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Following Sakurai, I know how the Cartesian components of a tensor transform under rotation, in classical physics and also in quantum physics. For example, the Cartesian components of a vector change as $$V^{\prime}_i= R_{ij} V_j$$ and the Cartesian components of the corresponding vector operator transform according to $$\hat{D}^\dagger(R)\hat{V}_i\hat{D}(R)=R_{ij}\hat{V}_j.$$ This can be generalized to Cartesian components of tensors of higher ranks. In particular, the Cartesian components of a tensor of rank-$k$ transform under the rotation as$$T_{i_1i_2...i_k}^\prime=R_{i_1j_1}R_{i_2j_2}...R_{i_kj_k}T_{j_1j_2...j_k}$$ and the corresponding Cartesian components of the tensor operator transform under the rotation according to $$D^\dagger(R)T_{i_1i_2...i_k} D(R)=R_{i_1j_1}R_{i_2j_2}...R_{i_kj_k}T_{j_1j_2...j_k}.$$

Sakurai tells us how the spherical tensor operators $\hat{T}^{(k)}_q$ in quantum mechanics. A spherical tensor of rank-$k$ is a set of $(2k+1)$ operators whose components transform under rotation as (Sakurai) $$\hat{D}^\dagger(R) \hat{T}^{(k)}_q\hat{D}(R)=\sum\limits_{q'=-k}^{k}D_{qq'}^{(k)*}\hat{T}_{q'}^{(k)}$$ or $$\hat{D}(R) \hat{T}^{(k)}_q\hat{D}^\dagger(R)=\sum\limits_{q'=-k}^{k}D_{q'q}^{(k)}\hat{T}_{q'}^{(k)}.$$ where these $\hat{D}(R)$'s are unitary representations of $R\in SO(3)$ in the appropriate ket space of the quantum system.

But what is a spherical tensor $T^{(k)}_q$ (not a spherical tensor operator, $\hat{T}^{(k)}_q$) in classical physics? For irreducible spherical tensors of rank-$k$, I would expect that under rotation the following should be the case: $$T^{(k)}_q\stackrel{R}{\rightarrow} \sum_{q'=-k}^{k}C^{(k)}_{qq'}T^{(k)}_{q'}.$$ But if this is the case, how are the coefficients $C^{(k)}_{qq'}$ related to $R_{ij}$?

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  • $\begingroup$ Perhaps you could state what you know them to be in quantum. Probably if you do that, you'll understand what they are in classical. $\endgroup$
    – DanielSank
    Jun 29, 2023 at 6:05
  • $\begingroup$ @DanielSank Please tell me if the question is clear. $\endgroup$ Jun 29, 2023 at 6:23

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According to the Peter-Weyl theorem, if $G$ is a compact topological group, then every strongly continuous unitary representation $G\ni g \mapsto U_g$ in a fixed Hilbert space ${\cal H}$ is the orthogonal direct sum of a suitable family of finite-dimensional irreducible unitary representation $G \ni g \mapsto U^{(j)}_g$.

The representations of this family are pairwise unitarily inequivalent.

In other words, ${\cal H}$ decomposes into an orthogonal sum made of finite dimensional subspaces ${\cal H}_k \subset {\cal H}$: $${\cal H} = \oplus_{k\in H} {\cal H}_k$$ (infinite if ${\cal H}$ is infinite dimensionall) with $U_g({\cal H}_k) \subset {\cal H}_k$ and $U_g|_{{\cal H}_k}= U_g^{(j_k)}$ (up to unitary equivalence). Notice that a representation $U^{(j)}$ may appear several times as we can have $j_k=j_{k'}$. Evidently if ${\cal H}_{k}$ and ${\cal H}_{k'}$ are unitarily isomorphic in that case and the abstract space where $U^{(j_k)}\equiv U^{(j_k)}$ acts is indicated by ${\cal H}^{(j)}$.

The finite dimensional representations $U^{(j)}$ (defined up to unitary equivalence) are the building blocks to construct all possible continuous unitary representations of $G$.

Now let us focus on a given $U^{(j)}$ and let us fix an orthonormal basis in the space ${\cal H}^{(j)} \equiv \mathbb{C}^{d_j}$. As every $U_g^{(j)}$ is unitary, it can be viewed as a unitary marix of coefficients $D^{(j)}(g)_{nm}$. These matrices can be viewed as matrices connecting different orthonormal frames in ${\cal H}^{(j)}$, since they are unitary matrices.

In the very special case of $G=SU(2)$, all elementary irreducible (finite dimensional) matrices $D^{(j)}(g)_{mn}$ are classified. As a matter of fact $j=0,1/2,1, 3/2, 2, \ldots$ and $d_j=2j+1$.

