2
$\begingroup$

I have several doubts regarding the nonlinear realization of a spontaneously broken symmetry and hope they are appropriate to be grouped, and I appreciate any insights.

Consider the group breaking pattern $G \to H$, where $\{T_i\}$ furnish a representation of (the Lie algebra of) $H$ and $\{X_\alpha\}$, an orthonormal completion of a basis for (the Lie algebra of) $G$.

1) Within the heart of the construction is the repeatedly stated fact that a general group element may be decomposed as $$ g=e^{\theta_\alpha X^\alpha}e^{u_i T^i}. $$

This does not seem to be formally true to me. Writing $g$ as an exponential and using BCH on the RHS, we obtain a finite system of infinite-degree polynomial expressions on $\theta_\alpha,~u_i$ for a specific fixed $g$. Even if we truncate the polynomials for small $\theta_\alpha,u_i$, it doesn't look obvious that there exists a solution to the system, even over $\mathbb{C}$. Is the common statement true? In which situations?

Indeed, the fact above should be exactly true in a neighborhood of the identity -- this is especially obvious because, as a theorem states, close enough to the identity, any group element could be written as a single exponential. Some texts make it clear that this is the case (it works only on a neighborhood of the identity) but why is this enough?

2) Another thing that is commonly stated when introducing the CCWZ construction is that a generic local field configuration is just a rotation of the favoured vacuum by the full group $$ \phi(x)=g(x)v. $$ Again, I don't understand why this is the case. Shouldn't $g$ only take $v$ to a general, possibly inequivalent, vacuum (with $h$ taking it to an equivalent one), and not to any field configuration?

3) An extra statement is one implying that a nonlinear realization is specially useful when there is strong coupling in the high energy theory, and that when there is weak coupling we may use a linear realization. I understand that a nonlinear realization is useful at low energies because then we may realize the full symmetry with a few modes integrated out (since the Goldstone bosons' transformations depend only on themselves); but I can't see how this is useful on a broken symmetry (since we only need the low energy effective theory to be symmetric by $H$ in this case) and what it has to do with weak or strong coupling.

$\endgroup$
2

1 Answer 1

0
$\begingroup$

I am not sure with what text I'd be shadow-boxing; I suggested a pretty good summary in the comments. It is understood you are completely, and hands-on familiar with the SO(4)→SO(3) σ model, which contains a σ and three π(ions); and have illustrated and worked out all relevant formulas that CCWZ abstracted and generalized into mathematese, but not intending to confuse students who did not have the straightforward σ-model under their belt. You would not be asking these questions had you bashed through the entirety of the model.$^\natural$

You have three ("unbroken") generators of isospin T in the sub algebra H and three (SSBroken) ones X, which do not close into a subalgebra, and commute with the Ts into themselves: they furnish a representation of isospin. (It might help you, or not, to think of the Xs as the three boosts of the Lorentz group and the three Ts as the rotation generators, the little group of the 4-vector with just a timelike component.)

  1. I have no idea what your problem is, and what you believe you are attempting to do. Indeed, $$ g=e^{\theta_\alpha X^\alpha}e^{u_i T^i}=e^{\phi_\alpha X^\alpha+ w_i T^i} , \\ \phi(\theta,u)_\alpha = \theta_\alpha -f^\alpha_{\beta j}\theta_\beta u_j /2+... , ~~~~~ w_i(\theta, u) = u_i+ ... $$ for real antisymmetric generators. The series in the exponent of the r.h.side converges: it is the foundation, the point!, of Lie theory, (for a finite number of generators, of course). In any case, you are supposed to already have done this as a homework problem for the SO(4) model! The parameterization provided is completely generic and holds to all orders. Tasteful reviews, and the original CCWZ papers have a proof. For non-pathological cases of physical interest, $(\theta, u)\mapsto (\phi, w)$ is invertible.

  2. H is the little group of v, i.e. leaves it unchanged, so isospin rotations don't do a thing to it, but the three axials X rotate it to a degenerate vacuum and parameterize the three pions (Goldstone bosons). The exponential $\exp (\vec \pi \cdot \vec X)$ pumps pions (goldstons) in and out of the vacuum v instead of leaving it invariant.

