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Let $$\rho_\theta \equiv \rho(R_\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\tag{1}$$ be a representation of $SU(2)$, and consider the tensor product representation $\rho_\theta \otimes \rho_\theta$. We want to write it in the basis $$E_1 = e_{11} + e_{22}, \quad E_2 = e_{11} - e_{22}, \quad E_3 = e_{12} + e_{21}, \quad E_4 = e_{12} - e_{21},\tag{2}$$ where $e_{ij} \equiv \mathbf e_i \otimes \mathbf e_j$, and $\{\mathbf e_1, \mathbf e_2\}$ is a standard cartesian basis. The book I'm reading (Blennow - Mathematical Methods for Physics and Engineering, Example 4.35) states that $$\rho_\theta \otimes \rho_\theta E_1 = E_1,\tag{3}$$ $$\rho_\theta \otimes \rho_\theta E_2 = \cos(2\theta)E_2 - \sin(2\theta)E_3,\tag{4}$$ $$\rho_\theta \otimes \rho_\theta E_3 = \sin(2\theta)E_2 + \cos(2\theta)E_3,\tag{5}$$ $$\rho_\theta \otimes \rho_\theta E_4 = E_4.\tag{6}$$ However, I get a different result for (4) and (5), namely $$\rho_\theta \otimes \rho_\theta E_2 = \cos(2\theta)E_2 + \sin(2\theta)E_3,\tag{7}$$ $$\rho_\theta \otimes \rho_\theta E_3 = -\sin(2\theta)E_2 + \cos(2\theta)E_3.\tag{8}$$ It looks like a mixup with active and passive rotations, but I don't think that should be applicable here, seeing as we are given the matrix (1) explicitly. Let me show you how I computed (7), and maybe you can spot a mistake? Here goes: $$\begin{split} \rho_\theta \otimes \rho_\theta E_2 &= \rho_\theta \otimes \rho_\theta(\mathbf e_1 \otimes \mathbf e_1 - \mathbf e_2 \otimes \mathbf e_2)\\ &= (\rho_\theta \mathbf e_1) \otimes (\rho_\theta \mathbf e_1) - (\rho_\theta \mathbf e_2) \otimes (\rho_\theta \mathbf e_2)\\ &= (\cos(\theta) \mathbf e_1 + \sin(\theta) \mathbf e_2) \otimes (\cos(\theta) \mathbf e_1 + \sin(\theta) \mathbf e_2)\\ &\quad - (-\sin(\theta) \mathbf e_1 + \cos(\theta) \mathbf e_2) \otimes (-\sin(\theta) \mathbf e_1 + \cos(\theta) \mathbf e_2)\\ &= \cos^2(\theta) e_{11} + \cos(\theta)\sin(\theta)(e_{12} + e_{21}) + \sin^2(\theta)e_{22}\\ &\quad - \sin^2(\theta) e_{11} + \sin(\theta)\cos(\theta)(e_{12} + e_{21}) - \cos^2(\theta)e_{22}\\ &= (\cos^2(\theta) - \sin^2(\theta)) e_{11} + 2\cos(\theta)\sin(\theta)(e_{12} + e_{21}) - (\cos^2(\theta) - \sin^2(\theta)) e_{22}\\ &= \cos(2\theta) e_{11} + \sin(2\theta)(e_{12} + e_{21}) - \cos(2\theta) e_{22}\\ &= \cos(2\theta) E_2 + \sin(2\theta) E_3. \end{split}\tag{9}$$ Could it be that the matrix representation in (1) is meant to indicate how contravariant components transform, and the basis vectors, being covariant, transform inversely? If so, what is a practical way of computing the matrix elements of $\rho_\theta \otimes \rho_\theta$ in the given basis, seeing as the inverse might not always be as obvious as it is here?

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    $\begingroup$ Your equations (7) and (8) are the right ones. To the contrary, equations (4) and (5) that is (4.109b) and (4.109c) in the textbook respectively are wrong. I don't think it's a typo. May be he was using inadvertently the inverse matrix for $\rho_\theta$. $\endgroup$
    – Frobenius
    Commented Apr 17, 2021 at 10:07
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    $\begingroup$ Thank you for answering! Yes, that possibility crossed my mind as well. However, there is a similar issue in Example 4.51, equation (4.196), and it seems less likely that the author would make the same mistake twice. Or at the very least, it seemed more likely that I was the one making an error. $\endgroup$
    – ummg
    Commented Apr 17, 2021 at 15:07

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I just heard from the author of the textbook who confirms that it is indeed a typo. So equations (7) and (8) in my question are the correct ones.

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  • $\begingroup$ Also, if anyone else is reading the same textbook, there is a similar mistake in eq. (4.196). The correct matrix is the transpose of what is written. $\endgroup$
    – ummg
    Commented Apr 23, 2021 at 14:53

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