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.

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 $T_m^{2}$?

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!

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: $T_{2k+1}^{2k+1} = (S_x+i S_y)^{2k+1}$?


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 $L_-$.

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  • $\begingroup$ Thanks for the response. I'm still a bit confused on the proportionality that you've written, $T_l^l \backsim (x+iy)^l$. Am I to take it then that for any Cartesian tensor defined by some $\vec{S}$ that I can just replace those position coordinates with the coordinates of my vector? So, in this case I'd have $T_2^2 \backsim (S_x+i S_y)^2$? If so, how is that equivalence between the position coordinates and the vector components popping out? Thanks $\endgroup$ – dsm Feb 13 '19 at 1:59
  • $\begingroup$ You don't want to use $\hat S_x$ etc because these operators do not commute. The $\hat x$ and $\hat y$ operators commute so you don't have to worry about ordering when taking powers. The key is writing the sph. harmonics in terms of Cartesian $\hat x$ etc. As the spherical harmonics transform correctly under rotation the corresponding operators will be the correct tensors. The trickiness is in the normalization: basically the spherical harmonics are normalized but the tensor operators "need not be" so that if you start with $k(x+iy)^\ell$ you get a valid tensor $T^\ell_\ell$ for any $k$. $\endgroup$ – ZeroTheHero Feb 13 '19 at 2:10
  • $\begingroup$ I suppose I understand your question a little bit different now. Basically $\hat S_+^\ell$ will be proportional to $T^\ell_\ell$ and you can get the other components through commutation with $L_-$. That will give you the tensor in terms of $\hat S_i$. $\endgroup$ – ZeroTheHero Feb 13 '19 at 2:13
  • $\begingroup$ When we have an even rank tensor, shouldn't that substitution still work for the highest subscripted element? Just explicitly not combining the terms, e.g. $T_2^2 = (S_x+i S_y)^2 = S_x^2-S_y^2+i(S_xS_y+S_yS_x)$, and then just lower it? Which will be annoying, but no problem. Or are you saying the failure to commute completely takes that equivalence out of the ball game, even for that highest element for an even rank? $\endgroup$ – dsm Feb 13 '19 at 3:01
  • $\begingroup$ For the highest component then proper symmetrization with every possible ordering should work but there might be some combinatorics subtleties. $\endgroup$ – ZeroTheHero Feb 13 '19 at 3:06

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