# Whence the $i$ in QM Poisson bracket definition?

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.

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

• Note that "QM Poisson bracket" is not a term that is in use today. The symbol $[u,v]$ is called a commutator, and while it is connected to the Poisson brackets of hamiltonian mechanics (see this question for the connection, and also possibly this one), no one calls it a Poisson bracket because it isn't really one. – Emilio Pisanty May 12 '17 at 17:35
• OK, I'll bite--why the downvote? There are two upvoted answers with +2 and +3. Can the question have been so devoid of interest? Really? – daniel May 13 '17 at 13:11

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.

Is it unavoidable: No.

Is it convenient: Yes.

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

• This answer would be a lot more helpful if it explained why it's convenient to turn the anti-hermitian thing into a hermitian thing. – DanielSank May 12 '17 at 17:38
• @DanielSank yep. – AccidentalFourierTransform May 12 '17 at 18:47