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

[Dahaka14]

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?

 

[malawi_glenn]

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.

 

[sir-pinski]

You are correct in noting that the \pi\^{+} and \pi\^{-} have different charges because they are composed of different quarks. The \pi\^{+} is composed of an up quark and an anti-down quark, while the \pi\^{-} is composed of a down quark and an anti-up quark. These two combinations result in opposite charges, which is why the \pi\^{+} has a charge of +e and the \pi\^{-} has a charge of -e.

On the other hand, the \pi\^{0} is composed of both an up quark and a down quark, but in a specific linear combination called an isospin state. This linear combination results in a net charge of 0, even though the individual quarks have charges of +e and -e. This is because the up and down quarks have opposite charges, canceling each other out in the \pi\^{0}.

So, to answer your question, the difference in charges between the \pi\^{+}, \pi\^{-}, and \pi\^{0} is due to the different combinations of quarks that make up each particle. This is a fundamental aspect of the Standard Model of particle physics and is one of the reasons why we have different types of particles with different properties. I hope this clarifies things for you.