minimal contribution

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    Michael Heines

    For alpha larger than zero, the sign is negative. This means we need to look for the maxima of 2 cos^2(x) – sin^2(x), which are x=0° and x=180° -> parallel

    For alpha smaller than zero, the sign is positive, so to obtain the lowest contribution we must find the minima of 2 cos^2(x) – sin^2(x) which are at x=90° and x=270° -> orthogonal

    For alpha equal to zero there is no quadrupole contributions so all orientations have (up to the quadrupole term) the same energy

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