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 $x>0$?

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If you're using the usual conventions, don't you mean $Ae^{\kappa x}$ on the left, not right? –  Dan Piponi Mar 29 '13 at 20:25

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.

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I think i get it but please correct me if i am wrong. On the interval $x > +\tfrac{L}{2}$ i can only increase $x$ towards $+\infty$. Soo i have to make sure that i have a negative exponent or else i ll get $\psi = \infty$! This is why on this interval i need only $Ae^{-\kappa x}$. It is vice versa for the interval $x<-\frac{L}{2}$ where i can only decrease $x$ towards $-\infty$. This means i have to take a part with a positive exponent so i won't get $\psi = \infty$. It is all about the signs of the exponents if i am not mistaken. –  71GA Mar 29 '13 at 20:39
Pretty much. You can go a step further and actually try integrating $\int_{-L/2}^\infty \ dx \ e^{\kappa x}$ and you will see that you will get $\infty$, so there isn't any hope of normalizing this to anything finite. –  DJBunk Mar 29 '13 at 20:50
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 $x\to \pm \infty$
@71GA You should be able to convince yourself that in general, the normalisation condition is violated for any function that does not vanish at $x = \pm \infty$. This means that the solution $e^{\kappa x}$ is not a valid solution on the right side of the box, and a similar argument applies to the left side. Have you actually tried evaluating the normalisation integral for the various possibilities to see what comes out? –  Mark Mitchison Mar 29 '13 at 20:36