A quantum field theory has finite zero-point energy if the sum over all boson modes $b$ of the $n$th power of the boson mass $ m_b^n $ equals the sum over all fermion modes $f$ of the $n$th power of the fermion mass $ m_f^n $ for $n= 0$, 2, and 4. The zero-point energy of a theory that satisfies these three conditions with otherwise random masses is huge compared to the density of dark energy. But if in addition to satisfying these conditions, the sum of $m_b^4 log m_b/mu$ over all boson modes $b$ equals the sum of $ m_f^4 log m_f/mu $ over all fermion modes $f$, then the zero-point energy of the theory is zero. The value of the mass parameter $mu$ is irrelevant in view of the third condition ($n=4$). The particles of the standard model do not remotely obey any of these four conditions. But an inclusive theory that describes the particles of the standard model, the particles of dark matter, and all particles that have not yet been detected might satisfy all four conditions if pseudomasses are associated with the mean values in the vacuum of the divergences of the interactions of the inclusive model. Dark energy then would be the finite potential energy of the inclusive theory.
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