Lecture 23 - Superconductivity II - Theory

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1 D() Lecture 23: Superconductivity II Theory (Kittel Ch. 10) F mpty D() F mpty Physics 460 F 2000 Lect 23 1 Outline Superconductivity - Concepts and Theory Key points xclusion of magnetic fields can be used to derive energy of the superconducting state eat Capacity shows there is a gap Isotope effect ow does a exclude B field? London penetration depth (1930 s) Flux Quantization ow we know currents are persistent! Cooper instability - electron pairs Bardeen, Cooper, Schrieffer theory (1957) (Nobel Prize for work done in UIUC Physics) (Kittel Ch 10 ) Physics 460 F 2000 Lect 23 2 Meisner ffect Magnetic field B is excluded B = + µ 0 M For type I s, µ 0 M = - for T < T c Perfect Diamagnetism! B - µ 0 M Meisner ffect (1934) A can actively push out a magnetic field - Meisner effect (For < c in type I s and < c1 in type II s) xcludes Magnetic Field Zero Field Cooled Field Cooled 0 0 c c T > T c T < T c T > T c T < T c Physics 460 F 2000 Lect 23 3 Physics 460 F 2000 Lect 23 4 From previous lecture From previous lecture ffect of a Magnetic Field Magnetic fields tend to destroy superconductivity c Note: = external applied field B = internal field B = + µ 0 M M = Magnetization T c Phase Transition SUPRCONDUCTING STAT IS A NW PAS OF MATTR T nergy : normal vs. superconducting The free energy F of the plus magnetic field is increased because magnetic field B is excluded The normal state energy is nearly independent of field Transition at c Physics 460 F 2000 Lect 23 5 Physics c 460 F 2000 Lect 23 6 From previous lecture F F N F S (0) F S () = F S (0) + 2 /2µ 0 1

2 nergy : normal vs. superconducting Therefore F S ( c ) = F S (0) + c2 /2µ 0 = F N (0) or F = F N (0) - F S (0) = c2 /2µ 0 Typical Values: F ~ 10-7 ev/electron! SMALL! F F N F S (0) Superconduc. F S () = F S (0) + 2 /2m 0 nergy : normal vs. superconducting ow do we understand the small values F ~ 10-7 ev/electron? Similar to the description of thermal energy F ~ D( F ) 2 ~ k 2 where is the region affected - as shown by the gap in the heat capacity - agrees with experiment D() mpty k Physics c 460 F 2000 Lect 23 7 F Physics 460 F 2000 Lect 23 8 Coherence Length The typical length associated with the mechanism of superconductivity is the feature associated with the Fermi surface is ξ = 1/ k = hv F /2 where is the region affected ow is a field excluded? What makes B = 0 inside superonductor? Supercurrents flowing on the boundary! asiest geometry - long thin rod B = 0 inside (Understood from the BCS theory see later) Typical values Al T c = 1.19K ξ = 1,600 nm Pb T c = 7.18K ξ = 83 nm k Current around boundary causes field inside that cancels the external field - A supercurrent that flows with no decay Physics 460 F 2000 Lect 23 9 Physics 460 F 2000 Lect Supercurrents J flowing on the boundary! Supercurrent state London Penetration Depth λ L Maxwell s q.: B = µ 0 j B = - 2 B = µ 0 j Also B = A (A not unique) state Physics 460 F 2000 Lect London PROPOSD that in the gauge A = 0, A normal = 0, j = - A/(µ 0 λ L2 ) so j = - B /(µ 0 λ L2 ) λ L Physics 460 F 2000 Lect

3 London quations ere we give a derivation of the London equations that gives physical insight and the expression for the penetration depth λ L The free energy for the system with a supercurrent and the penetrating B field is n s = superfluid density F = F 0 + kin + mag v(r) = velocity where mag = dr B 2 /8π and kin = dr ½mv 2 n s with j(r) = n s q v(r) Using B = µ 0 j we find F = F 0 + (1/8π) dr [B(r) 2 + λ L2 ( B(r)) 2 ], where λ L2 = ε 0 mc 2 /n s q 2 Varying the form of B(r) by adding δb(r) the change δf is δf = (1/4π) dr [B(r) δb(r) + λ L2 ( B(r)) ( δb(r)) ] Integration = (1/4π) dr [B(r) - λ L2 B(r)] δb(r) by parts At the minimum, δf = 0 for all possible δb(r) which requires that B(r) - λ L2 B(r) = B(r) + λ L2 2 B(r) = 0 Which leads to the London quation Physics 460 F 2000 Lect Therefore 2 B = B/ λ L 2 Solution: B decays into with the form B(x) = B(0) exp(-x/λ L ) xplains Meisner effect B vanishes inside the λ L state Physics 460 F 2000 Lect The superconducting state is a quantum state Landau and Ginsburg (before the BCS theory) proposed all the electrons act together to form a new state Ψ, with Ψ 2 = n s where n s is the superfluid density Ground state: Ψ G = n s - No current flowing Consider now Ψ = ( n s ) exp( iθ(r)) - the phase in a wavefunction corresponds to a current The velocity operator is v = p/m = (1/m)( - i h -(q/c)a) Thus j = q Ψ v Ψ=(n s q/m) (h θ -(q/c)a) and curl j = - (n s q 2 /mc) B Physics 460 F 2000 Lect The superconducting state is a quantum state - II This quantum state leads to a theory of the London penetration depth The equation curl j = - (n s q 2 /mc) B and the London proposal leads to curl j = - B /(µ 0 λ L2 ) λ L2 = ε 0 mc 2 / n s q 2 Agrees with experiment! See earlier slide for alternative derivation BUT what is m? What is q? ow do we really know it is quantum in nature? Physics 460 F 2000 Lect Quantized Flux The flux enclosed in a ring is quantized! Consider a line inside the The current j = 0 inside h θ - (q/c)a = 0 inside the Magnetic field threading ring Current only near surface Quantized Flux -II The line integral of θ is the change in θ around the loop = 2π x integer The line integral of A is the surface integral of B (See Kittel p 281) = total flux Φ enclosed in the ring Result: Φ = (2π hc/q) x integer -- quantized! Line integral on a closed contour inside the Physics 460 F 2000 Lect Result: Charge q = 2e - pairs! Physics 460 F 2000 Lect

