Two Hermitian operators $\hat{A}$ and $\hat{B}$ are such that they commute but don't anti-commute. In this case, even they commute their uncertainty product will not be zero, is it right?

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    $\begingroup$ (almost) any operator pair that commutes doesn't anticommute, so I'm not sure why you think that there is something special about operators that commute "but don't anti-commute." $\endgroup$ – Jahan Claes Apr 22 at 15:04
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    $\begingroup$ @Bob of course, any operator commutes with itself but doesn't anticommute with itself. Same with the identity, it commutes with any operator but it anticommutes with none. $\endgroup$ – user2723984 Apr 22 at 15:08
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    $\begingroup$ @Snpr_Physics, you should clarify your question, why do you think anticommutation should have any effect on uncertainty (I assume that by uncertainty product you mean the product of their standard deviations with respect to some state?) $\endgroup$ – user2723984 Apr 22 at 15:10
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    $\begingroup$ No, if they commute they have simultaneous eihenstates regardless of their anticommutator. $\endgroup$ – Jahan Claes Apr 22 at 15:48
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    $\begingroup$ To amplify the point of @Jahan Claes, the potentially nonvanishing term of the Robertson–Schrödinger inequality in fact vanishes by virtue of sharing eigenstates. $\endgroup$ – Cosmas Zachos Apr 22 at 16:34

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