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I am reading the book on group theory and stuck with a simple problem. Why $$(2\bigotimes2)\bigoplus(2\bigotimes1)\bigoplus(1\bigotimes2)\bigoplus(1\bigotimes1)=3\bigoplus1\bigoplus2\bigoplus2\bigoplus1$$ is true? Here the digits denote irreducible representations of $SU(2)$ group of given dimension. As far as I know, $m\bigotimes n=(m+n)\bigoplus(m+n-1)\ ... \bigoplus|m-n+1|\bigoplus|m-n|$. So it follows that $2\bigotimes2=4\bigoplus3\bigoplus2\bigoplus1\bigoplus0$ and etc. So my answer is: $$4\bigoplus3\bigoplus2\bigoplus1\bigoplus0\bigoplus3\bigoplus2\bigoplus1\bigoplus3\bigoplus2\bigoplus1\bigoplus2\bigoplus1\bigoplus0$$ Could someone please point out the mistake? Thanks in advance.

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  • $\begingroup$ Your formula following "as far as I know.." can't be right because the left side is a representation of dimension $mn$ while the right is a representation of dim $m+n + (m + n - 1) + ... + |m-n|$. So, e.g., when $m = n = 2$, you get $4 = 3 + 2 + 1$, which is obviously false. $\endgroup$ – Jason DeVito Jan 18 at 21:57
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We have $2 \bigotimes 2 = 3 \bigoplus 1$, $2 \bigotimes 1 = 1 \bigotimes 2 = 2$ and $1 \bigotimes 1 = 1$. Note how the identities would be valid if we hade ordinary times and plus (e.g. $2 \times 2 = 3 + 1$). This gives $$ (2 \bigotimes 2) \bigoplus (2 \bigotimes 1) \bigoplus (1 \bigotimes 2) \bigoplus (1 \bigotimes 1) = (3\bigoplus1) \bigoplus 2 \bigoplus 2 \bigoplus 1. $$

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  • $\begingroup$ Thanks a lot! It seems i confused one formula with another. Actually the formula for direct product of two representations is for number j labeling the representation, so the dimension is 2*j+1 $\endgroup$ – Ismail Gadzhiev Jan 18 at 22:10

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