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In a paper, "Nonlinear Realization of Global Symmetries" by Zhong-Zhi Xianyu, I see the line

"There are basically two ways of symmetry realization in Hilbert space. One is such that the vacuum state is invariant under the action of all symmetry transformations, which is known to be the Wigner-Weyl realization; the other contains degenerate vacua, which can transform into each other under some of the symmetry transformation. This is known as the Nambu-Goldstone realization."

My question is what does this really mean? What is the relation between nonlinear realization and the coset construction? Why Goldstone modes are identified with points in the coset space G/H where H is the linearly realized subgroup of the full symmetry G?

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  • $\begingroup$ Linked. So your coset space amounts to $S^2$, a basketball... Review in your question what you do know. $\endgroup$ Dec 27, 2022 at 14:20

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The answer by @Andrew is impeccable, and so is his bridge comment on illustrating the picture with the coset space SO(3)/SO(2) ~ S2, a basketball (or our earth!) I have been asking students in the past to make sure they can translate the mathematese (single orbit to transitive action of the group on the coset space, and the lot...) through this basketball for quite a while, so I thought I should summarize it for you, as it should be useful to anyone, in an extended footnote to his fine answer. Moving on to pious abstractions before mastering the obvious might be deprecated.

Take the generic scalar fields in some notional 3D isospace, $\vec \phi= (\phi_x, \phi_y, \phi_z)^T$ to remind you of undergraduate geometry: x,y,z are really isospin indices, in fact. Think of an SO(3)-invariant potential, $$ \lambda (\vec\phi^2-v^2)^2. $$

The SO(3) isorotations are generated by three generators L, $$ \vec\theta\cdot \vec L= \begin{bmatrix} 0&\theta_z&-\theta_y\\ -\theta_z&0&\theta_x\\ \theta_y&-\theta_x&0 \end{bmatrix}, $$ where each of the three angles $\vec \theta$ corresponds to each rotation generator about the obvious axis (earth's spin axis, and ascending from, e.g., Kigali and Darwin, respectively, for the sake of argument.) So, the infinitesimal rotations around each axis are $$ \delta ~\vec \phi= {\vec \theta} \times \vec \phi ~. $$

Arranging that the three fields have vanishing expectation values, and placing the vacuum at the minimum of the potential (SSB), we choose $\langle \phi_z\rangle =v$.

So, define $\sigma\equiv \phi_z-v$, such that $\langle \sigma\rangle=0$, just like $\langle \phi_x\rangle=0$ and $\langle \phi_y\rangle=0$. In this language, the minimum of the potential is now in this vacuum at zero, $$ \lambda \Bigl ( \phi_x^2+\phi_y^2+ \sigma^2+ 2v\sigma \Bigr )^2. $$

Now, the infinitesimal rotations are lopsided, $$\bbox[yellow]{ \delta \phi_x= \theta_y v +\theta_y \sigma - \theta_ z \phi_y, ~~~\leadsto \langle \delta \phi_x\rangle = \theta_y v \\ \delta \phi_y= \theta_z \phi_x-\theta_x v-\theta_x \sigma , ~~~\leadsto \langle \delta \phi_y\rangle = -\theta_x v \\ \delta \sigma= \theta_x \phi_y-\theta_y \phi_x ~~~~~~~~~~\leadsto \langle \delta \sigma \rangle = 0 }~~. $$ This is to say that $\langle \phi_z\rangle =v $ is invariant under $L_z$, polar, rotations. The subgroup generated by $L_z$, namely, SO(2), is the "little group", the "stabilizer subgroup", H, "the unbroken group", whatever: the spin of the earth. It just rotates $\phi_x,~\phi_y$ into each other, so transforms linearly, which one calls the Wigner-Weyl mode.

But the other two generators, $L_x, ~ L_y$, have a problem with the vacuum: they shift the fields $\phi_x,~\phi_y$ into and out of the vacuum, and are called SSBroken generators. (Hidden, actually, since the potential is still symmetric under their action, as it is under the full symmetry, and since they follow all commutation rules properly; only in less-than-transparent coordinates.)

They are said to be realized in the Nambu-Goldstone nonlinear mode. (Why nonlinear? Is a shift nonlinear? Yes, in SSB physics. Think of sending λ to infinity, and freezing the scalars at the bottom of the potential on a basketball, $\phi_z=\sqrt{v^2-\phi_x^2-\phi_y^2} $. Substituting the dependent variable $\phi_z$ in terms of the other two yields aggressively nonlinear functional dependence.)

It is evident from the potential that σ is massive, but $\phi_x,~\phi_y$ are massless. They each correspond to each SSBroken generator. They are called Nambu-Goldstone bosons, and, crucially, as you probably learned,

  • Their hallmark is the nonvanishing expectation values of their variation,

as emphasized in the second column of the variations above! There is a generic consequence of this: they correspond to the eigenvectors of the mass matrix, $$ \Large\langle\small \frac{\delta^2V}{\delta\phi_i\delta\phi_j}\Large\rangle ~~\small, $$ in our case diag$(0,0,8\lambda v^2)$. That is, they are its null vectors $\phi_x=(1,0,0)^T$ and $\phi_y=(0,1,0)^T$, while $\sigma =(0,0,1)^T$ is quite massive!

