Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

consider a transition for an electron in the Hydrogen atom from the ground state to the 1st excited state. Let's say this transition occurs through absorption of a photon of exactly the energy required to excite the electron.

what happens to the photon right after it is absorbed? is it annhilated when it is absorbed? does its wavefunction cease to exist? I'm curious about this process thanks.

share|improve this question
add comment

3 Answers

up vote 3 down vote accepted

Well, in the personificated picture that we often have of particles, it indeed ceases to exist. In a less dramatic language this can also be described on the level of fields: electron in the atom is a localized Dirac field while the incoming photon is an EM wave. This process is then morally no different from classical theory of interaction of waves with matter: some of the wave is reflected, some is scattered, some is absorbed.

Similarly, one could ask a question similar to yours about the creation of particle-antiparticle pairs. Are these brought into existence by some act on magic? It turns out that on the level of fields these are nothing else than quantum mechanical fluctuations that you can imagine as ripples and peaks on the sea surface. The sharp peaks then essentially have properties of particles. Indeed, this is how we think of particles in general: they are well-formed local portions of the field.

share|improve this answer
add comment

No, the wave function of a photon does not cease to exist. Technically a photon is an excited state of an oscillator. After absorption the oscillator wave function becomes a wave function of the ground state. The electron wave function, on the contrary, becomes an excited one from the ground state wave function.

The total wave function is a product of independent state wave functions. None of them ceases to exist while interactions.

Each state can also be described with "occupation numbers" $n_i (t)$. Physical processes are redistributions of the occupation numbers. Transition from $n_{ph} = 1$ to $n_{ph} = 0$ is absorption of a photon but not a photon wave function destruction.

share|improve this answer
add comment

I'm going to go way out on a limb and say that this is a hypothetical experiment that has never been carried out. I believe it is impossible to prepare a hydrogen atom in the ground state, expose it to light energy, and then observe it in the excited state. Never been done, as far as I know.

EDIT: This posting has been accumulating anonymous downvotes since it went up yesterday. I don't believe I've said anything controversial. Certainly, everyone "knows" that this it what happens when you expose hydrogen to light energy. I'm just questioning whether it's actually observable in an experimental sense.

I happen to believe that nothing of the sort happens. I think that when hydrogen is in the presence of an oscillating electromagnetic field of the right frequency, it is driven into a superposition of the s and p states. In this state, the charge distribution of the atom oscillates like a tiny antenna; and it interacts with the external field exactly as would a classical antenna with the same parameters. There is nothing in the mathematics of the interaction which requires it to end up in a pure p state, and nothing that requires the energy to be exchanged with the external field in specific quantities.

I know that my opinions would be considered to be verging on "crackpot", but that is not the point. I am simply asking people who think they know better than me: just what is the experimental evidence that their model of quantized energy exchange is superior to my model. I have already said that I do not believe that observation of the s-p transition of hydrogen provides any such evidence.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.