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1

The short answer is yes. One can convince oneself this is indeed the case by doing the dimensional counting as it was done by Everett You. However, it is by no means a proof. The problem is that the valence bond states are not linearly independent. Even though there are much more valence bond states than the number of singlets made from $N$ spin-one-half ...

0

A partial answer, is that supposing the gamma matrices, block-diagonal , as $\begin{pmatrix}A&\\&\epsilon A\end{pmatrix}, \begin{pmatrix}&A\\\epsilon A&\end{pmatrix}$, where $A$ is hermitian or anti-hermitian, and $\epsilon =\pm1$, give constraints on $A$ and $\epsilon$ due to $(\gamma^0)^2= \mathbb Id_4, (\gamma^i)^2= - \mathbb Id_4$. For ...

2

Yes, a spin-singlet state is also an RVB state. The valence bound states (singlet-product states) over-complete the Hilbert space of spin-singlet states.

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How about just testing the two different cases? I.e. if $\mu\not=0$ then the LHS becomes $$(\gamma^\mu)^\dagger= (\gamma^i)^\dagger= -\gamma^i$$ while the RHS becomes $$(\gamma^\mu)^\dagger=\gamma^0\gamma^i\gamma^0 = -\gamma^0\gamma^0\gamma^i=-\gamma^i~~~~~~~~ (\text{OK}).$$ For $\mu=0$, the case ...

3

First, if you take the fundamental representation (representation $N$) of $SU(N)$ made of $N$ objects $\varphi^i$, the transformation law is : $\varphi^i \to U^i{}_j \varphi^j$. By taking the complex conjugate, you get : $\varphi^{*i} \to (U^*)^i{}_j \varphi^{*j}= (U^\dagger)^j{}_i \varphi^{*j}$. Now, looking at the last expression with $U^\dagger$, one ...

3

Ok, I think there is a mistake here: A general tensor $\varphi^i$ transforms as: $$\varphi^i\rightarrow U^i_{\phantom{1}j}\varphi^j$$ whereas $\varphi_i$ transforms as: $$\varphi_i\rightarrow (U^\boldsymbol{\ast})_i^{\phantom{1}j}\varphi_j$$ Where did you find these equations? The unitary matrix element in the second line should not be a complex ...

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