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In physical three-dimensional space, a rotation about an arbitrary axies $\hat{\textbf{n}}$ through an angle $\phi$ can be represented by $$R(\hat{\textbf{n}},\phi)=e^{-i(\textbf{J}\cdot\hat{\textbf{n}})\phi}$$ which is an element of $SO(3)$. In this relation, $\textbf{J}$ is dimensionless, each $J_i$ ($i=1,2,3$) is antisymmetric and satisfies the commutation relation $$[J_i,J_j]=i\epsilon_{ijk}J_k.$$ This is a group theoretical relation and as far as I understand, it has nothing to do with classical or quantum.

The Hermitian matrices $J_i$'s represented the angular momentum operators in quantum mechanics (apart from a factor of $\hbar$). But in classical mechanics, we don't have operators or matrices associated with angular momentum. In classical mechanics, we only have numbers $\textbf{r}\times\textbf{p}$ associated with the angular momentum of a particle.

Question Then what do the antisymmetric matrices $J_i$'s represent in classical mechanics? Do they have anything to do with classical angular momentum?

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As you refer in your question the matrices $\:J_{1},J_{2},J_{3}\:$ are hermitian not antisymmetric as you call them in your last sentence. You'll find a representation of them in the footnote (1) of my answer here :Deriving the unitary operator U(R) associated with a rotation R using Wigner's theorem: \begin{equation} \mathcal{S}_{1}\equiv i \begin{bmatrix} 0&\hphantom{\!\!\!-}0&\hphantom{\!\!\!-}0\\ 0&\hphantom{\!\!\!-}0&\!\!\!-1\\ 0&\hphantom{\!\!\!-}1&\hphantom{\!\!\!-}0 \end{bmatrix} ,\quad \mathcal{S}_{2}\equiv i \begin{bmatrix} \hphantom{-}0&\hphantom{\!\!\!-}0&\hphantom{\!\!\!-}1\\ \hphantom{-}0&\hphantom{\!\!\!-}0&\hphantom{\!\!\!-}0\\ -1&\hphantom{\!\!\!-}0&\hphantom{\!\!\!-}0 \end{bmatrix} ,\quad \mathcal{S}_{3}\equiv i \begin{bmatrix} 0&-1&\hphantom{-}0\\ 1&\hphantom{-}0&\hphantom{-}0\\ 0&\hphantom{-}0&\hphantom{-}0 \end{bmatrix} \tag{01} \end{equation} These matrices have nothing to do and have no relation with angular momentum or any other quantity in classical mechanics.

In quantum mechanics if you suppose that a point particle possesses internal degrees of freedom then a first simple assumption is to represent its state not by a scalar wave function $\:\psi(\mathbf{x})\:$
but by a 3-vector wave function $\:\boldsymbol{\Psi}(\mathbf{x})$.Further, we assume that, when the state is rotated by $\:R(\mathbf{n},\theta)\:$, not only does $\:\mathbf{x}\:$ change into $\:\mathbf{x}'=R\mathbf{x}\:$ but also $\:\boldsymbol{\Psi}\:$ as a 3-vector changes into $\:\boldsymbol{\Psi}'=R\boldsymbol{\Psi}\:$. In this case we talk about a vector particle and the matrices represent the spin angular momentum $\:s=1\:$ (*) .


(*) "Quantum Mechanics",Leonard I.Schiff, 3rd edition 1968,McGraw-Hill : Spin of a vector particle (Section 27 ROTATION, ANGULAR MOMENTUM, AND UNITARY GROUPS), p.197-199.

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  • $\begingroup$ Note that since $R$ is orthogonal, $J_i$'s are antisymmetric. See ZeroTheHero's answer, for example. But yes. $J_i$'s are hermitian in quantum mechanics. @Frobenius $\endgroup$ – SRS Aug 8 '18 at 21:19
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The representation of the orbital angular momentum vector ${\bf L}$ in terms of derivatives does have a role in classical physics. The appearance of derivatives means that the quantity is only meaningful in a continuum field theory, so that there is a field defined at every spatial point, providing something that can be differentiated.

