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The generalized uncertainty principle says,

$$\sigma_A^2\sigma_B^2~ \ge ~\frac14\langle i[A, B]\rangle^2.$$

But the complex field is not ordered, i.e, inequalities like $\le$, $\ge$, etc are absurd. For instance, is $i>1$? Doesn't $i$ in there makes the whole inequation meaningless?

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    $\begingroup$ The commutator has its own factor of $i$ $\endgroup$ – Yuzuriha Inori Mar 10 '18 at 6:51
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    $\begingroup$ The more familiar, equivalent, formula is $\sigma_A^2\sigma_B^2~ \ge ~\frac14|\langle i[A, B]\rangle|^2.$. The absolute value arises form Cauchy-Schwartz' inequality...The factor $i$ can be therefore omitted as it shows up in the absolute value. Alternatively (J.Murray's answer below), as $i[A,B]$ is real you can keep it, dropping the absolute value. $\endgroup$ – Valter Moretti Mar 10 '18 at 10:33
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To elaborate on Yuzuriha's comment, if $A$ and $B$ are self-adjoint operators, then

$$\langle [A,B]\rangle = \langle\psi,AB\psi\rangle - \langle\psi,BA\psi\rangle = \langle A \psi,B\psi\rangle - \langle B\psi,A\psi\rangle$$

but

$$\overline{\langle [A,B]\rangle} = \overline{\langle A \psi,B\psi\rangle} - \overline{\langle B\psi,A\psi\rangle} = \langle B\psi,A\psi\rangle -\langle A \psi,B\psi\rangle = -\langle [A,B]\rangle $$

which implies that $\langle [A,B]\rangle$ is purely imaginary. The explicit factor of $i$ makes the whole thing make sense.


Of course, if you're in the mood for mindless pedantry, then complex numbers $z$ with $Im(z)=0$ are still complex numbers and therefore are members of an unordered set; in that case, you can define the obvious total order on the subset $R:=\{z\in\mathbb{C} | Im(z)=0\}$ and note that the objects in question belong to $R$.

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  • $\begingroup$ The step $\langle\psi,AB\psi\rangle - \langle\psi,BA\psi\rangle = \langle A \psi,B\psi\rangle - \langle B\psi,A\psi\rangle$ is true only for Hermitian operators. What about non-Hermitian observables? $\endgroup$ – Ayatana Mar 10 '18 at 7:15
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    $\begingroup$ All observables are Hermitian. $\endgroup$ – J. Murray Mar 10 '18 at 7:16
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    $\begingroup$ In fact, all observables are self-adjoint, which is a subtly stronger requirement. $\endgroup$ – J. Murray Mar 10 '18 at 7:17

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