# Recoil of an atom absorbing a photon

I am reading the article named Manipulating Atoms with Photons by Claude Cohen-Tannoudji and Jean Dalibard. On page no. 16, the following is said.

Consider the atom in its ground state $$g$$ and its center of mass initially at rest; suppose that a photon with wave vector $$\textbf{k}$$ is sent on this atom. If the atom absorbs the photon, it jumps to the excited state and it recoils with the momentum $$\hbar\textbf{k}$$.

My confusion

I am confused about the direction of the recoil momentum.

To be specific, let's say the photon's wave vector is $$\textbf{k} = k\hat{x}$$, where $$\hat{x}$$ is the unit vector. If the atom absorbs the corresponding momentum $$\textbf{p}_\text{photon} \equiv \hbar\textbf{k} = \hbar k \hat{x}$$, then I think the recoil momentum is $$\textbf{p}_\text{recoil} = -\hbar\textbf{k} = -\hbar k \hat{x}$$, so that the momentum of the atom is conserved before (it is $$0$$) and after the absorption ($$\textbf{p}_\text{photon} + \textbf{p}_\text{recoil} = 0$$). What am I missing here?

• > "then I think the recoil momentum is ... $-\hbar\mathbf k$" That makes no sense. The photon is absorbed, so its momentum lives on as momentum of the atom. There is always momentum $\hbar \mathbf k$. No body here has momentum $-\hbar\mathbf k$. Commented Jan 10, 2023 at 21:46

I am reading the article named [Manipulating Atoms with Photons][1] by Claude Cohen-Tannoudji and Jean Dalibard. On page no. 16, the following is said.

Consider the atom in its ground state $$g$$ and its center of mass initially at rest; suppose that a photon with wave vector $$\textbf{k}$$ is sent on this atom. If the atom absorbs the photon, it jumps to the excited state and it recoils with the momentum $$\hbar\textbf{k}$$.

I am confused about the direction of the recoil momentum. To be specific, let's say the photon's wave vector is $$\textbf{k} = k\hat{x}$$, where $$\hat{x}$$ is the unit vector. If the atom absorbs the corresponding momentum $$\textbf{p}_\text{photon} \equiv \hbar\textbf{k} = \hbar k \hat{x}$$,

...then I think the recoil momentum is $$\textbf{p}_\text{recoil} = -\hbar\textbf{k} = -\hbar k \hat{x}$$

No, that is wrong. You should have: $$\textbf{p}_\text{recoil} = \textbf{p}_\text{photon}=\hbar\textbf{k}$$

The total initial momentum is: $$0\;+\; \hbar\textbf{k}\;,$$ where $$0$$ is the initial momentum of the atom and $$\hbar\textbf{k}$$ is the initial momentum of the photon.

The total final momentum is: $$\hbar\textbf{k}\;+\; 0\;,$$ where $$\hbar\textbf{k}$$ is the final momentum of the atom and $$0$$ is the final momentum of the photon.

What am I missing here?

Conservation of momentum applies to the total/isolated system of atom plus photon: $$\textbf{P}_\text{total} = \textbf{p}_\text{atom} + \textbf{p}_\text{photon}$$

It is only $$\textbf{P}_\text{total}$$ that is necessarily conserved. Since $$\textbf{P}_\text{total}$$ started off as $$\hbar\textbf{k}$$ it must alway remain at that value regardless of how you apportion it between atom and photon.