- #1

Naveen345

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1. Where does this photon come from?

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

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In summary, electrons emit photons when they transition from one energy level to another. The photon comes from a three-way interaction vertex that allows electrons to emit or absorb them spontaneously. The transition is a gradual one, and the atom in a transition actually has an expectation value for a dipole moment.

- #1

Naveen345

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1. Where does this photon come from?

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

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- #2

AdrianTheRock

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Electrons have no currently known sub-structure, so it's impossible to say anything about "constituents" of them.

- #3

K^2

Science Advisor

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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

Science Advisor

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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

- #5

K^2

Science Advisor

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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.Bill_K said:

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. Itdoesjump, not fade away like a Cheshire cat. And the photon does not come out gradually, it comes out all at once.

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

mfb

Mentor

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<2p|1s>=0 as all eigenstates are orthogonal. That cross-term becomes relevant if you consider <x> or other observables, of course.K^2 said: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.

:)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

Science Advisor

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What I want is ψ

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

Science Advisor

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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?And the photon does, most certainly, get emitted over time.

- #9

K^2

Science Advisor

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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

Naveen345

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AdrianTheRock said:

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

Naveen345

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K^2 said:

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

andrien

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Naveen345 said: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.

it has nothing to do with it,if it exist even.Do the constituents of an electron have any role to play in this emission?

- #13

mfb

Mentor

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There is no "inside" and "outside" at electrons.Naveen345 said:1. Where does this vertex lie, inside or outside an electron? I mean is the photon emitted from inside an electron or outside.

Neither. The photon is produced (it is not present before), and the electron keeps its charge.2. Is it the electrical charge that gets converted into a photon or the photon already exists waiting to be emitted?

- #14

K^2

Science Advisor

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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.mfb said:Neither. The photon is produced (it is not present before), and the electron keeps its charge.

Why? Spontaneous emission just means it's not stimulated emission. It does not imply that emission is instantaneous.Naveen345 said:Does it mean spontaneity is a misnomer?

- #15

mfb

Mentor

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Ok. But then we should clarify that there is no photon "hidden in the electron", this is independent of the interpretation.K^2 said: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.

- #16

Naveen345

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K^2 said: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

cragar

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but charge is conserved

Or I could just bend an electrons path in a B field and cause it to emit light.

- #18

Naveen345

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Naveen345 said:Is the number of photons in this universe constant?

To Drakkith:

Please reply.

- #19

mpv_plate

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Naveen345 said: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

Naveen345

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mpv_plate said: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

andrien

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- #22

mfb

Mentor

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Neglecting general relativity, total energy in the universe is constant.Naveen345 said:Then, is mass + energy a 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

dennisfl8

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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? don't you need mass to create energy e=mc squared. is there an equation which relates a photon to energy?

- #24

andrien

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No.it does not have.One basic reason is gauge invariance.does a photon have mass?

two photons can create particle antiparticle pair.They have energy which is converted to mass.how can a photon create energy if it has no mass?

You may have written you need energy to create mass or vice versa.dont you need mass to create energy e=mc^{2}.

Every photon has energy,momentum,spin.Energy is simply E=his there an equation which relates a photon to energy?

- #25

dennisfl8

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isn't mass the the only source of energy

- #26

andrien

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energy and mass are same by E=mc^2

- #27

dennisfl8

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andrien said:No.it does not have.One basic reason is gauge invariance.

two photons can create particle antiparticle pair.They have energy which is converted to mass.

You may have written you need energy to create mass or vice versa.

Every photon has energy,momentum,spin.Energy is simply E=h^{-}ω

is a photon a force . like gravity, magnetic fields etc. are forces create by mass ?

- #28

agostino981

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dennisfl8 said:is a photon a force . like gravity, magnetic fields etc. are forces create by mass ?

Electromagnetic (mediated by photons), strong (mediated by gluons) and weak (mediated by W and Z bosons) interactions are mediated by virtual particles. Graviton is also a hypothetical boson that mediates gravity.

- #29

andrien

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electromagnetic interaction is mediated by photons.dennisfl8 said:is a photon a force . like gravity, magnetic fields etc. are forces create by mass ?

- #30

mfb

Mentor

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No, it is just one type of energy.dennisfl8 said:isn't mass the the only source of energy

Only if you use the concept of relativistic mass, which is done in ancient textbooks and bad TV documentations only.andrien said:energy and mass are same by E=mc^2

- #31

andrien

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mass of electron in natural units is 0.511 Mev.Only if you use the concept of relativistic mass, which is done in ancient textbooks and bad TV documentations only.

- #32

mfb

Mentor

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c=1 is a conventional choice, but this does not mean that my height is some nanometers.

- #33

andrien

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that is just a conventional way.To get back to original requires some work.And what you don't see.

- #34

K^2

Science Advisor

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You are missing the point. E=mc² is wrong. The correct formula is E² = p²c² + (mc²)²

- #35

andrien

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if you don't use the definition of relativistic mass then,right.Otherwise both are same.Energy is time component of four momentum and mass is invariant,so they are really different. if one avoids the notion of relativistic mass which is already abandoned then of course it is wrong.But that is just not what I say.It is a matter of definition.K^2 said:You are missing the point. E=mc² is wrong. The correct formula is E² = p²c² + (mc²)²

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