Question

If the wavefunction for a system is a superposition of wavefunction, the wavefunction of the system cannot be determined. Explain this statement.

Answer 1

Given two possible states of a quantum system corresponding to two wavefunctions ψa and ψb, the system could also be in a superposition ψ = αψa + βψb with α and β as arbitrary complex coefficients satisfying normalization.
This forms the soul of quantum mechanics!
Note that for a superposition state
ψ(x) = αψa(x) + βψb(x),
the probability density p(x) = |αψa(x) + βψb(x)|2 = |αψa(x)|2 + |βψb(x)|2 + α*βψ*(x)ψb(x) + αβ*ψa(x)ψ* a b (x) exhibits quantum interference aside from the usual addition of probability!
For example, let us consider ψ5 = ψ1 + ψ2 from our previous set of examples. Putting normalization aside, this looks like two distinct well-localized peaks. Each peak individually represented a particle that was localized at the position of the peak center. But now that there are two peaks, the particle is at neither position individual. It is not at both positions simultaneously, nor is it at no position at all. It is simply in a superposition of two states of definite position. The probability density of this superposition state will show no interference because when one of the component wavefunctions exhibits a peak, the other component wavefunction is zero, so their product is zero at all positions.
Similarly, ψ6 = ψ3 + ψ4 is a superposition of two states of definite momentum. It cannot be said that a particle in this state has one or the other momentum, nor can it be said that it has both or neither momenta. In contrast to the previous superposition example, though, the probability density will exhibit interference because the product of the two wavefunctions is not always zero as they are both sinusoidal waves.