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lectures accompanying the book: Solid State Physics: An Introduction, by Philip Hofmann (2nd edition 2015, ISBN-10: 3527412824, ISBN-13: 978-3527412822, Wiley-VCH Berlin. www.philiphofmann.net 1

Superconductivity

The discovery of superconductivity Resistance of Mercury at low temperatures (Heike Kamerlingh Onnes in 1911, Nobel price in 1913). At 4.2 K the resistance drops to an unmeasurably small value. This seems to be a phase transition with a transition range of around 10-3 K.

What about the low-temperature conductivity of metals? Metals should have a finite resistance at low temperatures. resistivitty (arb. units) 0 0 0 T C superconductor normal metal T (arb. units)

Periodic table of superconducting elements H Li Na K Rb Be Mg Ca Sr Cs Ba superconducting under normal conditions superconducting under high pressure He B C N O F Ne Sc Y La Hf Fr Ra Ac Rf Ti V Zr NbMo Ta Cr Mn Fe Co Ni Cu W Tc Re Ru Rh Os Ir Pd Pt Ag Au Db Sg Bh Hs Mt Ds Rg Uub Al Zn Ga Cd In Si Sn Hg Tl Pb P S Ge As Se Sb Bi Te Po Cl Br l At Ar Kr Xe Rn Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

Is the resistivity of a superconductor really zero or just very small? V I I R1 R2 Rsample V I R1 R2 Rsample I R3 Rsample R4 Rsample A two-point probe measurement will not work because the the contact resistance and the resistance of the wires dominates everything. A four point probe measurement can be tried but it does not work either: one is limited by the smallest voltage one can measure.

Is the resistivity of a superconductor really zero or just very small? F B I(t) = I 0 e t/ I A super-current can be induced in a ring. The decay of the current is given by the relaxation time (10-14 s for a normal metal). For a superconductor it should be infinite. Experiments suggest that it is not less than 100.000 years.

Destruction by magnetic fields external field B0 B C (0) B C (T) superconducting normal 0 0 temperature T T C B 0C (T ) = B 0C (0) [ 1 ( T T C ) 2 ]

Destruction by magnetic fields / current density j j C (B 0,T) T T C B C (0) B 0

Example: medical MRI scanner

The Meissner effect Superconductors (type I) are perfect diamagnets (with χm=-1) which always expel the external field. This so-called Meissner effect goes beyond mere perfect conductivity.

The Meissner effect

perfect conductor B=0 B=B a The Meissner effect T>T C

perfect conductor B=0 B=B a The Meissner effect T>T C I Edl = A d B dt

perfect conductor B=0 B=B a The Meissner effect T>T C A cool below T C I B B a B=B a Edl = d B dt

perfect conductor B=0 B=B a The Meissner effect T>T C A cool below T C I B B a B=B a Edl = d B dt B 0 B 0

perfect conductor B=0 B=B a The Meissner effect T>T C A cool below T C I B B a B=B a Edl = d B dt B 0 B 0 For a perfect conductor, the magnetic field inside the conductor cannot change. An enclosed field is possible.

The Meissner effect perfect conductor B=0 T>T C B=B a movement without friction m @v @t = qe @j @t = n sq 2 m E cool below T C B B a B=B a B 0 B 0

The Meissner effect perfect conductor B=0 T>T C B=B a cool below T C movement without friction m @v @t = qe @j @t = n sq 2 m E with Maxwell equation B B a B=B a curle = @B B 0 B 0 @ @t m n s q 2 curlj + B @t =0

The Meissner effect perfect conductor B=0 T>T C B=B a cool below T C movement without friction m @v @t = qe @j @t = n sq 2 m E with Maxwell equation B B a B=B a curle = @B B 0 B 0 @ @t m n s q 2 curlj + B @t =0 A @ @t Z m n s q 2 curljda + Z BdA = @ @t I m n s q 2 jdl + Z BdA =0.

The Meissner effect perfect conductor B=0 T>T C B=B a cool below T C movement without friction m @v @t = qe @j @t = n sq 2 m E with Maxwell equation B B a B=B a curle = @B B 0 B 0 @ @t m n s q 2 curlj + B @t =0 A mag. flux by j mag. flux by ext. field @ @t Z m n s q 2 curljda + Z BdA = @ @t I m n s q 2 jdl + Z BdA =0.

The Meissner effect perfect conductor superconductor B=0 B=B a T>T C B=0 T>T C B=B a cool below T C cool below T C B B a B=B a B B a B=B a B 0 B 0 B 0 B 0 The essence of the Meissner effect is that the superconductor ALWAYS EXPELS THE MAGNETIC FIELD (ideal diamagnet).

