# Angular momentum quantum mechanics operator

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. $${ {\bf \hat{L}} = -i \hbar \frac{1}{2} (\ \bf r \times \nabla - \nabla \times r )}?\tag{2}$$

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

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).

• 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. Commented May 31, 2018 at 20:26
• you're right, read too quickly, didn't see the half ^^. but it does stay the same anyway Commented May 31, 2018 at 20:30

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.

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.

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.