# Wave eigenfunction and eigenvalue for step potential

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?

You don't have to take account of normalization condition because if you have

$$\begin{cases} \psi_<(a)=\psi_>(a) \\ \psi_{<}^{'}(a)=\psi_{>}^{'}(a)\\ \psi(0)=0\\ \end{cases}$$

so it's easy to check that if you define $$\tilde{\psi}= \frac{\psi}{\int \psi^2 dx}$$ you obtain

$$\begin{cases} \tilde{\psi}_<(a)=\tilde{\psi}_>(a) \\ \tilde{\psi}_{<}^{'}(a)=\tilde{\psi}_{>}^{'}(a)\\ \tilde{\psi}(0)=0\\ \end{cases}$$

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?

You are right, for most values of $$E$$ there is no solution. However, for some special values of $$E$$ there is a solution. And these are the eigenvalues you are looking for.