mcas
- 22
- 5
- Homework Statement
- Find the coupling energy of a Cooper pair in a 1D lattice in tight binding (nearest neighbors), i.e. solve the Cooper Problem.
- Relevant Equations
- An equation to solve:
[itex]1=\lambda \sum_{\vec{k}} |w_{\vec{k}}|^2 \frac{1}{E-2\xi_{\vec{k}}}[/itex] for [itex]{\xi_\vec{k} \geq 0}[/itex]
where:
[itex]\lambda = const[/itex] is a part of a seperable potential [itex]V_{\vec{k}\vec{k'}}[/itex]
[itex]\xi_{\vec{k}} = \varepsilon_{\vec{k}}-\mu[/itex]
[itex]w_{\vec{k}}[/itex] is [itex]1[/itex] for [itex]\hbar w_D \geq \xi_{\vec{k}} \geq 0[/itex] and [itex]0[/itex] in any other case
We have a one dimensional lattice with a lattice constant equal to a (I'm omitting vector notation because we are in 1D). The reciprocal lattice vector is k_n=n\frac{2 \pi}{a}.
So to get the nearest neighbour approximation I need to sum over k = -\frac{2 \pi}{a}, 0, \frac{2 \pi}{a}.
If I understand everything correctly this would be the equation I need to solve for E:
1=\lambda ( |w_{\frac{2 \pi}{a}}|^2 \frac{1}{E-2\xi_{\frac{2 \pi}{a}}} + |w_{0}|^2 \frac{1}{E-2\xi_{0}} + |w_{-\frac{2 \pi}{a}}|^2 \frac{1}{E-2\xi_{-\frac{2 \pi}{a}}})
I'm not sure if this is the right approach. If so, would the values of |w_{i}| equal 1 in this equation? Do we know the values of \xi_{i}?
So to get the nearest neighbour approximation I need to sum over k = -\frac{2 \pi}{a}, 0, \frac{2 \pi}{a}.
If I understand everything correctly this would be the equation I need to solve for E:
1=\lambda ( |w_{\frac{2 \pi}{a}}|^2 \frac{1}{E-2\xi_{\frac{2 \pi}{a}}} + |w_{0}|^2 \frac{1}{E-2\xi_{0}} + |w_{-\frac{2 \pi}{a}}|^2 \frac{1}{E-2\xi_{-\frac{2 \pi}{a}}})
I'm not sure if this is the right approach. If so, would the values of |w_{i}| equal 1 in this equation? Do we know the values of \xi_{i}?