# Ground State Degeneracy in ferromagnetic Heisenberg model

I am reading the book "Lecture notes on Electron Correlation and Magnetism" by Patrik Fazekas.

It says, "The ground state (of Heisenberg FM model) is not unique. We have just found that the system has the maximum value of the total spin $S_{tot}= LS$. $S^{z}_{tot} = LS$ state which is maximally polarized in the z-direction. However the Hamiltonian is spin-rotationally invariant , hence turning the total spin in another direction does not change the energy: the ground state must be (2LS+1)-fold degenerate." where, L is the no. of lattice sites.

I don't understand why the ground state should be (2LS+1)-fold degenerate and not infinitely degenerate. Are we not considering a global rotation symmetry of the system? I understand it in the following way: All the spins in the system have the same quantization axis which is along z-direction. Now, if every spin is rotated by the same angle,it produces a new state but the energy of the system remains unchanged because it depends on the scalar product of spins in the Heisenberg Hamiltonian. Therefore, a global rotation chosen from any of the infinite no. of possible rotations should produce infinitely many possible states and leave the energy of the Hamiltonian invariant (equal to the ground state energy). "

Does the system has a single quantization axis(i.e. same quantization works for each spin)? or Do we need to consider a unique quantization axis for each spin? What happens to the quantization axis/axes of the system/spins as we consider a different state which has been globally rotated by some angle? If the quantization axis (z-axis) remains fixed, then the system no longer has the maximum value of $S^z$ in the new state obtained after a global rotation. But, if the quantization axis rotates with the system, then, why are there only, (2LS+1) possible rotations? In other words, how does L get into the degeneracy? Thanks

It is true both that the system has an infinite number of ground states, and that the ground state is $(2LS+1)$-fold degenerate, and there is no contradiction.
That is, in the $(2LS+1)$-dimensional space of ground states there are an infinite number of physically distinct states, each corresponding to a choice of the direction in which all of the spins are pointing, but these states are mostly not orthogonal to each other.
If you think about a single spin-$\frac12$ particle, this is already the case - there are an infinite number of states, but they form a finite dimensional space.
• Thank you for your reply. But, I don't understand why the space of ground states has to be $(2LS+1)$ dimensional. For example, in the case of 2 spin- $\frac{1}{2}$ particles, I understand that there are four possible states (3 triplet + 1 singlet) but there are only two states which have parallel spins(and therefore have the energy equal to the ground state energy) i.e. the two states from the triplet set with $m=−1$ and $m=+1$. But, in this case $(2LS+1)$-fold degenerate means 3-fold degenerate. – user1906035 Oct 28 '13 at 20:15
• Actually there are 3 states (or rather a 3-dimensional space of states) when the 2 spins are aligned - all of the triplet states where the total spin is 1. Asking for $m=\pm 1$ is too much, that means they are also aligned along a particular axis, but that is not important. – Rafael L. Greenblatt Oct 29 '13 at 10:30