Conductance and Thermopower of Ballistic Andreev Cavities Thomas Engl, Jack Kuipers, and Klaus Richter Institut fu¨r Theoretische Physik, Universita¨t Regensburg, D-93040 Regensburg, Germany (Dated: January 11, 2011) When coupling a superconductor to a normal conducting region the physical properties of the system are highly affected by the superconductor. We will investigate the effect of one or two superconductors on the conductance of a ballistic chaotic quantum dot to leading order in the total channel number using trajectory based semiclassics. The results show that the effect of one 1 superconductorontheconductanceisoftheorderofthenumberofchannelsandthatthesignofthe 1 quantumcorrectionfromtheDrudeconductancedependsontheparticularratiosofthenumbersof 0 channels of the superconducting and normal conducting leads. In the case of two superconductors 2 withthesamechemicalpotentialweadditionallystudyhowtheconductanceandthesignofquantum n correctionsareaffectedbytheirphasedifference. Asfarasrandommatrixtheoryresultsexistthese a are reproduced by our calculations. Furthermore in the case that the chemical potential of the J superconductors is the same as that of one of the two normal leads the conductance shows, under 0 certain conditions, similar effects as a normal metal-superconductor junction. The semiclassical 1 framework is also able to treat the thermopower of chaotic Andreev billiards consisting of one chaoticdot,twonormalleadsandtwosuperconductingislandsandshowsittobeantisymmetricin ] thephase differenceof thesuperconductors. l l a h I. INTRODUCTION tions with sufficiently large barrier strengths at the N-S - s interface that the differential conductance dI/dV van- e m Transport problems have always attracted a lot of ishes for voltages smaller than the superconducting gap attention in condensed matter physics. While the ∆/e. Inthisregimetheconductanceisdoubledcompared t. Landauer-Bu¨ttikerformalismwhichconnectsthe electri- to the conductance of the same normal conducting re- a m cal current to the quantum transmission probabilities of gionwithanormalconducting leadinsteadofthe super- a conductor is of key importance for transport through conducting one: an indication of the proximity effect6,7. d- nanosystems,similarformulaehavealsobeenderivedfor Whenincreasingthevoltagethedifferentialconductance n hybridstructuresconsistingofnormalconducting(N)re- has a peak at eV ≈ ∆ and finally approaches the con- o gions connected to superconductors (S)1–3 in which An- ductanceofthenormalconductingregionwithoutthesu- c dreev reflection4 plays a crucial role. perconductor. However,thetotalvalueofthecurrentfor [ Andreev reflection4 can occur whenever a normal high voltages exceeds that of a metallic junction by the 2 metal region is coupled to a superconductor. If an elec- so-calledexcess current. The earlyexperiments onN-I-S v tron hits the normal metal-superconductor (N-S) inter- junctions were in agreement with BTK-theory. However 5 face with an energy closely above the Fermi energy an later experiments8,9 found an enhancement of the differ- 8 additional electron-hole pair can be created, and the entialconductanceatV =0laterknownasthe zerobias 3 twoelectronsenterthesuperconductorformingaCooper anomaly. 3 2. pair. The hole however has to compensate the momen- Recently Whitney and Jacquod10 considered a some- tumoftheoriginalelectron,thereforeitretracestheelec- 1 what different type of setup. They considered a bal- tron path. Moreover the hole picks up a phase equal to 0 listic normal conducting region with a boundary giv- thephaseofthemacroscopicsuperconductingwavefunc- 1 ing rise to classically chaotic dynamics. Andreev reflec- : tion. v tionandinterferencebetweenquasiparticleswithslightly The early theoretical and experimental investiga- i different paths lead to a hard gap in the density of X tions of transport properties focused on the current states of such chaotic ballistic conductors coupled to a ar tnhorromuaglhmtehteal-inintseurflaactoer-osfupnoerrcmoanldumcettoarl-(sNu-pI-eSrc)oanndducSt-oNr-, superconductor11–13. In Ref. 10 such a chaotic Andreev quantum dotis coupled to two normal conducting and S junctions5. For these the BTK-theory1 applies, which one superconducting lead and its transport characteris- is based on the Landauer type equation tics was studied. Using a trajectory based semiclassical ∞ methodtheycalculatedtheaverageconductancebetween 2e I = Ω dǫ[1−R +R ][f(ǫ−eV)−f(ǫ)], (1) thetwonormalleadsofsuchchaoticAndreevbilliardsup 0 A h tosecondorderintheratioN /N whereN isthetotal −Z∞ S N S number of superconducting channels and N =N +N N 1 2 where I is the currentthroughthe N-S interface with an the sum of the number of channels in the normal leads. appliedvoltageV andΩameasureoftheareaofthejunc- If the superconducting chemical potential is the same as tion. In (1) R is the probability for normal reflection, thatofone ofthe two normalconducting leads(abbrevi- 0 R isthe probabilityforAndreevreflection,andf isthe atedto‘superconductinglead’anddepictedinFig.1(d)) A Fermi function. The BTK theory predicts for N-S junc- they found that the correction to the classical conduc- 2 Superconductor Superconductor N N N N 1 2 1 2 δτ S S 1 2 φ e e e* h S1, φ1 S2, φ2 e* e h* (a) (b) Superconductor Superconductor Bulk Superconductor h h* e Bulk Superconductor