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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.

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  • $\begingroup$ 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). $\endgroup$ – Qmechanic Jun 24 '11 at 22:01
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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.

There are probably enough distinct constructions that one would like to have some external motivation, either from Physics or from some other branch of Mathematics, for introducing a specific choice.

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  • $\begingroup$ 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 $\endgroup$ – BeauGeste Jun 25 '11 at 0:02

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