Photon hitting an atom

1. Feb 25, 2015

lucas_

When you hit a hydrogen atom with a photon, would it always register a hit 100% of the time? so do you have to compute for probability if it would hit it? I'd like to know if it is the photon wave function that interacts with the electron wave function so there should always be a hit all the time (because wave interacting with wave has larger scattering sections).

Or do you need to imagine it as a photon particle hitting the electron particle? Then it's like imagining a rocket aimed randomly at the earth/moon and calculating the probability the rocket would hit the moon, and you know the probability is so low. If the photon and atom's electron is like this. Hit would be rare. Or do you need to imagine it as a rocket wavefunction interacting with the moon wavefunction and there is a hit all the time because the wavefunction of moon is located at all space (this is just analogy of the electron wave function occupying the orbital all at ones)?.

2. Feb 25, 2015

jfizzix

What's a "hit"?

You can use quantum mechanics (and quantum electrodynamics) to describe what you are likely to detect given that you launch a photon at an atom (such as the probabilities for the photon to be detected at different scattering angles).

I'm not sure that there is a definite hit or miss going on. To get a complete picture, you'd have to describe the atom and the quantum electromagnetic field (initially in a 1-photon state) together. As a result, you can calculate what the state of atom-plus-field is at a later time. Because they interact, that joint state is entangled. Maybe that interaction counts as a kind of hit, but it can vary by degrees and isn't a definite yes or no situation. Conservation of momentum tells us that in this interaction, the total intial and final momenta of photon and atom when measured will be equal.

I imagine it as an interaction between photon and atom. The total state changes because of the "collision", and you can figure out what the new states and measurement probabilities are.

good question!

3. Feb 25, 2015

lucas_

I forgot all about the fields.. so this is quantum field theory all about? If the moon and rocket became fields.. their interaction is more probable than if there were particles and computing probabities of their hits? Or could we say the rocket and moon moves so fast all around that results in moonfield, rocketfield such that interactions is more likely? or it is not correct to imagine them as particles... in QFT, particles are momentum and energy of the fields so there are more interactions because they are more widespread (hence in a sense larger scattering cross sections)?

Is there probability of zero of photon interacting with an atom?

4. Feb 26, 2015

jfizzix

For any object, you can't completely describe it quantum mechanically without also describing everything it interacts with. You only get a single wavefunction when you are describing a closed system. The moon's an open system interacting with other objects, with, so you couldn't assign a single wavefunction to the moon, or really any macroscopic object unless they were under exceptional conditions.

Also quantum fields are only really assigned to describe particular kinds of particles, and not to large scale objects. One can talk about the quantum electron field, or a quantum quark field, but not a quantum baseball field (so far as I know).
That being said, you could hypothetically use quantum field theory to describe the collection of particles making up the moon and rocket, and see how they interact if there was nothing else interacting with them too. It is substantially easier just to show that the center of mass of these objects behave classically on average, and to just use Newton's laws for meaningful calculations.

5. Feb 26, 2015

Staff: Mentor

As jfizzix said, it depends what you call a "hit." By definition, you could call a it hit when a photon gets absorbed, so it becomes tautology. To the question "if a photon comes very close to the center atom, will it always get absorbed?", the answer is no: there is always a probability that the photon will just pass by.

It is also important to understand that it is not the electron that absorbs the photon. It is the system nucleus + electron that does the absorption.

6. Feb 26, 2015

lucas_

But in photoelectric effect, the photon directly interacts with the electron kinetically without any nucleus.. so is it not possible in the atom, it is the electron that is receiving the photon load? Are you saying it is the electric field in between electron and nucleus that receiving the photon momentum?

7. Feb 27, 2015

Staff: Mentor

What do you mean "without any nucleus"? You have a matrix of atoms, and their presence is essential.

No, it is the combined system that absorbs the photon. To keep things simple, to first approximation you need a dipole to absorb the photon. A free electron will not absorb a photon. See, e.g, https://www.physicsforums.com/threads/why-cant-a-free-electron-absorb-or-emit-photon.312121/

It is not the field, but the electron-nucleus dipole whose energy changes when interacting with electromagnetic radiation.

8. Feb 27, 2015

tijana

That collision I imagine like Compton disperse.
So what is happening when photon radiate E=hv to atom with some probability to do so? Would e gain E to separate from the atom or do the quant jump to another state? How e " know" which hv to absorb to make a quantum jump?
If I understood this well, cos I am from another field.

9. Mar 3, 2015

tijana

State of particle is give by specific wave function with characteristic quanta numbers,where the same value of E can suit more wave function or more states.
In term of resonance,at what E level will be e after absorbing photon specific hv?In which excitation state will be e? What is it depending on?