= Solution
The kinetic <matrix> is diagonal:
$$
\langle\phi_n|T|\phi_m\rangle
=\delta_{nm}\frac{\hbar^2\pi^2n^2}{2ma^2}.
$$
For the linear potential,
$$
\langle\phi_n|V|\phi_n\rangle=\frac{V_0}{2}
=\frac{9\hbar^2}{2ma^2},
$$
and
$$
\left\langle\phi_1\left|\frac xa\right|\phi_2\right\rangle
=2\int_0^1y\sin(\pi y)\sin(2\pi y)\,dy
=-\frac{16}{9\pi^2}.
$$
Thus, in units $\hbar^2/(ma^2)$,
$$
\mathcal H=
\begin{pmatrix}
(\pi^2+9)/2&-16/\pi^2\\
-16/\pi^2&(4\pi^2+9)/2
\end{pmatrix}.
$$
The lower <eigenvalue> gives the <two-mode variational bound for a linearly tilted square well>
$$
\boxed{
E_0\leq\frac{\hbar^2}{ma^2}
\left[
\frac{5\pi^2+18}{4}
-\sqrt{\frac{9\pi^4}{16}+\frac{256}{\pi^4}}
\right]}
\approx9.25936\,\frac{\hbar^2}{ma^2}.
$$
Back to article page
