Physics 498 SQD -- Lecture 3 -- Models and theories of superconductivity FINAL revised.pptx
- 1. Lecture 4: Ginzburg-Landau Theory Start discussion of how to understand the superconducting state Present four phenomenological approaches: 1. Thermodynamics picture --- motivated by early experiments 2. Order parameter picture --- the Gorter-Casimir “two-fluid” model 3. Electrodynamics picture --- the London equations 4. Non-local electrodynamics model --- the Pippard extension to the London equations Lecture 3: Models and theories of superconductivity Next Tuesday Today
- 2. THEORIES OF SUPERCONDUCTIVITY macroscopic problems (E&M) spatial variations (intermediate/mixed state) current applied macroscopic quantum phenomena 1st principle calculations → 𝑇𝑐 Thermodynamic details Excitations --- tunneling transport electrodynamics Phenomenological Thermodynamics Gorter-Casimir (two fluid) London Pippard Ginzburg-Landau Gorkov - Eliashberg (TDGL = Time-Dependent G-L) Gorkov Microscopic BCS + Anderson + Bogoliubov + Midgal Unconventional SC models Which is right? Which are useful? For superconductor device physics and quantum information applications? need some of both
- 3. Phenomena --- what were experiments saying? (1) SC was new (and unusual) – perfect conductivity and perfect diamagnetism (2) Sharp and reversible onset with temperature and magnetic field and current THERMODYNAMICS Compare “energy” of 𝑁 and 𝑆 states ⟹ 𝒰 = internal energy OK for isolated systems 𝐹 = 𝒰 − 𝑇𝑆 = Helmholtz free energy 𝐵 = constant 𝐺 = 𝒰 − 𝑇𝑆 − 𝑃𝑉 − 𝐻 ∙ 𝑀 = Gibbs free energy 𝐺𝑁 = 𝐺𝑆 at the phase transition (i.e. at Tc or Hc) New phase of matter with a phase transition which thermodynamic energy is appropriate? Must account for energy from field source since 𝐵 changes at transition (Meissner effect) Compare 𝑁 and 𝑆 for same applied field and require that OK for