h* h e* N1 N2 N1 N2 e*e e e* ee* h h* FIG. 2. The diagonal diagrams contributing to the conduc- (c) (d) tance up to third order in the number NS of channels of the superconductor. Here,e andh denoteelectron typeand hole typequasiparticlesandtheasteriskdenotesthatthepathen- N1 N2 N1 N2 ters thecalculations with the complex conjugated factors. S S S S 1 2 1 2 δτ φ φ density of states of Andreev billiards. Here we further extend this recent approachto the conductance. To this end a diagonal backbone is introduced which is given by (e) (f) apathanditscomplexconjugated. Thequantumcorrec- tioninleadingorderin1/N isthenobtainedbyattaching FIG.1. Schemaof variousAndreevbilliards considered here: anevennumberofsocalledtrees(orcomplexconjugated (a) Andreev interferometer with two superconducters, (b) double dot setup, (c) chaotic quantum dot coupled to one trees) as those used in Ref. 15. In this diagrammatical superconducting island. (d) the case of a superconducting language,in Ref. 10 the authors restricted themselves to lead with the same chemical potential as the right lead, (e) atmosttwotreesconsistingofjustonepathpair. There- theso-called “symmetrichouse”,(f)the“asymmetrichouse” fore their results arevalid only for small N /N and the S N where at lead 1 a neck is additionally inserted compared to validity of their results for larger N is not known. Un- S (e). like the results in Ref. 16, where the authers considered the distribution of the conductance of chaotic quantum dots with one open channel per lead, our results will be tance be huge (of order of N = N +N ) compared to valid for large numbers of channels in the normal leads. N S usualweaklocalizationeffects,inparticularitwasshown that the quantum correction may become negative or We will derive the conductance of the two setups in positivedependingontheratioN1/N2. Asimilarchange Ref. 10 - namely the setup with a superconducting is- inthesignofthequantumcorrectiontotheconductance land(seeFig.1(c))andwithasuperconductinglead(see may be caused by a change in the transparencies of the Fig. 1(d)) - to all orders in N /N . To this end we start S N leads14. inSects. II-IVbyconsideringthesemiclassicaldiagrams Using the same approach Whitney and Jacquod and their contribution to the transmission probabilities showed furthermore that to leading order in N /N the and thus to the conductance to leading order in 1/N. S N thermopowerofachaoticallyshapednormalmetalquan- In Sect. V we apply this approach to the setup with a tum dot with two normal leads and two superconduct- superconducting island. We show that our semiclassical ing islands (called a “symmetric house” and depicted in result for the conductance coincides with previous ran- Fig. 1(e)) is antisymmetric in the phase difference of dom matrix theory results17 existing for zero magnetic the superconductors. They also argued that the ther- field and temperature (though still with a phase differ- mopower vanishes if the two superconductors carry the ence φ = φ − φ between the superconductors). We 1 2 same amountof channels as long as no symmetry break- furthermoreconsiderthemagneticfieldandtemperature ing neck is inserted at one of the two superconductors dependence of the conductance of setup Fig. 1(c). (c.f. Fig. 1(f)). For the other setup of an Andreev billiard coupled to Herewecombinethetrajectorybasedsemiclassicalap- oneortwoseparatesuperconductingleads(Fig.1(d))we proaches of Refs. 10, 12, 13 and 15 and provide a com- will show as a main result in Sect. VI that the quan- prehensive calculation of the conductance and the ther- tum correction to the classical value of the conductance mopower of Andreev billiards. In Refs. 12 and 13 a changesits sign not only with the ratio ofthe number of method was developed for the systematic evaluation of channels in the two normal conducting leads N /N but 1 2 multiple sums over electron and hole type orbits arising also by tuning the ratio x = N /N . This sign change S N inasemiclassicalapproachtotheproximityeffectonthe wanotanticipatedinRef.10,sinceitrequiresananalysis 3 Superconductor Superconductor Superconductor Superconductor Superconductor h h h h* e e e e h* e h e* h h ee* e e* e h e e h* e ee* h h* e h e* e* h e* e* h* ee3I ee3II ee3III he3I he3II Superconductor Superconductor Superconductor Superconductor Superconductor e e h e e* e h h e h h h* h e h e h e* e e* e h* e h* e h* e* h* h* h e* h e* h he3III he3IV he3V he3VI he3VII FIG.3. Pairsofpathscontributingtothethird-orderterminx=NS/NN ofthetransmission. Electron (hole) pathsaregreen ′ (red). Thesolid(dashed)linesbelongtoγ (γ ). Atrajectorypairenteringfromtheleftandexitingtotherightcanconnectthe two normal conducting leads while a trajectory pair entering and exiting at the same side can only contribute if theincoming and outgoing channel belong both to thesame lead. tohigherordersinx. Thisconductancecorrectionisalso respectively. ζ and ζ′ are classical trajectories starting showntooscillatewiththephasedifferenceφbetweenthe at channel a and ending at channel b. The amplitudes two superconducting leads with period 2π. Finally, we A contain the stability of the trajectory ζ and S is its ζ ζ study the dependence of the conductance on an applied classicalaction. Moreover,T istheHeisenbergtimethe H magnetic field and temperature. The effects we observe