Why is it claimed that the spin of ${}^{60} \text{Co}$ is reversed under parity in Wu's experiment? Lee and Yang proposed Wu’s experiment to check whether parity is conserved during beta decay. According to Wikipedia, the experiment works because spin is reversed under a parity transformation.
However, it seems like spin shouldn’t be reversed under parity, because 
$$\vec{L}=\vec{r}\times \vec{p} = -(\vec{r})\times (-\vec{p}).$$
Why does Wikipedia, and some textbooks such as Griffiths, claim the spin is reversed under parity?
 A: This is a pretty unfortunate issue that come with some pop-science explanations of parity. Parity is the operation of 'reflection about a point', i.e. it maps
$$(x, y, z) \to (-x, -y, -z).$$
However, this is a bit complicated to imagine, so pop scientists instead talk about 'reflection in a mirror'. For example, if the mirror is the $xy$ plane, then
$$(x, y, z) \to (x, y, -z).$$
This is equivalent, because as far as we can tell, our world is rotationally invariant, and parity and mirror reflection differ by a $180^\circ$ rotation about the $z$ axis,
$$(x, y, z) \to (-x, -y, z).$$
So a theory not symmetric under parity is not symmetric under mirror reflection, and vice versa.
Under the standard conventions for parity, $\mathbf{L}$ stays the same, as you proved. We can get the same thing by thinking of parity as a mirror reflection plus a $180^\circ$ rotation about the axis perpendicular to the mirror. If $\mathbf{L}$ is perpendicular to the mirror, neither a mirror reflection nor the rotation change it, as you can check by drawing a picture. But when $\mathbf{L}$ is parallel to the mirror, both the mirror reflection and the rotation flip it. Wikipedia and Griffiths only consider the mirror flip, in this latter case.
A: In relativistic Quantum mechanics the spin in an arbitrary direction is not a conserved quantity or it doesn't commute with the Hamiltonian. That is for $S_z$ you cannot choose an arbitrary direction for $z$ and use it as a good quantum number as in case of the Schrodinger Hamiltonian where $S.\vec{n}$ is a conserved quantity for an arbitrary $\vec{n}$. 
What is conserved however in relativistic QM  is $$ S . \vec{p} $$. That is the component of spin along the direction of momentum. So consider the following reaction and its parity transformed reaction.
$$\Lambda \to p + \pi^{-}$$
Here is a diagram representing a parity non conserving decay.

the solid arrows represent spin direction. Notice that while spin is still aligned along the same direction but relative to the direction of momentum it has flipped. So $S.\vec{p}$ has changed its sign for the proton.
Now if you compute the probabilities for both these reactions then you would find that the first reaction is more probable than the second thus breaking parity symmetry.