Pumping Goldstone scalars into the vacuum alters it into a degenerate state, a new vacuum, as can be inferred from the "hallmark" test, above.

Observe the two generators $L_x, ~ L_y$ do not close into a subgroup. They are a set, "generating" the coset space SO(3)/SO(3). A coset is the set of group elements gH where $g\in G$, s.t the coset only monitors the broken generators. All gs yield the coset space G/H, 2d in terms of (broken) generators. The corresponding Goldstone fields are the (here, 2) "projective coordinates" parameterizing this (coset space) basketball. Any group rotation (around any axis) will trace a great-circle orbit on the globe.

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    $\begingroup$ Nice answer, very explicit :) $\endgroup$
    – Andrew
    Dec 27, 2022 at 19:43
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    $\begingroup$ @Cosmos Zachos Thank you so much for the explicit explanation. $\endgroup$ Dec 30, 2022 at 12:28
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A linearly realized symmetry is one where the fields transform linearly. For a generic set of fields $\phi_a$, under a linearly realized symmetry we will have a transformation $\phi_a\rightarrow \phi_a'$ where \begin{equation} \phi'_a = U_{ab} \phi_b \end{equation} Under a linearly realized symmetry, the vacuum is invariant under the symmetry. Additionally, correlation functions transform in an "obvious" way.

A non-linearly realized symmetry is one where the fields do not transform linearly. Typically, a non-linearly realized symmetry is associated with symmetry breaking and the presence of Goldstone bosons.

A simple example is a shift symmetry \begin{equation} \phi' = \phi + c \end{equation} In this case, the field $\phi$ itself can be thought of as a Goldstone boson, since the shift symmetry will prevent the existence of a mass term.

A more complex example is the breaking of a $U(1)$ symmetry. Under a linearly realized $U(1)$ symmetry, a complex scalar field $\Phi$ will transform as \begin{equation} \Phi' = e^{i\theta} \Phi \end{equation} But now suppose the field takes on a VEV $\bar{\Phi}$, which spontaneously breaks the $U(1)$ symmetry. We may reparameterize the field as $\Phi = \bar{\Phi} e^{i \varphi}$. Then in terms of $\varphi$, the $U(1)$ symmetry is now non-linearly realized \begin{equation} \varphi \rightarrow \varphi + \theta \end{equation} $\varphi$ is now the Goldstone boson associated with the broken $U(1)$ symmetry.

In more complicated symmetry breaking patterns, you will break from a group $G$ to a subgroup $H$. The remnant symmetry is associated with the generators of $H$, and is linearly realized. The Goldstone bosons are associated with the generators of $G$ that are not generators of $H$; this broken part of the symmetry group is non-linearly realized. The coset $G/H$ formalizes the idea of "the part of $G$ that is broken/non-linearly realized", and allows you to create a geometrical formalism (the coset construction) for understanding the allowed interactions of the Goldstone bosons.

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  • $\begingroup$ Thank you. My problem is actually in the last part what you said. Why the generators of H is realized linearly and the broken part is realized nonlinearly? And I am unable to follow the coset part. How the coset G/H is arising here? May be its too trivial but I am unable to get it. If you kindly elaborate.. $\endgroup$ Dec 26, 2022 at 15:20
  • $\begingroup$ "When G is broken into H, The set of ground states forms the coset space G/H" - How can I associate the ground states with elements of the coset space? $\endgroup$ Dec 26, 2022 at 15:28
  • $\begingroup$ @ArkapravaSil A canonical example is an $SO(N)$ scalar field theory. Let's just take it to be $SO(3)$ for simplicity. So we have three scalar fields, and in the unbroken phase the symmetry group is the 3d rotation group. Now we give the fields a vev, so the first two components of $\phi$ have a zero vev and the third component has a non-zero vev. You can think of the vev as "living on the z axis." Now rotations about the z axis leave the vev invariant -- so this 2 dimensional rotation is unbroken. The other two rotations will change the vev. Those two rotations form the coset $SO(3)/SO(2)$. $\endgroup$
    – Andrew
    Dec 26, 2022 at 15:53
  • $\begingroup$ I suspect you might have to add this illustration at the end of your answer in excruciating explicit detail. It is evident the OP and his teachers have skipped this all-crucial step, and wallow in generalities & abstractions. $\endgroup$ Dec 26, 2022 at 15:57
  • $\begingroup$ @Andrew Thank You. I have got it now. I had done the SO(3) case before, but missed some details due to my fault. That's why may be I could not connect. $\endgroup$ Dec 26, 2022 at 16:32

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