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I'm reading about the variational method in Shankar's Principles of Quantum Mechanics, page 433. The author states that if we have the trial ket $$|\psi\rangle = |E_0\rangle + |\delta\psi\rangle$$ where $|E_0\rangle$ is the eigenket corresponding to the ground state energy and we decompose $|\delta\psi\rangle$ into parts parallel and perpendicular to $|E_0\rangle$ such that $$|\psi\rangle = |E_0\rangle + \alpha|E_0\rangle + |\delta\psi_{\perp}\rangle$$ then we obtain $$\langle\psi|H|\psi\rangle = \frac{E_0|1+\alpha|^2 + \langle\delta\psi_{\perp}|H|\delta\psi_{\perp}\rangle}{|1+\alpha|^2 + \langle\delta\psi_{\perp}|\delta\psi_{\perp}\rangle}$$ $$\langle\psi|H|\psi\rangle = E_0 + O[(\delta\psi_{\perp})^2]$$ I don't know how the author arrives at the final line from the one above it. In particular, how do we know that $\langle\delta\psi_{\perp}|H|\delta\psi_{\perp}\rangle \sim O[(\delta\psi_{\perp})^2]$ ?

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Let $\Bbb I$ denote the identity on the Hilbert space so$$\langle\psi|H|\psi\rangle-E_0=\frac{\langle\delta\psi_\perp|H-E_0\Bbb I|\delta\psi_\perp\rangle}{|1+\alpha|^2+\langle\delta\psi_\perp|\delta\psi_\perp\rangle}.$$The numerator (denominator) is $O\left((\delta\psi_\perp)^2\right)$ ($O(1)$). Your question is about the numerator's behaviour. This $O$ characterization is correct in the sense that, if we scale $\delta\psi_\perp$, we scale $\langle\delta_\perp|M|\delta_\perp\rangle$ the same way for all operators $M$ from the Hilbert space to itself, including $\Bbb I$.

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  • $\begingroup$ Is this $O[(\delta\psi_{\perp})^2]$ characterisation due to the fact that we can write $\langle\delta\psi_{\perp}|M|\delta\psi_{\perp}\rangle = \int\delta\psi_{\perp}^*(x)M\delta\psi_{\perp}(x)dx$ ? $\endgroup$
    – Nitram
    Aug 24, 2021 at 6:20
  • $\begingroup$ @Nitram RE your comment: this is just a definition that you have written! $\endgroup$
    – Jacob A
    Aug 28, 2021 at 2:15

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