time dual to the mean level spacing. for some combinations of the ratios x and N /N turn Weareinterestedintheconductanceaveragedoverthe 1 2 out to be fairly similar to those found in the structures shapeofthe quantumdotoranenergyrangesmallcom- containing only one normal conducting lead. pared to the Fermi energy but large enough to smooth In Sect. VII we show how the methods derived before out fluctuations. Moreover we will take the semiclassi- canbeextendedtocalculatethetransmissioncoefficients cal limit ~ → 0. Therefore the energy dependent action oftwodotsconnectedtoeachotherbyaneckwehreeach difference in (2) causes fluctuations cancelled on aver- dothasonefurthernormalandonesuperconductinglead age unless it is of order of ~. Thus we have to pair (see Fig. 1(b)). The conductance of this setup is shown the trajectories in such a way that their action differ- to also be symmetric in the phase difference. The sign ence becomes sufficiently small. The easiest way to do of the quantum correction depends on the ratios x and this is to require that ζ = ζ′ which is known as the di- n = N /(N +N ), where N is the channel number of agonal approximation21. In Fig. 2 the trajectory pairs n 1 2 n the neck. contributing to the diagonal approximation are drawn In Sect. VIII we finally apply our calculations to the schematically for up to three Andreev reflections. The thermopower of the setup shown in Fig. 1(e) with both contributions of the diagonal pairs to the conductance equaland differentnumbers ofchannels as well asto the provide the classical conductance10 setupshowninFig.1(f). We findthatforthe symmetric N N house with different channel numbers and for the anti- g = 1 2 (3) cl symmetric house the thermopower is antisymmetric in N1+N2 the phase difference. if the superconductors are isolated and N (N +2N ) 1 2 S g = (4) II. CONTRIBUTING DIAGRAMS cl N +N +2N 1 2 S in the case of the superconducting leads. However as We will evaluate the quantum transmission between shownsemiclassicallyinRefs.12forthe densityofstates two normal conducting leads coupled to a classically and10fortheconductanceofAndreevquantumdotsone chaotic, ballistic quantum dot which is additionally con- hastogobeyondthe diagonalapproximationtofully ac- nected to superconducting leads such as depicted in count for quantum effects. The non-diagonal trajectory Fig. 1. In a trajectory-based semiclassical approach pairscontributingtotheconductanceofnormaljunctions the transmission probabilities may be written as18–20 in the limit ~ → 0 have been first considered in Ref. 22 Tij = T1H XbN=i1XaN=j1Xζ,ζ′qAζA*ζ′ei(Sζ−Sζ′)/~, (2) a2tahlrneed-sseogmfetanrtalrelarjreajeeclgictzoiteoordnyrystsoitsnrtherwitegcthhhciehcerhseso‘acrcnrdooeamsrrsseb’iiecntalro1cash/reyNotetoihvneeenrRacinenhfu.tmo2ht3bihs:eerrTre.-ghlsieoaronyef where the ∗ denotes complex conjugation. Here, a and whilethe remainingl onesavoidcrossing. Thedifference b label the channels in lead i ∈ {1,2} and j ∈ {1,2}, between a trajectory ζ and its partner ζ′ then leads to 4 S S ee3I ee3II ee3III he3I he3II he3III he3IV he3V he3VI he3VII FIG. 4. Diagrams corresponding to the trajectory pairs shown in Fig. 3. The full circles denote encounters while the empty circles denote Andreev reflections. Note that an encounter touching the superconductor is marked as Andreev reflection. An encountertouching a normal conducting lead is shown as an empty box. The solid (dashed) line represents ζ (ζ′). a small action difference as long as these stretches are Superconductor Superconductor close enough to each other. Such a region with l cross- h ing trajectorystretchesandl trajectorystretches‘avoid- e e e h* h* ing crossing’ will be referred to as l-encounter. Between e* e* h h these l-encounters two different trajectory stretches re- trace each other forming a path pair with vanishing ac- FIG. 5. If the e-h path pairs are cut off a diagonal type tion difference which will also be referred to as a link. diagram remains. Inwhatfollowswe willidentify the relevanttrajectory pairs contributing to the conductance beyond the diago- nalapproximationinleadingorderintheinversechannel number1/N. ThediagramswithtwoAndreevreflections maybe foundinRef.10. Howeverwewantto gobeyond second order in x=N /N . The trajectories contribut- S N (a) (b) (c) ing in third order in x, i.e. trajectories with three An- dreevreflections,areshowninFig.3. Thefirsttaskis to FIG. 6. Diagrams we neglect in leading order in 1/N due to find a structure in the diagrams contributing at leading ′ the formation of loops: (a) A non-diagonal ζ-ζ pair causes order in the channel number. To facilitate this we can theformationofaloop. (b)Aloopformedbyanoff-diagonal redrawoursemiclassicaldiagramsinaskeletonformand e-hpathpair. (c)Aloopformedduetothelackofadiagonal- represent encounters and path pairs by nodes and lines. typeζ-ζ′ path pair. For example the diagrams contributing to third order in x, shown in Fig. 3, can be redrawn as in Fig. 4. We first consider how to read of the channel number we haveto connectthis off-diagonalpartto the diagonal dependence from a given diagram, i.e. we use the dia- ‘backbone’byasecondζ-ζ′ pathpairthusformingaloop grammatic rules used in Ref. 23 disregarding an energy as indicated in Fig. 6(a). This loop however adds a link, andmagneticfielddependenceandanysignsforthe