# Wavefunction of Cooper Pairs

I tried to solve the wave-function of cooper pairs but i am stuck in an integral equation and have no idea how to solve for the wave-function. Before, i tell you how i got the integral in brief. Assuming two particle Schrodinger Equation and i converted the equation to relative (r) and center of mass coordinate (R). $$[-\frac{\hbar^2}{2\mu}\nabla_r^2-\frac{\hbar^2}{2m*}\nabla_R^2+V(\vec{r})]\Psi(\vec{r},\vec{R})=E\Psi(\vec{r},\vec{R})$$ $$\mu \to reduced\:mass$$$$m*\to total\:mass$$and taking only the relative coordinate equations(since potential is not depending on center of mass co-ordinate so both are independent) and transforming to momentum space (little calculation) one can get into this Integral. $$\Psi(\vec{k})(E-2e_k)=\int_{E_f-hw}^{E_f+hw} \frac{V(\vec{k}-\vec{k}')}{(2\pi^3)}\Psi(\vec{k'}) d\vec{k'^3}$$$$E_f$$is the Fermi energy. I assumed that $$V(k-k')\to -V_0$$ assuming this term is constant over the integrable range but i don't know how to integrate this. What is the form of $$\Psi(\vec{k'})$$ Any one have any idea how can i go from here? If you can show me the calculation it will be really helpful. Thanks in advance.For further informtion regarding the above calculation you can see it here https://portal.ifi.unicamp.br/images/files/graduacao/aulas-on-line/fen-emerg/lecture_notes_BCS.pdf

• What is the reason to consider $-V_0g(k)g(k')$ instead of $-V_0$? – Artem Alexandrov Jul 5 at 15:16
• In pdf-notes You can see that they defined modified wave-function as $\Delta(k)=(E-2\epsilon_k)\psi(k)$ and then write down $\psi(k')$ as $\Delta(k')/(E-2\epsilon_{k'})$. It is enough to obtain self-consistent equation (23) from (21) – Artem Alexandrov Jul 5 at 15:20
• Okay can you do it for -V? – Saptarshi Biswas Jul 5 at 15:53
• See my last comment – Artem Alexandrov Jul 5 at 20:58
• I don't understand why are You interested in wave-function? For observables, it seems useless. Sorry if I misunderstand Your goals – Artem Alexandrov Jul 6 at 15:07