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Within first order (or linear order) quantum perturbation theory, the Schrödinger equation (for a state $i$) can be written:

$$\delta H |\psi_i^0 \rangle + H^0 |\delta\psi_i\rangle=\delta\varepsilon_i|\psi_i^0\rangle+\varepsilon_i^0|\delta\psi_i\rangle $$

AKA Sternheimer's equation, where the exponent "$0$" denotes the ground-state, and the symbol $\delta$ corresponds to first order perturbations.

Now, I'm reading somewhere 2 things that confuse me:

  1. $\langle \psi_j^0|\delta\psi_i\rangle = \frac{\langle \psi_j^0|\delta H|\psi_i^0\rangle - \delta \varepsilon_i\langle\psi_j^0|\psi_i^0\rangle}{\varepsilon_i^0-\varepsilon_j^0}$

  2. $\langle \psi_j^0 | \delta\psi_i\rangle \sim \langle\psi_j^0|\delta H|\psi_i^0\rangle$

I don't understand how to obtain 1 from Sternheimer's equation, and I also don't get why the approximation 2 can be made.

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  • $\begingroup$ Isn't your point fully (and elegantly) addressed in the WP article you are citing? Which Sternheimer? $\endgroup$ – Cosmas Zachos Jul 25 '19 at 14:46
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In order to obtain 1, multiply from the left with $\langle \psi^0_j |$ and use that $$ \langle \psi^0_j | H^0 = \varepsilon^0_j\, \langle \psi^0_j | . $$ Note that 1 is only correct for $i \neq j$, so that the second term in the numerator of your expression vanishes ($\langle \psi^0_j | \psi^0_i \rangle = 0$).

Thus, for $i \neq j$, $$ \langle \psi^0_j | \delta \psi_i \rangle = \frac{\langle \psi^0_j | \delta H |\psi^0_i \rangle}{\varepsilon^0_i - \varepsilon^0_j} . $$ Your equation 2 is correct if $\sim$ is read as "is proportional to", but definitely not correct if $\sim$ is read as "is approximately equal to" (for dimensionality reasons alone).

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