The Cooper Problem in tight binding

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SUMMARY

The discussion centers on solving the Cooper problem in a one-dimensional lattice with a lattice constant of 'a'. The reciprocal lattice vector is defined as k_n=n(2π/a). The nearest neighbor approximation requires summing over k values of -2π/a, 0, and 2π/a. The equation to solve for energy E is established, confirming that the values of |w_i| should equal 1 and that the values of ξ_i can be derived from the problem statement or calculated from corresponding energies.

PREREQUISITES
  • Understanding of one-dimensional lattice structures
  • Familiarity with reciprocal lattice vectors
  • Knowledge of the Cooper problem in condensed matter physics
  • Ability to manipulate equations involving energy and wave functions
NEXT STEPS
  • Study the derivation of the Cooper problem in one-dimensional systems
  • Learn about the implications of the nearest neighbor approximation in solid state physics
  • Explore the calculation of ξ_i values from energy levels in quantum mechanics
  • Investigate the role of wave functions in lattice systems
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This discussion is beneficial for physicists, particularly those specializing in condensed matter physics, as well as students studying quantum mechanics and solid state theory.

mcas
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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}?
 
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Yes, this is the correct approach. The values of |w_{i}| should equal 1, and the values of \xi_{i} should be given in the problem statement or can be calculated from the corresponding energies.
 

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