Wave eigenfunction and eigenvalue for step potential

Question

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.

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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?

Answer 1

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} $

Answer 2

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.