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I am attempting two exercises from pages 20 - 21 of Srednicki's QFT book and for some reason cannot reproduce the required results (this is not homework and I am just working through the problems to try to learn more QFT). The task is show that

$$\Lambda=\exp(-i \theta S^{12}_{v}/\hslash),$$

where $\Lambda$ is the Lorentz matrix for rotation by an angle $\theta$ about the $z$ axis and $(S^{\mu \nu}_v)^{\rho}_{\tau}$ are matrices which form the vector representation of the Lorentz generators.

$$(S^{\mu \nu}_v)^{\rho}_{\tau}=\frac{\hslash}{i}(g^{\mu \rho}\delta^{\nu}_{\tau}-g^{\nu \rho}\delta^{\mu}_{\tau}).$$

Start by plugging in $1$ and $2$ into the definition of the matrices and we get:

$$S^{12}_v=\hslash \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}=\hslash \sigma_2 $$

$\sigma_2$ has eigenvalues $+1$ and $-1$, so $(\sigma_2)^2=1$. This means it is a self-inverse operator. Taylor expansion of a self-inverse operator gives us:

$$\exp(itO)=\cos(t) + i (\sin t)O$$

$$\exp(-i\theta S^{12}_{\nu}/\hslash) = \exp(-i \theta \sigma_2)= \cos \theta - i (\sin \theta) \sigma_2$$

In the solutions to this textbook it also quotes this result in terms of sinusoidals but then goes straight to the required result. If you do the explicit matrix multiplication we get:

$$\exp(-i\theta S^{12}_{\nu}/\hslash)=\cos \theta \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} -i \sin \theta \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}= \begin{bmatrix} \cos \theta & 0 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & \cos \theta \end{bmatrix} $$

This is not exactly what the rotation matrix should be, as in the top-left and bottom-right the $\cos \theta$s should be $1$s. I don't see where the $1$s come from as we are multiplying the whole identity matrix by $\cos \theta$, can someone clarify what I am missing here?

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This one is easy. Your matrix is not self-inverse. Just check it yourself, squaring it gives $$\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}, $$

which is not the identity operator.

Instead, you should consider the block submatrix for the $xy$ plane. Your trigonometric result is only valid on this 2d subspace. Then use the formula for the exponential of the block-diagonal matrix to reconstruct the correct result.

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