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)?


1 Answer 1


In this answer we will outline the proof given in Ref. 1 of the non-relativistic Goldstone theorem.

  1. 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 $$ f^0_i ~=~h_i{}^a \vartheta_a + e_i. \tag{6b}$$

  2. 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, $$ \hat{Q}_i | \Omega \rangle \left\{\begin{array}{rl} ~\neq~0 &\text{if $i$ is a broken generator in }\mathfrak{g/h}, \cr~=~0 &\text{if $i$ is an unbroken generator in }\mathfrak{h}. \end{array} \right. \tag{B} $$

  3. 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 $$\begin{align} \delta\phi^a~=~&\{\phi^a,Q_i\}\epsilon^i ~=~\epsilon^i h_i{}^a, \cr \delta \pi_a~=~&\{\pi_a,Q_i\}\epsilon^i ~=~ \epsilon^i \partial_a(f^0_i - h_i{}^b \pi_b)\cr \{Q_i,\pi_a-\vartheta_a\}~=~& \partial_ah_i{}^b (\pi_b-\vartheta_b),\cr \{{\cal H},Q_i\}~=~&0,\cr \delta {\cal L}_H~=~& \epsilon^id_tf^0_i. \end{align}\tag{4c}$$

  4. 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, $$\begin{align}\rho_{ij}~=~&\lim_{V\to \infty}\frac{1}{i\hbar V}\langle \Omega | [\hat{Q}_i,\hat{Q}_j] | \Omega \rangle\cr ~=~&\lim_{V\to \infty}\frac{1}{i\hbar}\langle \Omega | [\hat{Q}_i,\hat{J}^0_j] | \Omega \rangle.\end{align}\tag{2/9b} $$

  5. 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.

  6. 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 $n_{BG}=n_A+2n_B$.


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

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

  3. 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.


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