Optical Emission - due to Acceleration or Oscillation

what_are_electrons
Three Interconnected Questions:

1. When an atom is excited by a visible photon (KE=1 eV), does the probability (radial) density increase for the valence electrons or not?

2. If the probability density has increased outward, then what normally causes the excited state to begin the process of photon emission?

3. Once the excited atom has been "iritated" by some means, what is the cause of the photon emission - Choice A: electric dipole oscillation, or Choice B: some acceleration or deceleration process, or Choice C: another explanation which is...
 
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Let's take them slowly.First question:what is radial density...?Is it the PROBABILITY DENSITY...?If so,why would that happen??I'd say that the electron is ejected (just like in the photoelectric effect).

Daniel.
 
dextercioby said:
Let's take them slowly.First question:what is radial density...?Is it the PROBABILITY DENSITY...?If so,why would that happen??I'd say that the electron is ejected (just like in the photoelectric effect).

Daniel.
Sorry. Got my X-ray and visible wires crossed. Let's drop KE down to <1 eV. I edited my original post. Also changed radial to probability.
 
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The question is still the same...?What would the principles to account for an increase in (radial) probability distribution...?

Daniel.
 
When an atom absorbs a photon, it necessarily makes a transition to a more energetic state. Given that the average radius for an atomic state grows with energy, it most likely is the case that the radial probability's max does grow after absorbtion. But, the devil is in the details, and to be sure, you'll need to do some computations, or check in Condon and Shortley's The Theory of Atomic Spectra, a bible in its field. (The issue of photon absorbtion in multi-electron atoms is somewhat tricky.)
Regards,
Reilly Atkinson
 
what_are_electrons said:
Three Interconnected Questions:
1. When an atom is excited by a visible photon (KE=1 eV), does the probability (radial) density increase for the valence electrons or not?
Maybe you can make clear what you mean with this question.
The probability density for WHAT exactly will increase?

what_are_electrons said:
2. If the probability density has increased outward, then what normally causes the excited state to begin the process of photon emission?
That's a good question. I personally don't know how
one can explain spontaneous emission in an intuitive way.
To my knowledge the spontaneous emission is not contained in usual quantum mechanics. The process of spontaneous emission comes into play
if you quantize the electromagnetic field, that is you introduce the photons.

what_are_electrons said:
3. Once the excited atom has been "iritated" by some means, what is the cause of the photon emission - Choice A: electric dipole oscillation, or Choice B: some acceleration or deceleration process, or Choice C: another explanation which is...
As I stated above, I don't know why spontaneous emission occurs, that is I can't give you an intuitive explanation. All I know is that you can
describe the emission formally in second quantization by bosonic and fermionic annihilation and creation operators (maybe you know about that stuff).
 
Edgardo said:
Maybe you can make clear what you mean with this question.
The probability density for WHAT exactly will increase?
WHAT=electron

Edgardo said:
That's a good question. I personally don't know how
one can explain spontaneous emission in an intuitive way.
To my knowledge the spontaneous emission is not contained in usual quantum mechanics. The process of spontaneous emission comes into play
if you quantize the electromagnetic field, that is you introduce the photons.
I did not say "spontaneous" emission. My 3rd "interconnected" question uses the word "iritated" which defines the type of emission - ie stimulated by not in a coherent way.
 
reilly said:
When an atom absorbs a photon, it necessarily makes a transition to a more energetic state. Given that the average radius for an atomic state grows with energy, it most likely is the case that the radial probability's max does grow after absorbtion. But, the devil is in the details, and to be sure, you'll need to do some computations, or check in Condon and Shortley's The Theory of Atomic Spectra, a bible in its field. (The issue of photon absorbtion in multi-electron atoms is somewhat tricky.)
Regards,
Reilly Atkinson
Is the energetic state one of a near continuum of available quantum states or is it a restructuring of the original electron state?
 
dextercioby said:
The question is still the same...?What would the principles to account for an increase in (radial) probability distribution...?

Daniel.
That is the question isn't it. Does extra energy promote the electron that has "absorbed" the energy (for t <10(-18) sec or so) into a new quantum state which is farther from the nucleus, or does the energy directly modify the wave function (quantum state) of electron which stays at its original distance from the nucleus?
 
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  • #10
The last view is highly classical...Hence incorrect.If the electron acquires energy (from a photon or more),then its quantum state will be modified and therefore its initial radial probability distribution (r^{2}R_{nl}(r)) will be modified as well.Under such transition,the electron will certainly end up in a nonstationary state.

Daniel.
 
  • #11
dextercioby said:
The last view is highly classical...Hence incorrect.If the electron acquires energy (from a photon or more),then its quantum state will be modified and therefore its initial radial probability distribution (r^{2}R_{nl}(r)) will be modified as well.Under such transition,the electron will certainly end up in a nonstationary state.

Daniel.
OK. What is the nature of this nonstationary state?
 
  • #12
What do you mean...?Since the time-dependent SE is (for this case of interaction with external Em field) totally unsolvable,u cannot know it...There are other quantities which are computed in the theory of time-dependent perturbations (i'm assuming the time-dependent Hamiltonian to be of perturbative nature).

Daniel.
 
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