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Let's say we have a classical system with a Poisson bracket. We quantise this system to get a quantum theory where we choose some variable to operator replacement : $x\rightarrow X, p\rightarrow P$, such that the commutator $[X,P]\neq i\hbar$. Functions $f(x,p)$ get mapped to $f(X,P)$. Assume $X$ and $P$ are two arbitrarily chosen Hermitian operators on $L^2(R)$, with a continuous unbounded spectrum.

We now look at the space of functions $f(X,P)$, and define a Lie bracket $(f,g)$ as:

$$(f(X,P),g(X,P)) := i\hbar (\frac{\partial f}{\partial X}\frac{\partial g}{\partial P}-\frac{\partial f}{\partial P}\frac{\partial g}{\partial X})$$

where the differentiation is defined to act as if $f$ and $g$ were ordinary functions of $X$ and $P$.

The dynamics are defined using this new bracket:

$$\frac{dO}{da}=\frac{1}{i\hbar}(O,A)$$

where $A(X,P)$ is a generator like momentum, Hamiltonian or angular momentum.

The rest is the same: probabilities are given by the Born rule.

I want to know, this quantum theory is experimentally incorrect, but is there also anything logically inconsistent here?

And is there anything that logically motivates us to choose the commutator as the Lie bracket?

Ryder Rude
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    So $[X,P]=0$ ?? – Qmechanic Dec 11 '23 at 07:31
  • @Qmechanic "Assume X and P are two arbitrarily chosen Hermitian operators on L2(R), with a continuous unbounded spectrum.". The commutator isn't fixed. Is there anything logically inconsistent with this "quantisation"? – Ryder Rude Dec 11 '23 at 08:01

1 Answers1

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It seems like the choice of $[X,P]=0$ is also a logically consistent one, see KvN mechanics.

What's interesting is that we recover the usual commutator Lie bracket from the alternative Lie bracket anyway.

The dynamics of KvN mechanics can be given by the alternative Lie bracket in the post, which is just the definition of a Poisson bracket carried over to a Hilbert space.

However, if we then switch to the Schrodinger picture, we recover this Hamiltonian to describe the same dynamics:

$$H=-i\frac{\partial h}{\partial p}\frac{\partial }{\partial x} +i\frac{\partial h}{\partial x}\frac{\partial }{\partial p}$$

where $h$ is the Hamiltonian of the alternative Lie bracket. Switching to the Heisenberg picture then recovers the usual commutator Lie bracket:

$$\frac{dO}{dt}=\frac{1}{i} [O,H]$$

So we obtain a theory with two Lie brackets, both of which produce compatible evolution.

Interestingly, Quantum Mechanics's $[X,P]=i$ is the case where both these Lie brackets on the Hilbert space co-incide.

I have not investigated cases other than $[X,P]=i$ and $[X,P]=0$

Ryder Rude
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  • You are off... In phase-space quantization it is manifest that the commutator (Moyal bracket algebra) is distinctly different (non-isomorphic to) than the PB algebra. In fact, the Groenewold-VanHove theorem excludes a well-defined f map. – Cosmas Zachos Dec 11 '23 at 15:35
  • @CosmasZachos Thanks. Is the KvN part of the answer correct (because all operators commute in the alternative Lie algebra there)?. But then I also write "In Quantum Mechanics, both these Lie brackets (the commutator and the alternative one from the post) co incide". But note that the alternative bracket is similar in formula to the Poisson bracket, but it applies on operators, so that there ordering ambiguities in the operators. So it is not exactly the same as the Poisson bracket. – Ryder Rude Dec 11 '23 at 16:25
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    Wrong; you are getting faked out by Hilbert space. If you work in the (equivalent!) phase-space, the paradoxes and contradictions of KvN go away, and you distinctly see how Moyal brackets (the equivalent of commutators) are inequivalent to Poisson brackets. It is all but impossible to formulate or even discuss the G-VH theorem in Hilbert space. You might read the original. – Cosmas Zachos Dec 11 '23 at 17:41
  • @CosmasZachos thanks. i will read it – Ryder Rude Dec 11 '23 at 17:43
  • Linked. – Cosmas Zachos Dec 11 '23 at 17:48