mo- thus giving a factor 1/N compared to the same diagram ment. A path pair hitting lead j contributes a factor of without the loop, and therefore decreases the number channel number N . The path pair, or link, itself how- of Andreev reflections by at least one and therefore the j ever contributes a factor 1/N while each encounter con- contributiontotheconductanceissuppressedbyafactor tributes a factor N. From the trajectory pairs shown in of the order 1/N such that it would contribute to sub- S Fig. 3 we see that if we cut off all e-h and e∗-h∗ pairs leading order in the inverse channel number. Therefore we again get a diagonal like contribution as depicted in the ‘off-diagonal’ parts may only consist of e-h or e∗- Fig.5. Forexampleifwecutthe e-hpairatthe veryleft h∗ pairs. In the same way we may neglect loops formed ofee3Iwegetthediagonalcontributiontothesecondor- by e-h or e∗-h∗ path pairs as the one in Fig. 6(b). der in x in Fig. 2 since there are two Andreev reflections In terms of graphs, the ‘off-diagonal’ parts again be- andif we cutthe ‘off-diagonal’parts in sayee3III we get come rooted plane trees as in Refs. 13, 15. These trees a diagonal contribution at first order in x. startwithalink(root)whichconnectinganencounterto Staying at leading order in 1/N implies that the ‘off- the diagonal like backbone. From this encounter several diagonal’path pairs cannot consist of one ζ- and one ζ′- further links emerge all ending again at an encounter or stretchsinceeachofthosepath pairshasto be traversed at a superconducting channel. In contrast to the trees byζ andζ′ inthe samedirection. Thusinorderto come in Refs. 13, 15, the trees here - we will call them ‘side- backfromtheoff-diagonalpartstartingwithanζ-ζ′ pair trees’ - start at the ‘diagonal encounter’ such that their 5 rootdoesnottouchachannelbutthediagonalbackbone smallerthanthesuperconductinggap∆havetobetaken instead. Note that we draw the diagrams such that the into account ǫE ≪ ∆. This allows us to approximate T non-complex conjugated side trees are at the upper side exp[−iarccos(ǫ/∆)] ≈ −i such that the scattering ma- of the diagonal backbone while the complex conjugated trix of Andreev reflection becomes independent of the ones are on the lower side of the diagonal backbone. energy24. Thus the diagrammatic rules for the ζ-side The fact that the path pairs along the backbone are trees read13,23 lcooompps:osTedheotnwlyo otrfaζje-ζct′opriaeisrsζisanadgaζi′nmduuset tbootnhegstleacrttinagt • An e-h path pair contributes N 1+iǫ+b2 −1 leadj andendatleadi. Thereforethepathpairshitting • An l-encounter contributes −N(cid:2) 1(cid:0)+ilǫ+l2b(cid:1)2(cid:3) thenormalleadshavetobeζ-ζ′ pairsandso,ifthereisa ‘diagonal’ encounter entered by a ζ-ζ′ pair and left only • Ane-hpathpairhittingthesupe(cid:0)rconductorS (cid:1)con- j by e-h and e∗-h∗ pairs, there must be a corresponding tributes N . encounterenteredonlybye-hande∗-h∗ pairsandleftby Sj a ζ-ζ′ pair. Therefore we again would get a loop essen- • An l-encounter touching the superconductor Sj tially formed by one e-h and one e∗-h∗ pair as shown in contributes NSj. Fig. 6(c). • Each Andreev reflection at the superconducting Alltoldthediagramshavetoconsistofadiagonaltype leadj convertinganelectronintoaholecontributes ‘backbone’ consisting of ζ-ζ′ path pairs and encounters −ie−iφj. (whichmayalsotouchthesuperconductor)andζ-andζ′- sidetreesemergingfromthesediagonalencounters. Note • Each Andreev reflection at the superconducting that when pairing a ζ with a ζ′ stretch these stretches leadj convertingaholeintoanelectroncontributes have to be traversed by the same kind of quasiparticle, −ieiφj. i.e. it has to be an e-e∗ or a h-h∗ pair. This is related with b ∝ B/~ where B is the magnetic field applied to the fact that each encounter has an even number of to the system. The proportionality factor depends on entering and exiting path pairs. the actual system23. These diagrammatic rules have to However, there is still one possibility left we have not be complex conjugated for a ζ′-side tree and imply that mentioned yet but that needs a special treatment. If when exchanging electrons and holes we just have to re- the ‘diagonal’ part consists of only two path pairs and place φ ↔ −φ. Thus a side tree starting with a hole one 2-encounter with one ζ-side tree (a side tree formed by ζ) and one ζ′-side tree this encounter can be moved gives the same contribution of a side tree starting with an electron but with negative phase. Therefore we only into one of the normal conducting leads, say lead i. An need to evaluate side trees starting with electrons. exampleforan2-encountertouchingtheincomingleadis The evaluation of the side trees then follows essen- the trajectorylabelledby he3IVinfigures3 and4which tially those in Refs. 15, 13 and 25. However here the arises from the trajectory labelled by he3VII by moving root of the tree does not hit any channel and therefore theencounterintothelead. Howeverthisisonlypossible can not touch the superconductor which simplifies the if the trajectory connects lead i to itself and thus if the calculation. Moreover from the rules above for a path electroniscoherentlybackscattered. Inthiscasewehave pair hitting a channel in the superconductor S we get only one side tree and one complex conjugated side tree 1 a factor −ie−iφ/2N if an electron