# What is the spin rotation operator for spin > 1/2?

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:

$$R_\alpha(\textbf{n})=\cos\biggl(\frac{\alpha}{2}\biggr)-i\vec{\sigma}\cdot\textbf{n}\sin\biggl(\frac{\alpha}{2}\biggr)$$

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. –  Vašek Potoček Oct 28 '12 at 22:32
For example, in the case $s = 1$ with the representation @Arnold Neumaier suggested, see en.wikipedia.org/wiki/…;. You can identify terms involving $\cos\theta$, $\sin\theta$ and constants in the matrix elements. The result gets even more complex for higher spins. –  Vašek Potoček 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 $S_z = \mathord{\rm diag}(-s,-s+1,\ldots,s-1,s)$. –  Vašek Potoček 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 $X(a)b=a \times b$.
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. –  Vašek Potoček Oct 25 '12 at 22:30