How does spherical tensors act on a ket as an operator?

Question

I have this question when studying the 2021 edition of Sakurai's Advanced Quantum Machanics as a beginner. I read the answers in PSE:334091 and PSE:770079, but still need help on it. I wish for an answer that employs the methods in Sakurai's book so is presented in a self-contained manner. Thanks in advance for the patience in reading my question.

On P$233$, the spherical harmonics are $Y_l^m(\hat{n})$, where $\hat{n}$ is the directional vector. He replaces $\hat{n}$ by some vector $\mathbf{V}$, the result is a spherical tensor \begin{equation}\tag{3.458} T_q^{(k)} = Y_{l=k}^{m=q}(\mathbf{V})\,. \end{equation} In $(3.459)$, the explicit form of the tensors $T_{m}^{(l)}$ are given by replacing the three vector components $(\hat{n})_{x,y,z}$ with $V_{x,y,z}$ in the formula of spherical harmonics $Y^{m}_{l}(\hat{n})$.

( From this I reckon that the spherical tensor operator $T_q^{(k)}$ acts on a vector in 3-dimensional Euclidean space $\mathbf{V}$ to generate a list of numbers $T_q^{(k)}(\mathbf{V})$. But an operator in this book will act on a ket to produce another ket, as defined in (1.17) or (1.32), which is not quite consistent with this interpretation.)

Then he derives the transformation of spherical harmonics under rotation: \begin{equation}\tag{3.463} Y_{l}^m(\hat{n}') = \sum_{m'} Y_{l}^{m'}(\hat{n}) \mathscr{D}_{m'm}^{(l)}(R^{-1})\,, \end{equation} where \begin{equation}\tag{1} |\hat{n}'\rangle = \mathscr{D}(R) |\hat{n}\rangle\,,\quad \mathscr{D}_{m'm}^{(l)}(R^{-1}) = \langle l,m'|\mathscr{D}(R^{-1})|l,m \rangle\,. \end{equation} After which he writes that "If there is an operator that acts like $Y_{l}^{m}(\mathbf{V})$, it is then reasonable to expect " \begin{equation}\tag{3.464} \mathscr{D}^\dagger (R) Y_{l}^m(\mathbf{V})\mathscr{D}(R) = \sum_{m'} Y_{l}^{m'} (\mathbf{V})\mathscr{D}_{m'm}^{(l)}(R^{-1})\,. \end{equation} Then equation $(3.464)$ becomes the defining property of spherical tensors with rank $l$: \begin{equation}\tag{$3.464'$} \mathscr{D}^\dagger (R) T^{(l)}_m \mathscr{D}(R) = \sum_{m'} T^{(l)}_{m'} \mathscr{D}_{m'm}^{(l)}(R^{-1})\,. \end{equation} He gives an example of the $m = +2$ component of the spherical tensor of rank 2 as $$(U_x+i U_y)(V_x+iV_y)\,.$$

( From here, a spherical tensor of rank $l$ has $2l+1$ components: $T^{(l)}_m, m = -l,-l+1,\dots,l$, and each component is a spherical tensor operator that transforms under rotation as $(3.464')$. )

My question boils down to understanding this example:

1. Is $(U_x+i U_y)(V_x+iV_y)$ a tensor operator and how does it acts on an arbitary ket $|\alpha\rangle$ or the eigenket of $\{J^2,J_z\}$ $|j,m\rangle$?

2. If we consider $(U_x+i U_y)(V_x+iV_y)$ in $(3.464')$, some problems emerge. First, the other components of rank 2 tensor is not specified. Next, the left hand side of the equation is not defined, or it can only be understood as another expression of the right hand side.

In summary, I think there are some abuse of notation and terminology on spherical tensors in this book. In $(3.458)$, the spherical tensor $T_q^{(k)}$ seems to be a number that depends on $q,k$ and a vector $V$. But in $(3.464')$ the spherical tensor $T_m^{(l)}$ should be an operator that acts on kets.
If so, then how to make the notations consistent? If the tensor operators are written as $\hat{T}_q^{(k)}$ and $\hat{Y}_k^q$, then how to obtain the number $\hat{Y}_k^q(\mathbf{V})$ from this operator?
If $\hat{Y}_k^q(\mathbf{V}) = \langle\mathbf{V}|\hat{Y}_q^{(k)} |\mathbf{V}\rangle$, then $(3.464)$ follows naturally from $(3.463)$, and should be written as \begin{equation}\tag{2} \mathscr{D}^\dagger (R) \hat{Y}_{l}^m\mathscr{D}(R) = \sum_{m'} \hat{Y}_{l}^{m'} \mathscr{D}_{m'm}^{(l)}(R^{-1})\,. \end{equation} Is there a definition of tensor operators like this?

Answer 1

1. Yes it is a tensor operator and it acts in the usual way: $$ (U_x+iU_y)(V_x+iV_y)\vert n\ell m\rangle $$ is computed by distributing the operators. As your operator is an $\ell=2$ tensor, call it $T^{-2}_2$ and the results of this action will be $$ \sum_{n'\ell’}\vert n’\ell’ m-2\rangle\langle n’\ell’m-2 \vert T^{-2}_2 \vert n\ell m\rangle $$ where $\ell’$ is in the decomposition of $\ell\otimes 2$ and the sum over $n’$ is not restricted by the possible $\ell’$’s. For instance, if you work in an infinite spherical well with $(x+iy)(p_x+ip_y)$, the sum may well extend to all $n'$’s.

2. I don’t understand why you claim there’s a problem with the component. You have a product of two tensors with $m=-1$ so the only possible outcome is an $m=-2$ tensor, and the only possibility here is $\ell=2$ if we assume $U_x+iU_y$ and likewise for $V$ are vectors.

3. I don’t understand why you claim the left side is not defined either. If you rotate the $\ell=2,m=-2$ component of a tensor, you get a sum of $\ell=2$ tensor $$ R^{-1}T^{-2}_2 R =\sum_{m=-2}^2 T^m_{2} D^{2}_{m,-2}(R) $$ where I don’t want to use $D(R)$ for the rotation operator because I use $D^{\ell}_{mm’}(R)$ for the rotation matrix. Now you might have to compute the other components of $T^{m}_{2}$ through commutation, or in your case you can use CGs to combine the components of the vectors $T_{U;1}^{m’}T_{V;1}^{m’’}$, where $T^{-1}_{W;1}= (W_x+iW_y)$, but that’s technical not conceptual.

4. A tensor operator is NOT a number: it is an operator and it need not in general be a combination of vector operators. Now the tensor might not be irreducible, but you then reduce it so the parts act irreducible. Its matrix elements are numbers which are expressible in terms of a reduced matrix element common to all components, multiplied by a CG, and divided by a dimensionality factor. I completely fail to see how this is not clear since the tensors constructed from operators.

Various authors have various notations for tensor operators, but it is a stretch to suggest that there’s an abuse of notation. My go-to source on this material is the book by Gordon Baym “Lecture notes on quantum mechanics”.