hits the channel and but no ‘diagonal’ part. S1 −ieiφ/2N if a hole hits the channel, rather than just a Sinceweknowthestructureofthetrajectorypairscon- S1 factor of the numbers of channels, and equivalently for a tributing at leading order in the inverse channel number path pair hitting S . we canstartevaluating them. Becausethe contributions 2 of the encounters and the stretches are multiplicative23 Similar to Ref. 13 as long as the phase difference φ is zero and no encounter touches the superconductor the wemayfactorisethecontributionofagivendiagraminto contributionofasidetreewithcharacteristicv-whichis the contributions of side trees starting at the first en- counter with an α-type quasiparticle, Pα(ǫ,x), the first the vectorwhichl-thentryis the numberofl-encounters of the tree - but without the contribution of the path encounteritselfandthediagramremainingwhencutting pairs hitting one of the superconductors is the diagram after the first encounter as in Fig. 8. We will first evaluate the contribution arising from the sum- mation over a possible side tree. 1+iǫ+b2 −1V(v) 1+ilαǫ+lα2b2 α=1 (cid:0)1+iǫ+b2 2lα−(cid:1)1 (cid:0) (cid:1) Y III. SIDE TREE CONTRIBUTIONS = 1(cid:0)+iǫ+b2 −(cid:1)nV(v) 1+ilαǫ+lα2b2 . α=1 (cid:0) 1+iǫ+b2 lα(cid:1) We restrict ourself to sufficiently low temperatures (cid:0) (cid:1) Y (5) such that only energies ǫE (measured with respect to (cid:0) (cid:1) T the Fermi energy and in units of the Thouless energy Heretheencountershavebeenlabelledbyαandweused E = ~/2τ where τ is the mean dwell time) much that the side tree has to satisfy15 n = L(v)−V(v)+1, T D D 6 where n is the number of links touching the supercon- ductor, V(v) = v is the total number of encounters l l≥2 of the tree and LP(v) = lv. This is because every l- l l≥2 encounter creates 2l−1Padditional path pairs and each path pair has to end either in an encounter or at the superconductor. We then enumerate the number of l-encounters by x l and the number of l-encounters touching the supercon- ductor S at an odd numbered channel by z(i). An l- i o,l encounter touching the superconductor means that the l incoming trajectory stretches hit the superconductor at FIG.7. Iftheoddnumberedsubtreeshavezerocharacteristic one and the same channel. We look at the generating and hit the same superconductor the top encounter may be function f(x,z(o1),z(o2)) which counts the number of pos- slid into thesuperconductor. sible side trees and their encounter types and derive a recursion relation for it by cutting the side tree at its top encounter. If the top encounter is traversed by 2l the superconductor as indicated in Fig. 7. This is only stretchesanddoesnottouchthesuperconductorthetree possible if the odd numbered subtrees have zero charac- thenhasthecontributionofthetopencountertimesthat teristic and hit the same superconductor. By sliding the ofall2l−1subtreesgivingx flfˆl−1,wherefˆisthesame encounter into the superconductor the total number of l as f but with φ replaced by −φ accounting for the fact Andreev reflections do not change such that the phase that each even numbered subtree starts with a hole in- factor providedby the encounter touching the supercon- stead of an electron. If the top encounter however is an ductor is the same as the phase factor provided by the encounter traversed by 2l stretches touching S its con- odd numbered side trees we start from. For a side tree i tribution is z(i)fˆl−1. In total we therefore have startingwithanelectrontheoddnumberedsubtreeswith o,l zero characteristic provide one Andreev reflection con- N N vertinganelectronintoahole. Sincefromanl-encounter f =−i S1e−iφ/2−i S2eiφ/2 N N l odd numbered subtrees emerge the phase factor of an ∞ l-encounter touching Si is −ie−ilφi. + x flfˆl−1+ z(1)+z(2) fˆl−1 (6a) l o,l o,l N Xl=2h (cid:16) (cid:17) i zo(1,l) =(−i)l NS1e−ilφ/2r˜l−1, (7a) N N fˆ=−i S1eiφ/2−i S2e−iφ/2 N N N z(2) =(−i)l S2eilφ/2r˜l−1 (7b) o,l N ∞ + x fˆlfl−1+ zˆ(1)+zˆ(2) fl−1 (6b) The total power of a tree with 2n−1 Andreev reflec- l o,l o,l Xl=2h (cid:16) (cid:17) i tionsisagain l(l−1)vl =L−V =n−1. Thusinorder to get the required prefactor of (1+iǫ+b2)−n we can where the first two terms account for empty side trees P make the change of variables which consist of one link and one Andreev reflection at S or S and zˆ(i) is the same as z(i) but with φ replaced r 1 2 o,l o,l f =g(1+iǫ+b2), r˜= . (8) by −φ. 1+iǫ+b2 Due to the fact that the links of the side trees are tra- After making this change of variables and performing versed by one electron at energy +ǫ~/2τ and one hole D the summations in (6a,b) using geometric series we get at energy −ǫ~/2τ in opposite directions an l-encounter D in view of Eqs. (7a,b) consists of l electron-stretches traversing the encounter in the same direction and l hole-stretches traversing 1+iǫ+b2 g 2b2+iǫ rgˆg2 b2(1+rggˆ)rgˆg2 the encounter in the opposite direction. Thus we have + + 1−rggˆ (1−rggˆ)2 (1−rggˆ)3 x = − 1+ilǫ+l2b2 / 1+iǫ+b2 lr˜l−1 in line with (cid:0) (cid:1) (cid:0) (cid:1) l ix(1+y) ix(1−y) (5). The powers of r˜are included in order to keep track + + =0 of the or(cid:0)der of the tre(cid:1)es.