quantum mechanics - 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?

Answer

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

$$\begin{equation} \Delta p \ge \frac{\hbar}{\Delta x} \end{equation}$$ 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).