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I found this site to give a nice explanation of isospin. Apparently, the charged pions can be described by a superposition of $\pi^1$ and $\pi^2$.

Let us now consider the $\pi$'s in the context of isospin. There are three different charge states $\pi^+$, $\pi^0$ and $\pi^-$. They fit simply into a $I=1$ multiplet $\pi^a$, corresponding to $I_Z=+1,0,-1$ respectively, with

$\pi^\pm = \frac{1}{\sqrt{2}}(\pi^1\mp i \pi^2)$
$\pi^0=\pi^3$

Where $\pi^\pm$, $\pi^0$ are field operators. However, I fail to understand, what the $\pi^1$, $\pi^2$ and $\pi^3$ are.

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infinitezero
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  • It's currently unclear what exactly this question is asking without clicking on the link you provided. To make questions more accessible and guard against link rot, please include all relevant information, such as the explanation of notation or specific terminology used, in your question. – ACuriousMind Jun 24 '18 at 22:33
  • I did not find more information on that site, this is what spawned my question. – infinitezero Jun 24 '18 at 22:35
  • Forget the site. Look up these definitions in s a basic particle physics book. –  Jun 24 '18 at 22:53
  • I've read Griffiths for example, but it never came up. Can you recommend one? – infinitezero Jun 24 '18 at 22:54

1 Answers1

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Take a look at plain old vectors. Three dimensional space is spanned by ${\bf e}^1$, ${\bf e}^2$, and ${\bf e}^3$, also known as $\hat x$, $\hat y$, and $\hat z$. Of course:

$$ {\bf e}^i \cdot {\bf e}^j = \delta_{ij} $$

and

$$ ({\bf e}^i \times {\bf e}^j)\cdot {\bf e}^k = \epsilon_{ijk} $$

What are the eigenstate of rotations about the arbitrarily chosen $z$-axis? They are (normalized):

$$ {\bf e}^0 \equiv {\bf e}^3 $$

and

$$ {\bf e}^{\pm} = \frac 1{\sqrt 2}({\bf e}^1 \mp i{\bf e}^2)$$

You can show that the ${\bf e}^m$ with $m=(-1, 0, 1)$ are eigenvectors of a $z$-rotation by $\phi$ with eigenvalue:

$$ \lambda = \exp{(im\phi)}$$

This is known as the spherical representation of vectors, in contrast to the more familiar cartesian representation. (It's also called the fundamental representation of SO(3)). They transform exactly like a quantum spin-1 particle, aka, a vector boson.

Change the $\bf e$ to $\pi$, and replace spin with the name isospin, and you have 2 representations of pions.

The "cartesian" version isn't used, because the basis states aren't explicitly eigenstates of rotations about $I_3$--the equivalent $z$-axis in isospin space.

JEB
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  • Nit: can you explain what you mean exactly by saying that $e^{1} \pm e^{2}$ are eigenvectors of rotations? As far as I can tell, it is demonstrably not the case. – Prof. Legolasov Jun 25 '18 at 03:08
  • So they are just the isospin components of the pion field? Then I just find the notation confusing, because for neutrinos, the numbers refer to the mass eigenstates. Is it $\pi^1 = I_1 \pi$? – infinitezero Jun 25 '18 at 12:21
  • Solenodon I think he is missing an $i$ there. The same applies for basic vectors of circular polarisation vectors. – infinitezero Jun 25 '18 at 12:23
  • @SolenodonParadoxus infinite zero is right. I completely forgot the $\sqrt{-1}$, which comes from $e^{i\phi}=cos{\phi}+i\sin{\phi}\propto x +iy$. – JEB Jun 25 '18 at 23:22
  • Coming back to this, why do they need to be explicit eigenstates of rotations? – infinitezero Jun 15 '21 at 12:49