6-6  ①  We have, from Question 6-5, ψ(x) = a sin⁡kx, where k  = nπ/L. 

Therefore, substituting into our free particle energy expression from (6-4), we get: 

②  An increase in L corresponds to decreasing localisation of the particle. The energy expression shows that this also decreases the energies of the associated wavefunctions. This is consistent with the ideas we developed in Chapter 5, where we found that the bigger range of angular wavenumber (and therefore of momentum) needed to produce a superposition wavefunction with a narrower spatial range resulted in a kinetic energy price that must be paid for this localisation.

③  If n=0, we get ψ(x) = a sin(0)  =  0.

Therefore the wavefunction would be equal to zero at all places. This is not possible because we know from the previous chapter that ψ2(x) gives us the probability density of finding the particle in the region of x, and hence the integral of ψ2 over all values of x must equal 1. This condition could clearly not be met for n=0.

Consequently, the lowest energy state is n=1, which has energy E = h2/8mL2. This means that it is impossible for the particle to have zero energy. Even in its lowest energy state – the ground state – it has some energy, which we call the zero point energy.