Recoil of an atom absorbing a photon

Question

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?

Answer 1

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.