I am stuck on a question in Chapter 11 of Advanced Solid State Physics by Philip Phillips, which asks to do the Cooper instability calculation for triplet pairing.

I attempt to solve the Schroedinger equation

$ [-\frac{\hbar^2}{2 m} (\nabla^2_1 + \nabla^2_2)+V(r_1 - r_2)] \psi(r_1,r_2) = E \psi(r_1,r_2)$

with a antisymmetric wavefunction $\psi$ following the steps in the textbook.

First express the wavefunction in center of mass coordinate

$\psi(r_1,r_2) = \phi(r) e^{i Q \cdot R},$

where $r = r_1 - r_2, Q = k_1 + k_2$ and $R = (r_1 + r_2)/2$.

Expand $\phi(r)$ in a Fourier series

$\phi(r) = \displaystyle\sum_{k} \frac{e^{i k \cdot r}}{\sqrt{V}} \alpha_k$.

Spatial antisymmetry then requires that $\alpha_k = - \alpha_{-k}$, but I don't know how this requirement changes the subsequent calculations.

Any explanation and hint are appreciated.


1 Answer 1


After Fourier transform you get an equation

$\displaystyle (E-\frac{k^2}{m})\alpha_{\vec{k}}=\sum_{\vec{k}'}V_{\vec{k}\vec{k}'}\alpha_{\vec{k}'}$

Here $V_{\vec{k}\vec{k}'}\sim \int \mathrm{d} \vec{r} e^{i(\vec{k}-\vec{k}')r}V(\vec{r})$.

Now we need to make some simplifying assumption about $V$. For $s$-wave, usually we take $V_{\vec{k}\vec{k}'}=\text{constant}$ (and some cutoff by Debye frequency). We can expand $V$ into partial waves:


And write $a_{\vec{k}}=\sum_{lm}a_{lm}(k)Y_{lm}(\hat{k})$. Now we can work within a fixed angular momentum channel:

$\displaystyle (E-k^2/m)\alpha_{lm}(k)=\sum_{k'}V_{l}(k,k')\alpha_{lm}(k')$.

Unfortunately this is still not solvable and further assumption has to be made about $V_l(k,k')$.

  • $\begingroup$ Thanks! This is very helpful. Is there a usual assumption we make for triplet pairing? $\endgroup$
    – SlipStream
    Mar 29, 2015 at 22:44
  • $\begingroup$ Basically, in order to solve the equation (regardless of the pairing symmetry), you have to assume $V_l(k, k')=v_l(k)v_l(k')$. $\endgroup$
    – Meng Cheng
    Mar 29, 2015 at 22:53
  • $\begingroup$ Is $\nu$ the coherence factor for cooper pair? Could you give the explicit expression for $\nu_1$? $\endgroup$
    – SlipStream
    Mar 30, 2015 at 3:34
  • $\begingroup$ No. $v$ is just a parameter in this model interaction. $\endgroup$
    – Meng Cheng
    Mar 30, 2015 at 4:21

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