Given the Schrödinger equation:
$$-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2} + V(x)\psi = E\psi$$
where:
$$\left\{ \begin{array}{l} V(x) = V_0 \text{ for }x>a \\ V(x) = 0 \text{ for } 0\leq x \leq a \\ V(x) = \infty \text{ for } x<0 \end{array}\right.$$
and $V_0 > E$. Solving the schrodinger equation we get for $x\leq 0$:
$$\psi(x) = N_1\sin\left(\sqrt{\frac{2mE}{\hbar^2}}x+\phi\right)$$
And for $x>0$:
$$\psi(x) = N_2\exp\left({-\sqrt{\frac{2m(V_0-E)}{\hbar^2}}x}\right)$$
Where I neglected the other term $\exp\left({\sqrt{\frac{2m(V_0-E)}{\hbar^2}}x}\right)$ because the wave function should be normalizable.
The thing is, because we want $\psi(0)=0$, we fix $\phi = 0$. We are left with 2 Unknowns, while we have 3 conditions left. We want $\psi$ to be continuous at $a$ , We want $\psi'(x)$ to be continuous at $a $, and finally we want it to be normalized.
How is this possible?
