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When we are applying Group Theory to see whether the matrix element $\langle i|H'|f\rangle$ vanishes, we look at how the matrix element transforms. It can be shown that the matrix elements transforms as follows $$ \Gamma = \Gamma_i\otimes\Gamma_j\otimes\Gamma_f $$ where $\Gamma_i$ is the irreducible representation of the state $i$, $\Gamma_j$ is the reducible (can be irreducible) representation of the perturbed Hamiltonian and $\Gamma_f$ is the irreducible representation of the state $f$. It is said that the matrix element vanishes if $\Gamma$ does not contain the identity representation.

What is the intuition behind it? I have been told that we require the matrix element to transform as a scalar (essentially be invariant) under the symmetry operations which would mean that the matrix element corresponds to the identity representation. However, I don't find this answer satisfactory as we only require $\Gamma$ to contain the identity representation. $\Gamma$ can also contain different irreducible representations which means that it can also transform other than the identity representation. What am I missing ?

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This matrix element $\langle i\vert H'\vert f\rangle$ is evaluated as a mapping from your Hilbert space to the field of complex numbers.

Based on symmetry alone, the corresponding tensor product may have non-trivial transformation properties. However, the matrix element is evaluated as an inner product, and using the definition of this inner product, we conclude that all terms with non-trivial transformation properties must be zero as they do not belong to the set of complex numbers. We are then only left with terms transforming as the identity representation.

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  • $\begingroup$ What fo you mean by the matrix product ? Do you mean the tensor product of the corresponding matrices ? $\endgroup$
    – emir sezik
    Nov 25, 2021 at 9:56
  • $\begingroup$ We evaluate the matrix element as an inner product of the bra and ket vector and then argue based on the properties of this inner product. I will make an edit. $\endgroup$
    – B. Brekke
    Nov 25, 2021 at 10:05
  • $\begingroup$ Ah I see, thank you very much ! $\endgroup$
    – emir sezik
    Nov 25, 2021 at 10:16

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