Why do \pi\^{\pm} have different charges than \pi\^{0}?

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SUMMARY

The discussion clarifies the charge differences among pions, specifically \(\pi^{\pm}\) and \(\pi^{0}\). The \(\pi^{+}\) consists of one up quark and one anti-down quark, resulting in a charge of +1e, while the \(\pi^{-}\) contains one down quark and one anti-up quark, yielding a charge of -1e. In contrast, the \(\pi^{0}\) is a neutral particle formed from a linear combination of up and down quark pairs, leading to a net charge of 0. The confusion arises from misinterpretations of quark contributions to charge.

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Dahaka14
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How is it that [tex]\pi\^{\pm}[/tex] have charges [tex]\pm e[/tex] and [tex]\pi\^{0}[/tex] has a charge of 0? The [tex]\pi\^{+}[/tex] has one up quark and an anti-down quark, which doesn't add up to zero (same deal goes for [tex]\pi\^{-}[/tex], and the [tex]\pi\^{0}[/tex] has a linear combination that doesn't seem to add up to zero.

Where have I gone wrong?
 
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pi + has as you say, one up and one anti-down which gives you +2/3 + 1/3 = 1

You are saying (The [tex]\pi ^+[/tex]
has one up quark and an anti-down quark, which doesn't add up to zero ) But the pi + should have CHARGE = 1e! As you said in the sentence before this one.

Maybe try to be more consistent to yourself.. try again.
 

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