Representing a reducible Cartesian tensor as a spherical tensor

Question

I'm quite confused by this transformation, and am trying to gain fluency in moving back and forth between these objects. I understand that a second order dyadic Cartesian tensor can be represented as the sum of rank 0, 1, and 2 irreducible spherical tensors. I also know that the crucial point of extending this process to spherical tensors of higher rank is the fact that $r^lY_l^m$ is a homogeneous polynomial in $(x,y,z)$ of order $l$. My understanding of why this is the crux, while shaky, is because this polynomial is in fact a spherical tensor of rank $l$, proven by showing that it transforms properly under rotations. That is my background, here is where I'm trying to apply it.

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Consider the traceless symmetric Cartesian tensor

$$T_{ij} = \frac{S_iS_j+S_jS_i}{2}-\frac{\delta_{ij}}{3}\vec{S}\cdot\vec{S}$$

with the components of $\vec{S}$ satisfying the angular momentum commutation relations

$$[S_i,S_j] = \sum_{k=1}^{3}i\hbar\epsilon_{ijk}S_k.$$

How am I to construct the five components of the related spherical tensor $T_m^{2}$?

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Since the components of $S$ do not commute, I think it is appropriate to first determine $T_2^2$, and then use the commutation

$$[J_-,T_q^{(k)}] = \sqrt{(k-q+1)(k+q)}T_{q-1}^k$$

to determine the others. Would this be a valid approach to constructing the spherical tensor? If so, I'm a bit confused about determining what $T_2^2$ ought to be. I know that I'll be using the fact that $r^2Y_2^2$ is a homogeneous polynomial, but that connection is the main source of my confusion.

Thanks for any insight!

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EDIT - Add-on question

It has been clarified that $T_q^k = r^kY_k^q \backsim (x+iy)^k$ can be extended to generating spherical tensors from a vector operator $\vec{A}$, satisfying $[A_i,A_j]=0$, with $T_k^k = \alpha(A_x+i A_y)^k$. $\alpha$ is the normalization constant. From here, the lower elements can be generated using the stated lowering commutation identity. The add-on question here is the following:

For a vector operator $\vec{S}$ that satisfies the stated angular momentum commutation relations, does the above generating relationship still hold for the highest indexed element if the rank of the spherical tensor is even? In other words, is $T_{2k}^{2k} = (S_x+i S_y)^{2k}$ still the proper highest ordered element of the spherical tensor? How about if the rank is odd: $T_{2k+1}^{2k+1} = (S_x+i S_y)^{2k+1}$?

Answer 1

Yes your approach will work. Indeed since $T^{1}_1\sim (\hat x+i\hat y)$ you should be able to check that $T^{2}_2\sim (\hat x+i\hat y)^2$ and take it from there using the commutation with $J_-$, which is proportional to the Cartesian expression for $r^2Y^2_2(\theta,\phi)$. More generally, $\hat T^\ell_\ell\sim (\hat x+i\hat y)^\ell$. Of course a table of spherical harmonics, including expressions in Cartesian coordinates, is especially useful. The actual proportionality coefficient between $r^2Y^2_2(\theta,\phi)$ and $T^2_2$ actually depends on the normalization of your tensor operators.

If you want the tensors in terms of $\hat S_i$ etc then you can verify that $\hat S_+^\ell$ satisfies $$ [L_+,\hat S_+^\ell]=0\, ,\qquad [L_0,\hat S_+^\ell]=\ell \hat S_+^\ell $$ so this must be proportional to the highest component of the $\ell$ tensor $T^\ell_\ell$. You can get the remaining components in terms of $\hat S_i$ by lowering with the commutator of $L_-$.