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5

For the case $SU(n)$, $n>2$ the matrix and its inverse are not related by a similarity transform, so the representation where one acts with $g$ and with $g^{-1}$ are not isomorphic. For $SU(2)$ you can check that $$g^{-1}=EgE^{-1} $$ where $$E=\begin{vmatrix} 0 & 1 \\ -1 &0 \end{vmatrix}$$ This means there is no reason to act with $g^{-1}$. ...


4

Note that the finite transformation of: $$ W^a_\mu \to W^a_\mu + \frac{1}{g} \partial_\mu \theta^a + \epsilon^{abc} \theta^b W^c_\mu $$ is: $$ W^a_\mu t^a \to g W_\mu^a t^a g^{-1} + \frac{i}{g} \partial_\mu g \tag{1} $$ where: $$ g = \exp(-i \theta^a t^a) \;\;\; \text{and} \;\;\; [t^a,t^b] = i \epsilon^{abc} t^c $$ Thus, the first term on the right-hand ...


2

Consider a theory of fields $\phi:M\to T$ where $M$ is a manifold, and $T$ is a set. In physics, $T$ is often either a vector space or a manifold. We call $M$ the domain of the theory, and we call $T$ the target space. of the theory. We call a function from $M$ to $T$ a field configuration, and the set of all field configurations is denoted $\mathcal F$. ...


5

Using the Littlewood-Richardson (LR) rules for Young tableaux, one may show that $$ \begin{array}{c} [~~]\cr [~~] \end{array} \quad\otimes\quad \begin{array}{c} [a]\cr [b] \end{array} \quad=\quad\begin{array}{c} [~~]\cr [~~] \cr [a]\cr [b]\end{array} \quad\oplus\quad\begin{array}{rl} [~~]&[a]\cr [~~]&\cr [b] \end{array} ...


2

You need to work out the tensor product and will find a direct sum of different contributions \begin{multline} [(1/2, 0) \oplus (0, 1/2)] \otimes [(1/2, 0) \oplus (0, 1/2)] =\\ \big((1/2, 0) \otimes (1/2, 0)\big) \oplus \big((1/2, 0) \otimes (0, 1/2) \big)\oplus \quad \\\big((0, 1/2) \otimes (1/2, 0)\big) \oplus \big((0, 1/2) \otimes (0, 1/2)\big) = \\ (0, ...


0

short answer if $ \hat {S}^{-1} S = \mathbb{I}$ I can give you a general example of $\psi^\dagger\psi$ not being invariant. because for Dirac spinor $\psi$ whe have the following transformation rules $$\psi(x) \rightarrow S[\Lambda] \psi(\Lambda^{-1}x)=S[\Lambda] \psi(x^\prime) \\ \psi^\dagger(x) \rightarrow \psi^\dagger(\Lambda^{-1}x) ...


1

If we assume that $$\Psi {'} = \hat {S}\Psi$$ and $${\bar{\Psi}}{'} = \bar {\Psi} \hat {S}^{-1},$$ it follows that the product of the two transforms as $$(\bar{\Psi}\Psi)'={\bar{\Psi}}{'}\Psi {'}=\bar {\Psi} \hat {S}^{-1}\hat {S}\Psi=\bar{\Psi}\Psi,$$ which is a consequence of $$\hat {S}^{-1}\hat {S}=\mathbb{1}.$$



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