A vector $T^{(j)} \in {\cal H}^{(j)}\equiv \mathbb{C}^{2j+1}$ has components $T^{(j)}_{m}$, $m=1,2, \ldots, 2j+1$, such that, changing orthonormal basis connected by the matrices $D^{(j)}(g)$ transforms as $$T'^j_{m} = \sum_{n=1}^{2j+1} D^{(j)}_{mn}(g) T^j_n\:. \tag{1}$$

A spherical tensor of order $j$ is noting but a vector in ${\cal H}^{(j)}$ when the reference group is $G:=SU(2)$.

In view of this definition, its components transforms as in (1) when we change orthogonal basis and these basis are connected by the corresponding representation of $SU(2)$. We can also suppose to deal with a fixed basis. In that case (1) describes the active action of $SU(2)$ on these vectors.

An important point is that, if $j=0,1,2,...$ then $U^{(j)}$ is also a representation of $SO(3)$, since the covering map $\pi : SU(2) \ni g \mapsto R_g \in SO(3)$ has the nice property that, for $R\in SO(3)$ and $j\in \mathbb{N}$ $$SO(3) \ni R \mapsto V^{(j)}_R := U^{(j)}_{\pi^{-1}(R)}$$ is well defined and defines a group representation.

For $j=1$, ${\cal H}^{(1)} \equiv {\mathbb C}^3$ and the matrices $D^{1}$ can be chosen (working in a suitable orthonormal basis of $ {\mathbb C}^3$) as the well-known matrices which represent the rotations of spherical harmonics. In this picture ${\cal H}^{(1)}$ is a subspace of $L^2(S^2)$.

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Suppose you have a spherical tensor $T^{(k)}_q$. This object rotates under rotation $R$ as

$$ T'^{(k)}_q = \sum_{q'=-k}^kD(R)_{q, q'}^k T^{(k)}_{q'} $$

Where $D(R)^k_{q, q'}$ is the Wigner D-matrix for rotation $R$. The link provides explicit (albeit complicated) formulas relating $D(R)^k_{q, q'}$ to $R$. (There's a chance I've gotten $q$ and $q'$ swapped around in the subscripts).

A spherical tensor of rank $k$ is a group of $2k+1$ numbers which transform in the same way under rotations as the $2k+1$ spherical harmonic functions of rank $k$ transform amongst each other under rotation.

I'd suggest replacing the $\hat{D}(R)$ you are using with $U(R)$ or $\hat{U}(R)$. $R$ is the rotation matrix, $D(R)$ is the Wigner D-matrix corresponding to that rotation, and $U(R)$ is the unitary matrix which is a representation of that matrix on the quantum Hilbert space. These are three highly related, but distinct objects.

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  • $\begingroup$ Here is my problem. Can't we talk about spherical tensors without talking about quantum mechanics? If the definition is in terms of $D(R)$ (or $U(R)$, in your notation), suggests to me that we are using the knowledge of quantum mechanics to define a spherical tensor. This is weird! Certainly, for defining how Cartesian components of a tensor transform, we didn't need that. $\endgroup$ Jun 29, 2023 at 6:54
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    $\begingroup$ @Solidification No, $D(R)^k_{q, q'}$ does not require quantum mechanics to define. It does not require $U(R)$. $D(R)^k_{q, q'}$ can be defined in terms of transformation properties of spherical harmonics which are the family of solutions on the sphere of a certain differential equation which can arise in contexts unrelated to quantum mechanics $\endgroup$
    – Jagerber48
    Jun 29, 2023 at 7:00
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    $\begingroup$ @Solidification so it's a well known fact or definition (en.wikipedia.org/wiki/…) that $Y_l^{'m} = Y_l^{m'} D^l_{m', m}$ But it does remain to be seen if we can use this relation to "straight-forwardly" derive an explicit formula for $D^l_{m', m}$. I imagine it should be possible, but I don't know how to do it and I don't have a reference off-hand. Or at least not one that moves into group theory, hilbert spaces, and representation theory (which I think is to be avoided per the spirit of the question..) $\endgroup$
    – Jagerber48
    Jun 29, 2023 at 7:07
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    $\begingroup$ Using the formula I've given and orthogonal you can see that $D$ is directly related to the overlap integral between $Y_l^m$ and a rotated version of $Y_l^{m'}$. See e.g. physics.stackexchange.com/questions/383594/…. If this integral can be used to calculate the explicit formula on the linked Wikipedia page, I'm not sure. $\endgroup$
    – Jagerber48
    Jun 29, 2023 at 7:12
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    $\begingroup$ But I guess I'll emphasize that the definition of $D^k_{q, q'}$ as the matrix elements for rotations of spherical harmonics need not have ANYTHING to do with quantum mechanics. Also, once we have spherical harmonics, we could calculate the elements of $D$ numerically if needed. The fact that analytic methods/tricks used to give an analytic expression for $D$ us notations and intuitions from quantum mechanics doesn't mean we NEED quantum mechanics to understand $D^k_{q,q'}$ or spherical tesnors. $\endgroup$
    – Jagerber48
    Jun 29, 2023 at 7:32

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