Why levitation? The levitation stems from the same reason as ordinary diamagnetic levitation: a combination of gravitational force and a magnetic force due to the inhomogeneous field. The only difference is the different strength because of the much higher (negative) susceptibility.

Isotope effect 0.585 Sn log (T C / 1K) 0.575 0.565 ~M -1/2 remember: energy for an harmonic oscillator ω = γ M. 2.05 2.10 log (M / 1 amu) Changing the isotope of a material should have a very minor influence on the electronic states. The isotope effect suggests that the lattice vibrations are involved in the superconductivity.

London equations (1935) zero resistance @j @t = n sq 2 m s E 1. London equation

London equations (1935) zero resistance @ @t @j @t = n sq 2 m s E ms n s q 2 curlj + B 1. London equation =0 Z @ @t ms n s q 2 curljda + Z mag. flux by j BdA = @ @t I ms n s q 2 jdl + mag. flux by ext. field Z BdA =0

London equations (1935) zero resistance @ @t @j @t = n sq 2 m s E ms n s q 2 curlj + B 1. London equation =0 Z @ @t ms n s q 2 curljda + Z mag. flux by j BdA = @ @t I ms n s q 2 jdl + mag. flux by ext. field Z BdA =0 Meissner effect m s curlj + B =0 n s q2 2. London equation

London equations (1935) zero resistance @ @t @j @t = n sq 2 m s E ms n s q 2 curlj + B 1. London equation =0 Z @ @t ms n s q 2 curljda + Z mag. flux by j BdA = @ @t I ms n s q 2 jdl + mag. flux by ext. field Z BdA =0 Meissner effect I m s curlj + B =0 2. London equation n s q2 Z ms n s q 2 jdl + BdA =0

Field penetration m s now we have curlj + B =0 n s q2 and the Maxwell equation (inside the superconductor) curlb = µ 0 j

Field penetration m s now we have curlj + B =0 n s q2 and the Maxwell equation (inside the superconductor) curlb = µ 0 j combining this gives curlcurlb = µ 0 curlj = µ 0n s q 2 m B

Field penetration m s now we have curlj + B =0 n s q2 and the Maxwell equation (inside the superconductor) curlb = µ 0 j combining this gives curlcurlb = µ 0 curlj = µ 0n s q 2 m B

Field penetration m s now we have curlj + B =0 n s q2 and the Maxwell equation (inside the superconductor) curlb = µ 0 j combining this gives curlcurlb = µ 0 curlj = µ 0n s q 2 m B curlcurlb = graddivb B = µ 0n s q 2 m B

Field penetration m s now we have curlj + B =0 n s q2 and the Maxwell equation (inside the superconductor) curlb = µ 0 j combining this gives curlcurlb = µ 0 curlj = µ 0n s q 2 m B curlcurlb = graddivb B = µ 0n s q 2 m B = µ 0n s q 2 m B B

Field penetration m s now we have curlj + B =0 n s q2 and the Maxwell equation (inside the superconductor) curlb = µ 0 j combining this gives curlcurlb = µ 0 curlj = µ 0n s q 2 m B curlcurlb = graddivb B = µ 0n s q 2 m B = µ 0n s q 2 m B B This permits only an exponentially damped B field in the superconductor (exercise)

Field penetration B 0 superconductor 4B = µ 0n s q 2 m B λ L 4j = µ 0n s q 2 = m j L p m/µ0 n s q 2 The decay length is in the order of 1000 Å.

Ginzburg-Landau theory (1950) order parameter (r) = 0 (r)e i (r) n s = Ψ Ψ = Ψ 2 0 coherence length ξ (length over which can change appreciably) Ψ(r)

Key-ideas of the BCS theory of 1957 (Bardeen, Cooper, Schrieffer) The interaction of the electrons with lattice vibrations (phonons) must be important (isotope effect, high transition temperature for some metals which are poor conductors at room temperature). The electronic ground state of a metal at 0 K is unstable if one permits a net attractive interaction between the electrons, no matter how small. The electron-phonon interaction leads to a new ground state of Bosonic electron pairs (Cooper pairs) which shows all the desired properties.

The electron-phonon interaction/ Cooper pairs Polarization of the lattice by one electron leads to an attractive potential for another electron.

The electron-phonon interaction / Cooper pairs k Polarization of the lattice by one electron leads to an attractive potential for another electron.

