Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 1 / 18D (Quantum Mechanics)

The attractive potential has $\Delta >0$. Its time-independent Schrödinger equation isIntegrate from $-\epsilon$ to $\epsilon$. Continuity makes the delta term equal to $-{\hslash }^2\chi (0)/(2M\Delta )$, while the integral of $E\chi$ tends to zero as the interval shrinks. The second-derivative term integrates to the difference of first derivatives. HenceThis expresses the point interaction through a derivative jump while the wavefunction remains continuous.

For a negative energy write $E=-{\hslash }^2{\lambda }^2/(2M)$, $\lambda >0$. Away from the delta interaction the decaying solutions are $C{e}^{-\lambda x}$ for $x>0$ and $C{e}^{\lambda x}$ for $x<0$, with the same $C$ by continuity. The derivative jump is $-2\lambda C=-C/\Delta$, giving $\lambda =1/(2\Delta )$. Thus the bound state in a delta potential hasThe normalization follows from ${\int }_{\mathbb{R}}{e}^{-2\lambda |x|}dx=1/\lambda$. This is a genuine square-integrable negative-energy state.

For the symmetric double-delta potential, take an odd state at $E=-{\hslash }^2{\lambda }^2/(2M)$, $\lambda >0$. On $0<x<a$ odd parity selects $B\sinh (\lambda x)$. For $x>a$, decay and continuity give $B\sinh (\lambda a)e^{-\lambda (x-a)}$; extend to negative $x$ by oddness. At $x=a$ the jump condition isFor a nonzero state this means $\lambda \Delta [1+\coth (\lambda a)]=1$, and thereforeOddness makes the jump at $-a$ follow automatically: both its derivative jump and its required wavefunction value have the corresponding opposite signs.

There is a positive solution precisely when $\Delta <a$. To see this, rewrite the equation asThe right side decreases strictly from $a$ to zero as $\lambda$ increases from zero to infinity: its derivative has the sign of $(1+2a\lambda )e^{-2a\lambda }-1<0$. Hence for $0<\Delta <a$ there is exactly one positive solution, while $\Delta =a$ is a nonnormalizable zero-energy threshold and $\Delta >a$ gives no odd bound state. This verifies the stated sufficiently-small-$\Delta$ condition and identifies its exact boundary, consistent with the odd bound state of a symmetric double-delta potential.