Why does perturbating spins in the x axis not remove a degeneracy? Suppose we have a Hamiltonian proportional to two spin operators in the z axis:
$$
H_0 = (\vec{s}_{1} +\vec{s}_{2})^2
$$ 
Now suppose I have a perturbation proportional to a different component of spin (a  vector perturbation):
$$
H_1=\lambda(s_{1x}+s_{2x})
$$
I was told that this perturbation does not remove a degeneracy because it does not create a coupling in the degenerate eigenspace and does not couple eigenstates of different eigenspaces.
I do not understand why is it so.
 A: I would say that the perturbed Hamiltonian has the same eigenvectors of the non-perturbed one, but the degeneracy is broken.
We are dealing with a composition of two spins. Only spin variables appear, so we can work in the space $\xi=\xi_1\otimes \xi_2$, where $\xi_1$ is the spin state space associated to one of the particles: eigenstates of $\vec s_1^2$ and $s_{1x}$ provide a basis for this space. Analogous for $\xi_2$.
We know that starting from this two bases we can build the eigenstates common to $\vec S ^2=(\vec s_1 + \vec s_1)^2$ and $S_x=s_{1x}+ s_{1x}$, which constitute a basis for $\xi$. Let's call this eigenstates $∣s,m⟩$, so that:
$$
\begin{cases} \vec S^2∣s,m⟩=\hbar^2s(s+1)∣s,m⟩ \\ S_x∣s,m⟩=m\hbar∣s,m⟩\end{cases}
$$
The initial hamiltonian is $H=c\vec S^2$, its eigenstates are clearly $∣s,m⟩$, its eigenvalues are $c\hbar^2s(s+1)$, which have a degeneracy $2s+1$ for each possible value of $s$. But we said that  $∣s,m⟩$ are also eigenvectors for $S_x$ , so when you consider the hamiltonian $H'=\vec S^2+\lambda S_x$, $\quad$ $∣s,m⟩$ are still eigenvectors for $H'$. Concluding, the eigenvectors where not modified but the new eigenvalues are $c\hbar^2s(s+1)+\lambda\hbar m$, so now they depend both on $s$ and on $m$ and in general should not be degenerate.
