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Ground State Degeneracy in ferromagnetic Heisenberg model

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Oct26-13, 08:28 AM
P: 3
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 Stot = LS. Sztot = 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 Sz 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?
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