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In the context of solving the eigenvalue equation for an operator $C = A(1) + B(2)$ in terms of the eigenvectors of each of $A(1)$ and $B(2)$, which are the extended operators from the Hilbert spaces $\scr E_1$ resp. $\scr E_2$ to $\scr E = \scr E_1\otimes\scr E_2$, the author finds that $$ C|\varphi_n(1)\rangle|\chi_p(2)\rangle = (a_n + b_p)|\varphi_n(1)\rangle|\chi_p(2)\rangle = c_{np}|\varphi_n(1)\rangle|\chi_p(2)\rangle $$ where $a_n$ and $b_n$ are the eigen values of $A(1)$ and $B(2)$ to $|\varphi_n(1)\rangle$ and $|\chi_p(2)\rangle$ resp., assuming no degeneracy in them. In the case of degeneracy of the eigenvalues of $C$, the author comments that this may be the case if, e.g., there is two different pairs of indices such that $c_{mq} = c_{np}$, and in this case the eigenvector of $C$ corresponding to this eigenvalues is of the form $$ \lambda|\varphi_n(1)\rangle|\chi_p(2)\rangle + \mu|\varphi_n(1)\rangle|\chi_p(2)\rangle $$ which I think he meant (note the indices) $$ \lambda|\varphi_n(1)\rangle|\chi_p(2)\rangle + \mu|\varphi_m(1)\rangle|\chi_q(2)\rangle $$ Question: ...right ?

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    $\begingroup$ My French edition correctly reports the second equation, so that's probably a typo of the English edition. $\endgroup$ Dec 3, 2020 at 15:06
  • $\begingroup$ @MassimoOrtolano Thanks! this is the most important confirmation, since I know that the original is french $\endgroup$
    – Physor
    Dec 3, 2020 at 15:22

2 Answers 2

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The first equation in your answer should have a $a_{n} + b_{p}$, but I'm assuming that's a typo. For your question, you are correct. The eigenspace associated to a degenerate eigenvalue has as a basis the eigenvectors sharing it. [generalizes to $n$-fold degeneracies]

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  • $\begingroup$ Thanks, I've edited it: $b_n \to b_p$ $\endgroup$
    – Physor
    Dec 3, 2020 at 14:14
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I agree with you, otherwise we have

$$|\psi\rangle = \mu|\varphi_n(1)\rangle|\chi_p(2)\rangle+\lambda|\varphi_n(1)\rangle|\chi_p(2)\rangle = (\mu+\lambda)|\varphi_n(1)\rangle|\chi_p(2)\rangle,$$

which does not contemplate every possible eigenvector associated to this eigenspace, but rather only one.

PS: are you sure about the indices in the $c_{mp} = c_{nq}$?

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  • $\begingroup$ $(c_{mp}=c_{nq}) \to (c_{mq}=c_{np})$ $\endgroup$
    – Physor
    Dec 3, 2020 at 14:17
  • $\begingroup$ that what I thought, thanks! $\endgroup$
    – Physor
    Dec 3, 2020 at 14:20

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