弱耦合超导体中涡旋芯的解析电荷密度分布
Analytical Charge Density Profile of Vortex Core in Weak-Coupling Superconductor
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中文总结 AI 辅助
该研究阐释弱耦合超导体涡旋芯的电荷密度分布，通过分析Caroli-de Gennes-Matricon束缚态推导电荷分布的闭式表达，揭示其由束缚态与扩展态补偿产生的符号交替2k_F调制，明确其源于电中性与大动量转移下屏蔽低效的结论。
中文摘要 AI 辅助
自洽的Bogoliubov-de Gennes计算长期以来表明，求解超导涡旋内部的泊松方程会将单符号的电荷耗尽转变为周期为π/k_F的符号交替调制。本文对该结果给出基础阐释：针对阶梯型能隙中的Caroli-de Gennes-Matricon束缚态，证明束缚态旋量的归一化几乎与角动量无关，这使得模式求和可简化为闭式表达。在涡旋内部，涡旋的绕旋作用从完备性求和中移除一个贝塞尔通道，因此密度在涡旋线上消失，并呈现波矢为2k_F的Friedel型振荡；在涡旋外部，该求和给出随相干长度衰减的1/r包络，并带有残余波纹。束缚态电荷积分不为零，电中性要求扩展态精确补偿该电荷，这种补偿在长波长下是完整的，但在费米圆直径处失效，剩余的是仅被因子4k_F²/(4k_F²+k_TF²)缩小的符号交替2k_F调制，该因子对任意金属均介于1/2到3/4之间。因此该振荡并非微弱效应，而是电中性与大动量转移下屏蔽效率低下的结果：在屏蔽总电荷中，平滑项相互抵消，仅剩余波纹。
英文摘要
Self-consistent Bogoliubov-de Gennes calculations have long shown that solving the Poisson equation inside a superconducting vortex turns a one-signed charge depletion into a modulation that alternates in sign with period $π/k_F$. We give an elementary account of that result. Taking the Caroli-de Gennes-Matricon bound states in a step-like gap, we show that the normalization of the bound-state spinor is nearly independent of angular momentum, which collapses the mode sum into closed form. Inside the core the vortex winding removes one Bessel channel from a completeness sum, so the density vanishes on the vortex line and carries Friedel-like oscillations of wavevector $2k_F$; outside it the sum gives a $1/r$ envelope decaying over a coherence length, with a residual ripple. The bound-state charge does not integrate to zero, so neutrality obliges the extended states to compensate it exactly. That compensation is complete at long wavelength but fails at the diameter of the Fermi circle, and what survives is a sign-alternating $2k_F$ modulation reduced only by $4k_F^2/(4k_F^2+k_{TF}^2)$, a factor lying between one-half and three-quarters for any metal. The oscillation is therefore not a delicate effect but a consequence of neutrality and the inefficiency of screening at large momentum transfer: in the screened total the smooth terms cancel and only the ripple is left.
