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Let $F:\mathbb R ^n\to\mathbb R$ be a function that has a Taylor expansion, then it can be written (expanded at $a$) as $$ F(x)=\sum_{\alpha} \frac{(x_1 - a_1)^{\alpha_1}\dots(x_n - a_n)^{\alpha_n}}{\alpha_1 !\dots \alpha_n !}\frac{\partial^{|\alpha|}F(a)}{\partial_1^{\alpha_1}\dots \partial_n^{\alpha_n}} $$ where multi-index is used, i.e. $\alpha = (\alpha_1,\dots,\alpha_n) \in \mathbb N_0^n$ and $|\alpha| = \alpha_1 + \dots + \alpha_n$.

Now we take $F:\mathbb R^2 \to \mathbb R$ and the quantum mechanical operators $q$ and $p$ which do not commute, $[q,p] = i\hbar$, then (expanding at $a = 0$) there will be an ambiguity in the order (arrangement) of the powers of $q$ and $p$, since in general, for $m,n,j,k \in \mathbb N$, $$ q^m p^n q^j p^k \neq q^{m+j} p^{n+k} \qquad \iff \qquad [q^m p^n q^j p^k,q^{m+j} p^{n+k}] \neq 0 $$ So which one should be in the expansion ? or should they be symmetrised ?

Another question is how do we take partial derivatives of such function of operators. For example of the product $q^m p^n q^j p^k $, \begin{align} \partial_p(q^m p^n q^j p^k) &= n (q^m p^{n-1} q^j p^k) \quad ?\\ &=j (q^m p^n q^{j-1} p^k) \quad ?\\ &= \frac{1}{2}(n (q^m p^{n-1} q^j p^k) + j (q^m p^n q^{j-1} p^k)) \quad ?\\ & = ??? \end{align}

Cosmas Zachos
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Physor
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    There is an infinity of ordering prescriptions, all equivalent and interconvertible to each other. Some popular ones are Weyl ordering, normal ordering, anti-normal ordering, Born-Jordan ordering, etc... If you are taking derivatives w.r.t. noncommutative variables, you are being inefficient. There are consistent definitions of such, but there are better ways. You probably need to read up on the phase-space formulation. – Cosmas Zachos Sep 26 '21 at 16:20
  • "all are equivalent" means it doesn't matter which arrangement I choose, it gives the same operator $F(q,p)$ ? and what do you mean with "inefficient" ? – Physor Sep 26 '21 at 16:23
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    No, I mean you can systematically transform/translate from each one to any other one. Each prescription normally gives you different answers (amplitudes). But quantization prescription choice is subjective: *nobody* can dictate for you which one to choose. You decide which models your physics problem best. "Inefficient" means you are doing something "wrong": you are counting the centipede's legs and dividing by 100. The problem you might wish to solve this way most probably has a better solution within phase-space QM. – Cosmas Zachos Sep 26 '21 at 16:28
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    You might appreciate this one. – Cosmas Zachos Sep 27 '21 at 13:44

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