  The ground state energy per particle of a dilute, homogeneous,
two-dimensional Bose gas, in the thermodynamic limit is shown rigorously to be
$E_0/N = (2\pi \hbar^2\rho /m){|\ln (\rho a^2)|^{-1}}$, to leading order, with
a relative error at most ${\rm O} (|\ln (\rho a^2)|^{-1/5})$. Here $N$ is the
number of particles, $\rho =N/V$ is the particle density and $a$ is the
scattering length of the two-body potential. We assume that the two-body
potential is short range and nonnegative. The amusing feature of this result is
that, in contrast to the three-dimensional case, the energy, $E_0$ is not
simply $N(N-1)/2$ times the energy of two particles in a large box of volume
(area, really) $V$. It is much larger.
