Ballentine (Quantum Mechanics: A Modern Development 2nd edition, page 290) writes the attached in his introduction of the variational method. My question is about his very last line: why does $H$ being Hermitian imply that stationarity will then require the functional $\Lambda$ to be real at the stationary $\phi,\psi$, and why does this in turn require $\phi = \psi$? Is Ballentine perhaps saying that, in general, we use the variational method (10.110) in such a way that we are searching for minimization of expectation values, and so we enforce $\phi = \psi$ on (10.110)? I am not sure because Ballentine is seeming to suggest that this is something we arrive at as a conclusion, rather than as an assumption.
Edit: To clarify, essentially my concern is this. Ballentine begins with (10.110) and says "often we will want to extremize functionals $\Lambda(\phi,\psi)$ like this". Ballentine then goes through the derivation and shows that an equivalent condition in the space of pairs of functions $(\phi,\psi)$ to the extremization of $\Lambda(\phi,\psi)$ is that the functions obey a pair of eigenvalue equations, with the eigenvalue being related to the value of the functional at those functions $\Lambda(\phi,\psi$). But then, and here is my question, Ballentine seems to say that if $H$ is Hermitian, then we can restrict this search in the space of all pairs of functions to simply the subspace of single functions $(\psi,\psi)$. I don't understand why this claim is true. How do we know we're not missing possible extremization points of the form $(\phi,\psi)$ with $\phi \neq \psi$?
