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In classical mechanics the angular momentum is given as $\bf L = r \times p$ and when going to quantum mechanics you replace $\bf r$ and $\bf p$ by their respective quantum operators, namely $\bf \hat{r} = r$ and ${\bf \hat{p}} = -i \hbar \bf{\nabla}$, that is $${\bf \hat{L}}= -i \hbar {\bf r} \times \nabla.\tag{1}$$ Shouldn't one use a symmetrised form for $\bf\hat{L}$, i.e. $${ {\bf \hat{L}} = -i \hbar \frac{1}{2} (\ \bf r \times \nabla - \nabla \times r )}?\tag{2}$$

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    $\begingroup$ It is exactly the same thing, thanks to the $\frac12$ factor of your definition. $\endgroup$ May 31, 2018 at 20:29

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No, because by definition of the vector product, it is antisymmetric and thus your definition of L would be 2 times the "normal" L. (check it by computing each component!) L is defined by the correspondance principle (+ its commutation relations).

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  • $\begingroup$ They've included a half in their post, and their operator is just the usual one. What they're probably worried about is that if they hadn't explicitly anti-symmetrised it it may not be Hermitian, but in fact the cross product only mixes components that commute so the naive approach is fine. $\endgroup$
    – jacob1729
    May 31, 2018 at 20:26
  • $\begingroup$ you're right, read too quickly, didn't see the half ^^. but it does stay the same anyway $\endgroup$ May 31, 2018 at 20:30
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To give an explicit answer with a calculation, consider the product: $$\vec{r}\times\nabla=\hat{i}(y\partial_z-z\partial_y)+\hat{j}(z\partial_x-x\partial_z)+\hat{k}(x\partial_y-y\partial_x)$$

Notice that since $[r_i,\partial_j]=0$ for $i\neq j$, you can rewrite this: $$\vec{r}\times\nabla=\hat{i}(\partial_zy-\partial_yz)+\hat{j}(\partial_xz-\partial_zx)+\hat{k}(\partial_yx-\partial_xy) = -\nabla\times\vec{r}$$

So that finally: $$-i\hbar\vec{r}\times\nabla = -i\hbar\frac{1}{2}(\vec{r}\times\nabla-\nabla\times\vec{r})$$

So the two operators are actually the same.

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No, because when expanding $L_k$ you see that only components $X_k$ and $P_h$ with $h \neq k$ enter the formula, and they commute. So you can safely use the classical definition also in QM and the symmetrised formula would produce the same result.

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The canonical commutation relations (CCR) are $$[X_i, P_j] = i\hbar \delta_{ij}$$ so for example $X$ commutes with $P_Y$ and $P_Z$.

When expanding the cross product, its components are $$ L_k = \epsilon_{ijk} X_i P_j$$ but notice the Levi-Civita symbol vanishes when two indices are equal. Hence you never have terms $X_i P_i$, so you are never actually multiplying observables that do not commute.

Hence no need to symmetrize.

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