(cid:0) Now cons(cid:1)ider an l-encounter 2(1+x) eiφ/2+irgˆ 2(1+x) e−iφ/2+irgˆ touching S . According to the diagrammatic rules after 1 (cid:16) (cid:17) (cid:16) (cid:17) (9) extracting the factor (1+iǫ+b2)−n as in (5) the con- tribution of the encounter and the link connecting the andthesameequationwithgˆandg exchangedandφre- top encounter to the backbone is N /N. However we placedby −φ. Here we used N /N =x andintroduced S1 S N haveto include the phasefactors contributedby the An- the difference of the numbers of channels of the two su- dreev reflections. To evaluate this phase factor we look perconductors y =(N −N )/N such that y =0 cor- S1 S2 S at the l-encounter touching the superconductor as aris- responds to the case of equal numbers of channels and ing from an l-encounter inside the dot by sliding it into y =±1 to the case of just one superconductor. 7 In the case that the two superconductors provide the same number of channels (y=0) those two equations are the same implying gˆ = g and (9) is equivalent to an algebraicequationof7thorderin g. This increaseinthe order of the equation with respect to the same case for the density of states13 is due to the fact that in the case (a) of the density of states we had no normal leads. The contribution Pe of the side trees starting with an electron is then obtained by giving all trees the same weightbysettingr=1ing. Thecontributionoftheside (b) trees starting with a hole are then given by replacing φ by −φ in g or setting r = 1 in gˆ. After setting r = 1 FIG. 8. (a) A diagram contributing to X4 is split right af- andeliminating sayPh the contributionofthe side trees ter the first 4-encounter and decomposes into two separate starting with anelectron Pe is givenby a rather lengthy diagrams where the second one contributes to Teh. (b) To ij equation of in general 11th order which factorises in the sum over all diagrams starting with an l-encounter we can case y =0 such that Pe =Ph =P is given by remove a factor corresponding to the first encounter (and its side trees) and a sum again over thetransmission diagrams. −P7+(2iβ+iβx)P6 + −b2x+3+iǫx−b2+iǫ P5 + −iβx+2ib2β+2(cid:0)ǫβ+2ib2βx+2ǫβx−4iβ(cid:1)P4 We first order the sum over all diagrams contribut- (cid:0) +(−2iǫ−3−2iǫx(cid:1))P3 ing in leading order in the channel number with respect + 2ib2β+2ib2βx−2ǫβx−2ǫβ+2iβ−iβx P2 to the first encounter. Then the first summand is of (cid:0) + iǫx+b2+b2x+1+iǫ P +(cid:1)iβx=0. course the diagram corresponding to the upper left tra- jectory in Fig. 2. Next there are all diagrams whose (10) (cid:0) (cid:1) first encounter is a 2-encounter followed by all diagrams whose first encounter is a 3-encounter etc. Note that If no magnetic field is applied (b = 0) the equation may we also allow for the first encounter to touch the su- be factorised, and one has to solve an equation whose order is lowered by 2. perconductor or (if the first encounter is a 2-encounter IftheAndreevinterferometerconsistsoftwosupercon- and i=j) the normal lead. We denote the contribution ductorswiththesamenumbersofchannels(y=0)theside of the sum over all diagrams having an l-encounter as tree contributions only depend on β = cos(φ/2) rather their first encounter and contributing to Tieje in leading than on φ itself. Therefore in this case the side tree con- order in the number of channels by Xl. We may include tributions are symmetric in φ and the contribution of a the diagonal diagram without any encounter by setting sidetreestartingwithaholeisthesameasthatofaside X1 = NiNj/N. The transmission coefficients are then tree starting with an electron. In the most simple case given by Tieje = l≥1Xl. Now we fix l ≥ 2 and split of the absenceof a magnetic field, zerotemperature (i.e. all diagrams contributing to Xl right after the first en- P ǫ=0)andzerophase difference(10)reducesto asecond counterintoonepartconsistingofthefirstpathpairand order equation: thefirstencountertogetherwithitssidetreesandthere- maining part such as indicated in Fig. 8(b). Note that −P2+iP +iPx−x (11) the diagonal type path pair leaving the first encounter is completely included in the second part. Since the di- yielding agrammatic rules are multiplicative the contribution of a diagram is given by the product of the two parts and i P(0,x)= 1+x− 1+6x+x2 . (12) hence they all have a common factor which is given by 2 the first diagonal type link, the first encounter and the (cid:16) p (cid:17) Note that we take the solution satisfying P(0,0) = 0 side trees emerging from it. To sum over all diagrams since when there is no superconductor the correction of starting with an l-encounter we pull out this factor and leading order in the channel number has to be zero. are left with a sum over the transmission diagrams as depicted in Fig. 8(b). This sum runs over all possible diagramscontributing to Tee if the first encounter is left ij IV. TRANSMISSION COEFFICIENTS by an electron and to Teh if it is left by a hole. How- ij everinordertobe abletofully identify thesumoverthe We will now demonstrate how to calculate the trans- second parts as the transmissionwe have to reassignthe mission coefficients Tαβ for transmission from lead j to contributed number of channels N contributed by the ij j lead i while converting an α-type quasiparticle in a β- first path pair leaving lead j to the second part. Due to type one, using Tee as an example, as the evaluation of the twopossibilitiesofwhichtransmissioncoefficientthe ij the other transmission coefficients will be similar. remaining diagrams contribute