Physical meaning of the chiral condensate in QCD

Question

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 $Q_{a}^{A}(0)=\int d^3x \psi^{\dagger}\gamma_{5}T_{a}\psi(\vec{x},0)$.

Answer 1

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.