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Considering a three-dimensional state space spanned by the orthonormal basis formed by the three kets $|u_1\rangle,|u_2\rangle,|u_3\rangle $. In the basis of these three vectors, taken in order, are defined the operators $$H=\hbar\omega_0 \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right) \qquad B=b\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right) $$ with $\omega_0$ and $b$ real constants

My question is:

How do I give a basis of eigenvectors common to H and B?

We know that $H$ and $B$ commute,that is $$[H,B]=0$$

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  • $\begingroup$ One way is by finding eigenvectors of an arbitrary linear combination of $H$ and $B$, say $\alpha H + \beta B$. $\endgroup$ – secavara Jan 13 '18 at 19:05
  • $\begingroup$ $|u_1\rangle$ is a no brainer. For the others, try: $|u_2\rangle \pm |u_3\rangle$. Since, for $H$, $\lambda_2 = \lambda_3$, any linear combination of their eigenvectors is also an eigenvector. $\endgroup$ – JEB Jan 13 '18 at 19:06
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It is sufficient to find the eigenstates of $B$ in the subspace spanned by $\vert 2\rangle=\left(\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right)$ and $\vert 3\rangle=\left(\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right)$. The eigenstates of $B$ in that subspace will automatically also be eigenstates of $H$ because the similarity transformation $T$ that will diagonalize $B$ will be of the generic form $$ T=\left(\begin{array}{ccc} 1&0&0 \\ 0&T_{22}&T_{23} \\ 0&T_{32}&T_{33}\end{array}\right) $$ and so will commute with $H$ on that subspace that $H$ on that subspace is (up to a scalar) the unit matrix.

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Note that any two Hermitian operators $\hat{A}$ and $\hat{B}$ that commute share a eigenbasis.

Since the two matrices are Hermitian, your problem reduces to finding the eigenbasis of either one. This requires calculating the eigenvectors. See e.g. https://math.stackexchange.com/questions/1409350/what-is-an-eigenbasis-and-how-do-i-calculate-it-with-the-information-below.

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    $\begingroup$ Admittedly this answer was my first impulse and I posted it as a comment but we have to be careful about the fact that while they share an eigenbasis, not any eigenbasis will be shared. For instance, $|u_1\rangle,|u_2\rangle,|u_3\rangle$ are eigenvectors of $H$ but not $B$. The way I find a shared eigenbasis in general is by computing the eigenvectors of an arbitrary linear combination of both operators. $\endgroup$ – secavara Jan 13 '18 at 19:21
  • $\begingroup$ Thanks! I'll have to digest that, maybe I'll have to delete this answer. $\endgroup$ – Martin C. Jan 13 '18 at 19:27
  • $\begingroup$ No problem, do as you wish. I did that with my comment haha $\endgroup$ – secavara Jan 13 '18 at 19:36

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