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.


1 Answer 1


First of all the parity operator commutes with the spin, so that doesn't affect it, and you just have to think about the rotation around x.

Take the rotation operator written in the Pauli formalism: \begin{equation} exp\left(\frac{-i\sigma\cdot\textbf{n}\phi}{2}\right) \end{equation} 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.


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