Guys, when normalising the solution to the Schrödinger equation in the case of the infinite square well, we do the following:
$$
\int_0^a\vert A\vert^2\sin^2(kx)\,dx=\vert A\vert^2\frac{a}{2}=1,
$$
so $\vert A\vert^2=2/a$. Now Griffiths says that "it's simplest to pick the positive real root: $A=\sqrt{2/a}$ (the phase of $A$ cassies no physical significance anyway)." I don't understand where the phase comes into play, if we were to consider $A=-\sqrt{2/a}$. Any ideas?