Superconductor Ginzburg-Landau model

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SUMMARY

The Ginzburg-Landau model for superconductors is characterized by parameters where α < 0 and β > 0, indicating stability in the free energy formulation. The free energy expression is given by E = ∫ d³x (B²/2μ + (1/2m)|(-iħ∇ - eA)ψ|² + α|ψ|² + β|ψ|⁴). The parameter α is directly related to the critical temperature, while β ensures that the free energy is minimized at finite values of |ψ|. A simplified approach involves minimizing the free energy under the assumption of zero vector potential and negligible gradient of ψ.

PREREQUISITES
  • Understanding of the Ginzburg-Landau theory for superconductivity
  • Familiarity with variational principles in physics
  • Knowledge of thermodynamic potentials and phase transitions
  • Basic calculus and differential equations
NEXT STEPS
  • Study the implications of the Ginzburg-Landau parameters α and β on superconducting properties
  • Learn about critical temperature and its role in phase transitions in superconductors
  • Explore the mathematical techniques for minimizing functionals in variational calculus
  • Investigate the effects of vector potentials on the Ginzburg-Landau model
USEFUL FOR

Physicists, materials scientists, and students studying superconductivity and phase transitions will benefit from this discussion.

vidi
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Homework Statement



Why is it that for a superconductor \alpha&lt;0,\beta&gt;0 in the Ginzburg-Landau model wit free energy formulation
E=\int d^3x\,\, \frac{\vec B^2}{2\mu}+\frac{1}{2m}|(-i\hbar\nabla-e\vec A)\psi|^2+\alpha|\psi|^2+\beta|\psi|^4

Homework Equations



E=\int d^3x\,\, \frac{\vec B^2}{2\mu}+\frac{1}{2m}|(-i\hbar\nabla-e\vec A)\psi|^2+\alpha|\psi|^2+\beta|\psi|^4

The Attempt at a Solution



I have no idea how to start! perhaps it has to do with some critical temperature?
 
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\beta &gt;0, otherwise the free energy could be minimized by |\psi |\rightarrow \infty. This may not be obviously unphysical, but the Ginzburg-Landau model does after all aim to model the phase transition around |\psi |=0.

\alpha &lt;0, then. The parameter \alpha is indeed related to the critical temperature, but you don't need to look at the temperature dependence here. Just consider the simpler case with \vec{A}=0 and \nabla \psi \approx 0 and minimize the free energy. This will give you the sign. Of course, you'll probably have to extend the arguments a bit, to account for the inhomogeneous case and non-zero vector potentials, but that's the basic idea anyway.
 
Thanks, Hypersphere.
 

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