Due to the symmetry of the problem, the emitting waves on both sides must have equal amplitudes (A), but their phases (α) may be different. This gives a single extra parameter; however, gluing the two solutions at x = 0 requires two boundary conditions (for both the wave function and its derivative), so in general there is no solution. In particular, re-writing Ψ Ψ 3 (after translation by q 0 ) as a sum of a cosine and a sine of k x , each having a different factor that depends on k and α, the factor of the sine must vanish, so that the solution can be glued symmetrically to its reflection. Since the factor is in general complex (hence its vanishing imposes two constraints, representing the two boundary conditions), this can in general be solved by adding an imaginary part of k, which gives the extra parameter needed. Thus E will have an imaginary part as well.