- 4. 𝑑𝒰 = 𝑑𝑄 − 𝑊 THERMODYNAMICS 1ST Law of Thermodynamics 𝑑𝑄 = 𝑇𝑑𝑆 𝑊 = 𝑃𝑑𝑉 − 𝐻𝑑𝑀 𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉 + 𝐻𝑑𝑀 2nd Law of Thermodynamics (quasistatic) internal energy heat input work done BY system 𝐺 = 𝒰 − 𝑇𝑆 + 𝑃𝑉 − 𝐻𝑀 𝑑𝐺 = d𝒰 − 𝑇𝑑𝑆 − 𝑆𝑑𝑇 + 𝑃𝑑𝑉 + 𝑉𝑑𝑃 − 𝐻𝑑𝑀 − 𝑀𝑑𝐻 𝑑𝐺 = −𝑆𝑑𝑇 + 𝑉𝑑𝑃 − 𝑀𝑑𝐻 𝑑𝐺 = −𝑀𝑑𝐻 constant 𝑇, 𝑃 𝐻= applied magnetic field 𝑆= entropy
- 5. (1) Normal state: 𝐺𝑁 𝐻 = 𝐺𝑁(0) independent of magnetic field for a non-magnetic material (2) Phase transition: 𝐺𝑁 𝐻𝑐 = 𝐺𝑆(𝐻𝑐) (3) Superconducting state: 𝐺𝑆 𝐻 − 𝐺𝑆 0 = What do we know? 𝐺𝑠 𝐺𝑁 𝐺 ∆𝐺 𝑆 𝑁 𝐻𝑐 𝐻 ∆𝐺 𝐻 = 𝐺𝑁 𝐻 − 𝐺𝑠 𝐻 = 𝐺𝑛 0 − 𝐺𝑠 0 − 1 8𝜋 𝐻2 = 1 8𝜋 𝐻𝑐 2 − 𝐻2 “Condensation energy” (stabilization energy) Δ𝐺(0) = 1 8𝜋 𝐻𝑐 2 Depends on T 𝑀 = − 𝐻 4𝜋 − 0 𝐻 𝑀 𝑑𝐻 = − 1 4𝜋 0 𝐻 𝐻 𝑑𝐻 = 1 8𝜋 𝐻2 perfect diamagnetism SC is a lower energy phase that is weakened by applying magnetic field --- costs energy to expel the field = 1 2 𝜇𝑜𝐻𝑐 2 in MKS units 𝑑𝐺 = −𝑀𝑑𝐻
- 6. Independent of field Entropy 𝐻𝑐 𝑇 𝑇 − 𝜕𝐻𝑐 𝜕𝑇 𝑝, 𝐻 for constant 4 c c N S H H G S S S T T G S T 2 ( ) (0) 1 c c c T H T H T 2 (0) ( ) c c c c H H T T T T T ∆𝑆 𝑇𝑐 𝑇 2 2 2 (0) 1 c c c c H T T S T T T 𝑇𝑐
- 7. Normal state: 𝑆𝑁 𝑇 = 𝛾𝑇 𝛾 = 2 3 𝜋2 𝑁 0 𝑘𝐵 2 Comments: As 𝑇 → 0, 𝑆 → 0 (for both states) as required by the 3rd law 𝑆𝑆 < 𝑆𝑁 ⇒ 𝑆𝐶 is an ordered state (𝐵𝐶𝑆) Slope 𝜕𝑆 𝜕𝑇 discontinuous at 𝑇 = 𝑇𝑐 𝑆 𝑇 𝑆𝑁 𝑆𝑆 𝑇𝑐 (Sommerfeld constant) 0 c H T Superconducting state: 𝑆𝑆 𝑇 = 𝑆𝑆 𝑇 − ∆𝑆
- 8. Latent heat (at transition) • In zero magnetic field, no latent heat because the transition occurs at 𝑇 = 𝑇𝑐, where 𝐻𝑐 = 0 • In finite magnetic field for T<Tc(H), transition by an applied field or an increase in T gives a finite latent heat 1st order in finite field 2nd order phase transition in zero field (1st derivative of 𝐺 is discontinuous) 𝐻𝑐 𝑇𝑐 𝐻 𝑇 1st Order 2nd Order 4 c c T H H L T S T (1st derivative of 𝐺 is continuous)
