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

  • #1
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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?
 

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  • #2
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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?
 
  • #3
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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?
 
  • #4
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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.
 

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