Let us consider an infinitesimal rotation described by a vector
directed along the axis about which the rotation takes place and whose magnitude is the angle of the rotation. We then have

where
is the change in the position vector
of the particle due to such a rotation.
Let us now consider a function of position,
Substituting in the rotated coordinate and expanding to first order in
we obtain
![{\displaystyle \psi \left(\mathbf {r} +\mathbf {\delta } \mathbf {r} \right)=\left[1+\mathbf {\alpha } \cdot \left(\mathbf {r} \times \mathbf {\nabla } \right)\right]\psi \left(\mathbf {r} \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d99b2c31922a1803151291484289e7efaec5551f)
Figure 1: Illustration of the rotation considered here.
The expression,

may therefore be interpreted as an operator that performs an infinitesimally small rotation of position coordinates. Noticing that

we may write this infinitesimal rotation operator as

Note that this expression only applies to infinitesimal rotations. We may construct a rotation operator for finite rotations, however, by applying this operator
times and letting
go to infinity. Doing so, we get the rotation operator for finite angle

In this form, we recognize that angular momentum is the generator of rotation. And we can write the equation relating the initial vector before rotation with the transformed vector as

This expression of the rotation operator is also valid when the rotation angle is not infinitesimal. What's more, this equation also implies that if we have a scalar instead of
, it would be invariant. We can also calculate the effect of the unitary operator
on the states:


This is the wavefunction evaluated at a rotated point.
A sample problem