- #1

CAF123

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## Homework Statement

A general rotation through angle ##a## about the axis ##\underline{n}##, where ##\underline{n}^2 = 1## is given by $$R(a,\underline{n}) = \exp(-ia\underline{n} \cdot \underline{T}),$$ where ##(T_k)_{ij} = -i\epsilon_{ijk}##. By expanding the exponential as a power series in ##a##, and explicitly summing the resulting series, show that $$R_{ij}(a,\underline{n}) = \delta_{ij}\cos a + n_i n_j (1-\cos a) - \epsilon_{ijk} n_k \sin a$$ using ##(\underline{n} \cdot \underline{T})_{ij}^2 = \delta_{ij} - n_i n_j## and ##(\underline{n} \cdot \underline{T})_{ij}^3 = (\underline{n} \cdot \underline{T})_{ij}##, (which I derived earlier)

## Homework Equations

##\exp(ix) = \cos x + i\sin x## and cos and sin power series.

## The Attempt at a Solution

$$R(a,\underline{n})_{ij} = [\cos(a (\underline{n} \cdot \underline{T})) - i\sin(a (\underline{n} \cdot \underline{T}))]_{ij}$$ Reexpressing in terms of the Taylor series for sin and cos: $$\left[\sum_{k=0}^{\infty} (-1)^k \frac{(a (\underline{n} \cdot \underline{T}))^{2k}}{(2k)!} - i \sum_{k=0}^{\infty} (-1)^k \frac{(a (\underline{n} \cdot \underline{T}))^{2k+1}}{(2k+1)!}\right]_{ij}$$ Reorganising the even and odd terms gives $$\left[\sum_{k=0}^{\infty} (-1)^k \frac{(\delta_{ij} - n_i n_j)^k}{(2k)!} - i \sum_{k=0}^{\infty} (-1)^k\frac{[(\delta_{ij} - n_i n_j)^{2k} (-i\epsilon_{ijk}n_k)]}{(2k+1)!}\right]$$

I am not really sure how to progress. Thanks for any help.