# Mirror/Parity symmetry in spin

We just saw parity symmetry and we were told about the experiments to see the non parity symmetry of disintegration, in particular one involving the reaction:

$$^{60}Co\longrightarrow^{60}Ni+ e + \bar \nu$$

Now, we were asked to check that if we prepare the system so that the spin is parallel to the $z$ axis and make a mirror symmetry in the $x$ would flip the spin, going from $|jm\rangle$ to $|j-m\rangle$.

I can't prove.

We were given as a clue, to have in mind the fact that the transformation is:

$$\left(\matrix{ -1&0&0\\ 0&1&0\\ 0&0&1 }\right)= \left(\matrix{ -1&0&0\\ 0&-1&0\\ 0&0&-1 }\right) \left(\matrix{ 1&0&0\\ 0&-1&0\\ 0&0&-1 }\right)$$

In the decomposition, the first matrix is the parity matrix already studied, while the second one is a rotation of $\pi$ around the x axis.

How can I check that the spin actually flips after the reflection?

First edition: I thought I could prove it seeing that the wave function of the orbital angular momentum is $Y_l^m$, and that the spin operator is an angular momentum operator, so we can use the symmetry of those functions:

$$Y_l^m=\alpha e^{im\phi}P_l^m(\cos\theta )$$ And so we have that the mirror symmetry is $\phi\mapsto \phi+\pi$, and so $Y_l^m\mapsto Y_l^{-m}$. This doesn't convince me because spin is not about spacial properties of a particle.

Take the rotation operator written in the Pauli formalism: $$exp\left(\frac{-i\sigma\cdot\textbf{n}\phi}{2}\right)$$ For a $\pi$ rotation around $\textbf{x}$ the rotation matrix takes form \begin{pmatrix} 0 & -i \\ -i & 0 \\ \end{pmatrix} from which you can see that multiplying by any spinor you would get it flipped.