Unified Description of Nambu-Goldstone Bosons without Lorentz Invariance I am reading an article Unified Description of Nambu-Goldstone Bosons without Lorentz Invariance, arXiv:1203.0609, by Watanabe & Murayama. It gives a proof on the counting of Nambu–Goldstone bosons without Lorentz invariance. I am trying to derive all the equations to get a better understanding. But I have met some trouble with Eqs. (5) and (6). How does the functions $e_i$ are determined so that $\delta\mathcal{L}_{\mathrm{eff}}$ does not depend on the fields explicitly? Could anyone provide a derivation of (6) from (4) and (5)?
 A: 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 Lagrangian description has variables $\phi^a$ that parametrize the coset $G/H$.
We start from a Lagrangian density$^1$
$$\begin{align} {\cal L}~=~&\vartheta_a(\phi)\dot{\phi}^a 
+\frac{1}{2}\bar{g}_{ab}(\phi)\dot{\phi}^a\dot{\phi}^b \cr
&-\frac{1}{2}g_{ab}(\phi)\vec{\nabla}\phi^a \cdot\vec{\nabla}\phi^b-{\cal V}(\phi),\cr 
a,b~\in~&\{1,\ldots, n_{BG}\equiv\dim\mathfrak{g/h}\}. \end{align} \tag{4} $$
which is invariant under the spatial Euclidean group $E(3)$ and time translations. The momenta read
$$\begin{align} \pi_a~=~&\frac{\partial {\cal L}}{\partial \dot{\phi}^a}~=~\vartheta_a+ \bar{g}_{ab}\dot{\phi}^b, \cr 
a~\in~&\{1,\ldots, n_{BG}\equiv\dim\mathfrak{g/h}\}. \end{align}\tag{A}$$
We assume that the infinitesimal transformation
$$ \delta\phi^a~=~\epsilon^i h_i{}^a(\phi), \qquad i~\in~\{1,\ldots, \dim\mathfrak{g}\}, \tag{4a}$$
is a Killing symmetry for both the metrics $g_{ab}$ and $\bar{g}_{ab}$, and
is a quasi-symmetry of the Lagrangian density. This has the effect that the potential ${\cal V}(\phi)$ must be a constant. In particular, we cannot have a mass-term, in agreement with the fact that the Goldstone mode should be gapless. We calculate:
$$\begin{align} \epsilon^i d_t f^0_i~=~&\delta {\cal L}\cr
~=~&\epsilon^i\left( h_i{}^b\partial_b\vartheta_a \dot{\phi}^a + \vartheta_a d_t h_i{}^a \right) \cr 
~=~&\epsilon^i\left( h_i{}^b\omega_{ba} \dot{\phi}^a + d_t(h_i{}^a \vartheta_a) \right),
\end{align}\tag{6a}$$
where we have defined a pre-symplectic 2-form
$$ \omega_{ab}~:=~\partial_a \vartheta_b -\partial_b \vartheta_a. \tag{5a}$$
This leads to OP's sought-for equations
$$\exists e_i: \quad 
h_i{}^b\omega_{ba} \dot{\phi}^a~=~d_te_i
\quad \Rightarrow\quad 
 h_i{}^b\omega_{ba} ~=~\partial_a e_i \tag{5}$$
and
$$ f^0_i ~=~h_i{}^a \vartheta_a + e_i. \tag{6b}$$


*Noether's theorem. The 0-component of the bare Noether current is
$$ j^0_i~=~h_i{}^a \frac{\partial {\cal L}}{\partial \dot{\phi}^a} 
~=~h_i{}^a (\vartheta_a+\bar{g}_{ab}\dot{\phi}^b), \tag{7a}$$
so that the 0-component of the full Noether current is
$$ J^0_i ~=~ j^0_i-f^0_i~=~h_i{}^a \bar{g}_{ab}\dot{\phi}^b-e_i.\tag{7b} $$
The full Noether charge is
$$ Q_i~=~\int_V \!d^3 r ~ J^0_i. \tag{7c} $$
Quantum mechanically,
$$ \hat{Q}_i | \Omega \rangle \left\{\begin{array}{rl} ~\neq~0 &\text{if $i$ is a broken generator in }\mathfrak{g/h}, \cr~=~0 &\text{if $i$ is an unbroken generator in }\mathfrak{h}. \end{array} \right. \tag{B} $$


