In the section on spherical tensors in Sakurai, he introduces the idea of going from Cartesian tensors to irreducible spherical tensors. He states the following:

A spherical harmonic can be written as $Y_{l}^{m}(\hat{n})$ where the orientation of $\hat{n}$ is characterized by $\theta$ and $\phi$. We now replace $\hat{n}$ by some vector $V$. The result is that we have a spherical tensor of rank $k$ (in place of $l$) with magnetic quantum number $q$ (in place of $m$), namely $$T_{q}^{(k)} = Y_{l=k}^{m=q}(\mathbf{V}).$$ To see the transformation of spherical tensors lets see how $Y_{l}^{m}$ transforms under rotations. We then get $$Y_{l}^{m}(\hat{n}') = \sum_{m'}Y_{l}^{m'}(\hat{n})\mathcal{D}_{m'm}^{(l)}(R^{-1}).$$

He then states the following which he doesn't motivate other than stating that he uses the unitarity of the rotation operator to rewrite $\mathcal{D}_{m'm}^{(l)}(R^{-1})$:

If there is an operator that acts like $Y_{l}^{m}(\mathbf{V})$, it is reasonable to expect $$\mathcal{D^{\dagger}}(R)Y_{l}^{m}(\mathbf{V})\mathcal{D}(R) = \sum_{m'}Y_{l}^{m'}(\mathbf{V})\mathcal{D}^{(l)^{*}}_{mm'}(R)"$$


  • Can anyone see how exactly he derives this final equation comes from?
  • Also is $Y_{l}^{m}$ defined as irreducible since we can write it's rotation as a linear combination of all angular momentum eigenstates $|l,m' \rangle$: $\mathcal{D}(R)Y_{l}^{m} = \sum_{m''}\sum_{m'}|l,m''\rangle \langle l, m''| \mathcal{D}(R)|l,m' \rangle \langle l,m'|l,m \rangle = \sum_{m''}|l, m'' \rangle \mathcal{D^{(l)}_{m,m''}}(R)$?

2 Answers 2


The formula handling has been well addressed by ZeroTheHero, but I think there's a bit more to be said about your second question,

  • Also is $Y_{l}^{m}$ defined as irreducible since we can write it's rotation as a linear combination of all angular momentum eigenstates $|l,m' \rangle$: $\mathcal{D}(R)Y_{l}^{m} = [\ldots]= \sum_{m''}|l, m'' \rangle \mathcal{D^{(l)}_{m,m''}}(R)$?

This isn't quite correct. We don't just call the $Y_l^m$ irreducible - that's a sloppy use of language. The set that is really irreducible is the set of linear combinations of the $Y_l^m$ where $l$ is fixed: $$ V_l=\operatorname{span}\{Y_l^m : -l\leq m \leq l\} $$ We call this set irreducible for two reasons:

  • If you take a spherical harmonic $Y_l^m$ and you rotate it with an arbitrary $R\in\mathrm{SO}(3)$, then the rotated $Y_l^m$ will be some linear combination of the $Y_l^{m'}$ where $l$ does not get changed by the rotation. The same happens with any arbitrary linear combination $f_l = \sum_m a_m Y_l^m$.

  • Moreover, the sets $V_l$ are the smallest possible sets with this property: if I take any strict (nonzero) subspace $W<V_l$, then there will always be a member $w\in W$ and a rotation $R\in\mathrm{SO}(3)$ such that the action of $R$ takes $w$ out of $W$ and into $V_l\setminus W$.

In more formal language, what this means is that $V_l$ carries an irreducible representation of the rotation group: it's the representation itself that's irreducible, not the set.

