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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.

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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 $H\psi=E\psi\,.$

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  • $\begingroup$ I don't get how you use the cauchy schwarz inequality with to different vectors. $\endgroup$
    – Gonenc
    Feb 20, 2015 at 18:21
  • $\begingroup$ The Cauchy-Schwarz inequality says that $|\langle a,b\rangle|\le\|a\|\|b\|$ with equality iff $a=\lambda b$ where $\lambda$ is a scalar. $\endgroup$
    – JLA
    Feb 20, 2015 at 19:32
  • $\begingroup$ Indeed but what is your $a$ and $b$ in this case? $|\langle H\psi,\psi\rangle|^2\le \langle H\psi,H\psi\rangle$ $\endgroup$
    – Gonenc
    Feb 20, 2015 at 19:37
  • $\begingroup$ $a=H\psi\,,b=\psi\,.$ Note that $\langle H\psi,H\psi\rangle=\|H\psi\|^2\,.$ So we have $|\langle H\psi,\psi\rangle|\le\|H\psi\|\|\psi\|=\|H\psi\|\,.$ $\endgroup$
    – JLA
    Feb 20, 2015 at 19:57

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