9
$\begingroup$

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 > $\frac{1}{2}$ ?

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – The Vee
    Oct 28, 2012 at 22:32
  • $\begingroup$ 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. $\endgroup$
    – The Vee
    Oct 28, 2012 at 22:37
  • $\begingroup$ 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 $S_z = \mathord{\rm diag}(-s,-s+1,\ldots,s-1,s)$. $\endgroup$
    – The Vee
    Oct 28, 2012 at 22:41

2 Answers 2

6
$\begingroup$

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: http://arxiv.org/abs/1402.3541

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – The Vee
    Oct 25, 2012 at 22:30
  • $\begingroup$ 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 ! $\endgroup$
    – Tarek
    Oct 26, 2012 at 9:16
  • 1
    $\begingroup$ One cannot guess from your question what you want unless you write it down clearly. Maybe you wish to update your question. $\endgroup$ Oct 27, 2012 at 7:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.