3
$\begingroup$

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.

$\endgroup$

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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