 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2017/3/b/solution

Away from the delta potential, the decaying bound state obeys

E_{0} = - ℏ^{2} κ^{2} / (2 m)

. Integrating the Schrödinger equation across the attractive interaction gives

φ^{'} (0^{+}) - φ^{'} (0^{-}) = - \frac{2 m g}{ℏ ^{2}} φ (0) .

For an even exponential, the left side is

- 2 κ φ (0)

. Therefore

κ = \frac{m g}{ℏ ^{2}}, E_{0} = - \frac{m g ^{2}}{2 ℏ ^{2}} .

φ_{0} (x) = κ e^{- κ ∣ x ∣}

, since

2 κ \int_{0}^{\infty} e^{- 2 κ x} d x = 1

. The single positive solution for

κ

gives one bound state; an odd decaying wave cannot be both continuous and nonzero with this interaction.

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