Scalar, Vector, Tensor Operators

Motivating Discussion

$A^{'} = D^{†} (R) A D (R)$ .

$A is a scalar if $A^{'} = A$ . Since $A^{'} = A + \frac{i}{ℏ} d φ [\vec{J} \cdot \hat{n}, A] = 0$ .

For vectors, $V_{i}^{'} = \sum_{j} R_{i j} V_{j}$ , $V_{i}^{'} = D^{†} (R) V_{i} D (R) = V_{i} + \frac{i}{ℏ} d φ [\vec{J} \cdot \hat{n}, V_{i}] = V_{i} + d φ [\hat{n} \times \vec{V}]_{i}$ . So, $[\vec{J} \cdot \hat{n}, V_{i}] = - i ℏ [\hat{n} \times \vec{V}]_{i}$ . Then, $[V_{i}, J_{j}] = i ℏ ε_{i j k} V_{k}$ . ( ${\vec{V}}^{'} = \vec{V} + d φ \hat{n} \times \vec{V}$ )

Example, suppose $\vec{V} = \vec{J}$ . Then, $[J_{i}, J_{j}] = i ℏ ε_{i j k} J_{k}$ . Example, suppose $\vec{V} = \vec{P}$ . Then $[P_{i}, L_{j}] = [P_{i}, r_{x} P_{i} - r_{i} P_{k}] = i ℏ ε_{i j k} P_{k}$ .

For Tensors, $T_{i j k \dots} = \sum R_{i i^{'}} R_{j j^{'}} R_{k k^{'}} \dots T_{i^{'} j^{'} k^{'} \dots}$ . 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 $Q_{i j}$ , Inertia $I_{i j}$ , Permitivity $ε_{i j}$ . Examples: $N = 3$ , Electric acceptability $\vec{P} = χ^{(1)} \vec{E} + χ^{(2)} \vec{E} \vec{E} + χ^{(4)} \vec{E} \vec{E} \vec{E} = χ_{i j}^{(1)} E_{j} + χ_{i j k}^{(2)} E_{j} E_{k} + χ_{i j k l}^{(3)} E_{j} E_{k} E_{l}$ .

Example: $N = 2$ . $T_{i j} = u_{i} v_{j}$ such that $\hat{T} = \vec{u} \otimes \vec{v} ≐ (\begin{matrix} u_{1} v_{1} & u_{1} v_{2} & u_{1} v_{3} \\ u_{2} v_{1} & u_{2} v_{2} & u_{2} v_{3} \\ u_{3} v_{1} & u_{3} v_{2} & u_{3} v_{3} \end{matrix})$ . Let $u_{i} v_{j} = \frac{\vec{u} \cdot \vec{v}}{3} δ_{i j} - \frac{\vec{u} \cdot \vec{v}}{3} δ_{i j} + \frac{u_{i} v_{j}}{2} + \frac{u_{i} v_{j}}{2} + \frac{u_{j} v_{i}}{2} - \frac{u_{j} v_{i}}{2} = \frac{\vec{u} \cdot \vec{v}}{3} δ_{i j} + \frac{u_{i} v_{j} - u_{j} v_{i}}{2} + (\frac{u_{i} v_{j} + u_{j} v_{i}}{2} - \frac{\vec{u} \cdot \vec{v}}{3} δ_{i j})$ . The first term is a diagonal matrix, $\frac{\vec{u} \cdot \vec{v}}{3} I = \frac{Tr (T)}{3} I$ , which behaves like a scalar. The second term is an antisymmetric matrix, $T_{i j (a)} = \frac{u_{i} v_{j} - u_{j} v_{i}}{2}, {\hat{T}}_{(a)} ≐ \frac{1}{2} (\begin{matrix} 0 & T_{12} - T_{21} & T_{13} - T_{31} \\ T_{21} - T_{12} & 0 & T_{23} - T_{32} \\ T_{31} - T_{13} & T_{32} - T_{23} & 0 \end{matrix})$ , there are only 3 unique components the others are related by a minus sign. Remember $u_{i} v_{j} - u_{j} v_{i} = ε_{i j k} (\vec{u} \times \vec{v})_{k}$ . The last one is a symmetric matrix that has 5 components: $T_{11} - \frac{Tr (T)}{3}, T_{22} - \frac{Tr (T)}{3}, \frac{1}{2} (T_{12} + T_{21}), \frac{1}{2} (T_{13} + T_{31}), \frac{1}{2} (T_{23} + T_{32})$ . 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 $q = - k, - k + 1, \dots, k$ , is the irreducible $k$th order spherical tensor if $T_{q}^{(k)}$ transform under rotations as $D (R) T_{q}^{(k)} D^{†} (R) = \sum_{q^{'} = - k}^{k} D_{q^{'} q}^{(k)} (R) T_{q^{'}}^{(k)}$ . Note that we only need one rotation matrix, not 3 independent rotations.