*Hamiltonian formulation. We can repeat the above analysis in the Hamiltonian formulation, cf. e.g. this Phys.SE post. The Hamiltonian density reads
$$\begin{align} {\cal H}~=~&\frac{1}{2}\bar{g}^{ab}(\phi)(\pi_a-\vartheta_a(\phi))(\pi_b-\vartheta_b(\phi)) \cr
&+ \frac{1}{2}g_{ab}(\phi)\vec{\nabla}\phi^a \cdot\vec{\nabla}\phi^b+{\cal V}(\phi).\end{align} \tag{C} $$
The Hamiltonian Lagrangian density is
$$ {\cal L}_H~=~\pi_a\dot{\phi}^a - {\cal H},\tag{4b} $$
while the 0-component of the full Noether current is
$$\begin{align} J^0_i~=~&h_i{}^a(\phi) (\pi_a-\vartheta_a(\phi)) -e_i(\phi)\cr
~=~&h_i{}^a(\phi) \pi_a -f^0_i(\phi) . \end{align}\tag{7d} $$
The canonical Poisson brackets are
$$ \begin{align}
\{\phi^a(\vec{r},t), \pi_b(\vec{r}^{\prime},t)\}~=~&\delta^a_b \delta^3(\vec{r}\!-\!\vec{r}^{\prime}),\cr
\{\phi^a(\vec{r},t), \phi^b(\vec{r}^{\prime},t)\}~=~&0,\cr
\{\pi_a(\vec{r},t), \pi_b(\vec{r}^{\prime},t)\}~=~&0.
\end{align}\tag{D} $$
The infinitesimal transformation is
$$\begin{align} 
\delta\phi^a~=~&\{\phi^a,Q_i\}\epsilon^i
~=~\epsilon^i h_i{}^a, \cr
\delta \pi_a~=~&\{\pi_a,Q_i\}\epsilon^i
~=~ \epsilon^i \partial_a(f^0_i - h_i{}^b \pi_b)\cr
\{Q_i,\pi_a-\vartheta_a\}~=~& \partial_ah_i{}^b (\pi_b-\vartheta_b),\cr
\{{\cal H},Q_i\}~=~&0,\cr
\delta {\cal L}_H~=~& \epsilon^id_tf^0_i.
\end{align}\tag{4c}$$


*The Poisson matrix/commutator of Noether charges. Next we define the Poisson matrix
$$\begin{align} \{Q_i,J^0_j\}~=~& \left\{ \int_V \!d^3 r \left(h_i{}^a \pi_a -f^0_i \right) , h_j{}^b \pi_b -f^0_j \right\} \cr
~=~& - h_i{}^a\partial_a h_j{}^b\pi_b +h_i{}^a\partial_a f^0_j - (i \leftrightarrow j)\cr
~=~& - h_{[i}{}^a\partial_a h_{j]}{}^b(\pi_b-\vartheta_b) +h_i{}^ah_j{}^b\omega_{ba}.
\end{align}\tag{9a/10}$$
Ref. 1 claims on p. 1 that the rectangular matrix $h_i{}^a$ has maximal rank. Quantum mechanically,
$$\begin{align}\rho_{ij}~=~&\lim_{V\to \infty}\frac{1}{i\hbar V}\langle \Omega | [\hat{Q}_i,\hat{Q}_j] | \Omega \rangle\cr
~=~&\lim_{V\to \infty}\frac{1}{i\hbar}\langle \Omega | [\hat{Q}_i,\hat{J}^0_j] | \Omega \rangle.\end{align}\tag{2/9b} $$


