  By introducing a quantitative `degree of commutativity' in terms of the angle
between spin-observables we present two tight quantitative trade-off relations
in the case of two qubits: First, for entangled states, between the degree of
commutativity of local observables and the maximal amount of violation of the
Bell inequality: if both local angles increase from zero to \pi/2 (i.e., the
degree of local commutativity decreases), the maximum violation of the Bell
inequality increases. Secondly, a converse trade-off relation holds for
separable states: if both local angles approach \pi/2 the maximal value
obtainable for the correlations in the Bell inequality decreases and thus the
non-violation increases. As expected, the extremes of these relations are found
in the case of anti-commuting local observables where respectively the bounds
of 2\sqrt{2} and \sqrt{2} hold for the expectation of the Bell operator. The
trade-off relations show that non-commmutativity gives ``a more than classical
result" for entangled states, whereas "a less than classical result" is
obtained for separable states. The experimental relevance of the trade-off
relation for separable states is that it provides an experimental test for two
qubit entanglement. Its advantages are twofold: in comparison to violations of
Bell inequalities it is a stronger criterion and in comparison to entanglement
witnesses it needs to make less strong assumptions about the observables
implemented in the experiment.