to we split X into two l 8 We will now show that this indeed holds for all ‘di- agonal encounters’ entered by a diagonal type e-e∗ pair e h bystartingwithconsideringencountersnottouchingthe h e h e superconductor. Since an l-encounter connects 2l links e h e h to each other, each diagonal l-encounter, where 2 of the e* h* e* h* links belong to the backbone, provides in total (2l−2) side trees implying that if the number of ζ-side trees is h* e* even the number of ζ′-side trees is even too, or they are both odd. Furthermore each side tree provides an odd number of Andreev reflection and therefore a conversion of an electron into a hole or vice versa, since each of its FIG.9. Simpleexamplesforencounterstouchingasupercon- l-encounters is left by (2l−1) additional path pairs and ductor. The electron paths are shown green while the hole each path pair increases the number of Andreev reflec- paths are shown red. The solid lines belong to ζ while the tions by one (this is closely related to the fact that we ′ dashedonesbelongtoζ . Ifthequasiparticlesenteringanen- consider diagrams contributing at leading order in the counter touching the superconductor following an diagonal- numberofchannels). Thus,aslongasthefirstencounter type path pair the diagonal-type path pairs leaving it are doesnottouchthesuperconductor,the enteringelectron traversed by holes and vice versa. leaves the encounter as an electron if the number of side trees p˜built by ζ is even and as a hole if the number of side trees built by ζ is odd. parts However if the first diagonal l-encounter touches the X =AeTee+BeTeh, l l ij l ij superconductor the first side tree starts with a hole in- stead of an electron and is therefore left by an electron. where Ae is the contribution of the first e-e∗ pair and l Since the electron leaving the first side tree hits the su- the l-encounter the path pair enters together with all perconductor the second side tree again starts with a sidetreesandwiththeenteringandexitingquasiparticle hole. If one proceeds inductively one finds that every being the same. Be is the same but with the entering l side tree starts with a hole and is left by an electron and exiting quasiparticle being different. which after that undergoes again an Andreev reflection. The transmissioncoefficientsmaythereforebe written Therefore if the first encounter entered by an electron as touches the superconductor it is always left by a hole ∞ ∞ and we can view it as arising from an l-encounter with N N Tee = i j + AeTee+ BeTeh, (13a) an odd number p˜ of ζ-side trees slid into the supercon- ij N l ij l ij l=2 l=2 ductor as indicated in Fig. 10 and therefore contributes X X ∞ ∞ to Be. An l-encountermay then touch the superconduc- Tiejh = AhlTiejh+ BlhTieje. (13b) tor ilf the number of ζ-side trees p˜ is odd and the odd l=2 l=2 numberedζ-side trees, whicharethe side treestraversed X X by ζ after an odd number of traversals of the encounter, Ahl andBlh are the same as Ael and Ble, respectively, but aswellasthe oddnumberedζ′-sidetreeshavezerochar- with electrons and holes exchanged. Equation (13b) is acteristic (i.e. consist of just one link and one Andreev obtainedinthesamewayas(13a)butwiththeadditional reflection). Moreoverthe links of the odd numberedside condition that there is no diagram without any Andreev trees have to hit the same superconductor such that the reflection contributing to it since converting an electron channels can coincide. When sliding such an encounter to a hole requires at least one Andreev reflection and into the superconductor the channels at which the odd therefore one encounter. The formulae for Thh and The ij ij numberedside treeshit the superconductorcoincideand are the same but with e and h exchanged. thelinksvanish. Thereforebesidethediagonal-typepath Thenexttaskistofindoutwhatcausestheencounter pairs from such a diagonal l-encounter touching the su- whichisenteredbyanelectrontobeleftbyanelectronor perconductor p = (p˜−1)/2 even numbered ζ-side trees ahole. ThetrajectoriesinFig.3andtheircorresponding starting with a hole and [(2l−2−p˜)−1]/2= l−2−p diagrams in Fig. 4 indicate that, as long as the first en- evennumberedζ′-sidetrees,whichalsostartwith a hole counterdoesnottouchthesuperconductor,anencounter emerge. entered by an electron is left by an electron if the num- ber of side trees on each side of the diagonal backbone Thus if we denote the contribution of the first α-α∗ emergingfromthis encounter is even(suchas inthe dia- pair and of the first l-encounter inside the dot with p˜ gramsee3II andhe3V)andbyahole ifitis odd(suchas ζ-side trees by xα and the contribution of the Andreev l,p˜ inthediagramsee3IIIandhe3VII).Ifthefirstencounter reflections providedby the first l-encountertouching the however touches the superconductor the encounter is al- superconductorS createdbyslidinganl-encounterwith j ways left by a hole if it was entered by an electron. This originally p˜ ζ-side trees into the superconductor S by j is also indicated in Fig. 9 zα , we find l,p˜,j 9 Si Sj Si Sj (a) (b) FIG. 10. (a) An 3-encountermay touch thesuperconductor Si if theodd numberedside trees havezero characteristic and hit the same superconductor. The numberof Andreev reflections stays the same. If theencounter touches thesuperconductor an entering electron is converted into a hole. (c) A