Irreducible form of Spherical tensor operators

Question

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 $$\mathcal{D^{\dagger}}(R)Y_{l}^{m}(\mathbf{V})\mathcal{D}(R) = \sum_{m'}Y_{l}^{m'}(\mathbf{V})\mathcal{D}^{(l)^{*}}_{mm'}(R)"$$

Questions:

• 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 $|l,m' \rangle$: $\mathcal{D}(R)Y_{l}^{m} = \sum_{m''}\sum_{m'}|l,m''\rangle \langle l, m''| \mathcal{D}(R)|l,m' \rangle \langle l,m'|l,m \rangle = \sum_{m''}|l, m'' \rangle \mathcal{D^{(l)}_{m,m''}}(R)$?

Answer 1

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.

Answer 2

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.)

Answer 3

This is a detailed derivation of the final equation for OP's first question.

The spherical harmonics can be viewed as a function, with input $\mathbf{\hat{n}}$ and output $\langle{\mathbf{\hat{n}}}|{lm}\rangle$. The directional eigenvectors are in one-to-one correspondence with their three components in the Cartesian basis $(n_x,n_y,n_z)$, so we may as well use them for the function input: \begin{align} Y_l^m: (n_x,n_y,n_z)\to \langle{\mathbf{\hat{n}}}|{lm}\rangle\,. \end{align} We can extend the domain of this function by allowing the input to be vector operators, or equivalently the three components of them. The generalized definition is \begin{align} Y_l^m: (V_x,V_y,V_z)\to \langle{\mathbf{\hat{n}}}|{lm}\rangle\,,\quad\text{replace $n_{x,y,z}$ by $V_{x,y,z}$}\,. \end{align} The extension of $Y_l^m$ is well defined if we specify its expression with noncommutative terms of $n_{x,y,z}$. For a directional eigenket under rotation, \begin{align} \mathbf{\hat{n}}\cong\left[\begin{matrix} n_x\\n_y\\n_z \end{matrix}\right]\to \mathbf{\hat{n}}'\cong\left[\begin{matrix} n'_x\\n'_y\\n'_z \end{matrix}\right] = R \left[\begin{matrix} n_x\\n_y\\n_z \end{matrix}\right]\,. \end{align} Here we use $\cong$ to imply the representation under Cartesian basis. So the function is \begin{align} Y_l^m: R \left[\begin{matrix} n_x\\n_y\\n_z \end{matrix}\right] \to \langle{\mathbf{\hat{n}}'}|{lm}\rangle\,. \end{align} Denote \begin{align} \mathbf{V}' = \mathscr{D}^\dagger(R)\mathbf{V}\mathscr{D}(R)\,, \end{align} then by definition of vector operator, \begin{align} \mathbf{V}'\cong \left[\begin{matrix} V'_x\\V'_y\\V'_z \end{matrix}\right] = R \left[\begin{matrix} V_x\\V_y\\V_z \end{matrix}\right]\,. \end{align} For the input vector operator $\mathbf{V}'$, the function is \begin{align} Y_l^m: \left[\begin{matrix} V'_x\\V'_y\\V'_z \end{matrix}\right]\to \langle{\mathbf{\hat{n}}}|{lm}\rangle\,,\text{ replace $n_{x,y,z}$ by $V'_{x,y,z}$}\,. \end{align} which is equivalently \begin{align} Y_l^m: \left[\begin{matrix} V'_x\\V'_y\\V'_z \end{matrix}\right] = R \left[\begin{matrix} V_x\\V_y\\V_z \end{matrix}\right]\to \langle{\mathbf{\hat{n}}'}|{lm}\rangle = \langle{\mathbf{\hat{n}}}|{\mathscr{D}^\dagger(R)}|{lm}\rangle\,,\text{ replace $n_{x,y,z}$ by $V_{x,y,z}$}\,, \end{align} where we used the definition of the directional eigenket. So we have \begin{align} Y_l^m(\mathbf{V}') & = \langle{\mathbf{\hat{n}}}|{\mathscr{D}^\dagger(R)}|{lm}\rangle\quad \text{replace $n_{x,y,z}$ by $V_{x,y,z}$}\\ & = \sum_{m'}\langle{\mathbf{\hat{n}}}|{lm'}\rangle\langle{lm'}|{\mathscr{D}^\dagger(R)}|{lm}\rangle\quad \text{replace $n_{x,y,z}$ by $V_{x,y,z}$}\\ & = \sum_{m'}Y_l^{m'}(\mathbf{V})\langle{lm'}|{\mathscr{D}^\dagger(R)}|{lm}\rangle\\ & = \sum_{m'}Y_l^{m'}(\mathbf{V}){\mathscr{D}^{(l)^*}_{mm'}(R)}\,. \end{align} In the final step, use that $Y_l^m(\mathbf{V})$ is a homogeneous polynomial of the components of $\mathbf{V}$, which is proven on P371 of Gordon Baym Lecture notes on quantum mechanics. So we have \begin{equation} Y_l^m(\mathbf{V}') = Y_l^m(\mathscr{D}^\dagger(R)\mathbf{V}\mathscr{D}(R)) = \mathscr{D}^\dagger(R) Y_l^m(\mathbf{V}) \mathscr{D}(R)\,. \end{equation} It follows the final equation.