G-parity - where does the minus sign come from?

[Federica]

Hi all,

I have a question on G-parity. I know it's defined as $G = exp(-i\pi I_{y})C$, with $I_y$ being the second component of the isospin and $C$ is the C-parity. In other words, the G-parity should be the C-parity followed by a 180° rotation around the second axis of the isospin.

Now, if I apply C on the isospin doublet $\begin{pmatrix} u\\ d \end{pmatrix}$ I get $\begin{pmatrix} \bar{u}\\ \bar{d} \end{pmatrix}$, which is quite clear.

If I apply the G-parity on the same doublet I get: $-\begin{pmatrix} \bar{d}\\ \bar{u} \end{pmatrix}$. I understand $\bar{d}$ and $\bar{u}$ are now inverted because of the rotation around $I_y$, but where does the minus sign come from?

 

[Vanadium 50]

but where does the minus sign come from?

The definition.

I am not sure what kind of answer would satisfy you. Why isn't "the definition" a good answer?

 

[Federica]

The definition.

I am not sure what kind of answer would satisfy you. Why isn't "the definition" a good answer?

I understand why the third component of the isospin changes its sign, that's because of the rotation. But why should I put a minus sign in front of the whole doublet?

 

[Vanadium 50]

Sorry, I was confused about which minus sign you meant.

You are not in an eigenstate of G with the doublet you wrote. So -1 can't be the eigenvalue because there isn't an eigenvalue.