 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2018/2/c/solution

Away from the shell, the reduced radial equation is free. Regularity at the origin and the definition of the S-wave phase shift give

u_{0} (r) = {A sin k r, B sin (k r + δ_{0}), r < R, r > R .

The reduced radial wave function is continuous at

R

; otherwise its second derivative would contain a delta derivative not present in the equation. Integrating the equation across the shell gives the jump

u_{0}^{'} (R^{+}) - u_{0}^{'} (R^{-}) = \frac{2 m V _{0}}{ℏ ^{2}} u_{0} (R) .

Divide by the common shell value and substitute the two free solutions. This yields

k cot (k R + δ_{0}) - k cot (k R) = \frac{2 m V _{0}}{ℏ ^{2}} .

If the shell value vanishes, use the undivided matching equations or a continuous limit rather than this logarithmic-derivative division. For an explicit phase shift of a delta-shell potential, write

x = k R

and

g = 2 m V_{0} R / ℏ^{2}

tan δ_{0} (k) = - \frac{g sin ^{2} x}{x + g sin x cos x} .

The phase is defined modulo

π

; choose its branch continuously as energy or coupling varies. At zero coupling the physical phase shift is zero.

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