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For spin $\frac{1}{2}$, the spin rotation operator $R_\alpha(\textbf{n})=\exp(-i\frac{\alpha}{2}\vec{\sigma}\cdot\textbf{n})$ has a simple form:


What about spin > $\frac{1}{2}$ ?

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I'm pretty sure the general form for the exponential would be too complicated. The first observation is that the spectrum of $-i\alpha\vec\sigma\cdot\bf n$ in a system with spin $s$ is $(-i\alpha(-s),-i\alpha(-s+1),\ldots,-i\alpha(s-1),-i\alpha s)$ and thus goniometric functions of arguments $\frac12\alpha$ through $s\alpha$ for half-integer spins, or $0$ (i.e., a constant term) through $s\alpha$ for whole spins would be present. – The Vee Oct 28 '12 at 22:32
For example, in the case $s = 1$ with the representation @Arnold Neumaier suggested, see…;. You can identify terms involving $\cos\theta$, $\sin\theta$ and constants in the matrix elements. The result gets even more complex for higher spins. – The Vee Oct 28 '12 at 22:37
The reason the exponential has so simple form in the case $s=\frac12$ is that $\frac\alpha2$ is the only frequency allowed by the analysis I posted above. It helps here that the powers of $\vec\sigma\cdot\bf n$ are always either identity or the original matrix. In higher dimensions, the set $\{S_x^k\}^n$ trajects a $2s+1$-dimensional space in a non-periodic fashion. Consider $S_z = \mathord{\rm diag}(-s,-s+1,\ldots,s-1,s)$. – The Vee Oct 28 '12 at 22:41

The same, except that the $\sigma_k$ are now not Pauli matrices but the generators of a su(2) representation of the desired spin. For example, the $3\times 3$ matrices $$ \sigma_\ell:=(2\epsilon_{jk\ell})_{j,k=1:3}$$ define the spin 1 representation on 3-vectors. [Maybe the factor 2 should take a different value.] The corresponding explicit formula comes from the Rodrigues formula $$e^{X(a)}=1+\frac{\sin|a|}{|a|}X(a)+\frac{1-\cos|a|}{|a|}X(a)^2,$$ where $X(a)$ is the matrix mapping a vector $b$ to $X(a)b=a \times b$.

For higher spin, the corresponding formula will depend on how you write the representation. Numerically, one would just diagonalize the matrix in the exponent; then computing the exponential is trivial. I don't know whether for general spin if there is any advantage in having an explicit formula.

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Well, the exponential form is the same, but the exponential won't be computed in the same easy formula, using just one $\cos$ and one $\sin$ of the half angle. That was justified by $-i\frac\alpha2\vec\sigma\cdot\bf n$ having just two pure imaginary eigenvalues of opposite sign. The generators in higher-dimensional representations will have an accordingly higher number of eigenvalues, e.g., an additional $0$ in the case you used for an example. – The Vee Oct 25 '12 at 22:30
Of course I am asking about the analogous formula for the expansion of the exponential in terms of cosines and sines not about the spin matrix ! – Tarek Oct 26 '12 at 9:16
One cannot guess from your question what you want unless you write it down clearly. Maybe you wish to update your question. – Arnold Neumaier Oct 27 '12 at 7:59

What one has for the spin 1/2, \begin{equation} \exp(i\alpha \mathbf{J}\cdot\hat{\mathbf{n}}) = \cos(\alpha/2) + i\sin(\alpha/2)\mathbf{J}\cdot\hat{\mathbf{n}}, \end{equation} as well for spin 1 through Rodrigues formula, \begin{equation} \begin{aligned} \exp(i\alpha \mathbf{J}\cdot\hat{\mathbf{n}}) & = 1 + i\hat{\mathbf{n}}\cdot\mathbf{J}\sin\alpha + (\hat{\mathbf{n}}\cdot\mathbf{J})^2(\cos\alpha-1) \\ & = 1 + \left[2i\hat{\mathbf{n}}\cdot\mathbf{J}\sin(\alpha/2)\right]\cos(\alpha/2) + \frac{1}{2}\left[2i\hat{\mathbf{n}}\cdot\mathbf{J}\sin(\alpha/2)\right]^2, \end{aligned} \end{equation} is a spin representation of rotation operators as finite order polynomials of the rotation generators for $j=1/2,1$, where the coefficients are sines and cosines of half the angle of rotation. It was known that this could be extended for higher spin representations, but the exact polynomial expression for any spin $j$ remained unknown. Fortunately, in 2014 this general expression was found by Curtright, Fairlie & Zachos. I leave here their publication:

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