This has already been asked, but I still have some issues with it: It has been established in this question that the ordering prescription is not a function that maps operators to operators, but instead just a map from symbols to operators.

Does that mean, giving an ordering prescription just makes sense when you are given a function $ R \rightarrow R $, out of which you want to make an "Operator-function"? I would like to know if I understood correctly by giving an example here. Let's say $A$ is the space of all linear Operators acting on the Hilbert space. My wild guess is that the Hamiltonian then is a function $A \rightarrow A$, for example (I know this example stems from single particle QM) by

$$ H(\hat{p}, \hat{x})= \frac{(\hat{p}-f(\hat{x}))^2}{2m}.$$

Since I employ a function that is defined on real numbers (taking the square, or subtracting), the definition of $H$ is not well defined, and could yield different results (because real numbers commute, while operators don't). By fixing the ordering of the operators (for example by normal ordering), I remove any ambiguities. Is that the right way to see it?


1 Answer 1


Yes. In your example, you can rearrange the expansion of $(p-f(x))^2$ in multiple ways: \begin{align} (p-f(x))^2&=p^2+2p f(x) + f(x)^2\, ,\\ &= p^2 + f(x)^2 + p f(x) + f(x)p\, ,\\ &=p^2+f(x)^2 + \frac{1}{2}p f(x) +\frac{3}{2} f(x)p \tag{1} \end{align} etc. and even more complicated forms if you consider the series expansion of $f(x)$. For instance, imagine $f(x)=x^3$ then $$ px^3= xpx^3=x^2px $$ and so forth. All these expressions for $(p-f(x))^2$ are strictly the same when using classical variables, but produce different operators under the replacements $x\to \hat x$ and $p\to \hat p$ because of the non-commutativity of $\hat x$ and $\hat p$.

An ordering procedure would determine a unique polynomial in $\hat x$ and $\hat p$ (or in $\hat a$ and $\hat a^\dagger$) that would in turn determine a unique operator.

(Note that I've never seen something like Eq.(1) but it is possible in principle).


Please note that for polynomials of the type $p^kf(x)$ in the classical variables $p$ and $x$ with $k\le 2$, i.e. for polynomials at most quadratic in $p$, it is possible to find an ordering of the operators so that the quantum commutators is (up to $i$'s and $\hbar$'s), the classical bracket. This ordering is distinguished although not necessarily commonly used; the procedure is inductive on the degree of $p$ and $x$ and fails when the degree of $p$ and the degree of $x$ are both strictly greater than 2.

See Chernoff, Paul R. "Mathematical obstructions to quantization." Hadronic Journal 4 (1981) for details.


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