# Momentum and acceleration of an atom after emitting of a photon?

I'm a second year physics student and we've been talking about light and the fact that it carries momentum. I've been thinking about a situation where there is an excited atom that has an electron at a high energy level. When the electron jumps back down to a lower energy level, it will release a photon. This photon has momentum and according to the conservation of momentum, the atom must gain the same amount of momentum in the opposite direction. I understand that the light is released and it immediately is moving at c, with momentum p = h/λ. This means that the momentum of the atom must be mv=-h/λ.

What's bugging me, is that the photon isn't accelerated, it has this momentum immediately when it is emitted, which means the atom must also have this momentum (but negative) when the photo is emitted. Does this mean that the atom goes straight from 0 velocity to a non-zero velocity without accelerating? Or is something else going on that I don't know about?

• The fact that it is going from zero velocity to a non zero velocity is in fact Acceleration. But thats where it stops. The recoil velocity of atom is such that the momentum of the atom equals the momentum of the photon released – Prasad Mani Oct 19 '16 at 15:30
• I don't think quantum mechanics has an answer to that question. The question itself doesn't fit into quantum mechanics, as QM is silent on the time evolution of "quantum jumps". – garyp Oct 19 '16 at 15:45
• Sure, but it is pretty clear still on conservation of momentum in the end. So, yes, the atom as a change in momentum. This is how laser cooling works. – Jon Custer Oct 19 '16 at 16:05
• @PrasadMani What bugs me about that is that acceleration is defined as dv/dt, but the time interval is 0, so dv/dt would be undefined. Would this be possible? – Turner Garrow Oct 19 '16 at 16:06
• The thing is, special relativity says massless particle have to travel at the speed of light. That is photon cannot be accelerated or decelerated. Since it didnt even exist before the electron de-excited, we can take comfort from the fact that as and when the photon was created, it flew off with the light speed. And since atom has to conserve the momentum, it will have to recoil with some velocity. Now the energy time uncertainty principle allows for it to have less energy and hence less momentum for a very small time duration. It eventually does gain the velocity needed to conserve momentum – Prasad Mani Oct 19 '16 at 16:25

My introduction to this concept was David Bohm's book "Quantum Theory", where he suggests (not surprisingly) that questions like the above don't have any classical analog at any level, and that this is because they are all bound up with the central concept that he stresses in the first chapter of his book.

His point is that quantum processes, once begun, are deterministic in the sense that they cannot be stopped, or reversed, or otherwise dealt with in any classical manner. I write this, not as an answer, although I think it is the only answer we currently have, but in the hope that in the 60 plus year's since the book's publication, someone can tell me that this is, or is not, still experimentally verified.

Photon emission is not an instantaneous process. The time evolution of, say, an excited atom in a cavity is perfectly continuous. The electromagnetic field begins in the state $|0 \text{ photons}\rangle$, then becomes a superposition of $|0 \text{ photons}\rangle$ and $|1 \text{ photons}\rangle$. Similarly, the state of the atom smoothly transitions between $|\text{stationary, excited} \rangle$ and $|\text{moving, not excited} \rangle$. All of this happens on a timescale on the order of the lifetime $\tau$ of the excited state.

The idea that emission is instantaneous comes from taking a classical point of view -- classically, there can be zero photons or one photon at once, but not both, so we must have switched between those states at some instant in time. But in reality these are just two states of the (quantum) electromagnetic field, which can be superposed like any other pair of quantum states.

Here is my view, This intuitive that photon isn't accelerated, right. Why?, because the classically speaking emission of Photon is just fluctuation or variation in electric field per say. You must have heard of how photon or electromagnetic radiation are generated by accelerating electrons or any charged particle. Now come to atom, it does has mass and if you want to change some object's velocity which have rest mass non zero then we need force. So I think atom is accelerated, but then you will ask me, what's the force and what's the time of impact? If idealy I consider a single photon is released then force will be due to the energy or interactions involved in the process, and time is will be exactly the $\dfrac{1}{\nu}$ where $\nu$ is the frequency of photon released. Instead of acceleration better consider the impulse, $F\Delta t=\Delta P$ Because if intraction time is very small we calculate impulse instead of acceleration, same as hitting a ball with baseball bat.

Consider this, as in a gravity free space I kept a bomb enclosed inside a box, and when I exploded the box rips to half and moved in opposite direction with equal momentum. Same as your's atom situation. Here we consider the impulse not the accelaration of the box.

$Vivek$

The momentum depends on the wave length (energy). The time from the wave starts until it reaches its top phase will be the acceleration time. The acceleration amount depends on the mass of the atom.

What's bugging me, is that the photon isn't accelerated, it has this momentum immediately when it is emitted, which means the atom must also have this momentum (but negative) when the photo is emitted.

To understand this I recommend you take a look at pair production and Compton scattering, or should I say inverse Compton scattering. Also note that in atomic orbitals electrons exist as standing waves. We can diffract electrons. Because of the wave nature of matter.

See how the electron is represented by a circle? Think of it as a 511keV photon going around and around. In Compton scattering you effectively "take a slice" off the incident photon and slap it onto one side of the electron. This is no longer rotationally symmetric, and as a result the electron moves. It is rapidly accelerated. Meanwhile the photon is rapidly decelerated, but in the vector sense. It doesn't change speed, it changes direction and loses energy. The inverse Compton is this in reverse. Your atom is something similar, but with no initial photon.

Does this mean that the atom goes straight from 0 velocity to a non-zero velocity without accelerating?

It means it accelerates rapidly, in the time it takes to "take a slice".

• "In Compton scattering you effectively "take a slice" off the incident photon and slap it onto one side of the electron. This is no longer rotationally symmetric, and as a result the electron moves" This is not a correct description of this scattering experiment. In fact, saying that something moves as a result of it failing to be rotationally symmetric is rather absurd. You can find many non-symmetric things that stand perfectly still in your direct environment. – Danu Nov 4 '16 at 14:23
• @Danu : that's how it works. In atomic orbitals electrons exist as standing waves. They exist as standing waves outside of atomic orbitals too. Hermann Weyl was the first to say this, google it. Only sometimes they aren't quite standing. – John Duffield Nov 4 '16 at 14:33