 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/32d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

H = - \partial_{x}^{2} - 2 χ^{2} sech^{2} (χ x), A = \partial_{x} + χ tanh (χ x) .

L^{2} (R)

A^{†} = - \partial_{x} + χ tanh (χ x)

A^{†} A = H + χ^{2}, A A^{†} = - \partial_{x}^{2} + χ^{2} .

In particular,

A^{†} A

is nonnegative, so every eigenvalue of

H

satisfies

E \geq - χ^{2} .

Equality is attained precisely when

A ψ = 0

. Solving that first-order equation gives

ψ_{0} (x) = C sech (χ x),

which is square-integrable. With unit normalization,

C = χ /2

, and

H ψ_{0} = - χ^{2} ψ_{0} .

It remains to exclude another bound state. Since this Pöschl-Teller potential tends to zero at infinity, a bound-state energy must have

E < 0

. If

H ψ = E ψ

, then applying

A

A^{†} A ψ = (E + χ^{2}) ψ

gives

- \partial_{x}^{2} (A ψ) = E (A ψ) .

The free one-dimensional Schrodinger operator has no nonzero square-integrable eigenfunction with

E < 0

. Consequently

A ψ = 0

, and the state must be proportional to

ψ_{0}

. Thus there is exactly one bound state, with

E_{0} = - χ^{2} .

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