# Tag Info

## Hot answers tagged symmetry-breaking

24

In the case of electroweak force and electromagnetism there is an Higgs mechanism, which makes the $W^{\pm}, Z$ bosons massive, and preserves the photon $\gamma$ massless. But the symmetry relating the electric and magnetic fields is actually a Lorentz symmetry, which is global and remains unbroken. The difference between electric and magnetic fields emerges ...

16

No. Separating electric fields from magnetic fields would require the breaking of local Lorentz symmetry, not some gauge symmetry, and there is no reason why that is going to happen.

14

The question has two parts: (1) What does "tree level" mean, and why are tree-level effects said to be classical in nature? (2) Is the Higgs mechanism classical in nature? Tree level $\to$ classical Solving linear differential equations is easy. Solving nonlinear differential equations is hard, usually too hard. If the nonlinearities are small ...

9

I'll give a brief answer that is somewhat at odds with Chiral Anomaly's answer. When we do quantum field theory we expand about some 'saddle point' of the action, i.e. some place where the variation of the action vanishes and thus we can think of it as classical. When you are playing around with just the Lagrangian in quantum field theory, you are looking at ...

5

You are really asking about parity doubling of baryons. Fixing language: (N is a silly meretricious conflation of 2 and 3, since for N > 3 the quark masses completely invalidate any axial charge considerations, as per Gell-Mann's cynosure). For any multiplet except pseudoscalars, say, for baryons B, $$B \mapsto e^{i\theta_L^i T^i \frac{1-\gamma^5}{2} + i\... 4 This is indeed a very confusing question and I have spent a lot of time parsing the literature looking for an answer. The references I found most useful are section 3 of the paper by Harlow and Ooguri, a set of lectures notes, and a study of SSB in superconductors(1) by van Wezel and van den Brink. Below, I summarize my current understanding; any comments ... 4 In this answer we will outline the proof given in Ref. 1 of the non-relativistic Goldstone theorem. Lagrangian formulation and symmetries. On one hand the action should be G-invariant. On the other hand we need the broken symmetry directions \mathfrak{g/h} to parametrize the vacuum manifold. It seems natural to assume that a low-energy effective ... 4 This is something like the inverse fallacy. :) Goldstone's theorem tells you that for every broken symmetry generator, you must find one massless mode in the spectrum of the theory. It does not tell you for every unbroken symmetry generator, you must find one massive mode in the spectrum of the theory. If that were the case, then if you had chosen a symmetry-... 4 This statement is called in the literature Adler zero. Nambu-Goldstone boson couples to the associated Noether current with a strength parametrized by the decay constant f :$$ \langle 0 | J_\mu (x)| \phi(p) \rangle = -i p_\mu f e^{-i p x} $$The matrix element between physical states has a pole for p^2 \rightarrow 0 and the residue corresponds to the ... 4 For what it's worth, it is useful to rewrite OP's scalar field$$\phi~\in~\mathbb{H}~\cong~\mathbb{R}^4\tag{1}$$as taking values in the quaternions,$$ {\cal L}~=~|\partial\phi|^2+m^2|\phi|^2-\lambda(|\phi|^2)^2. \tag{2}$$The 6-dimensional global symmetry group is$$G~=~U(1,\mathbb{H})_L\times U(1,\mathbb{H})_R.\tag{3}$$Here$$\begin{align}U(1,\mathbb{H})...

4

First of all, Lorentz symmetry can be spontaneously broken. That phenomenon is beyond the scope of the present question, though, so for this answer I'll assume that Lorentz symmetry is not spontaneously broken. In this case, the vacuum state is Lorentz invariant: if $U$ is the unitary operator implementing a Lorentz transform, then $U|0\rangle=|0\rangle$. ...

4

The general rule is that in the broken phase, you don't get a non-trivial CFT. Indeed, the coset effective actions depend on something dimensionful like the pion decay constant so they are not conformal. You can continue flowing to the IR, in which case interactions involving powers of $f_\pi$ will all be integrated out. This will give you a CFT but it will ...

4

Symmetry breaking has little to do with it: it dictates the renaming/rearrangement of the fields so as to expand around the peculiar vacuum involved; it amounts to a writing of variables, here $\sigma\mapsto \sigma + v$. It does not eat up fields represented in the kinetic term! The text you are reading is notorious for pedagogical complacency and omissions. ...

