0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ No. It is typically reducible, cf. Clebsch-Gordan coefficients. $\endgroup$ – AccidentalFourierTransform Nov 6 '19 at 16:19
  • $\begingroup$ @AccidentalFourierTransform if we assume that F is a 1-dimensional representation of G then is the direct product an irreducible representation of G? $\endgroup$ – user243882 Nov 6 '19 at 16:34
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.