Emission of a photon

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  • #1
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When an electron jumps to a low energy level, it emits a photon.

1. Where does this photon come from?

2. Do the constituents of an electron have any role to play in this emission?
 

Answers and Replies

  • #2
The photon comes from a three-way interaction 'vertex' that allows electrons (or any other electrically charged particles) to emit or absorb them spontaneously.

Electrons have no currently known sub-structure, so it's impossible to say anything about "constituents" of them.
 
  • #3
K^2
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Keep in mind that electron does not actually "jump". It's just the explanation we give to chemists so that they stop bugging us about the discrete energy levels. The transition is actually a gradual one. Say, you observe emission in transition from 2p to 1s. In that case, during transition, the total state is going to be a superposition a|2p> + b|1s>, with a² + b² = 1, and this state will actually have an oscillating electric dipole. That electric dipole is responsible for producing electromagnetic radiation emitted due to transition. The reason you get precisely one photon out of it is that the energy difference between 2p and 1s is exactly equal to the energy of a photon with frequency equal to frequency of the dipole oscillation of the a|2p> + b|1s> superposition state.

P.S. This really should be in atomic. There isn't any particle physics going on here. Well, apart from second quantization of electromagnetic field, but we don't seem to be actually discussing that aspect of it.
 
  • #4
Bill_K
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Arrg, I'm shocked to hear you say something like that, K^2! :eek: The view that atomic transitions are in any way continuous went out in the 1930's. By the same reasoning a single U-238 nucleus slowly evolves into a Th-234 nucleus over a period of 4 billion years.

You know as well as I do that the only thing that evolves continuously in a transition is the probabilities. The superposition state, while mathematically correct, has no objective existence. At any time the electron is either in the ground state or the excited state. It does jump, not fade away like a Cheshire cat. And the photon does not come out gradually, it comes out all at once.
 
  • #5
K^2
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Arrg, I'm shocked to hear you say something like that, K^2! :eek: The view that atomic transitions are in any way continuous went out in the 1930's. By the same reasoning a single U-238 nucleus slowly evolves into a Th-234 nucleus over a period of 4 billion years.

You know as well as I do that the only thing that evolves continuously in a transition is the probabilities. The superposition state, while mathematically correct, has no objective existence. At any time the electron is either in the ground state or the excited state. It does jump, not fade away like a Cheshire cat. And the photon does not come out gradually, it comes out all at once.
Bill, you make me sad. Overlooking normalization factors, |ψ> = |2p> + |1s>. Probability density <ψ|ψ> = <2p|2p> + <1s|1s> + 2<2p|1s>. The probability density for the electron, which, by the way, determines electric field of the atom, is different than either that of 2p, 1s, or any linear combination of these states, because you can't get that <2p|1s> in either the excited or the ground state. Only in transition.

Furthermore, <2p|x|1s> ≠ 0. That means the atom in a transition actually has an expectation value for a dipole moment! How the hell do you reconcile this with your primitive notion of a jumping electron? Sure, it's the expectation value, but neither 2p nor 1s has a dipole moment. Yet in transition, it has a non-zero expectation. That's an actual observable, not some obscure phase.

And the photon does, most certainly, get emitted over time. The probability of detecting the photon increases over time. The probability density for photon is E²+B². That means electric field increases over time.

There are no jumps in quantum mechanical evolution. The only jumps happen due to collapse in measurement. And even that can all together be avoided with the right interpretation.
 
  • #6
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Bill, you make me sad. Overlooking normalization factors, |ψ> = |2p> + |1s>. Probability density <ψ|ψ> = <2p|2p> + <1s|1s> + 2<2p|1s>. The probability density for the electron, which, by the way, determines electric field of the atom, is different than either that of 2p, 1s, or any linear combination of these states, because you can't get that <2p|1s> in either the excited or the ground state. Only in transition.
<2p|1s>=0 as all eigenstates are orthogonal. That cross-term becomes relevant if you consider <x> or other observables, of course.

There are no jumps in quantum mechanical evolution. The only jumps happen due to collapse in measurement. And even that can all together be avoided with the right interpretation.
:)
 
  • #7
K^2
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Blargh. Yes, sorry. My brain foo-barred there for a bit. While pointing out my error there, you might as well have pointed out that <ψ|ψ> doesn't give me probability density either. It gives me total probability, which is unity.

What I want is ψ*(x)ψ(x). That picks up the 2p-1s cross-term which is most certainly non-zero.

So anyways, probability density for transition is still completely different than any linear combination of densities from ground state and excited state.
 
  • #8
Bill_K
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And the photon does, most certainly, get emitted over time.
If I hand you an atom that has been previously excited, can you tell me how much of the photon has been emitted so far?
 
  • #9
K^2
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If I hand you an electron that passed through a double-slit, can you tell me which slit it went through? What you should be able to do, however, is set up an experiment which shows that it did not just go through one or the other. It did, in fact, go through both slits. E.g. Delayed Choice Quantum Eraser.

Similarly, it is possible to set up an experiment that shows absolutely definitively that the atom is neither in the state having emitted zero nor one photon. It's somewhere in between.

Now, for the photon itself, you can try to interpret it in different ways. You can try to pretend that superposition is just some voodoo. But that doesn't work with the atom. As demonstrated prior, the actual charge density, which is an observable, of atom in transition is distinct from either ground state or excited state. As further demonstrated, the dipole moment is present and can be measured.

This, in turn presents an excellent opportunity to cut through the Gordian knot you present above. Say, rather than make measurement that tells me which energy state the atom is in, I do measure the electric dipole of the atom you handed me. Say I get a non-zero result. That would indicate that the state has collapsed into something that is neither the ground nor excited state. Having conducted that measurement, I can tell you exactly what fraction of the photon has been emitted. Naturally, this measurement is still a product of collapse, so I don't reveal information about what that fraction was originally. And nonetheless, I can make a measurement that gives me an answer that is a fraction.
 