*Low energy linearized equations. We are mostly interested in low energy/infrared/long wavelength field configurations
$$\phi^a~\sim~ 0, \qquad
\pi_a-\vartheta_a~\sim~ \dot{\phi}^a~\sim ~0, \qquad 
\vec{\nabla}\phi^a~\sim~ 0.\tag{E}$$
The linearized Euler-Lagrange (EL) equations
$$ \frac{\partial {\cal L}}{\partial\phi^a}~\approx~d_t\left(\frac{\partial {\cal L}}{\partial\dot{\phi}^a}\right)+\vec{\nabla}\cdot\left(\frac{\partial {\cal L}}{\partial(\vec{\nabla}\phi^a)}\right)\tag{F}$$
read
$$\dot{\phi}^b\omega_{ba} + \bar{g}_{ab} \ddot{\phi}^b~\approx~g_{ab} \vec{\nabla}^2\phi^b. \tag{G}$$
After a Fourier transformation
$$\tilde{\phi}^a(\vec{k},\omega)~=~\int\!dt~d^3r~e^{-i\vec{k}\cdot\vec{r}+i\omega t}\phi^a(\vec{r},t),\tag{H}$$
they become
$$\underbrace{(-i\omega\omega_{ab} + \omega^2\bar{g}_{ab})}_{\text{Hermitian matrix}} \tilde{\phi}^b~\approx~\vec{k}^2g_{ab} \tilde{\phi}^b. \tag{I}$$
We will assume that the metric $g_{ab}$ is positive definite. Let us diagonalize $g_{ab}$ and scale the fields $\phi^a$ so that
$$g_{ab}~=~\delta_{ab}.\tag{J}$$
For fixed $\omega$, eq. (I) becomes an eigenvalue equation with eigenvalue $\vec{k}^2$. Since the matrix on the LHS of eq. (I) is Hermitian, we can choose the $n_{BG}$ eigenmodes/normal modes orthogonal. This gives us $n_{BG}$ dispersion relations of the form $\vec{k}^2=\vec{k}^2(\omega)$, which we in principle can invert to $n_{BG}$ dispersion relations of the form $\omega=\omega(\vec{k}^2)$.
Next use orthogonal transformations to bring the real antisymmetric matrix $\omega_{ab}$ on $2\times 2$ block-diagonal form
$$ [\omega_{ab}]~=~O\begin{pmatrix}
0&-\lambda_1& \cr
\lambda_1&0& \cr
&&\ddots& \cr
&&&0&-\lambda_{n_B}& \cr
&&&\lambda_{n_B}&0& \cr
&&&&&&0&\cr
&&&&&&&\ddots& \cr
&&&&&&&&0&
\end{pmatrix}O^T.\tag{K} $$
Let $n_A:=\dim{\rm ker}\omega_{ab}$. We can diagonalize an $n_A\times n_A$ subblock of $$[\bar{g}_{ab}]~=~O\begin{pmatrix}
\begin{matrix}*&*&*&*\cr
*&*&*&*\cr*&*&*&*\cr
*&*&*&*\cr\end{matrix} &
\begin{matrix}*&*&*&*\cr
*&*&*&*\cr*&*&*&*\cr
*&*&*&*\cr\end{matrix} \cr
\begin{matrix}*&*&*&*\cr
*&*&*&*\cr*&*&*&*\cr
*&*&*&*\cr\end{matrix} &
\begin{matrix} c^2_1&&\cr
&\ddots& \cr
&&c^2_{n_A}\end{matrix}
\end{pmatrix}O^T.\tag{L} $$
using orthogonal transformations without destroying the $2\times 2$ block-diagonal form of $\omega_{ab}$.
Finally recall that the degrees of freedom (DOF) are the number of initial conditions needed divided by 2, cf. e.g. this Phys.SE post.


*Main conclusions:

*

*A zero-direction in $\omega_{ab}$ corresponds to 1 type $A$ Nambu-Goldstone mode with linear dispersion relation $\omega \propto |\vec{k}|$.


*A canonical pair/$2\times 2$ block corresponds to 1 type $B$ Nambu-Goldstone mode with quadratic dispersion relation $\omega \propto |\vec{k}|^2$.


*${\rm rank}\rho_{ij}={\rm rank}\omega_{ab}=2n_B$ and $n_{BG}=n_A+2n_B$.
References:

*

*H. Watanabe & H. Murayama, Unified Description of Nambu-Goldstone Bosons without Lorentz Invariance, Phys. Rev. Lett. 108 (2012) 251602, arXiv:1203.0609.


*H. Watanabe & H. Murayama, Effective Lagrangian for Non-relativistic Systems, Phys. Rev. X4 (2014) 031057, arXiv:1402.7066.


*H. Murayama, What’s new with Goldstone?, 2013 CERN talk, slides.
--
$^1$ This answer uses a slightly different notation than Ref. 1. We suppress for simplicity higher-order derivative terms as we're mainly interested in infrared/long wavelength modes. Symmetry under Galilean or Lorentz boosts would imply that the presymplectic potential $\vartheta_a$ vanishes. Lorentz symmetry would imply that $g_{ab}=c^2\bar{g}_{ab}$.
$^2$ Ref. 1 has the opposite sign convention for the Noether current and Noether charge.
