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# 物理代写|粒子物理代写Particle Physics代考|PGPH11099 Tensor decomposition. Irreducible tensors

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## 物理代写|粒子物理代写Particle Physics代考|Tensor decomposition. Irreducible tensors.

Tensor decomposition. Irreducible tensors. Consider a two-index tensor $T_{i j}$. It has nine components. It can be decomposed into a symmetric and traceless part $S$, an antisymmetric part $A$ and the trace $T_{k k}$, with five, three and one independent components, respectively, as follows
\begin{aligned} T_{i j} &=\frac{1}{2}\left(T_{i j}+T_{j i}\right)-\frac{1}{3} \delta_{i j} T_{k k}+\frac{1}{2}\left(T_{i j}-T_{j i}\right)+\frac{1}{3} \delta_{i j} T_{k k} \ & \equiv S_{i j}+A_{i j}+\frac{1}{3} \delta_{i j} T_{k k} \end{aligned}
The new objects $S, A$ and $T_{k k}$ have the following properties: (a) The trace $T_{k k}$ of $T$ is a scalar under $O$ (3). Indeed, $T_{k k}^{\prime}=R_{k l} R_{k m} T_{l m}=\left(R^T R\right){l m} T{l m}=\delta_{l m} T_{l m}=T_{m m}$.
(b) The $A_{i j}$ transform linearly among themselves under $O(3)$ and form a two-index antisymmetric tensor, which is equivalent to a pseudo-vector.

Indeed, defining $\mathcal{A}i \equiv \epsilon{i j k} A_{j k}$, and making use of the properties of the $\epsilon$-symbol, we verify that $\mathcal{A}i^{\prime}=(\operatorname{det} R) R{i k} \mathcal{A}k$. (c) The five independent quantities $S{i j} \equiv(1 / 2)\left(T_{i j}+T_{j i}\right)-(1 / 3) \delta_{i j} T_{k k}$, form a symmetric traceless two-index tensor. ${ }^{13}$

Thus, under the action of the rotation group the five components of $S$, the three components of $A$ and the one component $T_{k k}$ transform independently. We say that the generic two-index tensor decomposes into a sum of three irreducible tensors of ranks 2,1 and 0 respectively. We write
$$\mathbf{3} \otimes \mathbf{3} \cong \mathbf{5} \oplus \mathbf{3} \oplus \mathbf{1}$$
and, as will become clear soon, this is an example of the general reduction formula (5.58).

The same procedure applies to the decomposition of higher tensor quantities. The operations of symmetrisation, anti symmetrisation and tracing give seven-, nine-, etc. dimensional irreducible representations of the rotation group.

## 物理代写|粒子物理代写Particle Physics代考|The Lie algebra of SO(3)

The Lie algebra of $\mathbf{S O}(3)$. The group elements of $S O(3)$ depend on three independent variables, the three rotation angles and therefore the Lie algebra $\mathfrak{c o}_{(3)}$ has three generators. On the other hand we know from the Cartan classification that there exists only one independent three-dimensional Lie algebra, the one we studied in association with the group $S U(2)$. It follows that we have an isomorphism $\mathfrak{f u}(\mathbf{2}) \cong \mathfrak{f}(\mathbf{0})$. In Problem 5.3, we ask the reader to verify this by explicit calculation. This is another example of the statement that two different groups may have identical Lie algebras.
We have studied the representations of this algebra and have found that they are labelled by an index $j$ which can take integer, or half-integer, values. Which of these representations will give, by exponentiation, representations of $S O(3)$ ? The group $S O(3)$ is characterised by the fact that a rotation by $2 \pi$ is identical to no rotation. This property should be preserved by all its representations. This means that the rotation matrix in any representation should satisfy $\exp {2 \mathrm{i} \pi \boldsymbol{n} \cdot \boldsymbol{J}}=\mathbf{1}$. Applied to rotations around the $z$-axis, it implies that $J_3$ should have integer eigenvalues. Thus,of all representations of the algebra, only the ones corresponding to integer $j$ yield by exponentiation representations of $S O(3)$. In order to obtain this result we have to use the global property of $S O(3)$ and identify a rotation of $2 \pi$ with the identity.

## 物理代写|粒子物理代写粒子物理学代考|张量分解。不可约张量

\begin{aligned} T_{i j} &=\frac{1}{2}\left(T_{i j}+T_{j i}\right)-\frac{1}{3} \delta_{i j} T_{k k}+\frac{1}{2}\left(T_{i j}-T_{j i}\right)+\frac{1}{3} \delta_{i j} T_{k k} \ & \equiv S_{i j}+A_{i j}+\frac{1}{3} \delta_{i j} T_{k k} \end{aligned}

(b) $A_{i j}$在$O(3)$下彼此之间线性变换，形成一个双指标的反对称张量，它等价于一个伪向量 确实，通过定义$\mathcal{A}i \equiv \epsilon{i j k} A_{j k}$并利用$\epsilon$ -符号的属性，我们验证了$\mathcal{A}i^{\prime}=(\operatorname{det} R) R{i k} \mathcal{A}k$。(c)五个独立的量$S{i j} \equiv(1 / 2)\left(T_{i j}+T_{j i}\right)-(1 / 3) \delta_{i j} T_{k k}$，形成对称无迹双指标张量。${ }^{13}$

$$\mathbf{3} \otimes \mathbf{3} \cong \mathbf{5} \oplus \mathbf{3} \oplus \mathbf{1}$$
，很快就会清楚，这是一般约简公式(5.58)的一个例子

## 物理代写|粒子物理代写粒子物理学代考| SO(3)的李代数

$\mathbf{S O}(3)$的李代数。$S O(3)$的群元素依赖于三个自变量，三个旋转角度，因此李代数$\mathfrak{c o}_{(3)}$有三个生成器。另一方面，我们从Cartan分类中知道，只有一个独立的三维李代数，就是我们和$S U(2)$群一起研究的那个。由此可见，我们有一个同构$\mathfrak{f u}(\mathbf{2}) \cong \mathfrak{f}(\mathbf{0})$。在问题5.3中，我们要求读者通过显式计算来验证这一点。这是另一个例子说明两个不同的群可能有相同的李代数。我们研究了这个代数的表示，并发现它们由一个索引$j$标记，该索引可以接受整数或半整数值。通过求幂，哪一种表示法可以表示$S O(3)$ ?组$S O(3)$的特点是$2 \pi$的旋转等同于不旋转。这一财产应由其所有代表加以保全。这意味着任何表示形式中的旋转矩阵都应该满足$\exp {2 \mathrm{i} \pi \boldsymbol{n} \cdot \boldsymbol{J}}=\mathbf{1}$。应用于围绕$z$轴的旋转，它意味着$J_3$应该具有整型特征值。因此，在代数的所有表示中，只有整数$j$对应的表示通过$S O(3)$的幂表示产生。为了得到这个结果，我们必须使用$S O(3)$的全局属性，并用标识符标识$2 \pi$的旋转

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