Variational Principle to find Energy Eigenfunctions In Quantum Mechanics one can estimate an upper bound for the ground state energy with the following functional:
$$\mathcal{F}[\psi(x)]  \equiv \int_{-\infty}^\infty \psi^*(x)\hat{H}\psi(x) \,\, dx \geq E_{gs} $$
The trial wave functions must be square integrable. I thought that I can find the energy eigenfunctions by finding the extrema of this functional. However since it involves complex conjugation and complex functions, I cannot compute the ordinary functional derivative and set it equal to zero. (When I do that I get utter nonsense.) How can I find the extrema of this functional and (hopefully) the energy eigenfunctions?
PS: I know that solving the SE is probably much much easier than dealing with the functional, nevertheless I wanted to get a new look at the variational principle.
 A: You need to minimize this on the $L^2$ unit sphere, ie. over functions $\psi$ with $\|\psi\|_2=1\,.$ We just do Lagrange multipliers. We can compute the Frechet derivative of $\langle \psi,H\psi\rangle$ on the sphere: So take a $\psi$ such that $\|\psi\|_{2}=1$ and take a $\delta\psi$ such that $\langle \psi+\delta\psi,\psi+\delta\psi\rangle =1$ to first order, so that $\langle \psi,\delta\psi\rangle=0\,.$ Minimizing the functional over the sphere, then it must be the case that 
$\displaystyle \langle\psi+\delta\psi,H(\psi+\delta\psi)\rangle-\langle\psi,H\psi\rangle$ $\displaystyle=\langle\delta\psi,H\psi\rangle+\langle \psi,H\delta\psi\rangle=2\text{Re}\langle H\psi,\delta\psi\rangle=0$ for all $\delta\psi$ such that $\langle\psi,\delta\psi\rangle=0\,.$ 
Now let $\phi=H\psi-\langle H\psi,\psi\rangle\psi\,.$ We have that $\langle\phi,\psi\rangle=0\,,$ so letting $\delta\psi=\phi$ we have that $\displaystyle \langle H\psi,H\psi\rangle-|\langle H\psi,\psi\rangle|^2=0\,.$ But from the Cauchy-Schwarz inequality we have that $|\langle H\psi,\psi\rangle|^2\le \langle H\psi,H\psi\rangle$ with equality iff $H\psi=E\psi\,.$  
