**7. Giaever tunnelling**

When we bring two metals in proximity, separated by a thin-insulating barrier, apply a tiny voltage nd then the current will flow in the circuit. There is a thininsulating barrier, but electrons will tunnel through the barrier. Now, what will happen if these metals are replaced by a superconductor? These are a set of experiments carried out by Norwegian-American physicist Ivar Giaever who shared the Nobel Prize in Physics in 1973 with Leo Esaki and Brian Josephson "for their discoveries regarding tunnelling phenomena in solids". What he found was that if one of the metals is a superconductor, the electron cannot just come in, as there is an energy barrier of Δ, the superconducting gap. Your applied voltage has to be at least as big as Δ for tunneling to happen. This is depicted in **Figure 24**. Let us see why this is the case.

Recall in our discussion of superconducting state that we had 2*p* pockets of which *<sup>p</sup>* were empty. We had a binding energy of � <sup>4</sup>ℏ*d*<sup>2</sup> *Np*<sup>2</sup> *<sup>ω</sup><sup>d</sup>* . What will it cost to

Let us take lanthanum copper oxide *La*2*CuO*4, where lanthanum donates three electrons and copper donates two and oxygen accepts two electrons and all valence is satisfied. The *Cu* is in state Cu2+ with electrons in *d*<sup>9</sup> configuration [11]. *d* orbitals are all degenerated, but due to crystal field splitting, this degeneracy is broken, and *dx*<sup>2</sup>*<sup>y</sup>*<sup>2</sup> orbital has the highest energy and gets only one electron (the ninth one) and forms a band of its own. This band is narrow, and hence all *k* states get filled by only one electron each and form Wannier packets that localize electrons on their respective sites. This way electron repulsion is minimized, and in the limit *U*≫*t* (repulsion term in much larger than hopping), we have Mott insulator and antiferromagnetic phase. When we hole/electron dope, we remove/add electron to *dx*<sup>2</sup>*<sup>y</sup>*<sup>2</sup> band. For example, *La*<sup>2</sup>*xSrxCuO*<sup>4</sup> is hole doped as Sr has valence 2 and its presence further removes electrons from copper. *Nd*<sup>2</sup>*xCexCuO*<sup>4</sup> is electron doped as Ce has valence 4 and its presence adds electrons from copper. These extra holes/electrons form packets. When we discussed superconductivity, we discussed packets of width *ωd*. In d-bands, a packet of this width has many more *k* states as d-bands are narrow and only 1–2 eV thick. This means *N* (k-points in a packet) is very large and we have much larger gap Δ and *Tc*. This is a way to understand high *Tc*, d-wave packets with huge bandwidths. This is as shown in **Figure 26**. As we increase doping and add

*SQUID Magnetometers, Josephson Junctions, Confinement and BCS Theory of Superconductivity*

*DOI: http://dx.doi.org/10.5772/intechopen.83714*

*(A) Depiction on how in narrow* d *band the electron wave packet comprises all* k *points to minimize repulsion. (B) The wave packet in two dimensions with wave packet formed from superposition of k-points (along the k-*

*The characteristic phase diagram of a high* Tc *cuprate. Shown are the antiferromagnetic insulator (AF) phase*

*on the left and* dome*-shaped superconducting phase (SC) in the centre.*

**Figure 26.**

**Figure 27.**

**91**

*direction) inside the Fermi sphere.*

#### **Figure 24.**

*(A) Depiction on how a tiny voltage between two metals separated by an insulating barrier generates current that goes through an insulating barrier through tunneling. (B) If one of the metals is a superconductor, then the applied voltage has to be at least as big as the superconducting gap.*

**Figure 25.** *Depiction on how upon tunneling an extra electron, shown in dashed lines, enters the superconducting state.*

bring in an extra electron as shown in **Figure 25**? It will go in one of the empty pockets, and then we have only *p* � 1 pockets left to scatter to reducing the binding energy to � <sup>4</sup>ℏ*d*<sup>2</sup> *Np p*ð Þ �1 *<sup>ω</sup><sup>d</sup>* with a change <sup>Δ</sup> <sup>¼</sup> <sup>4</sup>ℏ*d*<sup>2</sup> *Np <sup>ω</sup><sup>d</sup>* . Therefore, the new electron raises the energy by Δ, and therefore to offset this increase of energy, we have to apply a voltage as big as Δ for tunneling to happen.
