# Calculating $\langle \psi_j^0 | \delta\psi_i\rangle$ in perturbation theory [closed]

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.

• Isn't your point fully (and elegantly) addressed in the WP article you are citing? Which Sternheimer? – Cosmas Zachos Jul 25 '19 at 14:46

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).