Inside the <infinite square well>, the stationary <Schrodinger equation> for unit mass is
$$
-\frac{\hbar^2}{2}\psi''=E\psi,
\qquad \psi(0)=\psi(\pi)=0.
$$
Its normalized eigenstates and energies are
$$
\boxed{\psi_n(x)=\sqrt{\frac2\pi}\sin(nx),
\qquad E_n=\frac{\hbar^2n^2}{2}},
\qquad n=1,2,\ldots.
$$
The general normalized time-dependent solution is
$$
\boxed{\Psi(x,t)=\sum_{n=1}^\infty c_n\psi_n(x)e^{-iE_nt/\hbar},
\qquad \sum_{n=1}^\infty|c_n|^2=1}.
$$

Immediately after removing the barrier, the normalized wavefunction is
$$
\Psi(x,0)=
\begin{cases}
\sqrt{4/\pi}\sin(2x),&0\leq x\leq\pi/2,\\
0,&\pi/2<x\leq\pi.
\end{cases}
$$
Its <Fourier sine series> coefficients in the full-well basis are
$$
c_n=\frac{\sqrt8}{\pi}\int_0^{\pi/2}\sin(nx)\sin(2x)\,dx.
$$
They are
$$
c_2=\frac1{\sqrt2},
\qquad
c_n=\frac{4\sqrt2}{\pi}\frac{\sin(n\pi/2)}{4-n^2}\quad(n\ne2).
$$
The <Born rule> therefore gives
$$
\boxed{
\mathbb P(E_n)=
\begin{cases}
\dfrac12,&n=2,\\[4pt]
\dfrac{32}{\pi^2(n^2-4)^2},&n\text{ odd},\\[4pt]
0,&n\text{ even and }n\ne2.
\end{cases}}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 3 / 16B / Solution

Expand $g$ in its <Fourier sine series>
$$
g(y)=\sum_{n=1}^{\infty}g_n\sin(k_ny),
\qquad
k_n=\frac{n\pi}{a},
\qquad
g_n=\frac2a\int_0^a g(y)\sin(k_ny)\,dy.
$$
By <separation of variables>, the decaying solution of <Laplace equation> in each half-strip has $n$th mode proportional to $e^{-k_n|x|}\sin(k_ny)$. Continuity at $x=0$ and integration of the equation across the <Dirac delta function> require the normal-derivative jump
$$
\phi_x(0^+,y)-\phi_x(0^-,y)=g(y).
$$
The jump of $C_ne^{-k_n|x|}$ is $-2k_nC_n$, so
$$
\boxed{\phi(x,y)=-\sum_{n=1}^{\infty}\frac{g_n}{2k_n}e^{-k_n|x|}\sin(k_ny)}.
$$

For $g(y)=\delta(y-c)$, its Fourier coefficients are
$$
g_n=\frac2a\sin\frac{n\pi c}{a}.
$$
Thus
$$
\phi(x,y)=-\frac1\pi\sum_{n=1}^{\infty}
\frac1n e^{-n\pi|x|/a}
\sin\frac{n\pi y}{a}\sin\frac{n\pi c}{a}.
$$
Translation in $x$ gives the <Dirichlet Green function for an infinite strip> with source $(b,c)$:
$$
\boxed{G(x,y;b,c)=-\frac1\pi\sum_{n=1}^{\infty}
\frac1n e^{-n\pi|x-b|/a}
\sin\frac{n\pi y}{a}\sin\frac{n\pi c}{a}}.
$$

Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 1 / 14B / Solution

Linearization about zero and expansion in the <Fourier sine series> give, for the $n$th mode,
$$
\ddot q_n+\left(f'(0)+\frac{n^2}{R}\right)q_n=0.
$$
Every mode is oscillatory exactly when the smallest coefficient, at $n=1$, is nonnegative. Since $f'(0)<0$,
$$
\boxed{R\leq R_c=-\frac1{f'(0)}}.
$$

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 331 / 2 / ii / a / Solution

A Fourier sine series expands a function on an interval in the eigenfunctions $\sin(n\pi x/L)$ of the Dirichlet Laplacian. It is the odd Fourier-series extension across the interval endpoints.


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