 2s initial state has an infinite lifetime? 



#16
Apr2909, 07:51 PM

P: 398

Thanks for the vote of confidence (?). But I hope you don't feel you need to take my word for it on the pulsating balloon. It's one of the superpositions that's fairly easy to visualize: hold the 1s state steady and applying a 1/2 cycle time evolution to the 2s. One way the charge is pushed inwards, and 180 degrees later the charge is pushed outwards.




#17
Apr3009, 03:24 AM

P: 981

Or you could just learn how to do timedependent perturbation theory? Fermi's Golden rule? That would tell you that the decay rate is proportional to the matrix element of the perturbation (in this case, an electric dipole) connecting the final and initial states. The difference in 3s>2p and 3d>2p is largely due to this difference.




#18
Apr3009, 10:00 AM

P: 398

Yes, that also works.



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