I'm working through problem 7.3 from Griffith's quantum mechanics 3rd edition:
Problem 7.3: Two identical spin-zero bosons are placed in an infinite square well. They interact weakly with one another, via the potential (where is a constant with the dimensions of energy, and a is the width of the well).
V(x1,x2) = -aVoδ(x1-x2)
(a) First, ignoring the interaction between the particles, find the ground state and the first excited state—both the wave functions and the associated energies.
It made sense to me that since the particles were both bosons (and thus had symmetric wavefunctions), that I would describe the ground state energy as |Ψ0> = |Ψx1,0> + |Ψx2,0>. But the answer gave by multiplying the two wavefunctions instead. Why is this? Don't we express a superpositon as a linear combination? What am I missing here?