# I Wigner-Eckart theorem

#### Silviu

Hello! I am a bit confused about the dimensionality of the vectors in Wigner-Eckart theorem. Here it is how it gets presented in my book. Given a vector space V and a symmetry group on it G, with the representation U(G) we have the irreducible tensors $${O_i^\mu,i=1,...,n_\mu}$$ (where $n_\mu$ is the dimension of the $D^\mu$ irreducible representation (irrep) of G) having the property that $$U(g)O_i^\mu U(g)^{-1}=O_i^\mu D^\mu(g)^j_i$$ for all $g\in G$. Now for a set of irreducible tensors $O_i^\mu$ and a set of orthonormal vectors $e_j^\nu$, vectors which span an invariant subspace of V, we have: $$O_i^\mu |e_j^\nu>=O_k^\mu |e_l^\nu>D^\mu(g)^k_i D^\nu(g)^l_j$$ which shows that the vectors $O_i^\mu |e_j^\nu>$ transform under the direct product representation $D^{\mu \times \nu}$. Then using Clebsch-Gordan coefficients, we can diagonalize $D^{\mu \times \nu}$ and implicitly decompose the vector space it acts on into invariant subspaces. Thus we get: $$O_i^\mu |e_j^\nu>=\sum_{\alpha,\lambda,l}|w^\lambda_{\alpha l}><\alpha,\lambda,l(\mu \nu)i,j>$$ where $<\alpha,\lambda,l(\mu \nu)i,j>$ are the CG coefficients and $|w^\lambda_{\alpha l}>$ are the basis vectors corresponding to the irrep $D^{\lambda}$ in the decomposition of $D^{\mu \times \nu}$. Lastly, to get the Wigner-Eckart theorem they calculate this product: $$<e^l_\lambda|O_i^\mu |e_j^\nu>$$ and here it is where I get confused. The vector (actually I think it becomes a one form, but anyway) $<e^l_\lambda|$ is in the vector space V (it has lot's of zeros as it is invariant under a certain subspaces, but still in V). However $O_i^\mu |e_j^\nu>$ is in the vector space associated with $D^{\mu \times \nu}$. Now the first one would have the dimension the same as V while the second one will have a dimension $n_\mu \times n_\nu$ which don't have to be equal. How can you take the dot product of 2 vectors with different dimensionalities? What am I missing here? For reference the book I am using is Group Theory in Physics by Wu-Ki Tung and this is presented towards the end of chapter 4. Thank you!

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