In such a theory field theory, the parameterization of field configurations in terms of eigenstates of ${\bf L}^{2}$ and $L_{z}$ is the same as the parameterization in terms of multipoles. The study of these multipoles is a major topic in electrodynamics, both in statics and in radiation theory. In the electrostatic case dipolar field has $\ell=1$; a quadrupole has $\ell=2$, and so on. With multipole radiation, there is a relationship, just as in quantum mechanics, between the $(\ell,m)$ eigenvalues describing the field and the orbital angular momentum that the field carries.

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  • $\begingroup$ If I understand it correctly then you're saying that in a classical field theory (e.g. classical electrodynamics) these differential operators measure the $l$ value of each multipole in the multipole decomposition of the vector potential? Am I correct? Am I close? @Buzz $\endgroup$ – SRS Aug 8 '18 at 21:44
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The antisymmetric matrices $J_i$ are actually (as in quantum mechanics) angular momentum matrices.

The neat way to see this is to consider $$ R\cdot R^t=1 $$ and then take the differential of this: $$ dR\cdot R^{t}+ R\cdot dR^t=0 \tag{1} $$ Call $dR\cdot R^T=A$, and note that (1) can be rewritten as $$ A+A^T=0 $$ which means that $A$ is antisymmetric. The most general antisymmetric matrix can be written as the linear combination $$ A= \vec \omega \cdot \vec J $$ where $\vec\omega=(\omega_1,\omega_2,\omega_3)$ and $J_i$ is antisymmetric. If you take a rotation about $\hat z$, then $\vec\omega=\omega\hat z$ and then you can easily verify that $$ R(\omega_3)=\left(\begin{array}{ccc} \cos\omega_3&\sin\omega_3&0 \\ -\sin\omega_3&\cos\omega_3&0\\ 0&0&1\end{array}\right) \tag{2} $$ and that $$ A=dR(\omega)\cdot R^{-1}= \omega_3 \left(\begin{array}{ccc} 0&1&0\\ -1&0&0\\ 0&0&0\end{array}\right) $$ is indeed antisymmetric, with $J_z$ the angular momentum generator about $\hat z$. Because $J_z^2=-I$, it’s not hard to verify that $e^{\omega_3 J_z}$ gives back (2), confirming that $J_z$ is the generator of rotation.

Doing the same with $J_y$ and $J_x$, you can verify that these matrices have the commutation relations (without the “i” of course) of the angular momenta operators.

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  • $\begingroup$ Whatever you wrote is known to me and it didn't answer my question. My question was what do $J_i$'s represent in classical mechanics. A group theoretical relation has nothing to do with classical or quantum, and must therefore also have an interpretation in classical mechanics. $\endgroup$ – SRS Apr 8 '18 at 20:59
  • $\begingroup$ @SRS I don't understand your reply. They are generators of rotations. This has nothing to do with classical or quantum as you point out. Maybe we're talking at cross-purposes... $\endgroup$ – ZeroTheHero Apr 8 '18 at 21:12
  • $\begingroup$ You said $J_i$'s represent angular momentum matrices. But angular momentum are matrices/operators only in quantum mechanics. Does it mean that the matrices $J_i$'s have no interpretation in classical mechanics? $\endgroup$ – SRS Apr 8 '18 at 21:19
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    $\begingroup$ @SRS Any transformation, such as a rotation, is defined by generators (obviously). The generators of rotation are traditionally known as angular momentum operators, in classical and in quantum mechanics. If you look at rotational dynamics, the transformation from body to lab frame is done by the "operator" $\vec \omega \times$ in the sense that $dA/dt$ in the lab frame is $dA/dt+\vec\omega\times \vec A$ in the body frame. The term $\vec \omega\times $ has the antisymmetric form of $\omega\cdot \vec J$ so that, for instance, $(\vec \omega\times \vec A)_z=\vec\omega\cdot J_z\cdot \vec A$. $\endgroup$ – ZeroTheHero Apr 8 '18 at 21:32
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    $\begingroup$ @SRS Sorry if I can't be more helpful. There are lots of operators in classical mechanics (mostly in Hamiltonian mechanics), to generate canonical transformations for instance. $\endgroup$ – ZeroTheHero Apr 8 '18 at 21:34

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