  • #10
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The photon comes from a three-way interaction 'vertex' that allows electrons (or any other electrically charged particles) to emit or absorb them spontaneously.

Electrons have no currently known sub-structure, so it's impossible to say anything about "constituents" of them.
1. Where does this vertex lie, inside or outside an electron? I mean is the photon emitted from inside an electron or outside.

2. Is it the electrical charge that gets converted into a photon or the photon already exists waiting to be emitted?
 
  • #11
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If I hand you an electron that passed through a double-slit, can you tell me which slit it went through? What you should be able to do, however, is set up an experiment which shows that it did not just go through one or the other. It did, in fact, go through both slits. E.g. Delayed Choice Quantum Eraser.

Similarly, it is possible to set up an experiment that shows absolutely definitively that the atom is neither in the state having emitted zero nor one photon. It's somewhere in between.

Now, for the photon itself, you can try to interpret it in different ways. You can try to pretend that superposition is just some voodoo. But that doesn't work with the atom. As demonstrated prior, the actual charge density, which is an observable, of atom in transition is distinct from either ground state or excited state. As further demonstrated, the dipole moment is present and can be measured.

This, in turn presents an excellent opportunity to cut through the Gordian knot you present above. Say, rather than make measurement that tells me which energy state the atom is in, I do measure the electric dipole of the atom you handed me. Say I get a non-zero result. That would indicate that the state has collapsed into something that is neither the ground nor excited state. Having conducted that measurement, I can tell you exactly what fraction of the photon has been emitted. Naturally, this measurement is still a product of collapse, so I don't reveal information about what that fraction was originally. And nonetheless, I can make a measurement that gives me an answer that is a fraction.
Does it mean spontaneity is a misnomer?
 
  • #12
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When an electron jumps to a low energy level, it emits a photon.

1. Where does this photon come from?
it is just created at the moment.it is not inside the atom.that's all.It comes from the complicated machinery of quantum electrodynamics where EM field is generally quantized in terms of harmonic oscillators.
Do the constituents of an electron have any role to play in this emission?
it has nothing to do with it,if it exist even.
 
  • #13
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1. Where does this vertex lie, inside or outside an electron? I mean is the photon emitted from inside an electron or outside.
There is no "inside" and "outside" at electrons.

2. Is it the electrical charge that gets converted into a photon or the photon already exists waiting to be emitted?
Neither. The photon is produced (it is not present before), and the electron keeps its charge.
 
  • #14
K^2
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Neither. The photon is produced (it is not present before), and the electron keeps its charge.
Depending on whether you think of the photon as the state or excitation of a state. People usually mean the later, but there is nothing wrong with thinking that the photons always exist, and which ones are excited and which aren't is what changes.

Naveen345 said:
Does it mean spontaneity is a misnomer?
Why? Spontaneous emission just means it's not stimulated emission. It does not imply that emission is instantaneous.
 
  • #15
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Depending on whether you think of the photon as the state or excitation of a state. People usually mean the later, but there is nothing wrong with thinking that the photons always exist, and which ones are excited and which aren't is what changes.
Ok. But then we should clarify that there is no photon "hidden in the electron", this is independent of the interpretation.
 
  • #16
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Depending on whether you think of the photon as the state or excitation of a state. People usually mean the later, but there is nothing wrong with thinking that the photons always exist, and which ones are excited and which aren't is what changes.
Is the number of photons in this universe constant?
 
  • #17
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no because I could annihilate an electron and positron to form 2 photons.
but charge is conserved
Or I could just bend an electrons path in a B field and cause it to emit light.
 
  • #18
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Is the number of photons in this universe constant?
To Drakkith:

Please reply.
 
  • #19
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Is the number of photons in this universe constant?
The number of photons is not constant in the universe. The total number of photons in the universe can both grow or decrease.

No particle number is actually contant. New particles can be created or annihilated of any known type: electrons and positrons, quarks and antiquarks, neutrinos, muons and antimuons, all sorts of bosons, etc.
 
  • #20
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The number of photons is not constant in the universe. The total number of photons in the universe can both grow or decrease.

No particle number is actually contant. New particles can be created or annihilated of any known type: electrons and positrons, quarks and antiquarks, neutrinos, muons and antimuons, all sorts of bosons, etc.
Then, is mass + energy a constant?
 
  • #21
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particle number is constant in non interacting field theories.It is not conserved only when interacting field are treated.Also mass is same as energy.
 
  • #22
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Then, is mass + energy a constant?
Neglecting general relativity, total energy in the universe is constant.
In general relativity, energy is locally conserved - but there is no meaningfull way to define "the current energy of the universe", so it is pointless to talk about global energy conservation.

Mass is just a type of energy.
 
  • #23
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what are you guys working on ur phd's or do you already have them!!!!!!
could we please get down to basics for a moment.
does a photon have mass? how can a photon create energy if it has no mass? dont you need mass to create energy e=mc squared. is there an equation which relates a photon to energy?
 
  • #24
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does a photon have mass?
No.it does not have.One basic reason is gauge invariance.
how can a photon create energy if it has no mass?
two photons can create particle antiparticle pair.They have energy which is converted to mass.
dont you need mass to create energy e=mc2.
You may have written you need energy to create mass or vice versa.
is there an equation which relates a photon to energy?
Every photon has energy,momentum,spin.Energy is simply E=h-ω
 
  • #25
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isn't mass the the only source of energy
 

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