# Goldstone modes $\hat a^\dagger(k) \left|0\right>$ for small $k$ or $\hat a^\dagger(0) \left|0\right>$

Consider the Hamiltonian: $$H=\sum_{\vec k} \varepsilon (\vec k)a_{\vec k}^\dagger a_{\vec k}$$ with $\varepsilon(\vec k) \rightarrow 0$ as $|\vec k|\rightarrow 0$. I know that this has gapless excitations and therefore Goldstone modes but I am confused about the actual definition of what counts as a Goldstone mode. Do the refer to the states $\hat a^\dagger(k) \left|0\right>$ for small/infinitesimal $k$ which do still have some energy or to the ground states $\hat a^\dagger(0) \left|0\right>$ which have no energy.

The massless field is called the Goldstone 'mode'. The term 'mode' does indeed make one think of a particular momentum mode and in that it is a misnomer. For simplicity lets take the $U(1)$ case with $\phi=|\phi| e^{i\theta}$. After symmetry breaking the Goldstone part of the Lagrangian is
$$\mathcal L = \frac{1}{2} \phi_0^2 (\partial \theta)^2$$
with $\theta \in [0, 2 \pi)$ and for some non-zero value of $\phi_0$ corresponding to the minimum of the potential. Symmetry breaking chooses one value of $\theta=\theta_0$ arbitrarily. Linearized excitations can be quantized and they change the value of $\theta$. However, strictly speaking $a^\dagger_0 |\theta_0 \rangle$ is not defined as it is not normalizable by itself and you cannot put it in a wavepacket while keeping the nerdy zero. This is related to the fact that one cannot change the vacuum of this system if the volume is infinite and it is really only in that limit that the symmetry is spontaneously broken as for finite volume one can take linear superpositions of different $\theta$ much like one can form a superposition of position for a particle living on a circle.