# Is momentum still conserved in non-phase-matched nonlinear optical processes?

To be efficient, a phase-matching condition has to be fulfilled in many nonlinear optical processes. For instance, the phase-matching requirement for second-harmonic generation is

$k_{2\omega}=2k_{\omega}$ or $\Delta k = k_{2\omega}-2k_{\omega}=0$

It is often said that this is equivalent to momentum conservation. However, even if $\Delta k \neq 0$, the process still takes place - although with lower efficiency and a finite coherence length $L = \frac{\pi}{\Delta k}$.

How can the conversion process still occur while momentum is not conserved? Is there momentum transfer to the medium? I guess not, because in many nonlinear processes only virtual photons participate. Do the photons 'borrow' momentum to make the jump? In other words, how does this work?

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There is typically considered to be an uncertainty which softens the matching condition. In the case of momentum, the momentum state is only as well defined as the spatial extent of the interaction allows it to be. If the interaction length is given by $L$, which we can take to be an approximate measure of the position uncertainty $\Delta x$, then the corresponding momentum uncertainty is $$\Delta p \ge \frac{\hbar}{\Delta x}$$ so that there is a corresponding uncertainty in $\Delta k = \Delta p / \hbar$, which gives (up to some factors) $\Delta k \sim (\Delta x)^{-1}$. The finitude of the system size, either in time or space, means that the determination of the Fourier coefficients has a certain wiggle room (in $\omega$ or $\mathbf{k}$, respectively).
@RVL I don't think you do need quantum mechanics to make this argument work. An uncertainty relation also holds for classical wavepackets, while the momentum carried by the wave is related to the wavevector in classical EM. If you work through the classical formalism you should get exactly the same physics out. If you want to see it even quicker than that, note that the momentum $p = \hbar k$, so in the uncertainty relation $\hbar \Delta k \geq \hbar/\Delta x$ the $\hbar$ cancels and everything still holds in the classical limit $\hbar \rightarrow 0$. –  Mark Mitchison Feb 20 '13 at 11:59
@Mark Good point. Could you remind me, I thought the momentum of a classical EM wave was the Poynting vector/c^2, but I lost the connection to the momentum $p$ you describe for the moment. Thanks to both of you! –  RVL Feb 20 '13 at 13:00
The difference is between the momentum per photon ($\mathbf{p}=\hbar \mathbf{k}$) and the momentum density ($\mathbf{p}=c^{-2}\int_A \mathbf{S}$, where the subscripted $A$ means the integral over some area), the difference between quantum and continuous treatments of the fields. –  KDN Feb 20 '13 at 15:37
@RVL The Poynting vector $\mathbf{S}$ is related to $\mathbf{E} \times \mathbf{B}$. (I will not bother to write all the constants.) Using Faraday's equation, you should be able to show that for plane waves, $\mathbf{k}\times\mathbf{E} = \omega\mathbf{B}$. Plug this into the expression for the Poynting vector, remembering that EM waves are transverse, and you'll get something like $\mathbf{S} \sim \mathbf{k}E_0^2$, where $E_0$ is the amplitude of the electric field. –  Mark Mitchison Feb 20 '13 at 16:08