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Considering the QCD Lagrangian in the chiral limit, where all the quarks masses are set to zero. Then the Lagrangian has the following chiral symmetry: $$ SU(L)_{V} \times SU(L)_{A} \times U(1)_{V} \times U(1)_{A} $$ As it is known, this symmetry group doesn't reflect in the actual existence of the eigenstates, that is we can classify eigenstates under multiplets of $U(L)_{V}$ generators, while instead $U(L)_{A}$ is spontaneously broken.

When a symmetry is not there anymore, it means we have to find something to characterize the new configuration of the system; this takes the name of "order parameter".

In the specific case of chiral symmetry in QCD, the order parameter is the chiral condensate, i.e. the following operator $$ \langle\Omega \lvert \bar{\psi}\psi(\vec{0},0) \rvert \Omega \rangle $$

In fact, it can be found that $$ \langle \Omega \lvert \left[Q_{a}^{A} (0),\bar{\psi} \gamma_{5}T_{b} \psi(\vec{0},0)\right] \rvert\Omega \rangle=-\frac{1}{L} \delta_{ab} \langle\Omega \lvert \bar{\psi}\psi(\vec{0},0) \rvert \Omega \rangle $$ where $Q_{a}^{A}$ is the conserved charge, as it follows from Noether's theorem.

This expression, when the chiral condensate is $\neq 0$, leads to $Q_{a}^{A} \rvert\Omega \rangle\neq0$, which identifies the spontaneous symmetry breaking. Then the chiral condensate is a good order parameter for SSB.

My question is: is there a reason why that commutator is what we need to compute in order to have a chiral condensate? In particular I am confused about the fact that the operators appearing in the commutator differ each other only for a $\gamma_{0}$, $Q_{a}^{A}$ being $Q_{a}^{A}(0)=\int d^3x \psi^{\dagger}\gamma_{5}T_{a}\psi(\vec{x},0)$.

Cosmas Zachos
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ottavio
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1 Answers1

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It's actually the other way around. That axial rotation of the pions ensures that, given its non-vanishing v.e.v., given by the condensate (assumed to be produced by QCD: a fact!), they therefore must be the Goldstone modes of the SSB of the axial charges. But, first, the chiral condensate is required so as to unleash all this.

Take L=2, for simplicity, and let's be schematic (~) about normalizations, which you may adjust to your satisfaction, in comportance with the conventions of your text.

Let us consider the relevant fermion bilinears and their representation of the $SU(2)_L\times SU(2)_R$. (By the way, the 3 axial $\vec{Q}_A$ do not close to an SU(2), as you wrote, as their commutators close to $SU(2)_V$ instead. Don't ever write this again... Also, $U(1)_A$ is broken explicitly by the anomaly, not spontaneously.) So, the 4 bilinears, $\bar{\psi} {\psi} $, $\bar{\psi}\gamma_{5} \vec{\tau} {\psi} $ form a quartet of this chiral group formally analogous of the $\sigma, \vec{\pi}$ quartet of the σ-model of the 60s; in fact, they are the QCD interpolating fields for this quartet.

(If you are insouciant about i s and the like, you may think of this scalar-pseudoscalars quartet as a 4-vector $t,\vec{x}$ of the Lorentz group, as a familiar mnemonic of the combinatorics/groupery to follow: The vector isorotations are analogous to the 3 rotations, and the 3 axials are analogous to the 3 boosts, acting on 4-vectors.)

You then see that $$ \left[\vec{Q}_{V},\bar{\psi}\psi\right]=0, $$ $$ \left[\vec{Q}_{A},\bar{\psi}\psi\right]\sim \bar{\psi} \gamma_{5} \vec{\tau} \psi ~, $$ $$ \left[Q_{a}^{V} ,\bar{\psi} \gamma_{5}\tau_{b} \psi \right] \sim \epsilon _{abc} \bar{\psi}\gamma_{5}\tau^c\psi ~, $$ and, crucially, the relation of interest, where you note the $\gamma_{0}\gamma_{5}$ makes all the difference in the combinatorics, as in the σ -model link, above, $$ \left[Q_{a}^{A} ,\bar{\psi} \gamma_{5}\tau_{b} \psi \right] \sim \delta_{ab} \bar{\psi}\psi ~. $$
So the σ is an isosinglet; the axials transform the σ by $\vec\pi$s; the vector transform of the $\vec \pi$s is an isorotation thereof; and the suitable "diagonal" axial transforms of the $\vec\pi$s takes them to the σ, the QCD analog of the Higgs of the EW interactions, the guy with the v.e.v.

Now take the v.e.v. $ \langle \Omega \lvert ... ... \rvert\Omega \rangle$ of each of the above. The r.h.sides of the first 3 must vanish; the v.e.v.s of the Goldstone modes are null!

The v.e.v. of the last one does not, but $=v\approx (250MeV)^3$, your relation of interest. QCD just achieves that, by dint of strong dynamics. So, over and above making it impossible for the axials to annihilate the vacuum, it identifies the 3 pions as the Goldstone modes of the 3 SSBroken $\vec{Q}_A$.

In point of fact, you actually see that $\vec{Q}_{A} \rvert\Omega \rangle \sim |\vec{\pi}\rangle$, that is the axial charges pump pions (chiral goldstons) out of the vacuum--- the precursor to PCAC.

A final loose end, lest you might still object that the third relation with vanishing r.h.side would then be moot, $\langle \vec{\pi} \lvert \bar{\psi} \psi \rvert\Omega \rangle=0$; but, no, the pion is orthogonal to the σ, just as it is to the original vacuum "chosen", $\langle \vec{\pi} \vert\Omega \rangle=0$. (Could insert $\vert\Omega\rangle \langle\Omega\vert$ above and factor out the order parameter, $\langle \vec{\pi} \vert\Omega \rangle v=0$ )

Further related (formally identical) questions might be 281696, and, needless to say, Gell-Mann and Lévy's timeless 1960 σ -model classic.

Cosmas Zachos
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