 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2017/4/d/solution

For

m \geq 0

, expand

e^{i ρ c o s θ} = exp [i ρ (e^{i θ} + e^{- i θ}) /2]

. The lowest power producing the Fourier harmonic

e^{im θ}

(i ρ /2)^{m} / m!

. Meanwhile

J_{m} (ρ) \sim (ρ /2)^{m} / m!

. Matching these lowest powers gives

a_{m} = i^{m}

For negative order use

J_{- m} = (- 1)^{m} J_{m}

and the equality of the positive and negative Fourier coefficients of this even function of

θ

. This gives

a_{- m} = (- 1)^{m} i^{m} = i^{- m}

e^{i ρ c o s θ} = m \in Z \sum i^{m} J_{m} (ρ) e^{im θ} = J_{0} (ρ) + 2 m = 1 \sum \infty i^{m} J_{m} (ρ) cos (m θ) .

The cosine form has multiplicity one for

m = 0

and two for every positive

m

. Those multiplicities are essential for matching the incident plane wave.

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