The electron-phonon interaction / Cooper pairs k k' Polarization of the lattice by one electron leads to an attractive potential for another electron.

The electron-phonon interaction/ Cooper pairs k k' Polarization of the lattice by one electron leads to an attractive potential for another electron.

What is the distance between the electrons forming Cooper pairs? The lattice atoms move much slower than the electrons. The polarization is retarded Establishing the polarization takes k 2 D 10 13 s the electrons 10 ms move with v F 10 6 ms 1 so distance between the electrons forming the Cooper pair is around 10 7 m

Coherence of the superconducting state The electrons paired in this way are called Cooper pairs. They have opposite k vectors (a total k=0) and (in most cases) opposite spins, such that S=0. Cooper pairs behave like Bosons and can condense all in a single ground state with a macroscopic wave function. This leads to an energy gain with a size depending on how many Cooper pairs already are in the ground state. Cooper pairs are example of quasiparticles appearing in solids (like holes or electrons with an effective mass or phonons) k k'

Ginzburg-Landau theory (1950) order parameter (r) = 0 (r)e i (r) n s = Ψ Ψ = Ψ 2 0 coherence length ξ (length over which can change appreciably) Ψ(r)

The BCS ground state normal metal E F unoccupied excited unpaired single electron electrons states occupied single electron states energy (arb. units)

The BCS ground state normal metal superconductor at T=0 E F unoccupied excited unpaired single electron electrons states occupied single electron states energy (arb. units) unoccupied single electron states occupied single occupied electron states single electron stats 2Δ C = 3.53k B T C

Hydrogen molecule anti-bonding E E bonding 1 2 interatomic distance (a 0 ) 3 4 5

The BCS ground state normal metal superconductor at T=0 E F unoccupied excited unpaired single electron electrons states occupied single electron states energy (arb. units) unoccupied single electron states occupied single occupied electron states single electron stats 2 C = 3.53k B T C

transition temperature The BCS ground state 10 m 1 T C = 1.13 D exp g(e F )V Debye temp. density of states el-ph interaction strength

transition temperature The BCS ground state Debye temp. 10 m 1 T C = 1.13 D exp g(e F )V density of states nal to the Debye temp el-ph interaction strength isotope effect e Θ D ω D M 1/2.

transition temperature The BCS ground state Debye temp. 10 m 1 T C = 1.13 D exp g(e F )V density of states nal to the Debye temp el-ph interaction strength isotope effect e Θ D ω D M 1/2. superconductor at T=0 number of electrons in Cooper pairs g(e F ) g(e F ) D C = 3.53k B T C unoccupied single electron states occupied single occupied electron states single electron stats 2Δ

transition temperature The BCS ground state Debye temp. 10 m 1 T C = 1.13 D exp g(e F )V density of states nal to the Debye temp el-ph interaction strength isotope effect e Θ D ω D M 1/2. superconductor at T=0 number of electrons in Cooper pairs g(e F ) g(e F ) D C = 3.53k B T C unoccupied single electron states occupied single occupied electron states single electron stats 2Δ total energy gain g(e F ) 2 g(e F ) 2 2 D

The gap at higher temperatures Gap size ( (T=0)) 1.0 0.8 0.6 0.4 0.2 In Sn Pb 0.0 0.0 0.2 0.4 0.6 Temperature T(T C ) 0.8 1.0

Detection of the gap: tunnelling Normal metal Insulator Superconductor

Detection of the gap: tunnelling Normal metal Insulator Superconductor U=0 normal metal superconductor at T=0 unoccupied excited unpaired single electron electrons states insulator unoccupied single electron states E F occupied single electron states occupied single occupied electron states single electron stats 2D

Detection of the gap: tunnelling Normal metal Insulator Superconductor U=0 normal metal superconductor at T=0 normal metal 0<U<D/e superconductor at T=0 unoccupied excited unpaired single electron electrons states insulator unoccupied single electron states unoccupied excited unpaired single electron electrons states insulator unoccupied single electron states E F occupied single electron states occupied single occupied electron states single electron stats 2D E F occupied single electron states occupied single occupied electron states single electron stats 2D

Detection of the gap: tunnelling Normal metal Insulator Superconductor U=0 normal metal superconductor at T=0 normal metal 0<U<D/e superconductor at T=0 unoccupied excited unpaired single electron electrons states insulator unoccupied single electron states unoccupied excited unpaired single electron electrons states insulator unoccupied single electron states E F occupied single electron states occupied single occupied electron states single electron stats 2D E F occupied single electron states occupied single occupied electron states single electron stats 2D normal metal U>D/e superconductor at T=0 unoccupied excited unpaired single electron electrons states insulator unoccupied single electron states E F occupied single electron states 2D occupied single occupied electron states single electron stats

