 Solution

The time-independent Schrodinger equation isIntegrating it over $(a-\epsilon ,a+\epsilon )$ and letting $\epsilon \to 0$ shows that ${\psi }^{\prime }$ has no jump because $V$ is finite. A jump in $\psi$ would create a delta term in ${\psi }^{\prime }$ and hence a derivative of a delta in ${\psi }^{\prime \prime }$, so $\psi$ is also continuous.

For a bound state $-V_0<E<0$, defineThen the proposed even wavefunction solves the equation away from the interfaces, andContinuity at $x=a$ gives $B\cos (ka)=A{e}^{-\eta a}$, while continuity of the derivative gives $-Bk\sin (ka)=-A\eta e^{-\eta a}$. Dividing yields the second required relationThus the lowest even state of the finite square well is the first-quadrant intersection, with $0<ka<\pi /2$, of the circle $k^2+{\eta }^2=2m{V}_0/{\hslash }^2$ and the curve $\eta =k\tan (ka)$.