0
$\begingroup$

Chirality is related to parity, correct? It is a form of 'parity'?

So, How can the strong force violate chirality symmetry, but only the weak force violates parity symmetry?

I am confused about the distinction.

$\endgroup$
4
$\begingroup$

Here's the distinction:

  • Parity converts left-handed chiral spinors to right-handed chiral spinors, and conversely: $\psi_L\leftrightarrow \psi_R$.

  • Chiral symmetry means symmetry under $\psi_L\to e^{i\phi_L}\psi_L$ and $\psi_R\to e^{i\phi_R}\psi_R$, where the phase-factors $e^{i\phi_L}$ and $e^{i\phi_R}$ can be different. (It can also refer more specifically to the transformation with opposite-sign phases: $\phi_L=-\phi_R$.) Chiral symmetry does not exchange the left- and right-handed parts with each other.

In QCD, the only thing that breaks parity symmetry is the so-called theta-term, which is (mysteriously!) zero as far as experiments have been able to tell.

Chiral symmetry is broken explicitly by quark masses. If quarks were massless, then chiral symmetry would be a symmetry of the net of observables in QCD. However, chiral symmetry would still be spontaneously broken, meaning that the vacuum state does not respect the symmetry, even though the net of observables does.

For more detail, see the other answers.

$\endgroup$
3
$\begingroup$

The first answer references the strong CP problem which is that there's a possible mechanism for breaking CP which QCD in nature appears not to make use of.

To address the other part of the question, I want to explain why this has nothing to do with chirality as usually defined in the QCD context. Of course parity and chirality are very overloaded terms in physics but they are not the same in this case. Instead of the discrete spacetime symmetry denoted P above, the chiral symmetry which QCD breaks is a continuous internal symmetry which rotates the quarks. The name is justified because even though there are six types of quarks, we know experimentally that they should be grouped in sets of three. So there is clearly no symmetry like $SU(6)$ that could rotate them all. Instead, we have three up-type quarks with positive charge and three down-type quarks with negative charge. So the best we can hope for is a chiral symmetry like $SU(3) \times SU(3)$ where each factor acts on a different set of quarks.

So why can't we rotate between them? Well first the symmetry is explicitly broken because the quarks have different masses even at very high energies. But even if that were not the case, it is believed that the dynamics of the theory still break chiral symmetry spontaneously at a lower energy scale. The evidence we have for this is very similar to the evidence we have for a Yang-Mills mass gap. In particular, it is encouraging that pions behave in a way that makes them very similar to Goldstone bosons. It is believed that they would be exactly the Goldstone bosons for chiral symmetry if it were not broken explicitly but only spontaneously.

$\endgroup$
3
$\begingroup$

Chirality is related to parity, correct? It is a form of 'parity'?

Chirality generators are odd under a parity transformation: A mirror reflection reverses chirality, left to right, and vice-versa. They are different operators which is why they have different names. A parity transformation is discrete, but a chiral transformation is continuous, effected by exponentiating such generators.

Weak interactions treat chiral enantiomorphs very differently, but strong interactions do not: QCD is vectorlike, so gluons treat left- and right-handed quarks identically. Strong interactions never violate parity.

So, How can the strong force violate chirality symmetry, but only the weak force violates parity symmetry?

Chiral symmetry is an ideal, notional symmetry among left- and, separately, right-handed fermions (and flavor rotations thereof, but let's leave this out for the moment): In practice, fermion masses (engendered by weak Yukawa couplings) break chiral symmetry, so it's not there, in fact. However, such masses are small for the light quarks, much smaller than the QCD scale of ~ 200 MeV, which creates very massive hadrons out of such light quarks. So, we may consider/imagine zero quark masses in a zeroth approximation, to better study the effects of QCD gluonic interactions in making fat (~1 GeV), heavy hadrons, like nucleons.

This zeroth approximation (chiral limit) has chiral symmetry, but realized in a very peculiar (Nambu-Goldstone, spontaneously broken) mode: for most practical purposes, this symmetry is invisible, and a chiral rotation of a hadron (or a "constituent quark") does not match that of a degenerate enantiomorph, and the effective lagrangian for these states has the fat masses for the nucleons mentioned, which thus does not look chirally symmetric—your "can be considered to violate chirality" hyper-coded message.

It would still have zero masses for pseudoscalar mesons (like the pions and kaons), but the perturbative corrections of small quark masses take you away from that chiral limit and give the pseudoscalars small masses (still much smaller than those of the rest of the hadrons).

It is in this sense that QCD breaks chiral symmetry spontaneously/dynamically. At every step, any and all strong couplings and reactions maintain their even-handedness on parity enantiomorphs: they treat all states$^\natural$ and their mirror images identically. Left- and right-chiral nucleons couple to each other big-time, (and even oscillate into each other, but let's not go there).

What about the weak interactions? These basically treat right-chiral states very, very differently than left-chiral states, so there never was any question of any chiral symmetry surviving/existing in that theory.

As a result, weak forces break parity (almost "maximally"). Through the magic of SSB, they get the left-chiral fermions to couple to right-chiral fermions, generating the small quark masses mostly ignored above.

The takeaway: Weak interactions break both parity and chiral symmetry explicitly. Strong interactions don't break parity (that we now, so far) and break chiral symmetry both explicitly (by a little) and spontaneously/dynamically (by a lot).


$^\natural$NB, Geeky: I'm only putting this in for completeness, with a heavy heart, as it's really technical.

The spontaneously broken chiral symmetry, however, is reflected in a parity imbalance in the spectrum of hadrons (but not their interactions, as emphasized above). It is the pseudoscalar mesons which are light, not the scalars, and the nucleons are not parity-doubled: they have parity +, not -. (At higher energies parity doubling returns.) That is, the Goldstone QCD vacuum is not blind to parity, bubbling with pions, and the chiral condensate is scalar, not pseudoscalar. Still, any amplitude or interaction you'd write in the strong sector preserves parity.

$\endgroup$
2
$\begingroup$

In our mathematical description QCD could violate CP-symmetry due to its non-abelian nature, which introduces the CP-symmetry violating theta-term into the action. It does neither contribute to perturbation theory nor to the equations of motion because it is a total derivative. In a naive quantization in the Schrödinger picture it influences the energy eigenvalues of the Hamiltonian of the YM-theory. This is because instantons with nonzero size only appear in non-abelian gauge theory and the Chern-Simons three-form is not invariant under large gauge transformations (instantons). To make the wave function invariant under all gauge transformations, we need to introduce a term into the action that cancels an additional phase. So far to the origin of the Theta-term.

The CP-violation is closely related to the Atiyah-Singer index theorem, because the first Pontryagin is equal to the signed number of zero modes of the Dirac operator in the instanton background. Therefore the chiral current is not conserved on the quantum level. One could also argue that this is due to the need of renormalization and regularization of the Dirac propagator in the instanton background. In reality however, we do not observe such violation, meaning that either there is something canceling the theta term, recovering the symmetry of the QCD Lagrangian, or the theta term is extremely small. Measurements of the dipol of the neutron set the value below $10^{-9}$ for theta, which is a fine-tuning problem and one of the most underestimated problems of modern physics according to Jackiw.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.