Suppose the quark neutral current weak interaction lagrangian: $$ \tag 1 L = \bar{q}\gamma_{\mu}(g_{V}+g_{A}\gamma_{5})qZ^{\mu}, \quad q = u,d $$ How to derive from this Lagrangian the neutral pion-Z-boson interaction?

It seems that it has to be $$ L\approx f_{\pi}\partial_{\mu}\pi^{0}Z^{\mu}, $$ but I can't derive this current.

It seems that for doing this I can use the relation $$ \tag 2 \langle \pi^{0}(p)|J^{\mu}_{5,3}(x)|0\rangle =f_{\pi}e^{ipx}p^{\mu}, \quad J^{\mu}_{5,3} = \bar{Q}\gamma^{\mu}\gamma_{5}\tau_{3}Q, $$ where $$ Q = (u,d)^{T}, \quad \tau_{3}=\text{diag}(1,-1) $$ The problem is that I can construct the current $J^{\mu}_{5,3}$ in $(1)$ and then use $(2)$, but there are also other summands distinct from it, which I can't simplify in the spirit of $(2)$.


1 Answer 1


It might be easier to consider the symmetry current version of (1) in the effective chiral lagrangian for pions, so the linear or chiral models, and look for the coefficient of the neutral pion in it. That is to say, fix the normalizations for $$ Z_\mu (g_V J^\mu_3(V) + g_AJ^\mu_3(5)) $$ where (V) denotes the isospin current and (5) the axial current, of relevance here. This is unspeakably sloppy, and you should take care to pin down the normalizations from the PDG mini-review, for instance.

You might, alternatively but equivalently, use the expression $$ \frac{e}{\sin \theta_W ~ \cos \theta_W} Z_\mu (J^\mu_3 (V-5)-\sin^2 \theta_W ~ J^\mu (e.m.))~, \tag{*} $$ so then keeping just the axial half of the L= (V-A)/2 current, since the e.m. and the V current have the wrong parity and cannot couple to a single pion (among other things: the charge is zero!).

But, in any case, you notice the felicitous linearity in the pion field of the axial current. In the linear σ model linked above, $$ \vec{J} ^\mu (5)= -2\sigma \overleftrightarrow{\partial} \vec{\pi} = -2 f_\pi \partial^\mu \vec{\pi} -2\sigma ' \overleftrightarrow{\partial} \vec{\pi} $$ where σ was shifted around its v.e.v. $f_\pi$ by σ' at the moment of (strong) chiral symmetry breaking; however, here, we are only interested in the linear term! You see how this comports with the first of your (2), up to my random normalizations.

Alternatively, in the chiral model, more popular today, the R - L current need only be expanded to linear order in π, $$ \vec{J} ^\mu (5)=i f_\pi^2 \operatorname {Tr} \frac{\vec{\tau}}{2} (U^\dagger \partial^\mu U - U\partial^\mu U^\dagger)= -2 f_\pi \partial^\mu \vec{\pi} +... $$ axial, yet again. Actually, you can be even lazier and simply focus on the plain left chiral (right-invariant) current featured in (*), $\vec{J} ^\mu (V-5)=i f_\pi^2 \operatorname {Tr} \frac{\vec{\tau}}{2} U\partial^\mu U^\dagger = f_\pi \partial^\mu \vec{\pi} +... $, ignoring the e.m. current which has no neutral pion piece.

This is quite analogous to the less messy charged W couplings of the sigma model, responsible for charged pion decay. But I suppose I should not call the coupling messy: Both the Z and the π0 have identical vanishing charges for all quantum numbers Y, T3, and hence e.m. charge.

  • 1
    $\begingroup$ By the way, $Z$ and $\pi^0$ also have the same "reduced witdh" $\Gamma/M^3$. (it can be told that it is mostly because of the still huge variance of the measurement of $\pi^0$ decay, but still it makes an amusing coincidence between QCD and Electroweak magnitudes) $\endgroup$
    – arivero
    Commented Feb 19, 2018 at 3:25

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