# What are spherical tensors?

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 $$R_{ij}$$?

• 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. Jun 29, 2023 at 6:05
• @DanielSank Please tell me if the question is clear. Jun 29, 2023 at 6:23

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.

• 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. Jun 29, 2023 at 6:54
• @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 Jun 29, 2023 at 7:00
• @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..) Jun 29, 2023 at 7:07
• 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. Jun 29, 2023 at 7:12
• 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. Jun 29, 2023 at 7:32

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 $$L^2(S^2)$$.