Detection of the gap: tunnelling Normal metal Insulator Superconductor U=0 normal metal superconductor at T=0 normal metal 0<U<D/e superconductor at T=0 unoccupied excited unpaired single electron electrons states insulator unoccupied single electron states unoccupied excited unpaired single electron electrons states insulator unoccupied single electron states E F occupied single electron states occupied single occupied electron states single electron stats 2D E F occupied single electron states occupied single occupied electron states single electron stats 2D E F normal metal unoccupied excited unpaired single electron electrons states occupied single electron states U>D/e insulator superconductor at T=0 unoccupied single electron states occupied single occupied electron states single electron stats 2D tunnelling current (arb. units) D/e tunnelling voltage U

Detection of the gap: optical hν hν For radiation below hν < 2Δ no absorption, light is transmitted through a thin sheet of superconductor. For radiation below hν > 2Δ absorption takes place. Relevant light wavelength in the infrared.

Detection of the gap: heat capacity 4 heat capacity (arb. units) 3 2 1 c s c n Al 0 0 1.0 2.0 T (K)

Zero Resistance

Zero Resistance All Cooper pairs are in the same state and have THE SAME total k. For k=0, no current. For k 0 supercurrent.

Zero Resistance All Cooper pairs are in the same state and have THE SAME total k. For k=0, no current. For k 0 supercurrent. For a normal metal, single electrons can be scattered and be lost for the current. For a superconductor, one can only change the k of all Cooper pairs at the same time. This is VERY unlikely.

Zero Resistance All Cooper pairs are in the same state and have THE SAME total k. For k=0, no current. For k 0 supercurrent. For a normal metal, single electrons can be scattered and be lost for the current. For a superconductor, one can only change the k of all Cooper pairs at the same time. This is VERY unlikely. Cooper pairs can be split up but this costs 2Δ and they might condensate quickly again. At high current densities, there is enough energy available to break them up. Also at high magnetic field densities.

Superconducting current in a ring F B I

Superconducting current in a ring F B = h p I

Superconducting current in a ring F B = h p Macroscopic coherence: apply Bohr-Sommerfeld quantization 2 r = n = n h p I n = 0,1,2... 2 rp = nh

Superconducting current in a ring F B = h p Macroscopic coherence: apply Bohr-Sommerfeld quantization 2 r = n = n h p I n = 0,1,2... 2 rp = nh I more general pdr = nh

Superconducting current in a ring I pdr = nh F B p qadr = hn I

Superconducting current in a ring I pdr = nh F B m s n s q p I qadr = hn jdr q I p = m s v j = n s qv Adr = nh I

Superconducting current in a ring I pdr = nh F B m s n s q p I qadr = hn jdr q I p = m s v j = n s qv Adr = nh I Adr = Stoke s integral theorem curlada = Bda = B

Superconducting current in a ring I pdr = nh F B m s n s q p I qadr = hn jdr q I p = m s v j = n s qv Adr = nh I Stoke s integral theorem Adr = curlada = Bda = B m s n s q 2 I jdr B = n h q

Superconducting current in a ring F B m s n s q 2 I jdr B = n h q I

Superconducting current in a ring F B m s n s q 2 I jdr B = n h q integration path inside the ring material I B = n h q

Superconducting current in a ring F B m s n s q 2 I jdr B = n h q integration path inside the ring material I B = n h q The magnetic flux through the ring is quantized. q turns out to have a value of -2e in the experiment. So one flux quantum is h 2e =y 2.067 10 15 Tm 2

movie of superconducting maglev train from IFW Dresden

Many open questions

Many open questions Why does the train follow the tracks?

Many open questions Why does the train follow the tracks? Why does the train not fall down when turned upside down?

Many open questions Why does the train follow the tracks? Why does the train not fall down when turned upside down? How does the train remember the distance it has to be above the tracks?

Type I and type II superconductors Type I 0 0 B C B int 0 0 0 B = 0 int B B = B C int 0 M External magnetic field B 0

Type I and type II superconductors Type I Type II =0 0 0 B C 0 0 B C1 B C2 B int 0 0 0 B = 0 int B B = B C int 0 B int 0 0 0 M B int = 0 B 0<B int<b 0 C1 B B int = B 0 C2 M External magnetic field B 0

type II pure supc. phase mixed phase =0 normal phase 0 0 B C1 B C2 B int 0 0 0 B int = 0 B 0<B int<b 0 C1 B B int = B 0 C2 M external magnetic field B 0

Vortices in type II superconductors Φ B =h/2e side view vortex j top view The field penetrates through filaments of normal material, the rest remains superconducting. These filaments are enclosed by vortices of superconducting current.

