Scalar, Vector, Tensor Operators

\(A' = \mathcal{D}^\dagger(R)A\mathcal{D}(R)\).

$A is a scalar if \(A'=A\). Since \(A' = A + \frac{i}{\hbar}d\varphi[\vec{J}\cdot\hat{n},A] = 0\).

For vectors, \(V_{i}' = \sum_j R_{ij}V_j\), \(V_i' = \mathcal{D}^\dagger(R) V_i\mathcal{D}(R) = V_i + \frac{i}{\hbar}d\varphi[\vec{J}\cdot\hat{n},V_i] = V_i + d\varphi [\hat{n}\times\vec{V}]_i\). So, \([\vec{J}\cdot\hat{n},V_i] = -i\hbar[\hat{n}\times\vec{V}]_i\). Then, \([V_i,J_j] = i\hbar\varepsilon_{ijk}V_k\). (\(\vec{V}' = \vec{V} + d\varphi\hat{n}\times\vec{V}\))

Example, suppose \(\vec{V} = \vec{J}\). Then, \([J_i,J_j] = i\hbar\varepsilon_{ijk}J_k\). Example, suppose \(\vec{V} = \vec{P}\). Then \([P_i,L_j] = [P_i,r_xP_i-r_iP_k] = i\hbar\varepsilon_{ijk}P_k\).

For Tensors, \(T_{ijk\cdots} = \sum R_{ii'}R_{jj'}R_{kk'}\cdots T_{i'j'k'\cdots}\). It has \(N\) indicies and to rotate it needs \(N\) rotation matricies. $N$-th rank Cartesian tensor. Number of components, \(3^N\) for a Cartesian tensor.

Examples: \(N=2\), Quardrapole moments \(Q_{ij}\), Inertia \(I_{ij}\), Permitivity \(\varepsilon_{ij}\). Examples: \(N=3\), Electric acceptability \(\vec{P} = \chi^{(1)}\vec{E} + \chi^{(2)}\vec{E}\vec{E} + \chi^{(4)}\vec{E}\vec{E}\vec{E} = \chi_{ij}^{(1)}E_j + \chi_{ijk}^{(2)}E_{j}E_{k} + \chi_{ijkl}^{(3)}E_jE_kE_l\).

Example: \(N=2\). \(T_{ij} = u_iv_j\) such that \(\hat{T} = \vec{u}\otimes\vec{v} \doteq \begin{pmatrix} u_1v_1 & u_1v_2 & u_1v_3 \\ u_2v_1 & u_2v_2 & u_2v_3 \\ u_3v_1 & u_3v_2 & u_3v_3 \end{pmatrix}\). Let \(u_iv_j = \frac{\vec{u}\cdot\vec{v}}{3}\delta_{ij} - \frac{\vec{u}\cdot\vec{v}}{3}\delta_{ij} + \frac{u_iv_j}{2} + \frac{u_iv_j}{2} + \frac{u_jv_i}{2} - \frac{u_jv_i}{2} = \frac{\vec{u}\cdot\vec{v}}{3}\delta_{ij} + \frac{u_iv_j-u_jv_i}{2} + \left(\frac{u_iv_j+u_jv_i}{2} - \frac{\vec{u}\cdot\vec{v}}{3}\delta_{ij}\right)\). The first term is a diagonal matrix, \(\frac{\vec{u}\cdot\vec{v}}{3}\mathbb{I} = \frac{\text{Tr}(T)}{3}\mathbb{I}\), which behaves like a scalar. The second term is an antisymmetric matrix, \(T_{ij(a)} = \frac{u_iv_j-u_jv_i}{2}, \hat{T}_{(a)}\doteq \frac{1}{2}\begin{pmatrix} 0 & T_{12}-T_{21} & T_{13}-T_{31} \\ T_{21}-T_{12} & 0 & T_{23} - T_{32} \\ T_{31}-T_{13} & T_{32}-T_{23} & 0 \end{pmatrix}\), there are only 3 unique components the others are related by a minus sign. Remember \(u_iv_j-u_jv_i=\varepsilon_{ijk}(\vec{u}\times\vec{v})_k\). The last one is a symmetric matrix that has 5 components: \(T_{11}-\frac{\text{Tr}(T)}{3}, T_{22} - \frac{\text{Tr}(T)}{3}, \frac{1}{2}(T_{12} + T_{21}), \frac{1}{2}(T_{13}+T_{31}), \frac{1}{2}(T_{23}+T_{32})\). These are called, \(T^{(0)}, \hat{T}_{(a)}, \hat{T}_{(s)}\), spherical tensors.

Definition of a Spherical Tensor: The $(2k+1)-$component operator \(\{T_{q}^{(k)}\}\), $k-$th rank with \(q=-k,-k+1,\cdots,k\), is the irreducible $k$th order spherical tensor if \(T_q^{(k)}\) transform under rotations as \(\mathcal{D}(R)T^{(k)}_q\mathcal{D}^\dagger(R) = \sum_{q'=-k}^k\mathcal{D}^{(k)}_{q'q}(R)T_{q'}^{(k)}\). Note that we only need one rotation matrix, not 3 independent rotations.

\(\mathcal{D}(R)T_q^{(k)}|jm\rangle = \sum_{q'}\mathcal{D}_{q'q}^{(k)}T_{q'}^{(k)}\sum_{m'}\mathcal{D}_{m'm}^{(j)}|jm'\rangle\) This implies that \(T_q^{(k)}|jm\rangle \Rightarrow |kq\rangle\otimes|jm\rangle\) so that the rotation will operate on each independent basis. Thus, \(\langle j'm'|T_q^{(k)}|jm\rangle\neq0\) if \(m'=m+q\) and \(j+k\geq j'\geq|j-k|\).

Relation to Spherical Harmonics.

\begin{align} Y_0^0 &= \frac{1}{\sqrt{4\pi}} \\ Y_1^0 &= \sqrt{\frac{3}{4\pi}}\cos\theta \\ Y_1^{\pm1} &= \mp\sqrt{\frac{3}{8\pi}}\sin\theta\exp(\pm i\varphi) \end{align}

We will transform the cartesian basis to the spherical basis, \((\hat{i},\hat{j},\hat{k})\Rightarrow(\hat{\epsilon}_1^{-1},\hat{\epsilon}_1^{0},\hat{\epsilon}_1^{1})\) \(\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}=-r_1^1\hat{\epsilon}_1^{-1}-r_1^{-1}\hat{\epsilon}_1^{1}+r_1^0\hat{\epsilon}_1^0\) with \(\hat{\epsilon}_1^0=\hat{k},\hat{\epsilon}_1^{1}=-\frac{\hat{i}+i\hat{j}}{\sqrt{2}},\hat{\epsilon}_1^{-1}=\frac{\hat{i}-i\hat{j}}{\sqrt{2}}\). Thus, \(r_1^1=-\frac{x+iy}{\sqrt{2}},r_1^{-1}=\frac{x-iy}{\sqrt{2}},r_1^0=z\).

So, \(\vec{r}=\sqrt{\frac{4\pi}{3}}r\sum_{q=-1}^1(-1)^qY_1^q\hat{\epsilon}_1^{-q}\).

Thus, for any vector, \(\vec{V} = V_x\hat{i} + V_y\hat{y} + V_k\hat{k} = \sum_{q=-1}^1(-1)^qV_q^{(1)}\hat{\epsilon}_1^{-q}\). Then, \(V_{\pm1}^{(1)} = \mp\frac{V_x\pm iV_y}{\sqrt{2}},V_0^{(1)}=V_z\).