  The spectrum of the Hermitian Hamiltonian ${1\over2}p^2+{1\over2}m^2x^2+gx^4$
($g>0$), which describes the quantum anharmonic oscillator, is real and
positive. The non-Hermitian quantum-mechanical Hamiltonian $H={1\over2}p^2+{1
\over2}m^2x^2-gx^4$, where the coupling constant $g$ is real and positive, is
${\cal PT}$-symmetric. As a consequence, the spectrum of $H$ is known to be
real and positive as well. Here, it is shown that there is a significant
difference between these two theories: When $g$ is sufficiently small, the
latter Hamiltonian exhibits a two-particle bound state while the former does
not. The bound state persists in the corresponding non-Hermitian ${\cal
PT}$-symmetric $-g\phi^4$ quantum field theory for all dimensions $0\leq D<3$
but is not present in the conventional Hermitian $g\phi^4$ field theory.
