# Tensor operator analogy in classical physics

In quantum mechanics, tensor operators are defined through their commutation with the operators of spherical angular momentum components, \begin{aligned} \ [L_3,T(k,q)] &= \hbar q\ T(k,q), \\ [L_\pm,T(k,q)] &= \hbar \alpha^\pm(k,q)\ T(k,q\pm1). \end{aligned} This can be equivalently rewritten in Cartesian coordinates, e.g., a vector operator $(V_i)_{i=1}^3$ needs to satisfy $$[L_j, V_k] = i\hbar\ \epsilon_{jkl} V_l.$$ This is nicer in that it only uses self-adjoint observables. What I am wondering about is what this corresponds to in classical Poissonian mechanics, where an analogous condition would be $$\{L_j,V_k\} = \epsilon_{jkl} V_l \label{1}\tag{1}$$ for three phase-space functions $(V_1, V_2, V_3)$. Of course we all know that components of position, momentum, or angular momentum satisfy this, but I want to see a general solution.

What I have tried is rewrite (\ref{1}) as $$\{\epsilon_{jmn}x_mp_n, V_k\} = \epsilon_{jmn}\frac{\partial V_k}{\partial p_m}p_n - \epsilon_{jmn}x_m\frac{\partial V_k}{\partial x_n} = \epsilon_{jkl} V_l,\quad \forall j,$$ multiply both sides by the $j$-th component of a generic normal vector $\vec n$ and a generic displacement $\chi$ and interpret the left-hand side as the first-order term in a Taylor expansion, $$V_k(x_n - \chi \epsilon_{jmn}n_j x_m, p_m + \chi \epsilon_{jmn}n_j p_n) = V_k(x_n, p_m) + \chi n_j \epsilon_{jkl} V_l(x_n, p_m) + O(\chi^2),$$ where $V_k(x_n,p_m)$ is a shortcut for $V_k(x_1,x_2,x_3,p_1,p_2,p_3)$, or $V_k(\vec x, \vec p)$. This can be exponentiated to get $$V_k\big(\exp(-\chi n_j \epsilon_{j\bullet\bullet})_{mn} x_m,\,\exp(\chi n_j \epsilon_{j\bullet\bullet})_{mn} p_n\big) = \exp(\chi n_j \epsilon_{j\bullet\bullet})_{kl} V_l(x_n,p_m).$$ The exponentials now denote rotational matrices about $\vec n$ by the angle $\pm\chi$, $$\exp(\pm \chi n_j \epsilon_{j\bullet\bullet})_{mn} = R(\vec n,\pm \chi)_{mn},$$ so we obtain that $\vec V = (V_1, V_2, V_3)$ must satisfy $$\vec V(R(\vec n,-\chi)^T \vec x, R(\vec n,\chi)\vec p) = R(\vec n,\chi) \vec V(\vec x, \vec p).$$ We also use the orthogonality of $R$, $$R(\vec n,-\chi)^T = R(\vec n,-\chi)^{-1} = R(\vec n,\chi) =: R,$$ so $$\vec V(R \vec x, R \vec p) = R \vec V(\vec x, \vec p).$$

Denoting $\vec{\tilde{x}} = R \vec{x}$ and $\vec{\tilde{p}} = R \vec{p}$, we get $$\vec V(\vec{\tilde{x}}, \vec{\tilde{p}}) = R \vec V(R^{-1} \vec{\tilde{x}}, R^{-1} \vec{\tilde{p}}).$$ Now the arguments in the right-hand side are $\vec x$, $\vec p$ computed back from $\vec{\tilde{x}}$, $\vec{\tilde{p}}$, fed into the untransformed $\vec V$, and covariantly (NB: no difference between covariant and contravariant in $SO(3)$) transformed to the tilde basis. This is exactly the full transformation rule of a vector field, so the RHS equals $\vec{\tilde{V}}(\vec{\tilde{x}}, \vec{\tilde{p}})$, or $$\vec V(\vec{\tilde{x}}, \vec{\tilde{p}}) = \vec{\tilde{V}}(\vec{\tilde{x}}, \vec{\tilde{p}}). \label{2}\tag{2}$$ This would mean that the functions $(V_k)_{k=1}^3$ satisfy (\ref{1}) if and only if treating them as a vector field and transforming to a rotated basis would be the same as only plugging the transformed coordinates into the original (untransformed) functions, or that their functional form is invariant under vector field rotation. This seems to be a quite strong condition that not many things I used to call vector fields would satisfy. For example, take a constant vector field, say, $\vec V(\vec r, \vec p) = (0, 0, -g)^T$. In a rotated basis the transform rules dictate $\vec{\tilde{V}}(\vec{\tilde{r}}, \vec{\tilde{p}}) = R\cdot(0,0,-g)^T$, which in general are different three constants than $0$, $0$ and $-g$. So (\ref{2}) is broken, which is consistent with the fact that a Poisson bracket of the form (\ref{1}) with a constant $V_k$ is zero, rather than $\epsilon_{jkl}V_l$.

