My question is regarding eq.(3.39) in the second edition of Georgi's book (for those who have the book:)). The section deals with tensor product states where the states comprising the product transform under different representations of a given group. Quick summary of the setting before the equation: we look at an element g of some group and the representation given by D(g). We assume we have some state $|i,x\rangle = |i\rangle \otimes |x\rangle$ where $|i\rangle$ transforms under the representation $D_1(g)$ and $|x\rangle$ under $D_2(g)$. Georgi states (I can follow so far) that under a group operation $D(g)$ the state transforms as: $$D(g)|i,x\rangle = |j\rangle \otimes |y\rangle \left[D_{1\otimes 2}(g)\right]_{jyix} = |j\rangle \otimes |y\rangle \left[D_1(g)\right]_{ji} \left[D_2(g)\right]_{yx}. \;\;\; (1)$$ Here, the Einstein sum convention was used (and is from now on implied) and the subscripts denote matrix indices ... so far so good. Now the problematic part: Taking infinitesimal $\alpha_a$, Georgi looks at the transformation properties near the identity using generators $J_a$ and writes:

$\begin{align} (1+ i \alpha_a J_a) |i,x\rangle &= |j,y\rangle \langle j,y| (1+ i \alpha_a J_a) |i,x\rangle \\ &= |j,y\rangle (\delta_{ji}\delta_{yx} + i \alpha_a\left[ J_a^{1 \otimes 2}(g)\right]_{jyix}) \\ &=|j,y\rangle \left( \delta_{ji} + i \alpha_a [J_a^1]_{ji} \right)\left( \delta_{yx} + i \alpha_a [J_a^2]_{yx} \right). \;\;\; (2) \end{align} $

I cannot wrap my head around the last line ... using eq.(1), my result is $$ |j,y\rangle \left( \delta_{ji}\delta_{yx} + i \alpha_a [J_a^1]_{ji}[J_a^2]_{yx} \right) $$ which seems to be wrong. I would appreciate hints on what I am doing/understanding wrongly.


1 Answer 1


The short answer is : when we are taking an infinitesimal transformation $1 + i\alpha_a J^a$, what we are actually doing is taking a derivative of $g\mapsto D[g]$ at $g = 1$. Since $D_{1\otimes 2}[g] = D_1[g]\otimes D_2[g]$ involves a product, it is only natural (by Leibniz rule) that its derivative $J^a_{1\otimes 2}$ involves a sum.

Now, in the abstract Lie group / Lie algebra, we write : $$g = 1 + i\alpha_a T^a$$ so that in the representation : $$D[g] = 1 + i\alpha_a J^a$$

For the tensor product, we have : \begin{align} 1 + i\alpha_a J_{1\otimes 2}^a &= D_{1\otimes 2}[1 + i\alpha_aT^a] \\ &= D_{1}[1 + i\alpha_aT^a] \otimes D_{2}[1 + i\alpha_aT^a] \\ &= (1 + i\alpha_a J_{1}^a )\otimes (1 + i\alpha_a J_{2}^a )\\ &= 1 + i \alpha_a (J_1^a + J_2^a) \end{align} discarding the terms of order $\alpha^2$.

Therefore, we see that : $$J^a_{1\otimes 2} = J^a_1 + J^a_2$$

  • $\begingroup$ I see, thank you very much. To me, this is clearer than the way it is written in the book. Cheers! $\endgroup$ Apr 14, 2022 at 16:36

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