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Given a wave equation, say for example $\Psi(x,y,z,t) = a \cos\left(\omega t -\vec{k}\cdot \vec{r} \right)$, what conditions should be met for $\Psi$ to represent a plane wave?

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This rather depends on your application. Your solution is a plane wave and, as you can see from your solution, the disturbance is infinite in spatial and temporal breadth. Real waves in causal systems have a definite beginning and end in time and are confined to finite volumes. Just a slight quibble: yours is a solution to the wave equation, not the wave equation itself. Your solution is a plane wave solution to the Helmholtz equation:

$$\left(\nabla^2 + \frac{\omega^2}{c^2}\right)\Psi=0$$

which is linear, so that an arbitrary linear superposition of waves of the form you give are also solutions. Indeed, given certain reasonable assumptions on $\Psi$, all solutions of this wave equations can be represented as such superpositions (see my answer here for a discussion on the limits to this statement).

The better question is "how plane is a certain disturbance?" and this is quantitatively answered by the Fourier transform: real waves are superpositions of your basic solution:

$$\Psi(\vec{r},\,t) =\int_{\mathbb{R}^3} \exp(i(\vec{k}\cdot\vec{r}-c\,|\vec{k}|\,t))\,\psi(\vec{k})\,{\rm d}^3 k$$

and you use the Fourier space function $\psi(\vec{k})$ that defines the superposition weights to measure how non-plane the true wave is. A plane wave has $\psi(\vec{k})$ tightly concentrated about one lone wavevector $\vec{k}_0$. Laser beams are roughly plane, but they are finite in spatial breadth and so they always comprise a spread of directions about the nominal direction of propagation. It is the presence of "off-axis", i.e. slightly skewed plane wave components that beat together and confine the superposition to a finite beam.

The spread in Fourier space and spread in untransformed space are bounded by the Heisenberg inequality (in quantum mechanics, the HUP arises because position and momentum space are Fourier transforms of one another). I talk more about this in this answer here, on why all laser beams diverge and also here for a more general discussion of the Heisenberg inequality.

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