In 1d spin model with periodic boundary condition with N sites, each site has a spin. Now I have a Hamiltonian for this model, and I want to restrict this Hamiltonian to the sector with quasimomentum $k=0$, magnetization $m_z =1/3$ and space inversion parity $p=1$. But these conserved quantities give me some problems: if quasimomentum $k=0$, meaning for a particular spin configuration, like only one spin up, I move this configuration one site to the left or one site to the right, the wavefunction should be the same, so wavefunction for quasimomentum $k=0$ should be something like $\frac{1}{\sqrt{N}}\sum_{all~one~spin~up~configuration~rearrangement} |\uparrow \downarrow \downarrow ..... >$ (all the coefficients are same), and if I restrict $m_z=1/3$, meaning I fix the number of spins up and spins down, leaving me only one wavefunction left, and the space inversion parity $p=1$ condition seems redundant. I don't expect after these restrictions, I only get one wavefunction, so what is the problem?
1 Answer
Restricting the magnetization doesn't necessarily give you one wavefunction. Say you have 7 sites and start with the state $|\uparrow\uparrow\uparrow\downarrow\downarrow\downarrow\downarrow\rangle$. You can turn this into a state with 0 quasimomentum $$\frac{1}{\sqrt{7}}\left[|\uparrow\uparrow\uparrow\downarrow\downarrow\downarrow\downarrow\rangle+|\downarrow\uparrow\uparrow\uparrow\downarrow\downarrow\downarrow\rangle+|\downarrow\downarrow\uparrow\uparrow\uparrow\downarrow\downarrow\rangle+\dots\right]$$ You can build a distinct 0 quasimomentum state with the same magnetization by starting with $|\uparrow\downarrow\uparrow\uparrow\downarrow\downarrow\downarrow\rangle$ instead. $$\frac{1}{\sqrt{7}}\left[|\uparrow\downarrow\uparrow\uparrow\downarrow\downarrow\downarrow\rangle+|\downarrow\uparrow\downarrow\uparrow\uparrow\downarrow\downarrow\rangle+\dots\right]$$ This second state doesn't have a well defined parity (the first state has parity +1) but we have a choice between parity $\pm 1$ combinations $$\frac{1}{\sqrt{14}}\left(\left[|\uparrow\downarrow\uparrow\uparrow\downarrow\downarrow\downarrow\rangle+|\downarrow\uparrow\downarrow\uparrow\uparrow\downarrow\downarrow\rangle+\dots\right]\pm\left[|\downarrow\downarrow\downarrow\uparrow\uparrow\downarrow\uparrow\rangle+|\downarrow\downarrow\uparrow\uparrow\downarrow\uparrow\downarrow\rangle+\dots\right]\right)$$ So all three conditions matter, and even after imposing them there is still not a unique wavefunction in general.
