# Is there any ambiguity when forming operators for classical observables from other operators? [duplicate]

This question already has an answer here:

There are some classical quantities, for which we know corresponding operators: e.g. position, momentum. For some others it's straightforward to compose these operators using expressions similar to their classical counterparts, e.g.

$$\vec L=\vec r\times\vec p\;\leadsto\; \hat{\vec L}=\hat{\vec r}\times\hat{\vec p}.$$

After checking that the resulting operator is Hermitian, we can use it as the operator for our quantity.

But in some cases we have to do some special considerations. E.g. one term of the Laplace-Runge-Lenz vector in classical mechanics, whose operator cannot be straightforwardly composed from simpler operators:

$$\vec p\times\vec L\;\not\leadsto\;\hat{\vec p}\times\hat{\vec L},$$

because such an operator wouldn't be Hermitian. Fortunately, we can easily form an appropriate Hermitian version of it by adding a symmetrizing term, so that the expression is equivalent to the original classically, but results in an Hermitian operator:

$$\vec p\times\vec L\;\leadsto\;\frac12\left(\hat{\vec p}\times\hat{\vec L}-\hat{\vec L}\times\hat{\vec p}\right).$$

But is there always a unique choice of an Hermitian combination of operators which would reduce to a given classical expression of corresponding observables? If not, what are some examples where one classical expression could be represented by more than one different Hermitian operators? Do we have any criteria of choosing one over others without need to do experiment?

## marked as duplicate by ZeroTheHero, Community♦Dec 21 '18 at 22:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.