# Scalar, Vector, Tensor Operators

## Motivating Discussion

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

$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, $$V_{\pm1}^{(1)} = \mp\frac{V_x\pm iV_y}{\sqrt{2}},V_0^{(1)}=V_z$$.

### Example - Hydrogen Atom

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

$$H_{int} = -\vec{d}\cdot\vec{E}\Rightarrow \mathcal{P} = |\langle f|H_{int}|i\rangle|^2$$. I.e. when is $$\mathcal{P}\neq 0$$.

$$|i\rangle = |n_i\ell_im_i\rangle,|f\rangle = |n_f\ell_fm_f\rangle$$.

$$H_{int} = -e\vec{r}\cdot\vec{E} = -eE_0\vec{\epsilon}\cdot\vec{r}$$. Let $$\vec{E}=E_0\hat{\epsilon}$$.

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

If $$\hat{\epsilon}\cdot\vec{r}=\mp\sqrt{\frac{4\pi}{3}}rY_{1}^{\pm 1}$$ is circularly polarized. If you take a superposition of $$Y_1^{\pm 1}$$ you could get linear polarization in $$x$$ or $$y$$ directions.

$$\langle n_f\ell_fm_f|\hat{e}\cdot\vec{r}|n_i\ell_if_i\rangle = \int R_{n_f}^{\ell_f*}(r)rR_{n_i}^{\ell_i}(r)r^2dr\cdot\int Y_{\ell_f}^{m_f*}(\theta,\varphi)Y_1^{q}(\theta,\varphi)Y_{\ell_i}^{m_i}(\theta,\varphi)\sin\theta d\theta d\varphi$$.

So, $$\langle \ell_fm_f|Y_1^q|\ell_im_i\rangle = 0$$ dictates forbidden transitions since the radial integral will never be zero.

From our previous selection rules, $$m_f=m_i+q$$ and $$\ell_i+1\geq \ell_f\geq |\ell_i-1|$$.

Created: 2024-05-30 Thu 21:17

Validate