- 9. Specific Heat + 𝑙𝑖𝑛𝑒𝑎𝑟 𝑇𝑐 𝑇 ∆𝐶 − 𝑐𝑢𝑏𝑖𝑐 2 ( ) (0) 1 c c c T H T H T S C T T 2 2 2 2 2 (0) 1 2 1 4 2 c c N S c c c H H T T C C C T S H T T T T T T 0 at 0 C T 1 0 at T 0.7 2 c c C T T 2 2 2 (0) at = 4 c c c c c c T T H H C T T T T SC specific heat is higher - cubic +linear
- 10. Normal state: 𝐶𝑁 𝑇 = 𝑎𝑇 + 𝑏𝑇3 phonons 𝐶 𝐶𝑠 𝑇𝑐 𝑇 𝐶𝑁 Preview of BCS: electrons Thermodynamic predictions: Linear at low 𝑇 Cubic at higher 𝑇 Jump at 𝑇𝑐 = 1 2𝜋 𝐻𝑐 2 𝑜 𝑇𝑐 Jump precisely determined by 𝐵𝐶𝑆 = 1.43𝛾𝑇𝑐 Exponential at low 𝑇 ⇒ energy gap Superconducting state: 𝐶𝑆 𝑇 = 𝐶𝑁 𝑇 − ∆𝐶
- 11. Two Fluid Model Gorter – Casimir (1934) (analogous to that in superfluid 𝐻𝑒4, but preceded it) Thermodynamics + two fluids ⇒ T-dependences of measurable quantities: 𝐺(𝑇) ⇒ 𝑆(𝑇), 𝐶(𝑇) and 𝐻𝑐(𝑇) Order parameter = fraction of SC electrons 1 𝑥 𝑇𝑐 ? NORMAL CARRIERS: 𝐺𝑁 𝑇 = −1 2 𝛾 𝑇2 from 𝑆 = 𝛾𝑇 = 𝐺𝑆 𝑇 = − 1 8𝜋 𝐻𝑐 2 (𝑇) Then, 𝑑𝐺 𝑑𝑥 = 0 ⇒ 𝐺𝑁 = 𝐺𝑆 ⇒ phase equilibrium (No information about 𝑥) 1st guess: 𝐺(𝑇) = 𝑥𝐺𝑆(𝑇) + (1 − 𝑥)𝐺𝑁 𝑇 2nd guess: 𝐺 𝑇 = 𝑥𝐺𝑆 𝑇 + (1 − 𝑥)1/2 𝐺𝑁 𝑇 Then, 𝑑𝐺 𝑑𝑥 = 0 ⇒ 𝑥 = 1 − 𝐺𝑁 2𝐺𝑆 2 = 1 − (2𝜋𝛾)2 𝐻𝑐 4 𝑇4 SC CARRIERS: Minimize 𝐺(𝑇) to get 𝑥(𝑇) 𝐺(𝑇) = 𝑥𝐺𝑆 + 𝑓(𝑥)𝐺𝑁 −𝜕𝐺 𝜕𝑇
- 12. Choose 𝑥 𝑇 = 1 − 𝑇 𝑇𝑐 4 ~ 𝑛𝑆 Calculate quantities: Significance: • 1st use of concept of an order parameter • Idea of a superfluid + normal excitations --- key concept of BCS and useful for transport and non-equilibrium effects • Focus on thermal properties --- gets right form for electrodynamics , e. g. 𝜆(𝑇) so that Interesting model but not very justifiable density of superconducting electrons 2 2 c c H T 0 c x T 1/2 1/2 ( ) 8 ~ 1 8 c N c N S S c T H T G T H G G G G 2 3 2 ( ) ~ S S d G C T T T dT Then,