4 Persistent Currents ow can the current stop flowing? Only if some of the flux Φ leaks out of the ring But the flux can only decrease by quanta! Two length scales in superconductivity London Penetration depth λ L2 = ε 0 mc 2 /nq 2 (particles of mass m, charge q) (Understood from the BCS theory that m and q are for an electron pair see later) Typical values Al T c = 1.19K ξ = 1,600 nm λ L = 160 nm ξ/λ L = 0.01 Pb T c = 7.18K ξ = 83 nm λ L = 370 nm ξ/λ L = 0.45 There is an energy barrier for the flux to go through the to escape - time for current to decrease can be ~ age of universe! Physics 460 F 2000 Lect The ratio determines type I (ξ/λ L <<1) and type II (ξ/λ L > ~1) s see later Other examples are given in Kittel Physics 460 F 2000 Lect Type II Type II s are ones where it is favorable to break up the field into quanta - the smallest posible unit of flux in each vortex shown - for c1 < < c2 Lattice of quantized flux units BCS theory ints: Must involve phonons, small energy scale First: Cooper instability If for some reason there were an attractive interaction between two electrons above the Fermi energy in a metal, they would form a bound pair below the Fermi energy no matter how weak the interaction! Magnetic flux penetrates through the by creating small regions normal metal applied Physics 460 F 2000 Lect Two electrons of opposite k and opposite spin form a bound state Fermi surface is unstable! Physics 460 F 2000 Lect BCS theory - II What could cause the attraction? - phonons! The Coulomb interaction is repulsive But phonons can cause the Mattress effect -one electron causes the lattice to distort - the second electron is attracted the the distortion even after the first electron has left! BCS theory - III The Cooper idea shows there is a problem for two electrons - but what do all the electrons do? This is the key advance of BCS - to construct a new quantum wavefunction for all the electrons Fundamental change only for electrons within a energy range near the Fermi surface Opens an energy gap - explains the specific heat Two electrons of opposite k and opposite spin form a bound state! Physics 460 F 2000 Lect Forms single quantum state Ψ separated by a gap from other states Physics 460 F 2000 Lect

Result D() F BCS theory - IV D() mpty Gap F mpty Physics 460 F 2000 Lect 23 25 transition T c BCS prediction: T c = 1.

5 Result D() F BCS theory - IV D() mpty Gap F mpty Physics 460 F 2000 Lect transition T c BCS prediction: T c = 1.14 Θ D exp(-1/ud( F )) where is the Debye temperature (measure of phonon energy), D( F ) is then density of states at Fermi energy, and U = typical electron-phonon coupling energy Fits experiments for ratio of energy gap to T c ard to actually predict T c! xperiment: Al 1.2 K g 4.6 K Pb 7.2 K Au < K - not found to be superconducting! Na 3 C K (1990) YBa 2 Cu 3 O 7 93 K (1987) Record today 140 K Physics 460 F 2000 Lect elements lements that have large electron-phonon coupling NOT the best metals, NOT the magnetic elements Super conducting Physics 460 F 2000 Lect What is the Order Parameter? If superconductivity is a new state of matter and there is a phase transition between the normal and superconducting states: What is the order parameter? (Analogous to magnetization vector M in a magnet) The wavefunction Ψ= ( n s ) exp( iθ(r)) c Two components: magnitude n s, phase θ The ground state is for θ = constant Variations in θ(r) describe T c higher energy current carrying states (analogous magnons in a magnet) Physics 460 F 2000 Lect T Summary Superconductivity - Concepts and Theory xclusion of magnetic fields can be used to derive energy of the superconducting state Shows very small energy F ~ D( F ) 2 ~ k 2 where the gap is consistent with heat capacity ow does a exclude B field? London penetration depth (1930 s) forms a quantum state Flux Quantization ow we know currents are persistent! Cooper instability - electron pairs Bardeen, Cooper, Schrieffer theory (1957) (Nobel Prize for work done in UIUC Physics) Physics 460 F 2000 Lect Magnetism Next time Physics 460 F 2000 Lect

Lecture 22 Metals - Superconductivity

Lecture 22 Metals - Superconductivity Lecture 22: Metals (Review and Kittel Ch. 9) and Superconductivity I (Kittel Ch. 1) Resistence Ω Leiden, Netherlands - 1911.1 4.6 K g sample < 1-5 Ω Outline metals Recall properties (From lectures 12,