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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.

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- Thread starter TeTeC
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- #1

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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.

- #2

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It is a law of nature.

- #3

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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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You can find a discussusion about this in e.g. Cohen-Tannoudji's book "atom-photon interactions"

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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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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.

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It is a little more complicated than a simple minimum energy principal. In Bethe and Salpeter "

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.

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),

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- #9

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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?

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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

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.

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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.

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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.

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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?

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

But then why does the 3s initial state have a lifetime of only 160 nanoseconds, when the 2s initial state has an infinite lifetime?Because the oscillations are spherically symmetric, like a pulsating balloon. There is no net radiation from such an antenna.

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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.

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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?

(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 have looked all through Bethe and Salpeter "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.

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Yes, that also works.

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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_rad. This interaction is alwaysjAbut it is small in many cases so oneoncanprepare 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).

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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.

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- #23

f95toli

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This is basically the situation described by the Jaynes-Cummings Hamiltonian.

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Ah, ok... Many thanks!

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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

This is basically the situation described by the Jaynes-Cummings Hamiltonian.

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