= Solution
With the physics convention $\langle\Phi|\Psi\rangle=\int\Phi^*\Psi\,\mathrm d\tau$, the <adjoint operator> is defined by
$$
\boxed{\langle\Phi|\hat A\Psi\rangle
=\langle\hat A^\dagger\Phi|\Psi\rangle.}
$$
This definition includes the operator domains and their <boundary conditions>. If $\hat A$ is a <Hermitian operator> on the common domain, then $\hat A^\dagger\Phi=\hat A\Phi$. Conjugate symmetry of the <inner product> consequently gives
$$
\langle\Phi|\hat A\Psi\rangle
=\langle\hat A\Phi|\Psi\rangle
=\langle\Psi|\hat A\Phi\rangle^*,
$$
which is the required integral identity.
The <boundary conditions> must make the boundary form from <integration by parts> vanish for every pair of allowed functions. For a Schrödinger kinetic operator in one dimension, this means
$$
\left[\Phi^*\Psi'-(\Phi')^*\Psi\right]_{\rm boundary}=0.
$$
Sufficient examples are decay of the functions and derivatives at spatial infinity, homogeneous hard-wall conditions $\Phi=\Psi=0$ at finite endpoints, or matching periodic conditions on both functions and their derivatives. The exact boundary form depends on $\hat A$; square integrability alone does not replace the domain requirement.
Back to article page
