= Solution
Varying the canonical scalar action gives $\Box\phi+V_{,\phi}=0$. In a homogeneous flat <Friedmann-Lemaître-Robertson-Walker metric> written in <conformal time>, $\sqrt{-g}=a^4$ and $g^{00}=a^{-2}$, so
$$
\Box\phi=\frac1{a^4}\partial_\tau(a^2\phi')
=\frac1{a^2}\left(\phi''+2\frac{a'}a\phi'\right).
$$
The quadratic potential therefore gives
$$
\boxed{\phi''+2\frac{a'}a\phi'+a^2m^2\phi=0.}
$$
Equivalently $\ddot\phi+3H\dot\phi+m^2\phi=0$ in cosmic time. The intrinsic roll/oscillation time is $m^{-1}$ and the damping time is $H^{-1}$. If $m\gg H_0$, the scalar has long since begun rapid oscillations: averaging over a quadratic oscillation gives equal kinetic and potential energies, $\langle p\rangle\simeq0$, so it behaves as matter rather than accelerating <dark energy>. To remain nearly frozen or only begin rolling today requires the <mass scale of quadratic quintessence>
$$
\boxed{m\lesssim H_0\sim2.3\times10^{-18}\,\mathrm{s}^{-1}
\sim1.5\times10^{-33}\,\mathrm{eV}.}
$$
If its energy is predominantly potential, equating $m^2\phi_0^2/2$ to $\rho_{X0}=3M^2H_0^2\Omega_X$, with $M$ the <reduced Planck mass>, gives
$$
\boxed{|\phi_0|\simeq\sqrt{6\Omega_X}\,M\frac{H_0}{m}.}
$$
Taking the natural onset estimate $m\sim H_0$ and $\Omega_X\sim0.7$ gives $|\phi_0|\sim2M\sim5\times10^{18}\,\mathrm{GeV}$, of order the <Planck mass>. More strongly frozen, negative-pressure evolution requires a smaller mass and correspondingly larger field. The density fixes the product $m|\phi_0|$ at this accuracy, rather than a unique mass and field separately.
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