Goldstone's theorem, symmetry breaking and the Heisenberg model

Question

I'm currently researching symmetry breaking and Goldstone's theorem for a project in my third year of my theoretical physics degree. So my knowledge isn't from a formal teaching but my own research.

I began with trying to understand Goldstone's theorem and from what I understand it's the idea that if a continuous symmetry is broken, you get massless scalar fields (Goldstone bosons). I've gone through the mathematics of this and it seems to make sense.

However, I'm looking into the Heisenberg model as a sort of real world example of Goldstone's theorem and I'm coming across issues. I guess I don't have a question per se but more looking to see if my understanding is correct. So the Heisenberg model says the Hamiltonian is made up from the spins of nearest neighbours in a lattice. Clearly this hamiltonian is symmetric under rotation (if you rotate all the spins by theta then the net energy will remain the same?). The ground state would be the state in which all the spins are pointing in the same direction and clearly there are an infinite number of these as they can point in any direction provided they're all pointing the same way. I have then heard that "choosing" a ground state spontaneously breaks this symmetry, is that because you've now collapsed from a infinite number of possible ground states which are invariant under rotation to a single state and hence if you rotate all the spins then it wouldn't be that specific state you selected? Furthermore, where does Goldstone's theorem come into this? I've heard something about spin waves, are these the Goldstone bosons in this circumstance?

I hope somebody can help answer my questions or point me in the right direction. I've tried to explain myself clearly, whether or not that was achieved is a different story.

Answer 1

The Heisenberg model is actually an example of an exception to the standard Goldstone theorem for relativistic QFT. In the standard case, we expect that each broken symmetry yields a gapless mode with linear dispersion at small momenta. This is not true generally for non-relativistic systems such as the Heisenberg model. Consider the Hamiltonian \begin{equation} H=-J\sum_{n=1}^N\vec{s}_n\cdot\vec{s}_{n+1}, \end{equation} where we assume periodic boundary conditions $\vec{s}_{N+1}=\vec{s}_1$. Here we are describing a spin chain rather than a lattice for simplicity. Since the Hamiltonian couples nearest neighbors through a dot-product, the Hamiltonian itself exhibits rotational symmetry about the $x$, $y$, and $z$ axes. The ground state of this system has all the spins pointing in the same direction, which we will choose to be the $+z$ direction; that is, \begin{equation} |0\rangle=|\uparrow,...,\uparrow\rangle. \end{equation} Thus the vacuum, or ground state, of the theory has chosen a preferred direction in space and has broken the original rotational symmetry. Rather, $|0\rangle$ is only invariant under rotations about the $z$-axis. You can verify this fact for yourself by acting on the ground state with the standard SU(2) rotation operators (one for each lattice site). We say that the original SO(3) symmetry has been broken to SO(2). We have two broken symmetries in this case, corresponding to rotations about the $x$ and $y$ axes.

Now consider the following state \begin{equation} |k\rangle=\sum_{n=1}^N e^{ikn}|...,\uparrow,\downarrow_n,\uparrow,...\rangle, \end{equation} which is a linear combination of states with the $n$-th spin flipped down. One may show that this state is an energy eigenstate with excitation energy (above the ground state) \begin{equation} \Delta_E=\hbar^2J(1-\cos(k)), \end{equation} which is gapless and quadratic in $k$ for small $k$. This excitation is known as a magnon. The curious thing here is that we broke two symmetries but found only a single excitation with quadratic rather than linear dispersion. The reason is that the two symmetries that we broke are not independent since the generators of those symmetries satisfy the familiar commutation relations \begin{equation} [\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k, \end{equation} so that in particular \begin{equation} [\sigma_x,\sigma_y]=2i\sigma_z. \end{equation} By performing rotations about the $x$ and $y$ axes, we can generate a rotation about the $z$-axis.

The number of broken generators is exactly equal to the number of Goldstone bosons if \begin{equation} \langle 0|[Q_i,Q_j]0\rangle =0, \end{equation} for all broken symmetry generators $Q$. Now you may wonder what is so special about Lorentz invariance. Recall that Noether's theorem tells us that conserved charges (i.e. symmetry generators) can be written as the integral of a conserved current \begin{equation} Q=\int j^0(x) d^3x, \end{equation} whereas Lorentz invariance demands that expectation values of objects with free Lorentz indices must vanish \begin{equation} \langle j^0(x)\rangle=0, \end{equation} thereby ensuring that the number of Goldstone bosons is always equal to the number of broken symmetry generators in a relativistic theory.

You may find the following reference useful (the source of most of this info). Spontaneous Symmetry Breaking

Answer 2

Yes, you have it right. In this case, when you choose a particular ground state, you are choosing a particular direction for all of the spins to point in. Excitations away from this ground state, i.e., waves in which the spins of the particles oscillate away from the chosen ground state direction, are the Goldstone Bosons.