1 Josephson Effect. dx + f f 3 = 0 (1)

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1 Josephson Effet In 96 Brian Josephson, then a year old graduate student, made a remarkable predition that two superondutors separated by a thin insulating barrier should give rise to a spontaneous zero voltage DC urrent, I s = I sin φ, where φ is the differene in phase aross the juntion. And that if a finite DC voltage were applied, an AC urrent with frequeny ω = ev/ would flow. I is alled the Josephson ritial urrent. There is a myth that Brian Josephson did his alulation 96 and won the Nobel prize 97 as part of the solution to a homework problem of Phil Anderson s. The truth is that Anderson was a leturer on sabbatial at Cambrige in 96-6, and he gave a series of letures in whih he mentioned the problem of tunneling between two superondutors, whih Josephson then promptly solved. The idea was opposed at first by John Bardeen, who felt that pairing ould not exist in the barrier region. Thus muh of the early debate entered on the nature of the tunneling proess, whereas in fat today we know that the Josephson effet ours in a variety of situations whenever two superondutors are separated by a weak link, whih an be an insulating region, normal metal, or short, narrow onstrition. Let s first onsider the last example as the oneptually simplest. The Ginzburg Landau equation appropriate for this situation may be written ξ d f dx + f f = where ξ = is the GL oherene length and fx Ψx/Ψ m at. Take L ξ, so the deviations of Ψ oming from the bulk value Ψ of the first SC is small, and vie versa for the seond SC. Changes of Ψ in the onstrition our over a length sale of L, so that the first term is of Oξ/L f f. So we must solve a Laplae equation for f, d f = dx Phys. Lett., 5 96 Physis Today, July In Je. 9 I reeived an from Brian Josephson orreting this version of the history: Date: Wed, Jun 9 9::5 + From: Brian Josephson <bdj@am.a.uk> To: pjh@phys.ufl.edu Subjet: the Josephson myth Dear Peter, While browsing I ame aross your mention of the myth that I disovered the effet beause of a problem set by Anderson. Your orretion is not ompletely orret either! It was Pippard, my supervisor, who drew my attention to Giaevar s tunnelling expts. and his theory, whih started me thinking espeially as to how one ould get away without using oherene fators. Anderson on the other hand told me of the Cohen/Faliov/Phillps alulation involving a single superondutor when it ame our in PRL, whih gave me the idea of how to do the two-s. ase. Previously I had got the broken symmetry idea whih was ruial from a number of papers inluding Anderson s pseudospin model, and also expounded in his leture ourse whih I went to. Best regards, Brian J.

2 with B.C. f =, fl = e i Φ. The solution will be f = x + x L L ei φ. The solution an be thought of as two terms, the first Ψ, beginning to leak into the onstrition from the left, the seond Ψ leaking into the onstrition from the right. The GL expression for the urrent will be j = e Ψ Ψ Ψ Ψ e m i m Ψ ΨA }{{} zero = e m i Ψ [ xl + xl e i φ L + L ei φ = e Ψ m L whih means that the urrent will be ]. sin φ, I = I sin φ, I = e Ψ A, 5 m L where A is the ross setion. Given that we have two weakly oupled QM systems, it is not unreasonable justified by mirosopi theory to write down oupled Shrödinger equations i Ψ t i Ψ t = E Ψ + αψ 6 = E Ψ + αψ 7 where H i Ψ i = E i Ψ i and E = E = E if the superondutors are idential. Take Ψ i to be the density of pairs in SC i Ψ i = n i e iφ i Ψ i = ṅ i e iφ i + i n i φi e iφ i n i ṅ + i n φ = ie n iα n e iφ φ n 8 ṅ + i n φ = ie n iα n e iφ φ. n 9

3 z Β y x Figure : If we take the real parts and use ṅ = ṅ we get ṅ n = α n sinφ φ ṅ = α n sinφ φ n ṅ = α n n sinφ φ, Note I ve put V=A=. Then the urrent is just j = eṅ. If we take the imaginary parts we have n φ = E n α n osφ φ n φ = E n α n osφ φ, and by subtrating and assuming n n let s ouple idential superondutors at first we get φ φ = E E = ev V where for the seond equality we used the fat that the potential differene between the superondutors shifts the pair energies by ev. So we see that a finite voltage differene leads to a time hanging phase differene φ whih means an AC urrent via Eq.. Magneti fields. Now put a flux through the juntion where the B field is along the ŷ diretion and A = Bxẑ. The phase of the wave funtion Ψ must hange by φ φ e ds A for the theory to be gauge invariant. Notie that φ is now spae dependent. So the Josephson

4 J Φ/Φ Figure : equations will read j = eα n n }{{} j ev V = t sin φ e φ φ e ds A, 5 ds A, 6 and sine S A = we will have J = L = Lj πφ/φ dxjx = d L dz Bx = Bxd, 7 dxj sin [ os Φ os Φ e φ + πφ Φ Bxd ], 8 where Φ = π. What is the maximum urrent through the juntion for all possible φ? e We have to alulate dj = whih leads to the relation d Φ tan Φ = ot πφ/φ 9 and with a bit of tedious trigonometry to sin πφ/φ J = Lj πφ/φ. This formula produes the Josephson Fraunhofer interferene pattern. DC SQUID Superonduting Quantum Interferene Devie

5 J Φ J Figure : We ignore resistane and apaitane for now. The inside SC thikness is assumed muh greater than λ and sine v s = we have φ = A/Φ. The flux will be Φ = ds A = ds A + Φ = Φ φ φ + Φ ds φ + φ φ + ds A + Φ ds A + ds φ ds A = Φ φ φ + ds A + Φ φ φ + ds A }{{}}{{} γ The Josephson urrent through the SQUID will be γ Φ = γ γ. J = J sin γ + sin γ = J sin γ + sinγ + Φ. Whih means that the urrent osillates with the flux. And as a result of that the SQUID an be a sensitive measure of magneti fields. In pratie we inlude the apaitane and resistane of the devie. 5