The second-order Born approximation for the scattering amplitude is given by

$$ f^{(2)}(k_f, k_i) = -\frac{m}{2\pi\hbar^2} \left<k_f\left| V \frac{1}{E-H_0+i\eta}V \right|k_i\right>, $$

where $H_0 = \frac{p^2}{2m}$ and $V$ is the Yukawa potential $V(r) = V_0 \frac{\exp(-\kappa r)}{r}$ or, in momentum space, $V(q) = -V_0\frac{2\pi}{q^2+\kappa^2}$. In the last expression, I used $q := k_f - k_i$.

I would like to calculate the expression for the scattering amplitude above but what I'm struggling with is the operator in the denominator of the fraction. How do I solve this problem?

  • $\begingroup$ Hint: $|k_i>$ is an eigenstate of $H_0$ $\endgroup$
    – user110373
    Commented Dec 31, 2017 at 18:18
  • $\begingroup$ @user110373 Sorry, this doesn't help me. I know that $|k_i>$ is an eigenstate of the Hamiltonian but this still leaves me with the problem what to do with the Hamiltonian in the denominator. $\endgroup$
    – MeMeansMe
    Commented Jan 1, 2018 at 13:20

1 Answer 1


I struggled with this as well, but from what I've seen, you have to work in the eigenstate basis, and then you just replace the operator with it's eigenvalue. $$\frac{1}{H_0}|E_i>=\frac{1}{E_i}|E_i>$$ This is the same for expressions like: $$e^{-iH_0t}|E_i>=e^{-iE_it}|E_i>$$

  • $\begingroup$ This answer is pretty mysterious. Can you provide a more detailed explanation so that college level students can understand? $\endgroup$
    – dturn805
    Commented Jan 21 at 1:24

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