  • $\begingroup$ Yes this is a very good summary. I had written a long and complicated spiel on this and I wisely chose not to include it as your explanation is clearer and shorter than mine. $\endgroup$ May 22, 2017 at 13:13
  • $\begingroup$ Still no substitute for the actual maths, though. $\endgroup$ May 22, 2017 at 15:42

To understand a tensor operator one first needs to be reminded that it's an operator. Whereas an (angular momentum) state $\vert \ell m\rangle$ transform as a column vector under rotation: \begin{align} \vert \ell m\rangle \to R\vert \ell m\rangle &= \sum_{m'} \vert\ell m'\rangle \langle \ell m'\vert R\vert \ell m\rangle \\ &=\vert\ell m'\rangle\, D^\ell_{m'm}(R) \tag{1} \end{align} an operator is a matrix so heuristically one needs one transformation $R$ for the rows and another $R^{-1}$ for the columns so that the component $m$ of the (angular momentum, or spherical) tensor operator $\ell$ transforms like \begin{align} T^\ell_m \to RT^\ell_m R^{-1} &= \sum_{m'} T^{\ell}_{m'}\, D^\ell_{m'm}(R)\, . \tag{2} \end{align} Basically, Eq.(2) is taken to be the definition of a tensor operator. It is designed to that the component $m$ of the tensor $\ell$ transform under rotation exactly as the state $\vert \ell m$ does under rotation.

Alternatively, for infinitesimal transformation we have \begin{align} \hat L_\pm \vert \ell m\rangle = \sqrt{(\ell\mp m)(\ell \pm m+1)} \vert \ell m\pm 1\rangle \quad &\rightarrow\quad [\hat L_\pm, T^\ell_m] = \sqrt{(\ell\mp m)(\ell \pm m+1)} T^{\ell}_{m\pm 1}\, ,\\ \hat L_z \vert \ell m\rangle = m \vert \ell m \rangle \quad &\rightarrow\quad [\hat L_z, T^\ell_m] = m T^{\ell}_{m }\, ,\\ \end{align} Another way to think about this is to realize that a tensor operator acting on states $\vert s \mu\rangle$ will transform this state to another state $\vert s \mu' \rangle$ and so can be written as the linear combination $$ T^{\ell}_{m}=\sqrt{\frac{2L+1}{2s+1}}\sum_{\mu\mu'}C^{s\mu'}_{S\mu;\ell m} \vert s\mu'\rangle \langle s\mu\vert\, , $$ where $C^{s\mu'}_{S\mu;\ell m}$ is a Clebsch-Gordan coefficient. This way of writing the tensor components makes it clear that, if $\vert \ell m\rangle \to R\vert \ell m\rangle$, then $\langle \ell m' \vert \to \langle \ell m' \vert R^{-1}$ so that, under rotation, $T^\ell_m$ does go to $RT R^{-1}$.

Under this rotation, we now have \begin{align} R T^{\ell}_{m}R^{-1}&=\sqrt{\frac{2L+1}{2s+1}}\sum_{\mu\mu'}C^{s\mu'}_{S\mu;\ell m} R \vert s\mu'\rangle \langle s\mu\vert\,R^{-1}\, ,\\ &=\sqrt{\frac{2L+1}{2s+1}}\sum_{\mu\mu'}\sum_{MM'}C^{s\mu'}_{S\mu;\ell m} \vert sM'\rangle \langle sM'\vert R \vert s\mu'\rangle \langle s\mu\vert\,R^{-1}\vert s M\rangle\langle s M\vert\, ,\\ &=\sqrt{\frac{2L+1}{2s+1}}\sum_{\mu\mu'}\sum_{MM'}C^{s\mu'}_{S\mu;\ell m} D^s_{M'\mu'}(R) D^s_{\mu M}(R^{-1}) \vert s M'\rangle\langle s M\vert\, . \end{align} With some dexterity the sum of products $\sum_{\mu\mu'}D^s_{M'\mu'}(R) D^s_{\mu M}(R^{-1})$ can be combined using the Clebsch $C^{s\mu'}_{S\mu;\ell m}$ to give a sum of type $$ \sum_{m'}D^\ell_{m'm}(R)C^{s\mu'}_{S\mu;\ell m'} $$ from which Eq.(2) follows.

Broadly speaking, irreducible under a set transformation means there is no subset that transforms only amongst itself under this set of transformations.

(The combination rules for $D$'s can be fond in Varshalovich, D. A., Moskalev, A. N., & Khersonskii, V. K. M. (1988). Quantum theory of angular momentum. World Scientific.)


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