I have been trying to prove variational theorem in quantum mechanics for a couple of days but I can't understand the logic behind certain steps. Here is what I have so far:
\begin{equation}
E=\frac{\langle \phi ^*|H|\psi\rangle}{\langle \phi ^*|\phi\rangle}
\end{equation}
consider a wavefunction close to the ground state,
\begin{equation}
\psi=\psi_0+\delta \psi
\end{equation}
\begin{equation}
E=\frac{\langle\psi ^*_0+\delta \psi^* |H|\psi _0+\delta \psi\rangle}{\langle \psi ^* _0 +\delta \psi ^*|\psi _0+\delta \psi \rangle}
\end{equation}
(here is where I begin to struggle … I don't understand the mathematical step).
\begin{equation}
=\frac{\langle \psi _0^*|H|\psi _0\rangle+2\langle\delta\psi^*|H|\psi _0\rangle+\langle\delta\psi^*|H|\delta\psi\rangle}{\langle\psi^*_0|\psi_0\rangle+2\langle\delta \psi^*|\psi_0\rangle+\langle\delta\psi^*|\delta \psi\rangle}
\end{equation}
\begin{equation}
=\frac{E_0\langle \psi _0^*|\psi _0\rangle+2E_0\langle\delta\psi^*|\psi _0\rangle+\langle\delta\psi^*|H|\delta\psi\rangle}{\langle\psi^*_0|\psi_0\rangle+2\langle\delta \psi^*|\psi_0\rangle+\langle\delta\psi^*|\delta \psi\rangle}
\end{equation}
\begin{equation}
=E_0 +O((\delta\psi)^2)
\end{equation}
I understand that the change in the wavefunction leads to an energy term that is second order, yet how can we tell that the best wavefunction is the one that minimises this energy? Is it because we ignore the $\delta ^2$ term?
There is another alternative proof here which I also can not follow.
http://www.nyu.edu/classes/tuckerman/quant.mech/lectures/lecture_3/node1.html
Best Answer
I don't find this proof a good one, since the notation is messy and not very clear (not to say wrong). One proof can be given in a similar way to the one you posted in the link. Suppose the spectrum of $H$ is discrete and the set of eigenstates $\{|\phi_n\rangle\}$ constitutes an orthonormal basis with eigenvalues $E_n$, such that $E_0\leq E_1\leq E_2\leq\dots$. So for any normalized state $|\psi\rangle$, we can expand it in this base: $$|\psi\rangle=\sum_nc_n|\phi_n\rangle$$ Then we have \begin{align}\langle\psi|H|\psi\rangle&=\left(\sum_mc_m^*\langle\phi_m|\right)H\left(\sum_nc_n|\phi_n\rangle\right)\\ &=\sum_{m,n}c_m^*c_n\langle\phi_m|H|\phi_n\rangle\\ &=\sum_{m,n}c_m^*c_n\langle\phi_m|E_n|\phi_n\rangle\\ &=\sum_{m,n}c_m^*c_nE_n\langle\phi_m|\phi_n\rangle \\ &=\sum_{m,n}c_m^*c_nE_n\delta_{mn}\\ &=\sum_n|c_n|^2E_n\\ &\geq \sum_n|c_n|^2E_0=E_0, \end{align} since, $\sum_n|c_n|^2=1$ and $E_n\geq E_0$, where $E_0$ is the lowest eigenstate of $H$. It can be proved that this theorem also holds in the case that there is a lowest eigenvalue $E_0<\sigma_{ess}(H)$ in the spectrum of $H$, even though the spectrum is not made only of eigenvalues.