The representation can still be used. Particularly if you have finite lattices with periodic boundary conditions. The situation is: This is not an approximation. What you are doing is to re-express the basis of lattice sites |r> of real space in another space, which, for example, can be |k> = \sum_r exp(i k r) |r>. As long as the set of |k>s spans the same space as the |i>s before, this is an completely equivalent basis.
If you are using periodic boundary conditions (in a finite lattice), then in fact you still have translational symmetry (which means that the Hamiltonian commutes with all translations T: [H,T]=0). Consequently, the eigenstates of the Hamiltonian can be made to transform according to irreducible representations of the translation symmetry groups. And these happen to be labelled by exactly the same k-vectors as in the infinite case -- it is just that you only have a finite number of them. This is most easily seen when you set up a tight-binding hopping Hamiltonian (-t, this emulates the kinetic energy operator) with periodic boudary conditions:
<br />
\mathbb{t} \equiv<br />
\left(\begin{array}{ccccccc}<br />
0 & -t & 0 & 0 & \ldots & 0 & -t<br />
\\ -t & 0 & -t & 0 & \ldots & 0 & 0<br />
\\ 0 & -t & 0 & -t & \ldots & 0 & 0<br />
\\ 0 & 0 & -t & 0 & \ldots & 0 & 0<br />
\\ \vdots & & & & \ddots &<br />
\\ 0 & 0 & 0 & 0 & -t & 0 & -t<br />
\\ -t & 0 & 0 & 0 & 0 & -t & 0<br />
\end{array}\right) <br />
In that case the eigenfunctions of that operator are still plane waves (i.e., the operator is diagonal in the plane wave basis), just as in the infinite case.
If you have a finite lattice but *no* periodic boundary conditions, then translational symmetry is gone, and the eigenstates do no longer exactly transform according to irreps of the translation group. You can still define the plane wave basis, however (as described above), and you will see that for example the kinetic energy operator still is very diagonal dominant in this basis (but not exactly diagonal anymore).
