# Proof of variational theorem for perturbation theory in QM

I am reading this lecture on the variational method to get the 0-state energy of a system with Hamiltonian $H$. What they do is basically to say that $E_0$ is a lower bound for $\langle H\rangle$. Then they minimize the functional $E[\psi] = \langle H\rangle$ in order to approximate $E_0$. What I don't understand is that they perform the variation $E[\psi + \delta \psi]$. They expand this in series of $\delta\psi$ and get the expression for $\langle H\rangle$ (page 2):

$$\frac{\langle \psi|H|\psi\rangle + \langle \delta\psi|H|\psi\rangle + \langle \psi|H|\delta\psi\rangle + \ldots}{\langle\psi|\psi\rangle + \langle\psi|\delta\psi\rangle + \langle\delta\psi|\psi\rangle + \ldots}.$$ I understand up to this point. Then they say that this is equal to $$\frac{\langle \psi|H|\psi\rangle + \langle \delta\psi|H|\psi\rangle + \langle \psi|H|\delta\psi\rangle}{\langle\psi|\psi\rangle}\left[1 - \frac{\langle\delta\psi|\psi\rangle}{\langle\psi|\psi\rangle}- \frac{\langle\psi|\delta\psi\rangle}{\langle\psi|\psi\rangle}\right].$$ I got completely lost here. Where did those minus signs come from?

• @AccidentalFourierTransform Wow, didn't see that :P Can you post this as an answer so that the question does not remain unanswered? – Vladimir Vargas Apr 6 '17 at 20:22
• Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic Apr 6 '17 at 20:58

## 1 Answer

From AccidentalFourierTransform's comment:

$$\frac{1}{\langle\psi|\psi\rangle + \langle\psi|\delta\psi\rangle + \langle\delta\psi|\psi\rangle} = \frac{1}{\langle\psi|\psi\rangle}\frac{1}{1 + \dfrac{\langle\delta\psi|\psi\rangle}{\langle\psi|\psi\rangle} + \dfrac{\langle\psi|\delta\psi\rangle}{\langle\psi|\psi\rangle}}\\ =\frac{1}{\langle\psi|\psi\rangle}\left(1 - \dfrac{\langle\delta\psi|\psi\rangle}{\langle\psi|\psi\rangle} - \dfrac{\langle\psi|\delta\psi\rangle}{\langle\psi|\psi\rangle} + \ldots\right).$$