Past exam of the options II course of the University of Oxford / 2010 / J / J2 / Solution

The stationary Schrödinger equation isA nonzero bound-state wave function has ${\int }_{\mathbb{R}}|\psi {|}^{2}dx<\infty$, so it can be rescaled to have integral one. Odd parity means $\psi (-x)=-\psi (x)$.

For $-V_0<E<0$, set $k=\sqrt{2m(V_0+E)}/\hslash >0$ and $\kappa =\sqrt{-2mE}/\hslash >0$. Oddness removes the interior cosine solution, and normalizability removes the growing exponentials outside. A continuous odd candidate can therefore be writtenContinuity of the derivative at $a$ gives $k\cos (ka)=-\kappa \sin (ka)$; oddness gives the other boundary condition. A nonzero solution cannot have $\sin (ka)=0$, since the derivative condition would then force its interior coefficient to vanish. Thus, with $\xi =ka$ and $\eta =\kappa a$,Physical bound states require $\xi ,\eta >0$. On $0<\xi <\pi /2$, the cotangent condition gives negative $\eta$, so no physical intersection occurs. On $(\pi /2,\pi )$, the curve $-\xi \cot \xi$ rises from zero to positive infinity, while the positive quadrant of the circle decreases. Their first intersection exists exactly when the circle radius $R=\sqrt{2m{V}_0{a}^2/{\hslash }^2}$ exceeds $\pi /2$.

In particular,There is also no normalizable odd state at equality: the putative intersection has $\eta =0$, hence $E=0$, and its exterior tail does not decay. Higher positive branches begin beyond this first interval and require still larger $R$.