414 - pr 18-9 - the London equation and the Meissner effect: Derive the differential equation [18.54] for the vector-potential A starting from the London equation [18.52] for a static magnetic field. 2 µ 0e 2 ρ s −e ρ s (r ) 1 2 (1.1) [18.52] → J e (r) = A; [18.54] → ∇ A = A = 2 A; m m λ L 2
Hints: you may need ∇ × (∇ × A) = ∇(∇ • A) − ∇ A , and the following simple geometry in xyz-space. Consider the half-space x < 0 to be vacuum, while x > 0 consists of superconducting-metal. You have a magnetic field B(x) = B( x) zˆ , with B( x) = B0 for x < 0. Calculate the form of the magnetic field in the superconductor x > 0.
(1.2) Big ansatz: let electric current density be decomposed into paramagnetic and diamagnetic contributions, as first done on SSP 06 – 23 – Eq. [1.98], ℏ
ℏ
( Ψσ (∇ Ψσ ) − (∇ Ψσ ) Ψσ ) = mV ∑ ( k + 2mi †
†
r
r
1 2
q) e
i ⋅q•r
∇
†
ak ,σ ak +q ,σ ≡ Jσ = [paramagneti c];
k ,q
Jσ A =
−q
m
AΨσ† Ψσ =
−q
mV
(1.3)
2
A ⋅ ∑ ei⋅q•r ak† ,σ ak +q ,σ =
−e ρ (r )
k ,q
m
A = [diamagnetic] ; J e (r, A) = Jσ ∇ (r, A) + Jσ A ;
The vector potential A comes from the external field. In the superconducting state, let the wavefunction be such that the paramagnetic response exactly cancels any applied diamagnetic field, ∇ ∇ ∇ ∃A, A′ : H < H C → J e (r , A ) = J e (r , 0) = J e (r, A′) = 0; → J e (r, A ) =
J σ∇ (r, A) + 0 = Jσ ∇ (r , A ) = [diamagetic only] = (− e2 ρ (r ) / m) A;
(1.4)
For a negligible displacement current (i.e., magnetostatics), and in the gauge such that ∇ • A = 0 , we compute this diamagnetic current (1.4) as, 2 1 1 1 −e ρ (r) ∇×B 2 2 A = Jσ ∇ (r, A) + 0 = J e = A A A) (1.5) +0= ∇ × (∇ × A) = ∇ ∇ • − ∇ = ∇ − ∇ ( ) ( ( 0 ) ( ) m µ0 µ0 µ0 µ 0
The vector potential of a uniform field is A = 12 B ⋅ ( yzˆ − zyˆ ) . We can also read off, µ0 e 2 ρ (r ) 1 2 A ≡ 2 A; ∇ A=
m
λ L
λ L
≡
m
µ0 e 2 ρ (r )
; ρ (r ) ≡ ρ S ;
(1.6)
2
Discuss what what happens happens in the boundary-laye boundary-layerr of thickness thickness λ L = m / ( µ0 ρ s e ) ≡ [the London-leng London-length] th] . Consider a
vector potential A x< 0 = 12 B ⋅ ( yzˆ − zyˆ ) in the space x < 0. Then the solution to (1.6) in the space x > 0 is given by the ansatz A x >0 = 12 B( x)( yzˆ − zyˆ ) ,
2
2
A x >0 = 12 B ( x)( yzˆ − zyˆ ) = λ L ∇ A x> 0 = =
λ L2 2
λ L2 2
⋅
( 0 + ∇ [ − B( x) z] yˆ + ∇ [ + B( x) y] zˆ ) 2
2
ˆ ) → A x >0 = Ax >0 yˆ + Ax >0 zˆ = 21 λL B′′ ( yzˆ − zyˆ ) = λ L ⋅ ([ B′′y + 0 + 0] zˆ − [ B′′z + 0 + 0] y y
z
2
2
B′′ B
(1.7) ⋅ A x <0 ;
For finite (and therefore physically-realistic) fields, the potential must be continuous across the boundary x = 0, so A x >0 (0+ , y , z ) ≡ A x< 0 (0 − , y , z ) , which makes (1.7) into the ODE λ L 2 B ′′( x) = B ( x) = B+ e + x / λ L + B− e− x / λ L , which, for a field that is finite at x → +∞ , requires B+ = 0 , meaning B ( x) = B− e − x / λ L , and B− = B . Hence, we have a decaying field across the boundary, which we plot as, BHxLêB1.0
0.8
0.6
0.4
0.2
-2
2
xêlL
4
(1.8)
How large is λ L for a typical superconducting-metal at T = 0 K, where all the electrons have condensed into 28
−3
the superconducting-condensate? For copper, you have a charge-carrier-density of ρ (r ) = 8.47 ×10 m , and
the electrons can be assumed nearly-free ( m ≡ m* = m0 = 9.11 ×10 −31 kg ) so the penetration depth for copper is, Cu L
λ
=
m 2
µ0e ⋅ ρ (r )
=
9.11×10 (4π ×10
−7 N ⋅ s2 2
C
)(1.602 ×10
−31
−19
kg 2
28
−3
C) (8.47 ×10 m )
=
−8
1.83 ×10 m = 18.3 nm
(1.9)
