- m-Selection Rule and Construction of Spherical Tensor

1. Another Way of m-Selection Rule in Spherical Tensor #

Another way to see the m-selection rule is to note the transformation property of $T_{q}^{(k)}\ket{\alpha,jm}$ under rotation, namely,

$$ \mathcal{D}T_{q}^{(k)}\ket{\alpha,jm} = \mathcal{D}T_{q}^{(k)}\mathcal{D}^{\dagger} \mathcal{D}\ket{\alpha,jm}.\tag{1} $$

If we now let $\mathcal{D}$ stand for a rotation operator around the $z$-axis, we get by eq.3 and eq.4

$$ \mathcal{D}(\hat{z},\phi)T_{q}^{(k)}\ket{\alpha,jm} $$

$$ = e^{-iq\phi} e^{-im\phi} T_{q}^{(k)}\ket{\alpha,jm},\tag{2} $$

which is orthogonal to $\ket{\alpha’,j’ m’}$ unless $q + m = m’$.

The rotation operator around the $z$-axis is given by

$$ \mathcal{D}(z,\phi) = \exp\left(-\frac{i}{\hbar} J_z \phi\right).\tag{3} $$

Similarly, the spherical tensor transforms under rotation as

$$ \mathcal{D} T_q^{(k)} \mathcal{D}^{\dagger} = \sum_{q’} \mathcal{D_{q’q}^{(k)}} T_{q’}^{(k)}.\tag{4} $$

From eq.2, we find

$$ \mathcal{D}(\hat{z},\phi)\ket{\psi} = e^{-i(q+m)\phi}\ket{\psi}.\tag{5} $$

By eq.3, we also have

$$ \mathcal{D}(\hat{z},\phi)\ket{\psi} = \exp\left(-\frac{i}{\hbar} J_z \phi\right)\ket{\psi}.\tag{6} $$

Hence, $\ket{\psi}$ is an eigenket of $J_z$, and we can write

$$ \ket{\psi} = C \ket{\alpha,j,q+m}.\tag{7} $$

Since, by eq.2,

$$ \ket{\psi} = T_q^{(k)}\ket{\alpha,jm},\tag{8} $$

we obtain the inner product relation

$$ \langle\alpha’,j’m’|T_q^{(k)}|\alpha,jm\rangle $$

$$ = C \langle\alpha’,j’m’|\alpha,j,q+m\rangle.\tag{9} $$

which shows that unless $m’ = q + m$, $T_q^{(k)}\ket{\alpha,jm}$ is orthogonal to $\ket{\alpha’,j’m’}$.

2. Construction of Spherical Tensor from Irreducible Tensors #

Theorem: Let $X_{q_1}^{(k_1)}$ and $Z_{q_2}^{(k_2)}$ be irreducible spherical tensors of rank $k_1$ and $k_2$, respectively. Then,

$$ T_q^{(k)} = \sum_{q_1} \sum_{q_2} \langle k_1 k_2; q_1,q_2 | k_1 k_2; kq\rangle $$

$$ X_{q_1}^{(k_1)} Z_{q_2}^{(k_2)}\tag{10} $$

is a spherical (irreducible) tensor of rank $k$.

Proof:

$$ \ket{k_1 k_2; kq} = \sum_{q_1,q_2} (\ket{k_1,q_1} \otimes \ket{k_2,q_2} $$

$$ \bra{k_1,q_1} \otimes \bra{k_2,q_2}) \ket{k_1 k_2; kq}.\tag{11} $$

where $(\bra{k_1,q_1} \otimes \bra{k_2,q_2}) \ket{k_1 k_2; kq}$ is the Clebsch-Gordan coefficient, $\langle k_1 k_2; q_1 q_2 |k_1 k_2; kq\rangle$.

Using the Clebsch-Gordan coefficients, we rewrite eq.11 as

$$ \ket{k_1 k_2; kq} = \sum_{q_1,q_2} \langle k_1 k_2; q_1 q_2 |k_1 k_2; kq\rangle $$

$$ \ket{k_1,q_1} \otimes \ket{k_2,q_2}.\tag{12} $$

Here we can represent $\ket{k_1 k_2; kq}$ as $T_q^{(k)}$ and $\ket{k_1,q_1} \otimes \ket{k_2,q_2}$ as $X_{q_1}^{(k_1)} Z_{q_2}^{(k_2)}$, using the property

$$ (\bra{x} \otimes \bra{y}) (\ket{\alpha} \otimes \ket{\beta}) $$

$$ = \psi_{\alpha}(x) \psi_{\beta}(y).\tag{13} $$

Thus, we identify

$$ T_q^{(k)} $$

$$ = \sum_{q_1,q_2} \langle k_1 k_2; q_1 q_2 |k_1 k_2; kq\rangle X_{q_1}^{(k_1)} Z_{q_2}^{(k_2)}. $$

$$\tag{14}$$

This shows that $T_q^{(k)}$ is an irreducible spherical tensor of rank $k$.