Understand Wigner-Eckart Theorem & Dimensionality of Vectors

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SUMMARY

The discussion centers on the dimensionality of vectors in the context of the Wigner-Eckart theorem as presented in "Group Theory in Physics" by Wu-Ki Tung. The key point is the transformation properties of irreducible tensors \(O_i^\mu\) under the symmetry group \(G\) and their relationship with orthonormal vectors \(e_j^\nu\). The confusion arises from the dot product between vectors from different dimensional spaces, specifically between \(\) in the space associated with the direct product representation \(D^{\mu \times \nu}\). The resolution lies in understanding the invariant subspaces and the application of Clebsch-Gordan coefficients for proper dimensional alignment.

PREREQUISITES
  • Understanding of group theory and symmetry groups in physics.
  • Familiarity with irreducible representations (irreps) and their properties.
  • Knowledge of Clebsch-Gordan coefficients and their role in decomposing representations.
  • Basic concepts of linear algebra, particularly vector spaces and inner products.
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  • Study the application of Clebsch-Gordan coefficients in quantum mechanics.
  • Explore the properties of irreducible representations in more depth.
  • Learn about the mathematical foundations of the Wigner-Eckart theorem.
  • Investigate the relationship between invariant subspaces and their dimensionality in group theory.
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Physicists, mathematicians, and students studying quantum mechanics or group theory, particularly those interested in the Wigner-Eckart theorem and its applications in theoretical physics.

Silviu
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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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