Whence the $i$ in QM Poisson bracket definition?

Question

On p. 87 of Dirac's Quantum Mechanics he introduces the quantum analog of the classical Poisson bracket$^1$

$$ [u,v]~=~\sum_r \left( \frac{\partial u}{\partial q_r}\frac{\partial u}{\partial p_r}- \frac{\partial u}{\partial p_r}\frac{\partial u}{\partial q_r}\right) \tag{1}$$

as

$$uv-vu ~=~i~\hbar~[u,v]. \tag{7}$$

I'm not worried about the $\hbar$ but if there is an (alternative) explanation of why the introduction of $i$ is unavoidable that might help.

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$^1$ Note that Dirac uses square brackets to denote the Poisson bracket.

Answer 1

The imaginary unit $i$ is there to turn quantum observables/selfadjoint operators into anti-selfadjoint operators, so that they form a Lie algebra wrt. the commutator.

Or equivalently, consider the Lie algebra of quantum observables/selfadjoint operators with the commutator divided with $i$ as Lie bracket.

The latter Lie algebra corresponds in turn to the Poisson algebra of classical functions, cf. the correspondence principle.

Answer 2

Is it unavoidable: No.

Is it convenient: Yes.

Why: because given two hermitian operators $A,B$, their commutator is anti-hermitian.