What happens with the Fock state after the Schwinger boson transformation? Let us imagine that initially, I had the following Hamiltonian.
$$
\hat{H} = \frac{\alpha}{4} (a^\dagger b - b^\dagger a )^2 + \frac{\beta}{2}(a^\dagger a -  b^\dagger b), \quad [a,a^\dagger] = 1, \quad [b, b^\dagger] = 1,
$$
here $\alpha$ and $\beta$ - some real constants. And the initial state was $|\psi_0 \rangle = | n,0 \rangle  = \frac{(a^\dagger)^n}{\sqrt{n!}}|0,0\rangle $.
After applying Schwinger transformation ($L_z \rightarrow \frac{1}{2}(a^\dagger a - b^\dagger b)$, $L_{+} \rightarrow a^\dagger b$, $L_{-} \rightarrow a b^\dagger$) the Hamiltonian transforms into
$$
\hat{H} = \alpha L_x^2 + \beta L_z,
$$
here $L_{x} = \frac{1}{2}\left( L_{+} + L_{-} \right)$.
What will be with the initial state under this transform?
 A: I assume you're looking for the components of the state in the angular momentum representation? The state itself is technically the same, but you want it in a different basis.
We need to think about the actions of $L^2$ and $L_z$ in the Fock space. Consider
$$L_z(a^{\dagger})^n |0\rangle = \frac{n}{2} (a^{\dagger})^n |0\rangle$$
$L_z=\frac{1}{2}(N_a-N_b)$ is just two number operators, so this is easy to evaluate.
$L^2$ takes a bit of work; first, you should define $L_y=\frac{i}{2}(L_--L_+)$. Doing some operator math, you should get that
$$L^2 = L_x^2+L_y^2+L_z^2 = L_z^2-\frac{1}{4}(L_--L_+)^2+-\frac{1}{4}(L_-+L_+)^2 = L_z^2-\frac{1}{4}(2L_+L_-+2L_-L_+)
=\frac{1}{4}(a^\dagger aa^\dagger a+b^\dagger bb^\dagger b-2a^\dagger ab^\dagger b + 2a^\dagger ba b^\dagger+2a b^\dagger a^\dagger b )
=\frac{1}{4}(N_a^2+N_b^2+2N_aN_b+2N_a+2N_b )=\frac{1}{4}(N_a+N_b)(N_a+N_b+2)$$
In particular,
$$L^2(a^{\dagger})^n |0\rangle = \left(\frac{n}{2}\right)\left(\frac{n}{2}+1\right) (a^{\dagger})^n |0\rangle$$
This is really nice, since this tells you that $(a^{\dagger})^n |0\rangle$ is an eigenstate of both $L_z$ and $L^2$ with quantum numbers $m_l=\frac{n}{2}$ and $l=\frac{n}{2}$, or just
$$\left|n,0\right\rangle \propto (a^{\dagger})^n |0\rangle \propto \left|\frac{n}{2},\frac{n}{2}\right\rangle_l,$$
up to some normalization constant that can be computed using the original state.
