in Isospin space there are two fundamental states called up and down quarks, which satisfy the following eigenvalue equations:

$I u = (1/2) u$, $I d = (1/2) d$ and $I_3 u = (1/2) u, I_3 d = (-1/2) d$. The antiparticles have inverted signs in their 3-component of the isospin. This also raises my first question: How are anti-d and u not the same state? They satisfy the same eigenvalue equations for $I_3, I$ and if I recall correctly the space of isospin-$1/2$ particles is 2 dimensional, i.e. there's no room left for the antiparticles in a sense. How do I solve his paradoxon?

Mesons are pairs of quarks and antiquarks. This means we can write the states of mesons as tensorproducts of an antiquark and a quark state. For example let's look at combinations of u and d quarks:

$u^{(*)} = |1/2, \pm1/2>$, $d^{(*)} = |1/2, \mp 1/2>$ (the star denotes the antiparticle. The first component is total isospin, the second is 3rd component of Isospin)

Using Clebsch-Gordan we can write the tensor of those two states as a direct sum:

$|1/2, \pm1/2> \otimes$ $ |1/2, \mp 1/2> = |1,1> \oplus $ $|1,0> \oplus$ $ |1,-1> \otimes $ $|0,0>$ where the triplet of $I = 1$ states corresponds to the three pions.

My second question is, how do I recover the ELECTRICAL charge of a pion state from these equations? In my script there is this operator given by $Q = e(I_3 + \frac{1}{2} id)$ which should return the electrical charge of an isospin state. This doesn't seem to give the right values as for example $Q |1,1> = e(1 + 1/2) |1,1>$ which doesn't give the charge of a pion.


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    $\begingroup$ This should really be broken up into two separate questions. Also, \lvert and \rangle are your friends ;) $\endgroup$
    – user10851
    Jul 9, 2014 at 8:55
  • $\begingroup$ Careful with the algebra: the total isospin operator follows the same algebra as the angular momentum operator. The operator acts on an eigenstate as $J^2 \lvert j m \rangle = j ( j + 1 ) \lvert j m \rangle$, not just $j \lvert j m \rangle$ (sic). $\endgroup$
    – jordix
    Jul 9, 2014 at 9:28
  • $\begingroup$ @jordix Of ourse, you're right! $\endgroup$
    – user17574
    Jul 9, 2014 at 10:15

1 Answer 1


About the supposed paradox: $u$ and $\bar d$ have the same isospin quantum numbers, but not all the other properties. If you restrict your problem to only study the isospin space, you will not see that they have different charge and other different quantum numbers.

About the charge: I don't know where your equation comes from, but it seems close to the Gell-Mann–Nishijima formula

$Q = \left( I_3 + \frac{Y}{2} \right)$,

where $Y$ is the hypercharge and is $0$ for pions. So, in your case, the pion charge and its 3rd component of isospin are equal. There must be a typo in your notes.

  • $\begingroup$ I see. So this formulation would make sense: The total wavefunction (if one only looks at space, spin and isospin) is basically a factor of three wavefunctions: $\psi_{tot} = \psi_{Space} \psi_{Spin} \psi_{Isospin}$. If one only looks at the wavefunction part of the up and down quarks which acts on the Isospin space only, u and d are the only eigenvectors of that space (since the dimension of that space is 2). However the $d^*$ and $u^*$ are linear combinations of u and d on that space, but give different values for the quantum numbers on the other two spaces in which $\psi_{tot}$ acts. $\endgroup$
    – user17574
    Jul 9, 2014 at 10:21
  • $\begingroup$ Is that correct like that? Also, thanks for the correct form of the $Q$ operator. $\endgroup$
    – user17574
    Jul 9, 2014 at 10:22
  • $\begingroup$ For the total wave function you would need to consider flavor (isospin in this simple case), spin, color and space components. Four components overall, plus you should make sure that the wave function is symmetric, which must be the case for all bosons. You can easily find an amazingly good explanation of the topic in the Cambridge university lectures by prof. Mark Thomson (hep.phy.cam.ac.uk/~thomson/partIIIparticles/handouts/…). He uses the proton as an example but, if you understand that, the pion case is simpler. $\endgroup$
    – jordix
    Jul 9, 2014 at 12:40

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