An excited electron looses energy in the form of radiations. The radiation constitutes photons which move at a speed $c$. But, is the process of conversion of the energy of the electron into the kinetic energy of the photon instantaneous. Is there a a simple way to visualize this process rather than math?


Imagine throwing a stone into water and asking the question how fast do the waves accelerate?

The answer is that the waves don't accelerate. Any given body has a range of possible waves that can propagate along its surface and the total energy of the surface is related to the amplitudes of all those possible oscillatory modes. In principle at least there are always waves present due to thermal excitations, though in practice the amplitude of the waves will be vanishingly small. When you thrown in the stone you are transferring energy to some of the already existant oscillatory modes. So you are not accelerating anything in the sense you accelerate an object - you are just transferring energy to waves that are already moving.

Now consider your atom generating light. As the electron falls to a lower energy level it transfers energy into the quantum field that constitutes light, just as the stone transfers energy to the water. This energy goes into exciting modes that are already travelling at $c$, so no acceleration is necessary.

I would guess that much of the confusion arises from considering light as a photon. Light is a quantum field, and the photon is an approximation for its behaviour that works well in some cases but not in others. As a general rule the particle approximation is useful when considering the transfer of energy but the wave approximation is better when considering the propagation of energy. In this case the photon is the transfer of energy from the electron to the photon field, but the subsequent motion of that energy through spacetime is best described as a wave. So the photon is never accelerated in the usual sense of the word.

  • $\begingroup$ What is the 'quantum field that constitutes light'? $\endgroup$ – Rajath Krishna R Jan 17 '14 at 11:22
  • $\begingroup$ Actually, this is a transfer-of-energy situation, where energy from an excited electron is transferred to the EM field. That's why it's valid to treat it as an electron emitting a photon. The question is essentially whether this transfer is instant. $\endgroup$ – MSalters Jan 17 '14 at 11:23
  • $\begingroup$ @MSalters: the question title asks Does a photon instantaneously gain c speed, though I'll grant that the body of the question then asks about the time for the transfer of energy. $\endgroup$ – John Rennie Jan 17 '14 at 11:26
  • $\begingroup$ @RajathKrishnaR: That would be the electro-magnetic (EM) field, as described in quantum mechanics (as opposed to Maxwell's equations, which describe a non-quantum EM field) $\endgroup$ – MSalters Jan 17 '14 at 11:26
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    $\begingroup$ @RajathKrishnaR: the interaction between electrons and light is described by Quantum Electrodynamics, which is an example of a quantum field theory. $\endgroup$ – John Rennie Jan 17 '14 at 11:27

You have to understand that this is Quantum Mechanics. A classical view is by definition just an approximation.

The Heisenberg uncertainty means that you cannot know both the exact time and energy of a given event. At macroscopic scales, this doesn't matter. But consider that the radius of an atom is a few attolightseconds. The energy involved is about 1eV. Planck's constant is 4 feV's. So, to the degree that you can say something meaningful, it's instant.

Note that this is just the photon's energy. Its speed is always c, the speed of the associated EM field. But that field isn't precisely localized in space (again, Heisenberg).

Edit: dj_mummy's answer reminds me of another point. The electron emission event is a quantum event. That means you also get the superposition of states. Like Schrödinger's cat, the atom will and will not have emitted the photon, at the same time. But once you see the photon, the wave function collapses and the no-photon state has zero probability. Also, all states in which the photon was emitted at a different time also have zero probability; only one photon can be emitted.

The important thing to understand is that this collapse happens only on observation. Until you see the photon, there are multiple states with non-zero probability, corresponding with different times at which the photon is emitted. So, you can't just talk about the time when it's emitted.

  • $\begingroup$ Suppose the wave function collapses and we see a photon has been emitted. But in reality there should be a time when this photon is emitted(even though we are not looking). Is it that the reality is all these probabilities? Why would nature hide from us what has actually happened? How does nature benefit by following the uncertainty principle? $\endgroup$ – Rajath Krishna R Jan 17 '14 at 11:27
  • $\begingroup$ Reality is weird. That's why Schroedinger came up with his cat example, to clarify how incredibly weird it is. So, yes, if you don't look, there is no precise time at which the photon is emitted. $\endgroup$ – MSalters Jan 17 '14 at 15:02
  • $\begingroup$ The concept of "collapse" is often rejected by physicists who are trying to imagine what really going on there. $\endgroup$ – Brian Haak Jun 7 at 19:22

The best way to visualize this is to think of a 'before' and an 'after'. The 'before' is the state the electron is prepared in (with a definite momentum and spin orientation). After some time (can be arbitrarily small BUT NOT ZERO) one can observe the system again, what is found is the 'after' state. Now in both these states total momentum, energy, charge etc. are conserved.

What happened in between these 2 states? It is irrelevant, since there was no observer in between.

