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I am reading a book introducing basic concept of laser. It is pretty shocking to me that people can generate beam with almost all photons in the same state. In the book, it said that a two-level atoms (with energy level is $\hbar\omega_0$) can absorb a photon from the laser so to excite the atom. The laser with wavevector $k$ and frequency $\omega$. Here is what confusing me. The laser not necessarily have the same frequency as the atom (i.e. $\omega_0 \neq\omega$), so how come the atom absorb a photon which is not of the frequency $\omega_0$? Can it still absorb photon from laser? If so, where is the extra energy ($\hbar\omega - \hbar\omega_0$)gone?

Also, if this is possible, so how does the momentum of the atom changed if it absorb the photon from the laser? I know that if the atom absorb photon of wavevector $k$, then the momentum change on the atom will be $\hbar k$, does it make sense? It seems someone wrong to me because I think the atom can only absorb photon of frequency $\omega_0$ such that the momentum change should be $\hbar k_0$.

In addition, the book introduce a concept of lifetime of the atomic upper state $\tau$. I am trying to understand the force exert on the atom by absorbing one photon. I know that the force is $\Delta p/\Delta t$. Can I say $\tau$ is just how long it takes to make a momentum change if absorb an photon of wavevector $k_0$ so the force the atom experience is $F = \hbar k_0/\tau$ ?

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1 Answer 1

I'll start from where you ended. The excited state of the atom has a finite lifetime $\tau$. Therefore, due to the energy-time uncertainty relation $\Delta E\Delta t\sim\hbar$ the energy of the excited state is not specified exactly but has a finite linewidth. Moreover the atom moves and sees the frequency of the photon Doppler-shifted and this Doppler shift is again smeared because of the uncertainty between position and momentum. The atom can also undergo collisions that lead to an exchange of energy that can channel the excess energy to the surroundings or take the missing energy from it. These are the main reasons the linewidth of the atomic transition is finite.

Mathematically, you can describe this in the perturbation theory, treating the atom-light interaction as a perturbation. Assuming only one-photon processes you can derive the Hamiltonian $H_\textrm{int} \propto a\sigma_+ +a^\dagger\sigma_-$, where $\sigma_-$ is the atomic lowering operator and $\sigma_+=\sigma_-^\dagger$. The solution of this interaction can be found in many quantum optics textbooks; it leads to oscillations of the atom between the ground and excited states. The amplitude depends on the detuning of the field and the atomic transition, i.e., the difference $\omega-\omega_0$. However, the excitation probability is finite as long as the detuning is not too large.

Finally, the absorbed photon will transfer its momentum to the atom. That is nevertheless typically much smaller than the momentum the atom has. To get the grip of it, the photon momentum is $\hbar k\sim \hbar/\lambda$ which, for visible light is of the order of $10^{-27} \textrm{kg m s}^{-1}$. If you have an atom of a gas moving typically at the speed of hundreds of meters per second, its momentum is at least to orders of magnitude higher. In a rigorous treatment of the perturbation the interaction represents, the momentum conservation can be treated more carefully.

EDIT

Simply put, before the interaction the atom has momentum $\mathbf{p}$. When it absorbs a photon with momentum $\hbar\mathbf{k}$ (i.e., the momentum is given by the frequency of the light, not the frequency of the atomic transition), its momentum changes to $\mathbf{p}_2 = \mathbf{p}+\hbar\mathbf{k}$. After a while it will emit another photon with (generally different) momentum $\hbar\mathbf{k}_2$ so that the resulting atomic momentum reads $\mathbf{p}_3 = \mathbf{p}_2-\hbar\mathbf{k}_2 = \mathbf{p}-\hbar(\mathbf{k}_2-\mathbf{k})$.

If you wanted to talk about the force the photon exerts on the atom, you would have to divide its momentum by the time it takes the atom to absorb the photon, not by the lifetime of the excited state. But I have never heard about anyone considering the force in such a scenario.

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Thanks for the explanation. It gives me some idea on laser though I am still confusing on some concepts. In the last part when you state the photon momentum, you said the momentum is $p=\hbar k$, so is that $k$ the $k$ given by the laser? So even the atoms absorb a photon from laser at frequency $\omega_0$, the momentum transferred to the atom is still $p=\hbar k$ instead of $p=\hbar k_0$? In the first part you explain, can I say the atom will stay in the excited state about the time $\tau$ so within that duration, the atom is experiencing momentum $p$? –  user1285419 Apr 6 '13 at 21:12
    
I am also thinking a picture like this, the atom absorb an photon so its momentum changed to $p$ but in a while the atom will emit the photon so it will take some momentum from the atom and averagely the atom doesn't gain net momentum so the net force the atom gain is zero, is that right? But if we only consider the time before the photon emitted, can I say it will see the force about $p/\tau$? –  user1285419 Apr 6 '13 at 21:14

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