$D (R) T_{q}^{(k)} | j m ⟩ = \sum_{q^{'}} D_{q^{'} q}^{(k)} T_{q^{'}}^{(k)} \sum_{m^{'}} D_{m^{'} m}^{(j)} | j m^{'} ⟩$ This implies that $T_{q}^{(k)} | j m ⟩ \Rightarrow | k q ⟩ \otimes | j m ⟩$ so that the rotation will operate on each independent basis. Thus, $⟨ j^{'} m^{'} | T_{q}^{(k)} | j m ⟩ \neq 0$ if $m^{'} = m + q$ and $j + k \geq j^{'} \geq | j - k |$ .

Relation to Spherical Harmonics.

\begin{aligned} (1) & Y_{0}^{0} & = \frac{1}{\sqrt{4 π}} \\ (2) & Y_{1}^{0} & = \sqrt{\frac{3}{4 π}} \cos θ \\ (3) & Y_{1}^{\pm 1} & = \mp \sqrt{\frac{3}{8 π}} \sin θ \exp (\pm i φ) \end{aligned}

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

So, $\vec{r} = \sqrt{\frac{4 π}{3}} r \sum_{q = - 1}^{1} (- 1)^{q} Y_{1}^{q} {\hat{ϵ}}_{1}^{- q}$ .

Thus, for any vector, $\vec{V} = V_{x} \hat{i} + V_{y} \hat{y} + V_{k} \hat{k} = \sum_{q = - 1}^{1} (- 1)^{q} V_{q}^{(1)} {\hat{ϵ}}_{1}^{- q}$ . Then, $V_{\pm 1}^{(1)} = \mp \frac{V_{x} \pm i V_{y}}{\sqrt{2}}, V_{0}^{(1)} = V_{z}$ .

Example - Hydrogen Atom

In the electric dipole approximation, what are the allowed radiative transitions?

$H_{i n t} = - \vec{d} \cdot \vec{E} \Rightarrow P = | ⟨ f | H_{i n t} | i ⟩ |^{2}$ . I.e. when is $P \neq 0$ .

$| i ⟩ = | n_{i} ℓ_{i} m_{i} ⟩, | f ⟩ = | n_{f} ℓ_{f} m_{f} ⟩$ .

$H_{i n t} = - e \vec{r} \cdot \vec{E} = - e E_{0} \vec{ϵ} \cdot \vec{r}$ . Let $\vec{E} = E_{0} \hat{ϵ}$ .

$\hat{ϵ} = (0, 0, 1)$ implies polarization aligned along $z$ .

If $\hat{ϵ} \cdot \vec{r} = \mp \sqrt{\frac{4 π}{3}} r Y_{1}^{\pm 1}$ is circularly polarized. If you take a superposition of $Y_{1}^{\pm 1}$ you could get linear polarization in $x$ or $y$ directions.

$⟨ n_{f} ℓ_{f} m_{f} | \hat{e} \cdot \vec{r} | n_{i} ℓ_{i} f_{i} ⟩ = \int R_{n_{f}}^{ℓ_{f} *} (r) r R_{n_{i}}^{ℓ_{i}} (r) r^{2} d r \cdot \int Y_{ℓ_{f}}^{m_{f} *} (θ, φ) Y_{1}^{q} (θ, φ) Y_{ℓ_{i}}^{m_{i}} (θ, φ) \sin θ d θ d φ$ .

So, $⟨ ℓ_{f} m_{f} | Y_{1}^{q} | ℓ_{i} m_{i} ⟩ = 0$ dictates forbidden transitions since the radial integral will never be zero.

From our previous selection rules, $m_{f} = m_{i} + q$ and $ℓ_{i} + 1 \geq ℓ_{f} \geq | ℓ_{i} - 1 |$ .