So the photon does not actually 'come out of the electron. It is just detected with an electron in the 'after' state. There is no continuously observable emission of the photon.

  • $\begingroup$ Why can't we observe in between the 'before' and 'after'? $\endgroup$ – Rajath Krishna R Jan 17 '14 at 9:54
  • $\begingroup$ Sure you can. Place a second observer in between. But this new observer is now the 'after' :D . $\endgroup$ – dj_mummy Jan 17 '14 at 9:55
  • $\begingroup$ Because the wavefunction collapses if you measure (That's the Shroedinger's cat thing). By measuring you confine the system to be in one of the two possible states (There is no undead cat!). $\endgroup$ – Neuneck Jan 17 '14 at 9:56
  • $\begingroup$ Consider a man who throws a ball. Before-the man has the ball,after-the man threw the ball. But, what about when he is throwing? We can observe it in the macroscopic world. What is the case with these subatomic particles then? Is it the uncertainty principle? $\endgroup$ – Rajath Krishna R Jan 17 '14 at 9:59
  • $\begingroup$ The 'before' state is the ball with a definite momentum. Almost immediately, a photon from the sun scatters from the ball and hits your eye thus producing the 'after' state. There are billions of such observations of the system by your eyes. Every pair of successive states forms a before and an after. $\endgroup$ – dj_mummy Jan 17 '14 at 10:05

Taking an atom as an example, the initial state must already be a state of the combined atom-field system. It cannot be an eigenstate, then the atom would not decay. This with the exception where the initial state involves the atomic ground state and the field vacuum state.

But how is such a state created? One possibility is a scattering situation with the atom in its ground state in the distant past and a field wavepacket moving towards it. Since c, the speed of light, is finite, there is initially, as $t{\rightarrow}-\infty$, no interaction. Finally, as the time $t{\rightarrow}+\infty$, we observe an atom in its ground state and a second field wavepacket. Note that the initial field wavepacket must be localized, a momentum eigenstate would extend through all space. This does not mean that momentum eigenstates play no role, they emerge if a Fourier decomposition (plane wave expansion) is made. Then the scattering event is described in terms of the scattering operator kernel S(k,k'). Energy conservation (the initial and final atomic states are both the ground state) tells us that k and k' have equal magnitude. In this case all information is contained in the operator S. The scattering formalism does not seem to allow to pose the original question nor does it lead to a simple answer.

  • $\begingroup$ Sir, I'm a high school student so, I'm not able to understand your answer. $\endgroup$ – Rajath Krishna R Jan 17 '14 at 10:57
  • $\begingroup$ @RajathKrishnaR Yes, I understand your problem. But you may have understood that quantum scattering theory, which, in my opinion, is the proper formalism, does not give rise to a simple description. $\endgroup$ – Urgje Jan 17 '14 at 11:06
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    $\begingroup$ Sorry, I'm going to disagree with the presumption that quantum scattering is the right formalism for this question, especially given that you admit it's unsuitable for posing the question in the first place. $\endgroup$ – MSalters Jan 17 '14 at 11:11
  • $\begingroup$ @MSalters. For the problem I sketched there is no real alternative. One might try to use a semi-classical model where the field is treated classically but this does not really alter the situation. $\endgroup$ – Urgje Jan 18 '14 at 11:02
  • $\begingroup$ @MSalters. One further step can be taken. The delay time operator is defined in terms of the scattering operator and its energy derivatives. It gives a measure for the extra time due to the interaction as compared to a freely moving photonic wavepacket. In case the system possesses strong resonances it can be appreciable. $\endgroup$ – Urgje Jan 18 '14 at 11:17

Quantum Mechanics tells us that electrons only lose or gain energy equal to the energy of an incoming or outgoing photon. And by default, all photons travel at speed c in vacuum. As I understand it, there is no "conversion" time for energy. Photons are energy and energy comes in photons. What we choose to call them is more a reflection of the state of the system than how much or what wavelength of photons are emitted. Hope this helps!

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    $\begingroup$ Electron energy levels in atoms are given by the eigenstates of the Hamiltonian. Only photons with the right energy can excite an electron. Also I would not say that photons are energy. Photons are particles (excitations of the electromagnetic field described by the four-potential) that have a number of characteristics (like spin, momentum, energy, polarization). As you can see, energy is just one of the characteristics of a photon. So photons are not "just energy". $\endgroup$ – mpv Jan 17 '14 at 10:04

MC Physics would suggest that photons are formed from the joining of 2 opposite electro-static charged mono-charges that are emitted from particles or atoms due to excessive vibration. Those joined mono-charges are (almost) instantly accelerated to c by surrounding electric-magnetic forces then any excess force is applied to rotate the real particle to frequency. This is further described at:


  • $\begingroup$ A photon is not a composite particle. $\endgroup$ – PM 2Ring Jan 13 at 8:13

protected by Qmechanic Oct 24 '16 at 4:42

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