What do the antisymmetric matrices $J_i$ represent in classical mechanics?

Question

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?

Answer 1

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 $\:s=1\:$ ^(*) .

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^{(*) "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.}

Answer 2

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.

Answer 3

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.