1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Hint: $|k_i>$ is an eigenstate of $H_0$ $\endgroup$ – user110373 Dec 31 '17 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 Jan 1 '18 at 13:20
3
$\begingroup$

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: $$e^{-iH_0t}|E_i>=e^{-iE_it}|E_i>$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.