How does the magnetic flux penetrate the type II superconductor The magnetic field enters in the form of flux quanta in normal-state regions. These regions are surrounded by vortices of supercurrent. This can be made visible by Bitter method, optical techniques, scanning tunnelling techniques... 25 x 25 μm (educated guess)

Vortices in type II superconductors The vortices repel each other. If they can move freely, they form a hexagonal lattice. But if there are defects which pin the vortices, they cannot change their position. 25 x 25 μm (educated guess)

Vortices in type II superconductors Here the vortices can move quite freely. Their number changes as the external field is changed. 25 x 25 μm For a very high field, the vortices move so close together that the superconductivity in between breaks up. 25 x 35 μm

Vortices in type II superconductors

Vortices in type II superconductors The superconductor used in the train movie is a type II superconductor.

Vortices in type II superconductors The superconductor used in the train movie is a type II superconductor. It pins the vortices very strongly. Therefore the magnetic field it experienced while cooling through the phase transition is frozen in.

Vortices in type II superconductors The superconductor used in the train movie is a type II superconductor. It pins the vortices very strongly. Therefore the magnetic field it experienced while cooling through the phase transition is frozen in. This can explain the effects we did not understand before (staying on the tracks, train upside down, train on walls...).

Current-induced vortex movement A current through the material causes the vortices to move perpendicular to the current and to the magnetic field. This is very reminiscent of the Lorentz force. Vortex movement leads to energy dissipation and therefore resistance at very small current densities! This can be avoided by pinning the vortices. j F = j B force on vortex

Characteristic length scales type I or type II superconductor The penetration depth λ The coherence length ξ B 0 superconductor λ L

High-temperature superconductors highest discoverd TC(K) 140 120 100 80 60 40 Hg Pb Nb NbN Nb 3 Sn Cs 2 RbC 60 Nb 0 3 Ge 1900 1920 1940 1960 1980 2000 20 HgBa 2 Ca 2 Cu 3 O 8+x Bi 2 Sr 2 Ca 2 Cu 3 O 10+x YBa 2 Cu 3 O 7-x La 1.85 Sr 0.15 CuO 4 MgB 2 Year Discovery of ceramic-like high-temperature superconductors

Structure Superconductivity in the Cu+O planes. Very anisotropic

Some properties Extreme type II behaviour (this is why the train worked). Low critical current densities (because of the anisotropy) Ceramic-like: difficult to turn into wires

What makes the high-temperature superconductors superconduct? One is reasonably sure that Cooper pairs are also responsible here. But there is a big dispute on how they are formed: interaction of electrons with phonons, magnetic excitations (magnons), collective excitations of the electron gas (plasmons) or something else?

High-temperature superconductors highest discoverd TC(K) 140 120 100 80 60 40 Hg Pb Nb NbN Nb 3 Sn Cs 2 RbC 60 Nb 0 3 Ge 1900 1920 1940 1960 1980 2000 20 HgBa 2 Ca 2 Cu 3 O 8+x Bi 2 Sr 2 Ca 2 Cu 3 O 10+x YBa 2 Cu 3 O 7-x La 1.85 Sr 0.15 CuO 4 MgB 2 Year BCS XYZ

Conclusions

Conclusions Very broad field. An exception to almost every rule we have discussed.

Conclusions Very broad field. An exception to almost every rule we have discussed. Ferromagnetism and superconductivity apparently mutually exclusive.

Conclusions Very broad field. An exception to almost every rule we have discussed. Ferromagnetism and superconductivity apparently mutually exclusive. But we have seem some similarities in the phase transition.

Conclusions Very broad field. An exception to almost every rule we have discussed. Ferromagnetism and superconductivity apparently mutually exclusive. But we have seem some similarities in the phase transition. Both are ways to lower the energy of the electrons at the Fermi level. The relative energy gain may seem small but this is only because the Fermi energy is so high.

Conclusions Very broad field. An exception to almost every rule we have discussed. Ferromagnetism and superconductivity apparently mutually exclusive. But we have seem some similarities in the phase transition. Both are ways to lower the energy of the electrons at the Fermi level. The relative energy gain may seem small but this is only because the Fermi energy is so high. There are also other mechanisms for this and interesting situations arise when several compete.