Derivation of eq. (N.10) on page 791 of Appendix N in Ashcroft's SSP.

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SUMMARY

The discussion focuses on the derivation of equation (N.10) from Appendix N in Ashcroft's "Solid State Physics" (SSP). The scalar product is defined using the notation involving operators A and B, and the wave functions Φ. A key conclusion is that the summation of the scalar product leads to the simplification where the operator A can be expressed as A = ∑_f A Φ_f^* Φ_f, given that ∑_f Φ_f^* Φ_f = 1. This establishes a foundational relationship for further derivations in quantum mechanics.

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Homework Statement
I want to derive equation (N.10) on page 791, i.e.:
$$\sum_f(\Phi_i, A\Phi_f)(\Phi_f,B\Phi_i)=(\Phi_i,AB\Phi_i)$$
where ##A,B## are operators, and ##\Phi_i, \Phi_f## are the initial and final wavefunctions before and after the scattering.
Relevant Equations
$$(\Phi_i,\Phi_f) = \int \Phi_i^* \Phi_f$$
Well as always start with the definition of scalar product:

$$\sum_f (\Phi_i,A\Phi_f)(\Phi_f ,B\phi_i) = \sum_f \int \Phi_i^*A\Phi_f \int \Phi_f^* B\Phi_i=\int \int \Phi_i^* \sum_f A\Phi_f \Phi_f^* B\Phi_i$$

How to continue from the last equality?

Thanks.
 
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I think I got it, ##\sum_f A \phi_f^* \phi_f = A## since ##\sum_f \phi_f^*\phi_f = 1##.

Am I correct?
 

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