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The Gell-Mann matrices $\lambda^\alpha$ are the generators of $SU(3)$.

Applying an SU(3) - transformation on the triple $q = ( u , d, s )$ of 4-spinors looks like this:

$$ q \rightarrow q' = e^{i \Phi_\alpha \lambda^\alpha / 2} q.$$

So far I can follow and I also understand why the expression $\bar{q}q$ is invariant under this transformation.


Now my book defines axial transformations as $ q \rightarrow q' = e^{i \Phi_\alpha \lambda^\alpha / 2 \gamma_5} q$ and states that the expression $\bar{q}q$ is not invariant any longer under this transformation.

What confuses me is the fact that the $\lambda$ generators of $SU(3)$ and $\gamma$ matrices are being multiplied in the exponent, even though the $\lambda$ have 3 and the $\gamma$ have 4 dimensions.


Maybe this is not a matrix product but some sort of tensor product? In that case, how should the exponential expression be understood? I suspect $\lambda$ and $\gamma$ commute as they act on different vector spaces.

Or maybe it is a typo?

Or maybe the $\gamma_5$ is not 4-dimensional in this context?


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A Dirac spinor, as your $q$ is, has four components, corresponding to one left-handed and one right-handed Weyl (two-component) spinor, $$q = q_L + q_R.$$ $\gamma_5$ is the $4\times4$ matrix that is $1$ on the right-handed part and $-1$ on the left-handed part. The expression $$q\mapsto q' = \exp(i\Phi_a \lambda^a /2 \gamma_5)q$$ means $$q_{R(L)} \mapsto q'_{R(L)} = \exp(\pm i\Phi_a \lambda^a/2) q_{R(L)} \tag{1}$$ that is, that the left- and right-handed parts of $q$ transform differently.

The operator $A = \Phi_a \lambda^a \gamma_5$ is indeed a tensor product. Write the field $q$ can most explicitly be written $q_{f\alpha}$ where $f = u,d,s$ is a flavor index and $\alpha$ is a spinor index. Then $A$ is the product of an operator $\Phi_a \lambda^a$ acting on the flavor index, and an operator $\gamma_5$ acting on the spinor index.

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  • $\begingroup$ Thank you for this nice explanation. Especially the idea of replacing the gamma5 - Operator by its eigenvalue when acting on a left/right - handed spinor will help me in future, and I think it is giving me a very practical idea of an "axial" symmetry. $\endgroup$ – Konstantin Schubert Aug 10 '14 at 0:49
  • $\begingroup$ The observation about $\gamma_5$ is particularly useful when you come across $(1\pm\gamma_5)/2$, because what this does is pick out the right- or left-handed part. $\endgroup$ – Robin Ekman Aug 10 '14 at 13:45
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I) An axial (vector) symmetry transformation acts opposite (the same) under the left-handed and right-handed parts of a Dirac spinor, cf. chiral symmetry.

II) The full symmetry group is the product group $G=SU(3)_F\times SO(3,1)$. The quark $q$ transforms in the representation $\underline{3} \otimes \underline{4}$, i.e. under the fundamental representation $\underline{3}$ of the flavor symmetry group $SU(3)_F$ and under a Dirac-spinor representation $\underline{4}$ of the Lorentz group. The transformation act in the natural way, i.e. the Gell-mann matrices $\lambda^{\alpha}$ act on $\underline{3}$ and the $\gamma_{\mu}$ matrices act on the Dirac spinor $\underline{4}$. OP is correct that there is an implicitly written tensor product between the two matrices: $\lambda^{\alpha}\otimes \gamma_5$ in the exponential.

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  • $\begingroup$ Concerning product groups and tensor product representations, e.g. this Phys.SE post has another example from the standard model. $\endgroup$ – Qmechanic Aug 9 '14 at 19:47
  • $\begingroup$ I have accepted the other answer as the best answer since it is easier to understand. Your answer provided me with additional insight, thank you very much. $\endgroup$ – Konstantin Schubert Aug 10 '14 at 0:53

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