Excited state to ground state

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Hello,

Non-relativistic 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.
 

Answers and Replies

  • #2
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
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Hello,

Non-relativistic 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.
I suspect that quantum electrodynamics explains it, though I don't know too much about it.
 
  • #4
f95toli
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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. Cohen-Tannoudji's book "atom-photon interactions"
 
  • #5
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I bought this book a few weeks ago, but I still have to read it. Thanks !
 
  • #6
alxm
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You'd need to pose that question more specifically.

Is the question: "How can that transition occur, in non-relativistic 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
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Hello,

Non-relativistic 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.
It is a little more complicated than a simple minimum energy principal. In Bethe and Salpeter "Quantum Mechanics of One and Two Electron Atoms", there is a Table on page 266 showing the theory and calculated transition rates and lifetimes for all the levels in hydrogen up to 6h.

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
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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
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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.
But why doesn't the hydrogen 2s state decay to the 1s state by the same method all the other states decay?
 
  • #10
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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.
It is wrong and right at the same time.

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 non-relativistic 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
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But why doesn't the hydrogen 2s state decay to the 1s state by the same method all the other states decay?
Because the oscillations are spherically symmetric, like a pulsating balloon. There is no net radiation from such an antenna.
 
  • #12
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It is a little more complicated than a simple minimum energy principal. In Bethe and Salpeter "Quantum Mechanics of One and Two Electron Atoms", there is a Table on page 266 showing the theory and calculated transition rates and lifetimes for all the levels in hydrogen up to 6h.

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.
It is not forbidden but highly suppressed. It is still possible - via multi-photon radiation. The corresponding probability is rather small due to extreme symmetry mentioned in the previous posting (balloon oscillations).

Bob.
 
  • #13
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(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?
Because the oscillations are spherically symmetric, like a pulsating balloon. There is no net radiation from such an antenna.
But then why does the 3s initial state have a lifetime of only 160 nanoseconds, when the 2s initial state has an infinite lifetime?
 
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  • #14
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(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?
I haven't worked out all the combinations, but I think the superposition of 3s and 2p would probably have a strong oscillating dipole moment.
 
  • #15
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(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?
I haven't worked out all the combinations, but I think the superposition of 3s and 2p would probably have a strong oscillating dipole moment.
I have looked all through Bethe and Salpeter "Quantum Mechanics of One and Two Electron Atoms", but I have not found any reference to pulsating balloons, but I believe the 3s decays via the 2p state. The 3s to 2p is a delta N=1, delta L=1 transition, the same as 3d to 2p. Both have (approximately) the same transttion energy. But the 3s has a 160 nanosecond lifetime, and the 3d has a 15.6 nanosecond lifetime (from Bethe and Salpeter). Neither can go directly to the 1s, because that would be delta L = 0 or 2 (forbidden). So maybe the pulsating balloon theory is correct.
 
  • #16
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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
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Or you could just learn how to do time-dependent 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
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Yes, that also works.
 
  • #19
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It is wrong and right at the same time.

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 non-relativistic 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.
However there must be some further dissipation process involved if the atom remains in the ground state, no? Under unitary evolution the entire system (atom+EM-field) must return after a certain period of time to its original state. The period "t" must only satisfy exp(-itH)=unity. This is always possible if the Hilbert space one is working with is finite dimensional the eigenvalues of H are rational numbers, say e1=n1/d1, e2=n2/d2, .... elast=nlast/dlast, then you simply have t=2*pi*n1*n2*...*nlast. Unfortunately I have no idea what happens if one has infinite dimensional Hilbert spaces and/or non-rational eigenvalues... My feeling is that exp(-itH)=unity can still be fulfilled with arbitrary precision, even though the period might be huge... On one hand one can always approximate the non-rational numbers by rational ones with arbitrary precision, and on the other hand for such a simple process the effective Hilbert space is usually low-dimensional even if you allow for virtual processes...

Do you know if some dissipational process is responsible for the atom to remain in the ground state? Or is there something wrong with my argument about unitary evolution?
 
  • #20
f95toli
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Do you know if some dissipational process is responsible for the atom to remain in the ground state? Or is there something wrong with my argument about unitary evolution?
Interaction with environmental degrees of freedom (which can in turn be modelled as e.g spin-bath) will eventually lead to the atom staying in its ground state.
In the simplest models with just a constant loss of energy this leads to an exponential decay of the probability to find the atom in its excited state; the time constant turns out to be identical to the T1 time that is introduced (phenomenologically) in the Bloch equations (the other time constant is T2, which is the dephasing time; T1 and T2 are what is measured in e.g. NMR).
 
  • #21
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However there must be some further dissipation process involved if the atom remains in the ground state, no? Under unitary evolution the entire system (atom+EM-field) must return after a certain period of time to its original state. The period "t" must only satisfy exp(-itH)=unity. This is always possible if the Hilbert space one is working with is finite dimensional the eigenvalues of H are rational numbers, say e1=n1/d1, e2=n2/d2, .... elast=nlast/dlast, then you simply have t=2*pi*n1*n2*...*nlast. Unfortunately I have no idea what happens if one has infinite dimensional Hilbert spaces and/or non-rational eigenvalues... My feeling is that exp(-itH)=unity can still be fulfilled with arbitrary precision, even though the period might be huge... On one hand one can always approximate the non-rational numbers by rational ones with arbitrary precision, and on the other hand for such a simple process the effective Hilbert space is usually low-dimensional even if you allow for virtual processes...

Do you know if some dissipational process is responsible for the atom to remain in the ground state? Or is there something wrong with my argument about unitary evolution?
Yes, there is a natural dissipation process. It is still the same electron-oscillator interaction but with other, low-frequency oscillators. Factually the electron gives away its energy not to one oscillator but to infinitely many ones. Their frequency band is continuous, so there is no selected periods. It is as if a particle collided with many particles and given away its energy to the whole ensemble with different timings. It is impossible to reverse the process. In addition, we have to consider the initial low-frequency oscillators as excited, with unknown state (determined with the environment temperature T, for example).

Bob.
 
  • #22
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Ok, so this is the case if the EM-field can transfer energy to other surrounding systems... However, my question is: what happens if the entire universe consists only of one excited atom and the QED-vacuum?
 
  • #23
f95toli
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It will oscillate forever with a frequency given by the coupling strength (times 2 to be precise).
This is basically the situation described by the Jaynes-Cummings Hamiltonian.
 
  • #24
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Ah, ok... Many thanks!
 
  • #25
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It will oscillate forever with a frequency given by the coupling strength (times 2 to be precise).
This is basically the situation described by the Jaynes-Cummings Hamiltonian.
The 3d-2p and the 3s-2p transition energies in hydrogen are about 1.9 eV, so their transition lifetimes (15 to 160 nanoseconds) are very long compared to the uncertainty time (h-bar is about 6 x 10-16 eV-sec). The oscillation between the two states is occurring in vacuum, which even in inter-galactic space has a characteristic impedance of 377 ohms. Vacuum is "dissipative" in the sense that once the oscillating dipole radiates into vacuum, the photon moves away at about 3 x 1010 cm/sec and can no longer couple to the atom. Once the oscillating electron's EM field couples to free space and releases a single 1.9 eV photon, the 2p electron will never go back to the 3d (or 3s) state (it will go to the 1s state in about 1.6 nsec).
 

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