Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 1 / 15A (Quantum Mechanics)

The squared sine has integral $a$ over the interval of length $2a$, since it contains an integer number of half-wavelengths. A real positive phase convention therefore givesInside the infinite square well, the Time-independent Schrodinger equation is $-{\hslash }^2{\psi }_n^{\prime \prime }/(2m)=E_n{\psi }_n$. The wavenumber is $\pi n/(2a)$, soThe stationary states vanish at both walls and form an orthonormal basis of the square-integrable state space. A common arbitrary phase on each basis function does not alter any energy or measurement probability.

Choose the prepared wavefunction real and positive on its support. Its normalization isIn the ramp-state expansion in a symmetric infinite well, write $\psi ={\sum }_{n\ge 1}{c}_n{\psi }_n$. Orthogonality givesFor $\kappa =n\pi /(2a)$, a primitive is $-x\cos (\kappa (x+a))/\kappa +\sin (\kappa (x+a))/{\kappa }^2$. Evaluating both endpoints givesAn equivalent parity-separated form isThese formulas with ${\sum }_n{c}_n{\psi }_n$ are the requested explicit expansion. The convergence is in the square-integrable norm; at the idealized discontinuity at the right wall, a pointwise boundary value does not change the quantum state or its probability amplitudes.

The original PDF threshold contains $\hslash$, not $h$. It is $8E_1$, so the energies strictly above it are exactly those with $n\ge 3$. By the Born rule, the required probability is $1-|c_1{|}^2-|c_2{|}^2$. The two excluded amplitudes areConsequentlyThus, in increasing powers of $1/\pi$,The threshold lies strictly between $E_2$ and $E_3$, so no equality-at-threshold convention affects the answer.

The Parseval identity for the orthonormal stationary-state expansion is ${\sum }_{n\ge 1}|c_n{|}^2=\Vert \psi {\Vert }^2=1$. Splitting the amplitudes into even and odd indices givesMultiply by ${\pi }^2/3$ to obtain the requested identity. Its integer constants areThe series converges absolutely, since both parity contributions are $O(p^{-2})$. This derivation uses the normalization of the prepared state and completeness of the well stationary states, rather than a separate summation trick.