Finite square well I know how and why we use this form of stationary Schrödinger equation for finding $\psi$ outside the finite square potential well:
$$\frac{d^2 \psi}{dx^2}=\kappa^2 \psi$$
I Also know that the general solution to this equation is:
$$\psi = Ae^{\kappa\, x} + Be^{-\kappa\, x}.$$
But why do we use only left part $\psi = Ae^{\kappa \, x}$ for the left outer side $x<0$, and the right part $\psi = B e^{-\kappa\, x}$ on the right outer side part $x>0$?
 A: I think you are asking for a finite well of width L that is from $ -L/2< x< L/2$. Why do we only use 
$\psi(x) = A e^{+\kappa x} $ for $ x<-L/2 $ 
and 
$\psi(x) = B e^{-\kappa x} $  for $ x>+L/2 $. 
The reason is we want to be able to interpret the wavefunction as the probability density for finding the particle at x. For this to make sense the probability of finding the particle anywhere should be 1 or
$\int_{-\infty}^{-\infty}dx \ |\psi(x)|^2 = 1$.
If you have solutions of the form 
$\psi(x) = A e^{+\kappa x} + B e^{-\kappa x} $
this is only possible if $A = 0 $ in the $x > +L/2$ region and $B = 0 $ in the $x < +L/2$ region.
A: You need to ensure that the wavefunction is normalisable: $$\int\limits_{-\infty}^{\infty} \mathrm{d}x\, |\psi(x)|^2 = 1. $$
This ensures that the wave function yields a valid probability distribution upon application of the Born rule. If you use the normal convention of taking positive $x$ values to the right and negative $x$ values to the left, this implies that you can only keep $e^{-\kappa x}$ for the right outer side, etc. Otherwise the value of the wavefunction grows without bound as $x\to \pm \infty$
