# Spin operator eigenstate in Fock space

I am creating an operator group from representation of spin 1 operators $$J_{x} = \frac{1}{\sqrt{2}}\left(\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array} \right)\ \text{, } J_{y} = \frac{i}{\sqrt{2}}\left(\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{array} \right)\ \text{, } J_{z} = \left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array} \right)$$ $$\hat{J}_{\mu} = \sum\limits_{m=-1,0,1}(J_{\mu})_{mn}\hat{a}^{\dagger}_{m}\hat{a}_{n},$$ and I want to find eigenstate of $\hat{J}_{z}$ (or other two) in Fock space with fixed number of particles $N$. For example, I have found that the eigenstate of $J_{z}$ should have the form $$\left| J_{z}\right> = \sum\limits_{k=1/2(N-M)}^{N-M}C_{k}\left|N-k, M-N+2k,N-M-k \right>$$ with eigenstate $M$ and coefficients $C_{k}$. Due to $SU(2)$ structure of the spin operators I suppose that $C_{k}$ are not random numbers, but there is some theory behind them. Is there a generall way to determine those numbers (something similar to Clebsch-Gordan coefficients?) or to find the eigenstates?

• I am not sure what the three numbers in your ket state represent. How could a spin-1 N-particle state have only 3 components? In the mean time, please see my proposed answer and see if I have understood you correctly. – Pooya Nov 20 '14 at 4:01
• Also see a similar question. – Pooya Nov 20 '14 at 18:57

To find the spin eigenstates corresponding to a multi-particle state, all one needs to do is build the appropriate multi-particle spin operator using the direct product, e.g. $$J_z^{(2)}=J_z \otimes 1 + 1\otimes J_z \\ J_z^{(3)}=J_z \otimes 1 \otimes 1 + 1\otimes J_z \otimes 1 + 1\otimes 1\otimes J_z$$ and then diagonalize the resulting operator to find the eigenstates. For the case of $n=2$ one has $$J_z^{(2)}=diag(2,1,0,1,0,-1,0,-1,-2)$$ which happens to be already in diagonal form, therefore the eigenstates are trivial: $\langle j \vert \psi_i \rangle = \delta_{ij}$. In fact, because $J_z$ is diagonal, $J_z^{(n)}$ will also be diagonal for any $n$, therefore the corresponding eigenstates would be $\langle j \vert \psi_i \rangle = \delta_{ij}$, where $1 \le i,j \le 3^n$.