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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?

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  • $\begingroup$ Maybe math.stackexchange? $\endgroup$ – imallett Oct 23 '14 at 6:32
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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.

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  • $\begingroup$ Alright, I managed to get the answer by doing that. Although the calculations were quite annoying, so I'm not sure if it was that much faster that just brute forcing the original hamiltonian. $\endgroup$ – Spine Feast Oct 22 '14 at 14:32
  • $\begingroup$ Probably it will give the same effort as directly diagonalizing $H$ and verifying that they are also eigenvectors of $P$. $\endgroup$ – Mateus Sampaio Oct 22 '14 at 15:02
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Hint: When an eigenvalue for an operator $P$ is degenerate, there are more than one way to chose a set of eigenvectors. If the other commuting operator $H$ lifts that degeneracy, there will be a preferred choice of common eigenvectors.

More generally, a set of diagonalizable operators commutes if and only if the set is simultaneously diagonalizable.$^1$


$^{1}$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

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  • $\begingroup$ For the general case, because of domain issues, we usually define that two self-adjoint operators commute if their spectral projections do (or equivalently, their resolvents), which is the analogous case of the finite dimensional case pointed out in the link. $\endgroup$ – Mateus Sampaio Oct 22 '14 at 15:09
  • $\begingroup$ It is interesting to note that the theorem talks about diagonalizable operators (rather than selfadjoint operators, although admittedly the latter is what is relevant for QM). Being diagonalizable is a generic property. Being selfadjoint is a special property. $\endgroup$ – Qmechanic Oct 22 '14 at 21:01
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    $\begingroup$ If we restrict to the case of diagonalization via unitary operators $U$, that is $A=U^*DU$, then $A$ is normal, because $A^*A=U^*D^*UU^*D^*U=U^*D^*DU=U^*DD^*U=U^*DUU^*D^*U=AA^*$. The result is then the same in the infinite case, as there are also a spectral theorem for normal operators and we define commutativity in the same way as for self-adjoint ones. $\endgroup$ – Mateus Sampaio Oct 22 '14 at 21:43
  • $\begingroup$ $\uparrow$ Agree. $\endgroup$ – Qmechanic Oct 22 '14 at 21:44

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