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In Dirac's Principles of Quantum Mechanics (pg. 86 eq. 5), the quantum commutator is motivated by looking for a bracket that satisfies the same properties of the Poisson bracket. When deriving the commutator, he first calculates

$$ \{u_1 u_2 , v\}_{P.B.} $$ however, ensuring that the ordering of $u_1$ and $u_2$ is preserved. My confusion is that he allows $u_1$ and $u_2$ to commute with $v$ in his derivation. Is this because he is viewing the Poisson bracket in the abstract sense i.e.

$$ F\times F \rightarrow F $$ where $F$ is the set of all functions on phase space?

Replacing the coordinates with ones that are Grassmann-odd valued, would I pick up a minus sign whenever $u_1$ or $u_2$ is commuted with $v$, or would I let the commute exactly like the derivation of Dirac's? I'm assuming I would get the anticommutator but I feel like I'm missing something important.

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  1. Ref. 1 is only discussing bosonic/Grassmann-even variables, so the classical variables trivially commute, but Dirac has secretly the corresponding operator identity for the commutator in mind when he writes down the Poisson property/Leibniz rule (5) for the Poisson bracket, and then the order of operators is of course crucial.

  2. The generalization to supernumber-valued (operators) reads $$ [u_1u_2,v]~=~[u_1,v]u_2(-1)^{|v||u_2|}+u_1[u_2,v],\tag{5}$$ where $|\cdot|$ denotes the Grassmann-parity. See also e.g. Wikipedia and this related Phys.SE.

References:

  1. P.A.M. Dirac, Principles of QM, 4th ed, 1958; $\S21$ p. 86.
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  • $\begingroup$ So is the explicit definition of the Poisson Bracket on pg 85 eqn 1 not used in Dirac's derivation of the quantum commutator? As in, has he just used the fact that the quantum commutator will also satisfy the Leibniz rule and correspondingly just written down the only way you can do that in a way that preserves the order of $u_1$ and $u_2$? $\endgroup$
    – DIRAC1930
    Jan 12, 2022 at 18:56
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic ♦
    Jan 12, 2022 at 19:06
  • $\begingroup$ How does one go about deriving the commutative property of the super-Poisson bracket for odd Grassmann variables, namely $\{F,G\} = \{G.F\}$? I understand that the quantum bracket must obey the Leibniz rule (which is easily generalized for Grassmann numbers) otherwise it wouldn't be the quantum analogue of the Poisson bracket, but it's unclear why the fermionic Poisson bracket must be commutative in its arguments. $\endgroup$
    – DIRAC1930
    Jan 13, 2022 at 23:00
  • $\begingroup$ This is because the super-Poisson bracket inherits its properties from the super-commutator, cf. the semiclassical limit $\hbar\to 0$. $\endgroup$
    – Qmechanic ♦
    Jan 13, 2022 at 23:10

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