3

Since it seems that you're dealing with this in a statistical mechanics context, I'll use the Ising model as an example. Spontaneous symmetry breaking is a phenomenon in which the Hamiltonian or Lagrangian of your system has a certain symmetry, but some relevant state of your system does not. Take the nearest neighbor 2D Ising model $$H=-J\sum_{\langle i,j\... 3 Symmetry breaking is, in fact a topic within field theory. To understand it means understanding the mathematics of field theory. Certainly we can handwave and speak in analogies, but this does not convey accurately what the topic is really about or what it does for us. I will say this though. There is a difference between what we call symmetry and what might ... 3 The dimension of G/H is n_G-n_H, with a minus, not a plus. This is a quotient, not a product. The space G/H is naturally a manifold. The claim is that its tangent space is ismomorphic to V. Both the tangent space and V are vector spaces, they both have infinitely-many elements but a finite basis. The claim is that the bases for both have the same ... 3 How many Goldstone bosons? A system with a spontaneously broken symmetry is one where the vacuum is not invariant under the full continuous, semisimple symmetry group G of the theory, but only a subgroup H\subset G. We say that the symmetry G is spontaneously broken to H. In general, the equation defining the mass matrix of the field content in ... 3 The simplest way to prove this statement (actually, it gives you more, namely an upper bound on the 2-point function that decays with a power law) is the argument by McBryan and Spencer. You can actually prove much more, namely the full rotation invariance of the limiting Gibbs state. In my opinion, the simplest proofs are due to Dobrushin and Shlosman and ... 3 In the most formal sense (contrasting with the informal usage), a symmetry means that if you know something about one state, you can infer information about another state. The most trivial example is that if you've seen that gravity works in the past, you may assume it will work in the future. This is an example of a time-translation symmetry. The laws of ... 3 One can get extended SUSY by combining two supersymmetric multiplets into one with the appropriate choice of coupling constant. For example - in order to get \mathcal{N} = 2 SYM (Super Yang-Mills) one combines the gauge multiplet with gluon A_\mu and gluino \lambda_\alpha with the chiral multiplet with a fermion \psi_\alpha and a scalar \phi, all ... 3 I very much recall that example (many years ago!), and at the time it did confuse me too. The most important thing to take from that example, is the fact that in the case of the force being applied in the longitudinal direction of the rod (from top pressing down) as in diagram (c) you have above, is that the rod can bend in any direction. That is, there are ... 3$$ \langle \Omega|[J^\mu(y),\phi_n(x)]|\Omega\rangle=\langle \Omega|J^\mu(y)\phi_n(x)-\phi_n(x)J^\mu(y)|\Omega\rangle \\=\int d\Pi\ \langle\Omega|J^\mu(y)|p,\lambda\rangle \langle p,\lambda|\phi_n(x)|\Omega\rangle-\langle\Omega|\phi_n(x)|p,\lambda\rangle \langle p,\lambda|J^\mu(y)|\Omega\rangle \\d\Pi=\frac{d^3 p}{(2\pi)^3}\frac{1}{2E_p(\lambda)} $$where ... 3 You can't close the top loop. Every Higgs vertex has a left-handed and a right-handed top coming out of it. So any top "triangle" would have two top lines with the same chirality touching. 3 All the answers here and in the other question do not address the important difference between superconductivity and superfluidity: namely that the Nambu-Goldstone modes in superconductors are not gapless. The latter is an assumption for the validity of the Mermin-Wagner-Hohenberg-Coleman theorem, and therefore it does not apply. The question of whether ... 3 We do the exact same thing in the standard model when we break the electroweak symmetry. There are 3 broken generators \delta_-/2, \sigma_1/2, \sigma_2/2, and a leftover preserved U(1)_{EM} generator, \delta_+/2, where$$ \delta_\pm = \frac{1}{2}(\mathbf{1}\pm \sigma_3) $$are linear combinations of the EW gauge group generators. You can verify that ... 3 There are (at least) three options: We can retain the field's zero-mode \phi_0\equiv \int d^3x\ \phi(x) as an observable and retain the term \Pi_0^2 in the Hamiltonian. Then the model does not have a vacuum state at all. The spectrum of the Hamiltonian does have a finite lower bound, as required by the principles of QFT, but technically the Hilbert ... 2 If you work out the mass term using your second H, you will see that it is of the form$$m^2(\chi_1+\chi_2)^2 Only the combination $\chi_1+\chi_2$ has a mass, and in particular the combination $\chi_1-\chi_2$ is massless.

2

Well, the v.e.v. of the SM is about a quarter of a TeV, but that hardly matters, since the EM coupling varies very slowly there, So you wish to study the results of OPAL 2006, namely

2

I will try a simple explanation. Try putting a cylindrical pencil standing upright on a ball. Eventually, for a second, you will succeed, but then the pencil will probably fall. What tells the pencil in which direction to fall? At the start (pencil upright on the ball) there is cylindrical symmetry to the whole thing. The answer would be that some ...

2

The source $S^1$ is, as is usual in the homotopical classification of defects, coming from considering a loop around a defect line (in this case a cosmic string). If no such nontrivial loops would exist, then you can always deform the string to a "pointlike" object, i.e. it is not a stringy (one-dimensional) defect.

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