quantum mechanics - These two operators commute...but their eigenvectors aren't all the same. Why?

The Hamiltonian

$$H = \left[ \begin{array}{cccc} a & 0 & 0 & -b \\ 0 & 0 & -b & 0\\ 0 & -b & 0 & 0\\ -b & 0 & 0 & -a \end{array} \right] $$

commutes with the qubit exchange operator

$$ P = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{array} \right] $$

So I would expect the two to have the same eigenvectors. The eigenvectors of $P$ are easily seen to be $(1,0,0,0)^T; (0,0,0,1)^T; (0,1,1,0)^T ; (0,1,-1,0)^T$. The latter two are also eigenvectors of $H$, but the first two are not. Why? I thought commuting operators shared the same eigenbasis?

Answer

Call $u_1, u_2, u_3, u_4$ the eigenvectors described by you, respectively. Your claims are all right, but realize that both $u_1$ and $u_2$ share the same eigenvalue, that is $1$, i.e., $Pu_1=u_1$ and $Pu_2=u_2$. Hence, any linear combination of $u_1$ and $u_2$ will also be eigenvectors with the same eigenvalue $1$. Try to find eigenvectors of $H$ of the form $\alpha u_1+\beta u_2$, with $\alpha$ and $\beta$ being constants.