Wave equations for two intervals at Potential step

Lets say we have a potential step as in the picture: In the region I there is a free particle with a wavefunction $\psi_I$ while in the region II the wave function will be $\psi_{II}$.

Let me take now the schrödinger equation and try to derive $\psi_I$:

\begin{align} &~~W \psi = -\frac{\hbar^2}{2m}\, \frac{d^2 \Psi}{d\, x^2} + W_p \psi ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nonumber \\ &~~W \psi = -\frac{\hbar^2}{2m}\, \frac{d^2 \Psi}{d\, x^2}\nonumber \\ &\frac{d^2 \Psi}{d\, x^2} = -\frac{2m W}{\hbar^2}\,\psi \nonumber\\ {\scriptsize \text{DE: }} &\boxed{\frac{d^2 \Psi}{d\, x^2} = -\mathcal L\,\psi}~\boxed{\mathcal{L} \equiv \sqrt{\tfrac{2mW}{\hbar^2}}} \nonumber\\ &~\phantom{\line(1,0){18.3}}\Downarrow \nonumber\\ {\scriptsize \text{general solution of DE: }} &\boxed{\psi_{I} = C \sin\left(\mathcal{L}\, x \right) + D \cos \left(\mathcal{L}\, x \right)}\nonumber \end{align}

I got the general solution for the interval I, but this is nothing like the solution they use in all the books: $\psi_{I} = C e^{i\mathcal L x} + D e^{-i \mathcal L x}$ where $\mathcal L \equiv \sqrt{{\scriptsize 2mW/\hbar^2}}$. I have a personal issue with this because if $x= -\infty$ part $De^{-i \mathcal L x}$ would become infinite and this is impossible for a wavefunction! I know that i would get exponential form if i defined constant $\mathcal L$ a bit differently as i did above:

\begin{align} {\scriptsize \text{DE: }} &\boxed{\frac{d^2 \Psi}{d\, x^2} = \mathcal L\,\psi}~\boxed{\mathcal{L} \equiv -\sqrt{\tfrac{2mW}{\hbar^2}}} \nonumber\\ &~\phantom{\line(1,0){18.3}}\Downarrow \nonumber\\ {\scriptsize \text{general solution of DE: }} &\boxed{ \psi_{I} = C e^{\mathcal L x } + D^{-\mathcal L x} }\nonumber \end{align}

This general solution looks more like the one they use in the books but it lacks an imaginary $i$ and $\mathcal L$ is defined with a - while in all the books it is positive. Could anyone tell me what am i missing here so i could understand this?

• Why do you think $\mathrm{e}^{-i\mathcal{L}x}$ would go to infinity? $\mathrm{e}^{i\theta}$ has constant magnitude $=1$ for all real $\theta$. The complex exponentials are just a way of rewriting the solution in terms of $\cos$ and $\sin$. In your second form the minus sign should go under the square root -- $\mathcal{L}$ is imaginary in that case. – Michael Brown Apr 6 '13 at 11:03
• I would appreciate if you could show me how to transform $C \sin(\mathcal L x) + D \cos (\mathcal L x)$ using euler formula. Afterall i don't have $i\sin()$. – 71GA Apr 6 '13 at 12:44

The solution is $A\sin(ax) + B\cos(ax)$ when $\psi'' = -(a^2\psi)$
but $Ce^{bx} + De^{-bx}$ when $\psi'' = (b^2\psi)$ where $b=ia$
$\psi(0)_{ exponentialsoln. } = \psi(0)_{cos \& sin soln.}$
$Ce^0 + De^0 = B$