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# 物理代写|量子力学代写Quantum mechanics代考|PHYS3001 Tensors and spherical harmonics

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## 物理代写|量子力学代写Quantum mechanics代考|Tensors and spherical harmonics

It is also useful to understand that the $Y_{l m}$ ‘s represent tensors of rank $l$, constructed from the unit vector $\boldsymbol{\Omega}$. For example, neglecting the overall normalization one can verify that for $l=1$ the $Y_{1 m}$ are rewritten as
\begin{aligned} Y_{10} & \sim \cos \theta=\frac{z}{r}=\mathbf{\Omega}3 \equiv \mathbf{\Omega}_0 \ Y{1, \pm 1} & \sim \frac{1}{\sqrt{2}} \sin \theta e^{\pm i \theta}=\frac{x \pm i y}{r \sqrt{2}}=\frac{\mathbf{\Omega}1 \pm i \mathbf{\Omega}_2}{\sqrt{2}} \equiv \boldsymbol{\Omega}{\pm} \end{aligned}
Thus, one may think of $Y_{1 m}$ as the 3 components of the vector $\boldsymbol{\Omega}I$ taken in the basis $I=(+, 0,-)$ instead of the conventional cartesian basis labelled by $1,2,3$. The relation between the bases $(1,2,3)$ and $(+, 0,-)$ is given by the unitary transformation $U$ that satisfies $U U^{\dagger}=1$ as follows \begin{aligned} & \left(\begin{array}{c} \boldsymbol{\Omega}{+} \ \boldsymbol{\Omega}0 \ \boldsymbol{\Omega}{-} \end{array}\right)=\left(\begin{array}{ccc} \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} & 0 \ 0 & 0 & 1 \ \frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}} & 0 \end{array}\right)\left(\begin{array}{l} \boldsymbol{\Omega}1 \ \boldsymbol{\Omega}_2 \ \boldsymbol{\Omega}_3 \end{array}\right) \ & \left(\begin{array}{l} \boldsymbol{\Omega}_1 \ \boldsymbol{\Omega}_2 \ \boldsymbol{\Omega}_3 \end{array}\right)=\left(\begin{array}{ccc} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \ -\frac{i}{\sqrt{2}} & 0 & \frac{i}{\sqrt{2}} \ 0 & 1 & 0 \end{array}\right)\left(\begin{array}{l} \boldsymbol{\Omega}{+} \ \boldsymbol{\Omega}0 \ \boldsymbol{\Omega}{-} \end{array}\right) \end{aligned}

## 物理代写|量子力学代写Quantum mechanics代考|Radial & angular equations in d-dims

It is possible to generalize all the results to $d$-dimensions by following the general operator approach. In $d$-dimensions the generator of rotations in the $(I, J)$ plane is written as
$$L_{I J}=r_I p_J-r_J p_I, \quad I, J=1,2, \cdots d .$$
The $L_{I J}$ commute with all dot products constructed from $(\mathbf{r}, \mathbf{p})$. The commutators of these operators close into the same set (see problem)
$$\left[L_{I J}, L_{K L}\right]=\delta_{J K} L_{I L}-\delta_{I K} L_{J L}-\delta_{I L} L_{J K}+\delta_{J L} L_{I K} .$$

This set of commutation rules is the Lie algebra for $S O(d)$. Note that the $3-$ dimensional case is a special case that permits the rewriting in terms of the familiar $L_{1,2,3}$ as $L_{I J}=\epsilon_{I J K} L_K$. The quadratic Casimir operator is defined by
\begin{aligned} L^2 & =\frac{1}{2} \sum_{I, J=1}^d L_{I J}^2 \ & =\mathbf{r}^2 \mathbf{p}^2-(\mathbf{r} \cdot \mathbf{p})(\mathbf{p} \cdot \mathbf{r})-2 i \hbar \mathbf{r} \cdot \mathbf{p} \end{aligned}
Since it is constructed from dot products it must commute with all $L_{I J}$; this result may also be verified abstractly by only using the commutation rules of the $S O(d)$ Lie algebra. The combination $(\mathbf{r} \cdot \mathbf{p})(\mathbf{p} \cdot \mathbf{r})+2 i \hbar \mathbf{r} \cdot \mathbf{p}$ may be rewritten in terms of the Hermitian operator $p_r=\sum_I\left(\frac{1}{r} r_I p_I+p_I r_I \frac{1}{r}\right)$ and then solve for $\mathbf{p}^2$ in the form (see problem)
$$\mathbf{p}^2=p_r^2+\frac{1}{r^2}\left[L^2+\frac{\hbar^2}{4}(d-1)(d-3)\right] .$$

## 物理代写|量子力学代写Quantum mechanics代考|Tensors and spherical harmonics

$$Y_{10} \sim \cos \theta=\frac{z}{r}=\boldsymbol{\Omega} 3 \equiv \boldsymbol{\Omega}0 Y 1, \pm 1 \quad \sim \frac{1}{\sqrt{2}} \sin \theta e^{\pm i \theta}=\frac{x \pm i y}{r \sqrt{2}}=\frac{\boldsymbol{\Omega} 1 \pm i \boldsymbol{\Omega}_2}{\sqrt{2}} \equiv \boldsymbol{\Omega} \pm$$ 因此，人们可能会想到 $Y{1 m}$ 作为向量的 3 个分量 $\Omega I$ 采取的甚础 $I=(+, 0,-)$ 而不是标记为的传统笛卡尔甚 $1,2,3$. 碱基之间的 关系 $(1,2,3)$ 和 $(+, 0,-)$ 由酉音抬给出 $U$ 满足 $U U^{\dagger}=1$ 如下

## 物理代写|量子力学代写Quantum mechanics代考|Radial \& angular equations in d-dims

$$L_{I J}=r_I p_J-r_J p_I, \quad I, J=1,2, \cdots d .$$

$$\left[L_{I J}, L_{K L}\right]=\delta_{J K} L_{I L}-\delta_{I K} L_{J L}-\delta_{I L} L_{J K}+\delta_{J L} L_{I K} .$$

$$L^2=\frac{1}{2} \sum_{I, J=1}^d L_{I J}^2 \quad=\mathbf{r}^2 \mathbf{p}^2-(\mathbf{r} \cdot \mathbf{p})(\mathbf{p} \cdot \mathbf{r})-2 i \hbar \mathbf{r} \cdot \mathbf{p}$$

$$\mathbf{p}^2=p_r^2+\frac{1}{r^2}\left[L^2+\frac{\hbar^2}{4}(d-1)(d-3)\right] .$$

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