- 13. London Equations F. London, H. London (1935) Focus on describing the electrodynamics Variation of 𝐽 with time, space (screening currents) 1. Forces: 𝐹 = m𝑎 = m 𝑑𝑣 𝑑𝑡 = 𝑒𝐸 → 𝑑𝑣 𝑑𝑡 = 𝑒 𝑚 𝐸 𝐽 = 𝑛𝑒𝑣 → 𝑑𝐽 𝑑𝑡 = 𝑛𝑒 𝑑𝑣 𝑑𝑡 = 𝑛𝑒2 𝑚 𝐸 𝑚 𝑛𝑒2 𝑑𝐽 𝑑𝑡 = 𝐸 1st London Equation Λ 𝑑𝐽 𝑑𝑡 =𝐸 Λ = 𝑚 𝑛𝑒2 = London Parameter Fritz London Heinz London Return to the superconductor as a perfect conductor:
- 14. 2. Magnetic Fields: 𝛻 × 𝐵 = 4𝜋 𝑐 𝐽 𝛻 × 𝐵 = 4𝜋 𝑐 𝐽 = 4𝜋 𝑐Λ 𝐸 𝛻 × 𝛻 × 𝐵 = 4𝜋 𝑐 𝛻 × 𝐽 = 4𝜋 𝑐Λ 𝛻 × 𝐸 = 4𝜋 𝑐2Λ 𝐵 Λ 𝛻 × 𝐽 = − 1 𝑐 𝐵 ⇒ 𝛻2 𝐵 = + 4𝜋 𝑐2Λ 𝐵 1-D geometry 𝑥 𝑆𝐶 𝑑2𝐵 𝑑𝑥2 = + 4𝜋 𝑐2Λ 𝐵 𝐵 𝑥 = 𝐵 𝑥 𝑒−𝑥 𝜆 𝜆 = 𝑐2 Λ 4𝜋 1 2 excluded from bulk (expected for perfect conductor) 𝐵 Meissner effect ⇒ 𝐵 excluded Λ 𝛻 × 𝐽 = − 1 𝑐 𝐵 2nd London Equation 1st London Equation 𝐵
- 15. LONDON EQUATIONS 𝑑 𝑑𝑡 Λ𝐽 = 𝐸 𝛻 × Λ𝐽 = − 1 𝑐 𝐵 𝑐𝑔𝑠 𝑑 𝑑𝑡 Λ𝐽 = 𝐸 𝛻 × Λ𝐽 = − 𝐵 𝑀𝐾𝑆 Λ = 𝑚 𝑛𝑒2 Λ = 𝑚 𝑛𝑒2 Can these be justified? Start with the “canonical momentum” 𝑝 = 𝑚𝑣 + 𝑒 𝑐 𝐴 Assume same at finite field. 𝑣 = − 𝑒 𝑚𝑐 𝐴 𝐽 = 𝑛𝑒 𝑣 = − 𝑛𝑒2 𝑚𝑐 𝐴 Λ𝐽 = − 1 𝑐 𝐴 𝑑 𝑑𝑡 Λ𝐽 = − 1 𝑐 𝑑𝐴 𝑑𝑡 = 𝐸 𝛻 × Λ𝐽 = − 1 𝑐 𝛻 × 𝐴 = − 1 𝑐 𝐵 We expect 𝑝 = 0 in the ground state at zero field. London equations
- 16. Corrections: 𝐽 ⟶ 𝐽𝑠 superfluid part only 𝐸 ⟶ 𝐸′ = −𝛻𝜙 = 𝐸 − 1 𝑒 𝛻𝜇 𝑛 ⟶ 𝑛s 𝑇 Normal fluid: 𝐽𝑛 = 𝜎𝑛𝐸 does not obey London Comments: (a) 𝐽 ∝ 𝐴 not gauge invariant Valid only for London gauge: 𝛻 ∙ 𝐴 = 0 = 𝛻 ∙ 𝐽 = 0 𝐽⊥ = 0 ( at surface) (b) “Rigidity” of wavefunction electrons not affected by a magnetic field (c) 𝑝 = 0 suggests “condensation” into 𝑝 = 0 state ⇒ Bose pairing of electrons Going back to the “two-fluid picture: London equation applies only to the superfluid part superfluid density includes chemical potential
