One commonly finds this interaction Lagrangian in phenomenology (ignoring constants):

$$\mathscr{L} \propto \bar\psi \gamma^5\gamma^\mu \vec\tau\psi \cdot \partial_\mu \vec\pi \tag{1}$$

Evaluating the scalar product yields

$$\vec\tau \cdot \vec \pi = \begin{pmatrix}\pi_3 & \pi_1 -i \pi_2\\ \pi_1 + i \pi_2 & -\pi_3 \end{pmatrix} = \begin{pmatrix}\pi_0 & \sqrt{2}\pi^+ \\ \sqrt{2}\pi^- & -\pi_0 \end{pmatrix} \tag{2}$$

Here we've used the eigenstates of $I_z$, the generator for rotations around the z-axis, i.e. $\pi^\pm = \frac{1}{\sqrt 2}(\pi_1 \mp i \pi_2)$, $\pi_0 = \pi_3$

This introduces a factor of $\sqrt{2}$ at the vertices with charged pions and a factor of $-1$ for the neutron-neutron interaction. I am curious where these factors come from, because I can not reproduce it with Clebsch-Gordon coefficients.

Naive Attempt

Supposing I want to write down an interaction for a $0^-$ scalar particle with a derivative coupling, I could do it as such

$$\mathscr{L} \propto \bar\psi \gamma^5\gamma^\mu \psi \partial_\mu \phi$$

Now without knowing about the first Lagrangian, how would I introduce isospin? For $\phi = \pi$ and a nucleon, I can think of the following interactions:

$$ \begin{align} \pi^0 p &\rightarrow p \tag{4a}\\ \pi^- p &\rightarrow n \tag{5a}\\ \pi^0 n &\rightarrow n \tag{6a}\\ \pi^+ n &\rightarrow p \tag{7a} \end{align} $$

So naively I could write down for Lagrangians

$$ \begin{align} \mathscr{L} &\propto \bar\psi_p \gamma^5\gamma^\mu \psi_p \partial_\mu \pi^0\tag{4b}\\ \mathscr{L} &\propto \bar\psi_n \gamma^5\gamma^\mu \psi_p \partial_\mu \pi^-\tag{5b}\\ \mathscr{L} &\propto \bar\psi_n \gamma^5\gamma^\mu \psi_n \partial_\mu \pi^0\tag{6b}\\ \mathscr{L} &\propto \bar\psi_p \gamma^5\gamma^\mu \psi_n \partial_\mu \pi^+\tag{7b} \end{align} $$

but now the isospin factors are missing. How can I derive them? Just "guessing" $\tau_a \pi_a$ seems a bit contrived.

My attempt

Writing this down in braket notation, I would find the interaction $\pi^-p\rightarrow n$ to be something like

$$ \begin{align} \langle f| O |i \rangle &= \left\langle \tfrac{1}{2}, -\tfrac{1}{2}\right| O \left|1, -1; \tfrac{1}{2}, \tfrac{1}{2}\right\rangle \tag{8}\\ &= \sum \limits_{M=-\tfrac{1}{2}}^{\tfrac{1}{2}} \left\langle \tfrac{1}{2}, -\tfrac{1}{2}\right| O \left|\tfrac{1}{2},M\right\rangle\underbrace{\left\langle\tfrac{1}{2},M\,\right|\left.1, -1; \tfrac{1}{2}, \tfrac{1}{2}\right\rangle}_{=C^*}\tag{9}\\ &= -\sqrt{\tfrac{2}{3}}\left\langle \tfrac{1}{2}, -\tfrac{1}{2}\right| O \left|\tfrac{1}{2}, -\tfrac{1}{2}\right\rangle\tag{10} \end{align} $$

Here in $(9)$ we have added a 1 as a sum over the compound state, which gives us the clebsch-gordan coefficient $-\sqrt{\tfrac{2}{3}}$.

But I do not really know how to continue.

  • $\begingroup$ pion charge in (7b) is off. $\endgroup$ Commented May 6, 2022 at 22:10

1 Answer 1


You are looking down on the 2nd column of the Clebsch matrix, up to an over-all common normalization of $-1/\sqrt 3$, ignored here.

Rewrite (1),(2) as the sum of (4b,5b,6b,7b), $$ \bar\psi \gamma^5\gamma^\mu \vec\tau\psi \cdot \partial_\mu \vec\pi \\ = \bar\psi_p \gamma^5\gamma^\mu \psi_p \partial_\mu \pi^0 + \sqrt{2} \bar\psi_n \gamma^5\gamma^\mu \psi_p \partial_\mu \pi^- - \bar\psi_n \gamma^5\gamma^\mu \psi_n \partial_\mu \pi^0 + \sqrt{2} \bar\psi_p \gamma^5\gamma^\mu \psi_n \partial_\mu \pi^+ , $$ And you are trying to connect the relative coefficients with this second column of the Clebsch matrix, namely the projection onto isospin 1/2 of the $1\otimes 1/2$, nucleon-pion state; the respective coefficients of the terms in this order being $$ \langle \tfrac{1}{2}, \tfrac{1}{2} | 1, 0 \tfrac{1}{2}, \tfrac{1}{2} \rangle = -1/\sqrt{3}; \\ \langle \tfrac{1}{2}, -\tfrac{1}{2} | 1, -1; \tfrac{1}{2}, \tfrac{1}{2} \rangle =-\sqrt{2/3}; \\ \langle \tfrac{1}{2}, -\tfrac{1}{2} | 1, 0; \tfrac{1}{2}, -\tfrac{1}{2} \rangle =1/\sqrt{3} ; \\ \langle \tfrac{1}{2}, \tfrac{1}{2} | 1, 1; \tfrac{1}{2}, -\tfrac{1}{2} \rangle = -\sqrt{2/3}. $$ The last Clebsch may need rephasing, but you see that the relative coefficients comport with the top expression you see in books. This is where the relative strengths and phases of (4a,5a,6a,7a) come from.

  • $\begingroup$ Thanks! I discovered this myself last night, but I agree, that the last Clebsch-Gordan is actually positive. Can we actually rephase the CGC or would we need to rephase the positively charged pion field? $\endgroup$ Commented May 7, 2022 at 6:59
  • $\begingroup$ Not sure… I’d leave the pion alone and play with CGC conventions… one hasn’t used the isospin 3/2 sector which is bound by the Condon-Shortley convention; perhaps there is play room there, if one departed from that convention...? $\endgroup$ Commented May 7, 2022 at 14:25
  • $\begingroup$ I'm not sure that would work. I know for a fact, that the sign difference between the two neutral pions is needed so that the born terms in photo production ($\gamma N \rightarrow \pi N$) is gauge invariant. This essentially makes the s and u channel have the same sign, which are both partly cancelled by the t-channel. (an additional contact term is needed to cancel the rest of the t-channel). Presumably, this does not play a role for the charged pions and one can ignore the phase factor, because it does not appear in coherent sums. $\endgroup$ Commented May 7, 2022 at 14:32
  • $\begingroup$ Thinking about this further, I think this problem lies in the definition of the pions. There are of course infinitely many eigenstates to $I_3$. As such, writing $\vec\tau \cdot \vec\pi$ only works for appropriately chosen eigenstates. Otherwise, this shorthand notation does not produce the correct clebsch gordan coefficients. $\endgroup$ Commented May 8, 2022 at 10:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.