Help with Time-Independent Perturbation Theory "Good" States Proof

[AT_saavedra]

Homework Statement

Please refer to the attached file.
I am trying to make sense of the last part of the "good" states proof.

Relevant Equations

Please refer to the attached file.

(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.)
I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of Ψ_{a}^{0} and Ψ_{b}^{0} but right before that the proof just showed that they are orthogonal states. Just from the very basics of linear algebra these two statements seem incompatible to me.

I know there must be some error in my interpretation since the contradiction is too clear not to have been caught by the third version of the book, but despite reviewing the chapter once again, as well as my linear algebra book, I am still completely stuck. If someone could help me get out of this hole I have made for myself, I would deeply appreciate it.

 

[julian]

If $\gamma = \mu$, then you don’t necessarily have $\langle \psi_a^0 , \psi_\gamma (0) \rangle = 0$.

 

[AT_saavedra]

Thank you for your answer. I now see what I was misinterpreting. If I could ask a quick follow-up question, am I correct in assuming that $\psi_\gamma(0)$ is a linear combination of $\psi_a^0$ and $\psi_b^0$ because $\psi_a^0$ and $\psi_b^0$ are two distinct eigenvectors with distinct eigenvalues so they must span the space? (since the theorem is dealing with 2nd order degeneracy)

 

[julian]

To show that $\psi_a^0$ and $\psi_b^0$ span the space, we must prove they are linearly independent. Suppose

\begin{align*}
\kappa \psi_a^0 + \zeta \psi_b^0 = 0.
\end{align*}

Applying $A$ gives

\begin{align*}
\kappa \mu \psi_a^0 + \zeta \nu \psi_b^0 = 0.
\end{align*}

Subtracting $\mu (\kappa \psi_a^0 + \zeta \psi_b^0) = 0$ from this yields

\begin{align*}
\zeta (\nu - \mu) \psi_b^0 = 0.
\end{align*}

Since $\mu \ne \nu$, it follows that $\zeta = 0$, and hence $\kappa = 0$.

 

[AT_saavedra]

To show that $\psi_a^0$ and $\psi_b^0$ span the space, we must prove they are linearly independent. Suppose

\begin{align*}
\kappa \psi_a^0 + \zeta \psi_b^0 = 0.
\end{align*}

Applying $A$ gives

\begin{align*}
\kappa \mu \psi_a^0 + \zeta \nu \psi_b^0 = 0.
\end{align*}

Subtracting $\mu (\kappa \psi_a^0 + \zeta \psi_b^0) = 0$ from this yields

\begin{align*}
\zeta (\nu - \mu) \psi_b^0 = 0.
\end{align*}

Since $\mu \ne \nu$, it follows that $\zeta = 0$, and hence $\kappa = 0$.

Thank you very much, your answer was very thorough