- 17. Applications (1) Steady-state: 𝐸′ = 0 (2) Field expulsion: 𝛻 × 𝐵 = 4𝜋 𝑐 𝐽 𝛻2𝐵 = 4𝜋 Λ𝑐2 𝐵 = 1 𝜆𝐿 𝐵 𝜆𝐿 = 𝑚𝑐2 4𝜋𝑛𝑠𝑒2 1 2 = 𝑚 𝜇0𝑛𝑠𝑒2 1 2 𝑐𝑔𝑠 𝑀𝐾𝑆 𝐵 𝑥 = 𝐵0 𝑒 −𝑥 𝜆𝐿 𝐽 𝑥 = 𝑐𝐵0 4𝜋𝜆𝐿 𝑒 −𝑥 𝜆𝐿 𝐽0 Β0 𝐵, 𝐽 𝑥 𝜆𝐿 Superconductor 𝜆𝐿 50 nm 40 nm 85 nm 200 nm NbTi 300 nm 65 nm 140 nm YBCO (c) 700 nm Al Pb Nb PbBi Nb3Sn YBCO (a,b) Temperature dependence: 𝜆𝐿 𝑇 = 𝜆𝐿 0 1 − 𝑇 𝑇𝑐 4 −1 2 ~ 1 𝑛𝑠 𝑇 1 2 𝑛𝑠 𝜆𝐿 𝑇𝑐 Consistent w/ two-fluid model 𝑇
- 18. (3) Field applied to a finite thickness plate 𝐵0 𝐵0 0 −𝑎 𝑎 𝑆𝐶 Width = 2𝑎 𝑑2 𝐵 𝑥, 𝜆𝐿 𝑑𝑥2 = 1 𝜆𝐿 2 𝐵 𝑥 𝐵 𝑥, 𝜆𝐿 = 𝐵1𝑒−𝑥/𝜆𝐿 + 𝐵2𝑒+𝑥/𝜆𝐿 𝐵. 𝐶. 𝐵 ±𝑎 = 𝐵0 𝐵 𝑥, 𝜆𝐿 = 𝐵0 𝑐𝑜𝑠ℎ 𝑥 𝜆𝐿 𝑐𝑜𝑠ℎ 𝑎 𝜆𝐿 B vs. 𝜆𝐿 For different 𝑥 B vs. x For different 𝜆𝐿
- 19. (4) Current flow in a strip 𝐼 𝑡 2𝑎 (a) Current density (uniform): 𝐽 = 𝐼 2𝑎𝑡 But this will be modified to prevent field penetration into the sample 𝑥 −𝑎 +𝑎 𝑥 𝑩 Self-fields from currents will be screened by SC
- 20. (b) Self-consistent London solution (𝑡 ≪ 𝑎) Solve for current flow 𝐽 𝑥 : 𝛻2𝐽 = 1 𝜆𝐿 2 𝐽 𝑥 𝐽 𝑥 = 𝐴 𝑒 𝑥 𝜆𝐿 + 𝑒 −𝑥 𝜆𝐿 𝐼 = −𝑎 𝑎 𝐽 𝑥 𝑡𝑑𝑥 current conservation 𝐽 𝑥 = 𝐼 2𝑥𝑡 𝑐𝑜𝑠ℎ 𝑥 𝜆𝐿 𝑠𝑖𝑛ℎ 𝑎 𝜆𝐿 For uniform 𝜆 ≫ 𝑎, 𝐽 𝑥 → 𝐼 2𝑎𝑡 For 𝜆𝐿 ≪ 𝑎, 𝐽 𝑥 → current piles up on edge 𝜆𝐿 ~ 𝑎 𝑥 𝑩 𝜆𝐿 ≫ 𝑎 𝑥 𝑱 𝜆𝐿 ≪ 𝑎 𝜆𝐿~𝑎 𝜆𝐿 ≫ 𝑎 𝜆𝐿 ≪ 𝑎
- 21. (5) Current flow at N-S interface 2𝑎 𝑁 𝑆 𝑥 𝑦 2-D problem inside 𝑆 𝛻2 𝐽𝑥 = 1 𝜆𝐿 2 𝐽𝑥 𝛻2 𝐽𝑦 = 1 𝜆𝐿 2 𝐽𝑦 Δ ∙ 𝐽 = 0 = 𝑑𝐽𝑥 𝑑𝑥 + 𝑑𝐽𝑦 𝑑𝑦 𝐽𝑦 𝑦 = 0 = 𝐽𝑁 𝐼 = −𝑎 𝑎 𝐽𝑦 𝑥 𝑡 𝑑𝑥 = 2𝑎 𝑡 𝐽 𝐽𝑥 𝑦 → ∞ → 0 𝐽𝑦 𝑥, 𝑦 = −𝐽 𝑎 𝜆𝐿 cosh 𝑥 𝜆𝐿 sinh 𝑎 𝜆𝐿 − 𝑛=1 ∞ 2 −1 𝑛 𝑎 𝜆𝐿 2 cos 𝑛𝜋𝑥 𝑎 𝑎 𝜆𝐿 2 + 𝑛𝜋 2 𝐽𝑥 𝑥, 𝑦 = −𝐽 𝑛=1 ∞ 2 −1 𝑛 𝑛𝜋 𝑎 𝜆𝐿 2 𝑎 𝜆𝐿 2 + 𝑛𝜋 2 sin 𝑛𝜋𝑥 𝑎 exp − 1 𝜆𝐿 2 + 𝑛𝜋 𝑎 2 1 2 𝑦 exp − 1 𝜆𝐿 2 + 𝑛𝜋 𝑎 2 1 2 𝑦 Boundary conditions: Current constraint:
- 22. Beyond London ….. limitations of the London equations: (3) Predicts no dependence on impurities (contradicts experiment) Sn + 3% In doubled ℓ DIRTY CLEAN Determined from microwavwe wave absorbtion measurements of in doped samples: (1) Describe superfluid response – must put in normal response separately ⇒ two-fluid model (2) Applies to weak magnetic fields ⇒ cannot handle vortices, intermediate state, inhomogeneous materials, … (4) Local equations: 𝐽 𝑟 = − 1 𝑐Λ 𝐴 𝑟
