When constructing the effective Lagrangian, we parametrize the Goldstone bosons (such as the pions $\pi_a$) by $U = \exp(i \pi_a \tau_a/2 f)$, where $\tau_a$ are the Pauli matrices. (See e.g. here). Under transformations by $SU(N_f)_L\times SU(N_f)_R$, this transforms as $U \rightarrow L U R^\dagger$. We then construct the most general Lagrangian using invariant traces of this., such as $Tr(\partial_\mu U \partial^\mu U^\dagger)$.

If we want to include other fields, background or dynamic, we add source terms to the underlying QCD Lagrangian, for example $v_\mu \bar q \gamma^\mu q$ for a vector current. We then demand that the theory is invariant under local $SU(N_f)_L\times SU(N_f)_R$, and enforce this by treating the source terms as gauge fields. This gives them transformation rules under $SU(N_f)_L\times SU(N_f)_R$, and we can add them to the effective, chiral Lagrangian, while still maintaining the symmetry.

To my question, when we have a dynamic photon field, the QCD lagrangian will have the vector current $v_\mu = e Q A_\mu$, where $Q $ is the quark charge matrix. As with other vector currents, for example a chemical potential, this is included in the effective theory by the covariant derivative, $\nabla U = \partial_\mu U - i [v_\mu, U]$. However, in addition, we also treat give the charge matrix transformation properties, $Q_R \rightarrow R Q_R R^\dagger$, and similar for $Q_L$. This leads to the term $Tr(QUQU^ \dagger)$. See here, or here.

Why do we give $Q$ these transformation properties? It seems unmotivated. All other transformation properties are derived by how they appear in the QCD Lagrangian, even the spurion field which gives the quarks their mass. The vector field already has the transformation law $v_\mu + a_\mu \rightarrow R(v_\mu + a_\mu + i \partial_\mu) R^\dagger$, ad similar for $L$. Does not the additional law for $Q$ destroy this?

  • $\begingroup$ ? Q of course breaks SU(N) violently. For low energy QCD, it breaks flavor SU(3) violently, so both left and right SU(3)s. You understand, and should illustrate with real particles, that Q does not commute with all Gell-Mann λs. What are you talking about? $\endgroup$ Feb 21 at 20:50
  • $\begingroup$ Yes. However, one can break SU(N) by including external currents. For example, the mass is included by a $Tr(m^\dagger U + U^\dagger m)$-term, and a isospin chemical potential by including a vector current $v_\mu = \mu_I \delta^\mu_0 \tau_3/2$. (SU(2)). The transformation rules of these are deduced directly from the full QCD lagrangian including external currents. As far as I can see, there is no way to deduce the transformation of $Q$ in the same way. This is the crux of my question, how can you know this rule from only the underlying QCD Lagrangian? $\endgroup$ Feb 21 at 21:14
  • $\begingroup$ I'm confused: You know how you have aligned your particles, quarks, etc... in the Us, and since you know their extraneous charges, you know how to parameterize charge: Q= e diag (2,-1,-1)/3. What is it you are after? $\endgroup$ Feb 21 at 21:23
  • $\begingroup$ I am sorry if I am being a bit obtuse. I am after a way to deduce the transformation properties of $Q$ as a spurion field, $Q_R \rightarrow R Q_R R^\dagger$ and $Q_L \rightarrow L Q_L L^\dagger$ $\endgroup$ Feb 21 at 21:31
  • $\begingroup$ Yes, however the form of these terms can be deduced from the QCD Lagrangian. I have not seen anyone make that argument for the $Tr(QUQU)$-term. $\endgroup$ Feb 22 at 8:18

1 Answer 1


I a not quite sure of what is bothering you, except perhaps getting lost in the abstract notation and pulling the thread from the wrong (axiomatic) end. Take two quarks, so SU(2)×SU(2) for simplicity, and generalize trivially to N=3 in the end. The connection of charge action on quarks to such on pseudoscalar pions and thus U is here.

The electric charge commutes with neither isospin (vectors) nor axials in QCD, of course, $$ Q=\operatorname{diag}~( 2,-1)/3 = \tau_3/2+ 1\!\! 1 /6, $$ but only commutes with the third component of the isospin and axial charges, respectively, $$ \vec I = \vec \tau/2 \sim \int\!\! d^3x ~\bar q \vec \tau \gamma_0 q ~~\leadsto ~~\propto f^2 \int\!\! d^3x ~\operatorname {Tr} \vec \tau (U^\dagger \partial_0U + U\partial_0 U^\dagger), \\ \vec A =\gamma_5 \otimes \vec \tau/2 \sim \int\!\! d^3x ~\bar q \vec \tau \gamma_0 \gamma_5 q ~~\leadsto ~~\propto f^2 \int\!\! d^3x ~\operatorname {Tr} \vec \tau (U^\dagger \partial_0U - U\partial_0 U^\dagger), \\ \\ Q = I_3 + 1\!\! 1 /6, $$ where I have been cavalier with signs and normalizations, since they do not affect the flow of logic.

Isospin is realized linearly (so its charges are bilinear in the pseudoscalars), but the axials, and hence also the Ls and Rs, are realized nonlinearly, in the Goldstone mode (so they start with the term linear in them!).

Q is manifestly a vector charge!

In terms of the pion (Pauli vector "angle") of the logarithm of U, $$ \vec I \sim \int\!\! d^3x ~ \vec \pi \times \partial_0 \vec \pi + ..., \\ \vec A \sim \int\!\! d^3x ~ f \partial_0 \vec \pi + ... . $$

The charge operator then inherits the elegantly repackaged transformation properties of the left/right/isospin/axial charges, given $$ U\to e^{i(\vec \alpha \cdot \vec I - \vec \beta \cdot \vec A)} U ~e^{-i(\vec \alpha \cdot \vec I + \vec \beta \cdot \vec A)} ~~~~\leadsto \\ \int\!\! d^3x ~\overline{q_R} \gamma_0 \left( \frac{\tau_3}{2}+ \frac{1\!\!1 }{6}\right )q_R \to \int\!\! d^3x ~\overline{q_R} \gamma_0 ~ R^\dagger\! \left( \frac{\tau_3}{2}+ \frac{1\!\!1 }{6}\right ) R~q_R , $$ and likewise for the L piece. Do check this!

Likewise, you may express the action of the electric charge on the Us (and their h.c.) as a restricted isospin transformation, $$ U\to e^{i \alpha_3 I_3 } U ~e^{-i \alpha_3 I_3 } \leadsto i\alpha_3 [Q,U]+ O(\alpha^2_3), $$ since the identity part cancels out from left to right. Make sure you check this explicitly on the logarithm of U, and not just trust the logic, which you didn't trust in the first place. I am unsure what further motivation you expect. The term Tr$(QUQU^ \dagger)$ you are looking at is one of the two resulting from the seagull coupling to two photons.

Extension to three light quarks effortlessly shadows the above, given the full Gell-Mann - Nishijima formula, $Q= (\lambda_3+\lambda_8/\sqrt{3})/2 $.

  • $\begingroup$ I thank you for taking the time to explain. I can't say that I fully understand it yet, but I will look further into it with the help of your comments. Let me perhaps ask one final question---do the photon field still transform its usual way under $U(1)_{EM}$, $A_\mu \rightarrow A_\mu + i \partial_\mu \alpha/ e$, when included as a dynamical field in chpt? Because I have some difficulty consolidating this with the transformation rule of right/left currents in chpt, $r_\mu \rightarrow U_L r_\mu U_L^\dagger + i U_L \partial_\mu U_L^\dagger$ (similar for $l_\mu$), as $r_\mu = eQ_R A_\mu$. $\endgroup$ Feb 24 at 17:22
  • $\begingroup$ Of course! the EM U(1) is unbroken and gauged, even though the SU(2)xSU(2) is explicitly (and parts of it spontaneously/dynamically) broken. In fact, the covariant derivatives, which produced your fancy term (seagull with a photon bilinear) are such as to ensure full gauge invariance, given the photon transformation... $\endgroup$ Feb 24 at 17:31

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