I'm reading Griffiths' section on the infinite square well defined as having zero potential between 0 and a on the x-axis and being infinite everywhere else, and am confused about the following part when discussing the general solution inside the well. The bolded part is what confuses me, the rest is just context. Continuity of [tex]\psi(x)[/tex] requires that [tex] \psi(0)=\psi(a)=0,[/tex]so as to join onto the solution outside the well. What does this tell us about A and B? Well, [tex] \psi(0)=A\sin(0)+B\cos(0)=B,[/tex]so B=0, and hence [tex]\psi(x)=A\sin(kx).[/tex]Then [tex]\psi(a)=Asin(ka),[/tex]so either A=0 in which case we're left with the trivial-non-normalizable--solution [tex]\psi(x)=0,[/tex] or else [tex]\sin(ka)=0,[/tex] which means that[tex] ka=0\pm\pi\pm 2 \pi\pm 3\pi, ...[/tex]But k=0 is no good-again, that would imply [tex] \psi(x)=0[/tex] and the negative solutions give nothing new, since [tex]\sin(-\theta)=-\sin(\theta)[/tex] and we can absord the minus sign into A. My question here is, why do the negative solutions even matter? There are restrictions on k (E>0, mass is always for this case going to be nonzero, hbar is obviously nonzero and positive), and the general solution for the well that is used above is only applicable for the portion of the x-axis where x is always positive or zero at one of the boundaries. The restrictions on kx above make for a product of two positive real numbers, giving you another positive real number. I'm tempted to swear like a sailor because the argument will *never* be negative for that sine function, right? How does it matter? PS: Is there any way to put LaTeX inline on PF rather than just having it come out in separated chunks? I feel like the equations jumping every line feels like Christopher Walken reading a Physics textbook.