- 23. Brian Pippard (Cambridge, 1953) 𝐽 𝑟 depends on a weighted-average of 𝐴 over a range 𝜉 = coherence length Pippard Non-local Modification to 𝐽 𝐴 𝜉 𝑟 LONDON PIPPARD 𝐽𝑠 𝑟 = − 1 𝑐Λ 3 4𝜋𝜉0 𝑅 𝑅 ∙ 𝐴 𝑟′ 𝑅4 ℓ −𝑅 𝜉 𝑑3 𝑟′ 𝐽𝑠 𝑟 = − 1 𝑐Λ 𝐴 𝑟 𝑟 𝑅 𝑟′ 𝜊 𝑅 = 𝑟′ − 𝑟 How did he get this form? CHAMBERS expression for non-local resistivity 𝐽𝑛 𝑟 = 𝜎 3 4𝜋ℓ 𝑅 𝑅 ∙ 𝐸 𝑟′ 𝑅4 ℓ −𝑅 ℓ 𝑑3𝑟′ (replacing 𝐽𝑛 𝑟 = 𝜎𝐸 𝑟 ) Here, range of influence is ℓ due to memory over time between scattering events
- 24. Range of non-locality for SC: 𝜉 = Pippard coherence length 𝜉 = 𝜉0 𝜉 ℓ DIRTY CLEAN 𝜉0 ~ℓ Values of 𝜉0 𝐴𝑙 1600 𝑛𝑚 𝑆𝑛 230 𝑛𝑚 𝑃𝑏 83 𝑛𝑚 𝑁𝑏 38 𝑛𝑚 𝑃𝑏𝐵𝑖 20 𝑛𝑚 𝑁𝑏3𝑆𝑛 4 𝑛𝑚 𝑌𝐵𝐶𝑂 𝑎, 𝑏 1.5 𝑛𝑚 𝑌𝐵𝐶𝑂 𝑐 0.4 𝑛𝑚 DIRTY SC ℓ ≪ 𝜉0 1 𝜉 = 1 𝜉0 + 1 ℓ ~ ℓ ℓ 𝜉0 𝜉0 + ℓ 𝜉 ℓ = CLEAN SC (ℓ > 𝜉0)
- 25. 𝜆 vs. 𝜉 ⇒ LOCAL (London) vs. NON-LOCAL (Pippard) 𝐴 𝑥 𝜆 𝜉 𝐴 𝑥 𝜆 𝜉 𝐽 = − 1 𝑐Λ 𝐴 𝐽~ − 1 𝑐Λ 𝜆 𝜉 𝐴 Averages current from higher 𝐴 ⇒ effective increase in 𝐽 by 𝜉 𝜆 𝐽 = 𝑐 4𝜋 𝛻 × 𝐵 = − 1 𝑐Λ 𝜆 𝜉 𝐴 𝛻2 𝐵 = − 4𝜋 𝑐2Λ 𝜆 𝜉 𝐵 = − 1 𝜆L 2 𝜆 𝜉 𝐵 = − 1 𝜆2 𝐵 1 𝜆2 = 𝜆 𝜆L 2𝜉 Non-locality: 𝜆L > 𝜉 𝜆L < 𝜉 𝐽 = 𝑐 4𝜋 𝛻 × 𝐵 = − 1 𝑐Λ 𝐴 𝛻2 𝐵 = − 4𝜋 𝑐2Λ 𝐵 = − 1 𝜆L 2 𝐵 𝜆 = 𝜆L 2𝜉 1 3 = 𝜆L 𝜉 𝜆L 1 3 𝜆 = 𝜆L 𝜆 = 3 2𝜋 1 𝜉 𝜆L 𝜉 𝜆L 1 𝜉 > 𝜆L ~0.65
- 26. In CLEAN LIMIT, can be either London or Pippard limit: 𝜆𝐿 > 𝜉0 ⇒ 𝜆𝑃 = 𝜆𝐿 𝜆𝐿 < 𝜉0 ⇒ 𝜆𝑃 = 𝜆𝐿 𝜉0 𝜆𝐿 1 3 > 𝜆𝐿 LONDON PIPPARD In moderately DIRTY limit 𝜆𝑃 = 𝜆𝐿 ℓ 𝜆𝐿 1 3 ≥ 𝜆𝐿 𝜆𝑃 < 𝜉~ℓ < 𝜉 In very DIRTY limit 𝜉~ℓ < 𝜆𝐿, 𝜉0 Always in London (local) limit Pippard expression: ℓ −𝑅 𝜉 ⇒ ℓ− 𝑅 ℓ 𝐽~ − 1 𝑐Λ ℓ 𝜉0 𝐴 ⇒ 𝜆𝑃 = 𝜆𝐿 𝜉0 ℓ 1 2 > 𝜆𝐿 𝜆 ℓ DIRTY CLEAN 𝜉0 𝜉0 > 𝜆𝐿 𝜉0 < 𝜆𝐿∟ 𝜆𝑃 = 𝜆𝐿 𝜉0 𝜆𝐿 1 3 𝜆𝐿 Non-locality over the coherence length modifies the penetration depth depending on the impurity concentration: