$\rho\approx\frac{TeV^8}{m_{Pl}^4}$This is almost exactly the same seesaw game with the scales like the neutrino seesaw game. In the case of the neutrinos, we assume the right-handed SU(5)-neutral neutrinos to acquire the GUT scale masses - which is almost the same thing as the Planck scale above - and the unnaturally small value of the observed neutrino masses comes from the smaller eigenvalue(s) of the matrix ((mGUT, mEW), (mEW,0)).

Note that again, the neutrino masses are of the same order as the mass scale associated with the vacuum energy. This means that if the vacuum energy is, in some appropriate sense, a "2 x 2" matrix

$\left(\begin{array}{cc}0&+m_{SB}^4\\ +m_{SB}^4&-m_{Pl}^4\end{array}\right)$then the eigenvalue closer to zero will be of the desired order. Let me propose one more mechanism to generate the seesaw for the cosmological constant. The vacuum energy of free fields may be computed as the integral of "sqrt(k^2+m^2)/2" (the zero-point energy of a harmonic oscillator) over the three-dimensional momentum space. This integral is quartically divergent. Normally we consider the following expression to be a result in the dimensional regularization

- m
^{4}

- m
_{Pl}^{4}

$abs(m^4+im_{Pl}^4) = sqrt(m^8 + m_{Pl}^8)$If it is so, you can Taylor-expand because "m_{Pl}" is so much larger. The term "m_{Pl}^4" cancels between bosons and their fermionic superpartners while the first uncancelled terms will be of order

$\frac{m_{boson}^8-m_{fermion}^8}{m_{Pl}^4}$which is again of the right order of magnitude to match the observed vacuum energy because the numerator is of order "m_{SB}^8" where "m_{SB}" is the supersymmetry breaking scale. Yes, the calculation above does not really agree with the usual supergravity calculus but I think that we don't know whether the calculus is correct after SUSY breaking anyway.

**Update:**As we discuss here, our colleague from Brookhaven was brave enough to transform these ideas into a preprint based on the Wheeler-DeWitt equation.

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