# generalizing spin rotations

I have a question about the relation: $\exp(-i \vec{\sigma} \cdot \hat{n}\phi/2) = \cos(\phi/2) - i \vec{\sigma} \cdot \hat{n} \sin(\phi/2)$.

In my texts, I see $\phi\hat{n}$ always as c-numbers. My question is whether or not this relation can be generalized for $\hat{n}$ being an operator?

If so how exactly would the expression be different?

Thanks.

• Sure. What commutation relations do you want to impose on $[n^i\phi, n^j\phi]$ ? If they are still zero, then the relation remains the same (up to minor notational issues, assuming for simplicity that $n^i\phi$ isn't an operator in the $\sigma^j$ Lie algebra sense, but even that possibility can be dealt with). – Qmechanic Jun 24 '11 at 22:01

In your expression, $\hat n$ is a unit vector. For a general vector $\vec{n}$ of length $|\vec{n}|$, the expression would be $$\exp{(-i\vec{\sigma}\cdot\vec{n}\phi/2)}=\cos(|\vec{n}|\phi/2)-i\frac{\vec{\sigma}\cdot\vec{n}}{|\vec{n}|}\sin(|\vec{n}|\phi/2).$$
For a matrix case, a reasonably natural assumption would be that $\vec{\sigma}\cdot\vec{N}$ is a matrix constructed as a sum of tensor products, $\vec{\sigma}\cdot\vec{N}=\sigma_1\otimes N_1+\sigma_2\otimes N_2+\sigma_3\otimes N_3$. Other constructions are possible, but the end result will presumably be a matrix except in special cases. The matrix obtained would not necessarily be Hermitian, in which case the expansion of $\exp{(-i\vec{\sigma}\cdot\vec{N}\phi/2)}$ would include $\cosh$ and $\sinh$ components as well as $\cos$ and $\sin$ components. For a matrix $M$ that has eigenvalues $m_i$, $\exp{(M)}$ has eigenvalues $\exp{(m_i)}$, in the same basis, which can be used to expand the matrix $\exp{(M)}$ with $M=-i\vec{\sigma}\cdot\vec{N}\phi/2$.
The expression for the vector $\hat{n}$ comes out so nicely because the matrix $\vec{\sigma}.\hat{n}$ only has eigenvalues $\pm 1$. If one ensures by one's choice of matrices $\vec{N}$ and by one's choice of the construction $\vec{\sigma}\cdot\vec{N}$ that the resulting matrix only has eigenvalues $\pm 1$, one would have the same happy simplicity.
• Thanks @PeterMorgan. Yes, I see now that the type of operator would be important. I am thinking of two spins so the matrix would be another spin operator so $\vec{S} \cdot \vec{I}$ where S and I commute but the eigenvalues would not be +/-1 – BeauGeste Jun 25 '11 at 0:02