Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 4 / d

Table of contents
- Solution d

Solution

Part (b) shows that the transform of

g_{n}

is

q_{n} (k) e^{- k^{2} /2}

with

q_{n}

of degree

n

. Part (c) shows that it solves the same eigenvalue equation as

g_{n}

, with the same

μ_{n}

. The assumed uniqueness of the polynomial solution of each degree therefore forces

q_{n} = c_{n} p_{n}

; hence

g_{n} = c_{n} g_{n}

.

The constants can also be determined. Substituting

g_{n} = p_{n} e^{- x^{2} /2}

into the equation gives the Hermite differential equation

p_{n}^{''} - 2 x p_{n}^{'} - (1 + μ_{n}) p_{n} = 0.

Comparison of the leading coefficient yields

μ_{n} = - (2 n + 1)

. If the leading coefficient of

p_{n}

is

a_{n}

, part (b) gives leading coefficient

2 π (- i)^{n} a_{n}

for

q_{n}

. Thus

g_{n} (k) = 2 π (- i)^{n} g_{n} (k), c_{n} = 2 π (- i)^{n} .

Thus Hermite functions are Fourier eigenfunctions with the paper's unnormalized Fourier transform; dividing that transform by

2 π

gives the unitary eigenvalues

(- i)^{n}

.

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