# Which is pion(0) makeup?

1. Mar 18, 2006

### maverick6664

In my book (Greiner's Quantum Mechanics, vol2, symmetries) says after calculation with Clebsch-Gordan coefficients,

$$|\pi^0\rangle = \frac 1 2 (u\uparrow \overline{u} \downarrow + d\uparrow \overline{d} \downarrow - u\downarrow \overline{u} \uparrow - d\downarrow \overline{d} \uparrow ),$$

And I confirmed it.

Ignoring spin, It's $$|\pi^0\rangle = \frac 1 {\sqrt{2}} (u \overline{u} + d \overline{d})$$

However, some sites denote, it's the same: the sum (ex, this one, but others denote it's different; minus sign (ex. this one).

I wonder which is correct. I konw Internet resource is sometimes incorrect. And the latter is wiki... Or they mean the same? because $$u \overline{u}$$ and $$d \overline{d}$$ are orthogonal.

But if I do $$\hat{T_-}|\pi^+}\rangle = \hat{T_-}u\overline{d} = \frac 1 {\sqrt{2}} (u\overline{u} + d\overline{d}),$$ only plus is correct. (phase is ignored and each hand is normalized.)

So will anyone give me any hint which is correct or both are correct? I think at least plus is correct.

Last edited: Mar 19, 2006
2. Mar 19, 2006

### Meir Achuz

That sign is different in different books.
Many times the sign doesn't matter in a particular calculation if yhou are consistent throughout.
I think the minus sign is more common.
That is related to the common usage that charge conservation turns the d into -dbar (if I remember it correctly).
The Ispin doublet for the antiquarks is (ubar,-dbar).
Just be consistent in whatever you do.
I think the sign is only important in getting the G parity right.

Last edited: Mar 19, 2006
3. Mar 19, 2006

### maverick6664

Thank you for the reply. I don't understand exactly what you mean right now (why ispin doublet for the antiquarks is $$(\overline u,-\overline d)$$. In my understanding, it's $$(\overline u, \overline d)$$ and it must make difference), but I'll proceed keeping it in mind, because I don't have a book denoting minus explicitly...

Thanks.

Last edited: Mar 19, 2006