An example of $(V_k)_{k=1}$ that does conform to (\ref{1}) are the canonical projections \begin{aligned} V_1(x, y, z, p_x, p_y, p_z) &= x, \\ V_2(x, y, z, p_x, p_y, p_z) &= y, \\ V_3(x, y, z, p_x, p_y, p_z) &= z \\ \end{aligned} (as is well known), and indeed transforming the radius vector $\vec r = (V_1, V_2, V_3) = (x,y,z)$ gives $\vec{\tilde{r}} = (\tilde x, \tilde y, \tilde z)$, whose components are the first three elements of the tuple $(\tilde x, \tilde y, \tilde z, \tilde p_x, \tilde p_y, \tilde p_z)$, so \begin{aligned} \tilde V_1(\tilde x, \tilde y, \tilde z, \tilde p_x, \tilde p_y, \tilde p_z) &= \tilde x = V_1(\tilde x, \tilde y, \tilde z, \tilde p_x, \tilde p_y, \tilde p_z), \\ \tilde V_2(\tilde x, \tilde y, \tilde z, \tilde p_x, \tilde p_y, \tilde p_z) &= \tilde y = V_2(\tilde x, \tilde y, \tilde z, \tilde p_x, \tilde p_y, \tilde p_z), \\ \tilde V_3(\tilde x, \tilde y, \tilde z, \tilde p_x, \tilde p_y, \tilde p_z) &= \tilde z = V_3(\tilde x, \tilde y, \tilde z, \tilde p_x, \tilde p_y, \tilde p_z) \\ \end{aligned}

So the "classical vector observable" is something stricter than a "vector field" over phase space, requiring that the functions defining its components stay intact under rotations (or $GL$ transforms, as could be shown analogously). What are such special vector fields called? Or am I missing something in my derivation?

The requirement $\tilde{V}(x,p)=V(x,p)$ is simply that the vector field be rotationally symmetric. Rotational symmetry clearly holds for $\vec{x}=(x,y,z)$ (just picture is in your head), and you can convince yourself that it holds for angular momentum in phase space.
In Hamiltonian mechanics, the Poisson bracket makes the smooth functions $C^\infty(M)$ on phase space into a Lie algebra. The angular momenta then furnish a Lie algebra representation of the rotation group $\{L_i,L_j\}=\epsilon_{ijk}L_k$. You can check that the action of the $L_i$ on functions $f\in C^\infty(M)$ is just the infinitesimal version of the $O(3)$ representation $f\to f\circ R^{-1}$, where I've used $R(x,p)=(Rx,Rp)$ for short. This should, of course, not be confused with the $O(3)$ action on vector fields $V^i\mapsto R^i_j \cdot(V^j\circ R^{-1})$.
Lastly, note that 'proper' quantum vector fields carry a position index $V^j=V^j(x)$. They $\textbf{don't}$ satisfy $[J_i,V_j(x)]=i\epsilon_{ijk} V_k(x)$ unless $x=0$. The usual vector operators in quantum mechanics correspond not to vector fields, but to single vectors.
• How simple – I'm feeling very stupid right now :-) Would one still use some variation on the term "rotationally symmetric" if this property is held (as it is the case for $\vec r$ transformed as a contravariant or $\vec p$ as a covariant quantity) for a general $S \in GL(3)$? I can't think of any. Mar 1 '18 at 13:48
• Suppose we have some $V(x)$ such that $V(x)=A V(A^{-1}x)$ for any $A\in\text{GL}(3,\mathbb{R})$. Then, we have $V(\hat{z})=V(R(\theta)\hat{z})=R(\theta)V(\hat{z})$ where $R(\theta)$ is any rotation in the $x$-$y$ plane. This gives $V(\hat{z})=\lambda \hat{z}$ for some scalar $\lambda$. From this one obtains $V(x)=\lambda x$ in general. These fields are not just rotationally symmetric, they're proportional to $\vec{x}$.