# Possible error in Tannoudji, Vol. 1, p. 153

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 ?

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

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]

• Thanks, I've edited it: $b_n \to b_p$ Dec 3, 2020 at 14:14

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 $$c_{mp} = c_{nq}$$?

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