# Exponentials and linear operators

I don't quite understand how the following works, could anyone please explain?

Let $O_i,i=1,2,3$ be linear operators on a space of functions.

They satisfy the commutation relation $[O_a,O_b]=O_aO_b-O_bO_a=i\epsilon_{abc}O_c$

Why then is $$\exp(-i\phi O_2)O_1\exp(i\phi O_2)=O_1\cos\phi-O_3\sin \phi$$?

I don't understand how to deal with the exponentials. I thought that maybe it had to do with Euler's formula, but I don't quite know how that works with the operator.

Thanks in advance!

I believe it is related to angular momentum operators in quantum mechanics.

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I think that what you need to use here first is that the exponential of a linear operator $A$ is given by $\exp(A)=1+A+A^2/2+\dots$ then play around with terms –  tibL Nov 19 '12 at 12:46

## 1 Answer

The operators $O_1,O_2,O_3$ form a representation of the lie algebra $\mathfrak{so}(3)$. These are also the Pauli spin matrices.

The exponentials generate rotations in $\mathbb{R}^3$ so that $e^{i\phi O_2} \in SO(3)$. Any rotation in 3D is is specified by an axis and angle of rotation. Conjugation by another rotation, changes the axis by rotation, but doesn't change the angle.

\begin{eqnarray} e^{i\theta O_3}O_1 e^{-i\theta O_3} &=& O_1\cos \theta - O_2 \sin \theta \\ e^{i\theta O_3}O_2 e^{-i\theta O_3} &=& O_1\sin \theta + O_2 \cos\theta \\ e^{i\theta O_3}O_3 e^{-i\theta O_3} &=& O_3\end{eqnarray}

This is similar to how two permutations conjugate one another $\sigma (1234)\sigma^{-1} = (\sigma(1)\sigma(2)\sigma(3)\sigma(4))$.

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Thank you, John! –  Darren Nov 20 '12 at 3:03