  3. I'm not sure I understand what you are saying. If you are discussing chiral symmetry, at low energy only pions dominate, and non-pions are heavier hadrons suppressed by QCD scale powers in the effective lagrangian... if you are discussing QCD; at higher energies, inside hadrons, the full chiral symmetry is restored. Is that what you are talking about? What is the firm statement, as opposed to notional implications? An effective lagrangian normally involves both the σ and the three π s and exhibits all symmetries, broken & unbroken, in different, but transparent ways. The unbroken symmetries are an accounting aid (i.e. the action has to be an isosinglet), but the broken axial ones are the crucial ones dictating the highly restricted shift-invariant form of the effective action of goldstons: The goldstons must then couple with derivatives, (so their amplitudes have Adler zeros: vanish at zero momenta). Not so for the other particles (like the σ) lacking shift invariance.

When in doubt about the CCWZ method, many turn to Weinberg's QFT text for completeness. You might like this summary.


$^\natural$ Recall the fundamental of SO(4) is the quartet $$ (\phi_1, \phi_2, \phi_3, \phi_4)= ( \pi_1,\pi_2, \pi_3,\sigma), \\ (T_i)_{jk}= \epsilon _{ijk},~~~~~(X_\alpha)_{jk}= (\delta_{j\alpha} \delta_{k4}-\delta_{k\alpha}\delta_{j4}), $$
so the Ts rotate the three pions among themselves and leave the σ alone; while the Xs send the σ to the three pions, and any of the three pions to the σ. (Crucially, the Lie/current algebra here has a parity isomorphism dictating $w(-\theta, u)= w(\theta, u)$, but $\phi(-\theta, u)= -\phi(\theta, u)$. This reduces the explicit calculation to the mere composition law of rotations!)

Shifting the σ by a constant (v.e.v.) term v converts these latter three axial transformations to nonlinear ("shift") ones, $e^{\theta\cdot X} ~\vec \pi =\vec \pi+\vec \theta v+ ... $, so the gradient of the infinitesimal increment to the goldstons vanishes! This is the essence of SSB.

$\endgroup$
10
  • $\begingroup$ Thank you for the answer. I'm studying these matters by myself and the questions I pose emerged after reviewing (to the end) several texts regarding non-linear realizations/CCWZ/non-linear sigma models. I apologize if what I enunciate is simple enough (or nonsensical) and I should have understood it through said literature, but I haven't. Regarding 1.: what I tried to do is exactly what you wrote (which I have not seen in any source I have studied), and my doubt is how can we know this system of truncated polynomials have a solution? $\endgroup$
    – GaloisFan
    Commented Jun 25, 2022 at 22:46
  • $\begingroup$ Furthermore, this is supposed to work only near the identity -- in which case we are only able to recover any vacuum element close enough to the favoured point. Isn't that a problem? $\endgroup$
    – GaloisFan
    Commented Jun 25, 2022 at 22:47
  • $\begingroup$ 2. I understand that... my point is that we should be writing a general field, not just parametrizing the vacuum manifold. 3. Yes, only the low-energy degrees of freedom are present on the low energy theory, and because of that we realize the symmetry non-linearly (we don't have a complete basis for a linear realization). I'm asking why do we care about the full symmetry at all, and not only about the unbroken part, at low energies. $\endgroup$
    – GaloisFan
    Commented Jun 25, 2022 at 22:51
  • $\begingroup$ They are not truncated polynomials. It is an infinite series, which can be gotten exactly in special cases. Not just near the identity .... it works over the entire sectors continuously connected to the identity, e.g. a sphere... Lie's third theorem! Try a good book on Lie groups, such as Gilmore's, Wybourne's, Varadarajan's ... An effective lagrangian represents general fields, both Goldstone, and nongoldstone degrees of freedom. Only the goldstons survive at low energies, by dimensional analysis. $\endgroup$ Commented Jun 26, 2022 at 0:52
  • 1
    $\begingroup$ Clarified answer. He is a mathematician looking for pathological loopholes in a structure that works in normal constructions in physics. $\endgroup$ Commented Jun 26, 2022 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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