 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2018/s18/2/b/solution

Expand

G (x, x^{'})

in a complete orthonormal eigenbasis in its first argument. Applying the operator yields

\sum_{n} (E_{n} - E) c_{n} φ_{n} (x) = δ (x - x^{'})

\sum_{n} φ_{n} (x) φ_{n}^{*} (x^{'})

, so

c_{n} = \frac{φ _{n}^{*} ( x ^{'} )}{E _{n} - E}, G (x, x^{'}) = n \sum \frac{φ _{n} ( x ) φ _{n}^{*} ( x ^{'} )}{E _{n} - E} .

Consequently

lim_{E \to E_{n}} (E_{n} - E) G = φ_{n} (x) φ_{n}^{*} (x^{'})

For the explicit formula, use

E_{n} - E ≃ - (ℏ^{2} k_{n} / m) (k - k_{n})

and

sin ka ≃ (- 1)^{n} a (k - k_{n})

. Also

sin [k_{n} (a /2 - x)] = (- 1)^{n + 1} sin [k_{n} (x + a /2)]

. The resulting limit is

φ_{n} (x) φ_{n}^{*} (x^{'}) = \frac{2}{a} sin \frac{nπ ( x + a /2 )}{a} sin \frac{nπ ( x ^{'} + a /2 )}{a} .

The expression is symmetric in

x, x^{'}

and holds on either side of the source. The usual complex-analysis residue with respect to

E - E_{n}

has the opposite sign, since the source requests the limit with

E_{n} - E

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