I was studying group theory yesterday and i had this question. If i have an irreducible representation D that belongs in G and another irreducible representation F that belongs in G, is it right to say that the direct product $D {\otimes} F$ is also an irreducible representation and if yes why? I thought of angular momentum, i.e $$|j_1,m_1>{\otimes}|j_2,m_2>=|j_1,m_1>|j_1,m_1>$$ which is an irreducible representation.
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2$\begingroup$ No. It is typically reducible, cf. Clebsch-Gordan coefficients. $\endgroup$– AccidentalFourierTransformCommented Nov 6, 2019 at 16:19
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$\begingroup$ @AccidentalFourierTransform if we assume that F is a 1-dimensional representation of G then is the direct product an irreducible representation of G? $\endgroup$– user243882Commented Nov 6, 2019 at 16:34
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In general the tensor product of two irreducible representations is reducible. The best example is the coupling of two spin-1/2 states, which give $$ \frac{1}{2}\otimes \frac{1}{2}=0\oplus 1\, . $$
If one of the representation is 1-dimensional, then the result will usually remain irreducible. For instance, the alternating representation ${\cal A}$ of $S_n$, when tensored with an irrep $\{\lambda\}$ of $S_n$, will give the irrep conjugate to $\{\lambda\}$. Clearly the trivial irrep tensored with any irrep will give this same irrep back and nothing else.
answered Nov 6, 2019 at 16:38