more complicated diagram with two diagonal encounters that may touch the superconductor. Note that additionally the fourth side tree may also touch the superconductor but this does not affect the diagonal encounterbut is instead included in theside treerecursion. l−1 Ae = xe (Pe)p Ph p (Pe)∗ l−1−p Ph ∗ l−1−p, (14a) l l,2p Xp=0 (cid:0) (cid:1) (cid:2) (cid:3) h(cid:0) (cid:1) i l−2 Be = xe (Pe)p+1 Ph p (Pe)∗ l−1−p Ph ∗ l−2−p+ ze Ph p Ph ∗ l−2−p , (14b) l  l,2p+1 l,2p+1,j  Xp=0 (cid:0) (cid:1) (cid:2) (cid:3) (cid:16)(cid:0) (cid:1) (cid:17) Xj (cid:0) (cid:1) h(cid:0) (cid:1) i l−1  Ah = xh Ph p(Pe)p Ph ∗ l−1−p (Pe)∗ l−1−p, (14c) l l,2p Xp=0 (cid:0) (cid:1) h(cid:0) (cid:1) i (cid:2) (cid:3) l−2 Bh = xh Ph p+1(Pe)p Ph ∗ l−1−p (Pe)∗ l−2−p+ zh (Pe)p (Pe)∗ l−2−p , (14d) l  l,2p+1 l,2p+1,j  Xp=0 (cid:0) (cid:1) h(cid:0) (cid:1) i (cid:0) (cid:1) Xj (cid:2) (cid:3)   where we have used that p˜ has to be even for Aα and an electron and the last one also does. Thus there are l thus replaced p˜=2p and odd for Bα with p˜=2p+1. (p˜+1)/2e-stretchestraversedinpositivedirectioninthe l The next and final step is to find the contribution of encounter. In the same way one finds that the number the encounters. For that we would like to recall the dia- of e∗-stretchesare (2l−2−p˜)/2 and [(2l−2−p˜)+1]/2, grammatic rule for an l-encounter traversedby trajecto- respectively. Since the diagonal path pair is traversed ries with energies ±ǫ and in presence of a magnetic field by ζ and ζ′ in the same direction the directions of the b from Ref. 23: e∗-paths are also positive. Since the holes retrace the electron paths their directions in the encounters is nega- • An l-encounter inside the dot contributes a factor tive. Thusinbothcasesonefindsthatη =µ=(p˜−l+1). −N 1+ηiǫ+µ2b2 . So Here η is(cid:0)the difference (cid:1)between the number of traver- sals of e-stretches and the number of traversals of e∗- xl,p˜=− 1+(p˜−l+1)iǫ+(p˜−l+1)2b2 . (15) stretches and µ is the difference between the number of h i ζ-stretches traversedin a certaindirection and the num- For the contributionz˜α whicharises by sliding an l- l,p,j ber of ζ′-stretches traversedin the same direction. Since encounterintothesuperconductor15,asshowninFig.10 every electron path of the side tree is retraced by a hole we remember that the number p˜ of ζ-side trees emerg- every second stretch connected to a ζ-side tree is an e- ing from it is odd and the odd numbered ζ-side trees as stretchandtheyarealltraversedinthesamedirectionwe well as the odd numbered ζ′-side trees consist of only choose arbitrarily as ‘positive’. Therefore if the number one path pair and one Andreev reflection (i.e. they have ofζ-sidetreesiseventhenumberofe-stretchestraversed zero characteristic). Moreoverthe Andreev reflections of inpositive directionissimply p˜/2. Ifp˜is oddwe haveto the odd numbered side trees have to be all at the same account for the fact that the first ζ-side tree starts with superconductor. The contribution of the encounter it- 10 self and the first path pair is then NSj/N. However we eachprovideafactor−ie−iφi. HenceintotaltheAndreev also include the factors contributed by the Andreev re- reflections of the odd numbered ζ-side trees provide a flections in zα , too, which are stated in Sect. III. As factor (−i)p+1e−i(p+1)φi. Analogously the Andreev re- l,p˜,j forthesidetrees,thesephasefactorsmaybedetermined flections of the odd numbered ζ′-side trees contribute a by looking at the odd numbered side trees before sliding factor il−p−1ei(l−p−1)φi. Thus in the case of two super- the encounter into the superconductor since the number conductorswithphasesφ =−φ =φ/2thephasefactor 1 2 ofAndreevreflectionsofthe ζ- andζ′-trajectorycannot included in ze is given by (−i)pil−p−2e−i(2p−l+2)φ/2. l,p˜,1 changewhenslidingtheencounterintothesuperconduc- We thus have tor. Consider the p+1 = (p˜+1)/2 odd numbered side N trees which have zero characteristic and hit say Si: The zl,p˜,1 = S1il−p˜−1e−i(p˜−l+1)φ/2. (16) N Andreev reflections provided by these side trees convert an electron into a hole and thus the Andreev reflections For ze we have to exchange φ ↔ −φ and replace N l,p˜,2 S1 byN . Moreoverzh =ze | . Thereforewehave S2 l,p˜,j l,p˜,j φ→−φ l−1 Ae =− 1+i(2p−l+1)ǫ+(2p−l+1)2b2 (Pe)p Ph p (Pe)∗ l−p−1 Ph ∗ l−p−1 (17a) l Xp=0h i (cid:0) (cid:1) (cid:2) (cid:3) h(cid:0) (cid:1) i l−2 Be =− 1+i(2p−l+2)ǫ+(2p−l+2)2b2 (Pe)p+1 Ph p (Pe)∗ l−p−1 Ph ∗ l−p−2 l " Xp=0 (cid:16) (cid:17) (cid:0) (cid:1) (cid:2) (cid:3) h(cid:0) (cid:1) i x(1+y)e−i(2p−l+2)φ/2 −iPh p i Ph ∗ l−p−2 x(1−y)ei(2p−l+2)φ/2 −iPh p i Ph ∗ l−p−2 − − . 2(1(cid:0)+x) (cid:1) (cid:16) (cid:0) (cid:1) (cid:17) 2(1(cid:0)+x)(cid:1) (cid:16) (cid:0) (cid:1) (cid:17) # (17b) where we again used y = (N − N )/N . The case The transmission coefficients necessary for calculating S1 S2 S N = N is then obtained by setting y = 0 while the conductance may be calculated by evaluating the S1 S2 the case of just one superconducting lead corresponds side tree contribution by solving (9), inserting this into to y = ±1. Since exchanging electrons and holes corre- (17a,b) and performing the sums and finally inserting sponds to replacing φ by −φ, Ah and Bh are obtained into(13a,b)andsolvingforthe transmissioncoefficients. l l by the same formulae but with φ replacedby −φ includ- ing an exchange Pe ↔Ph. The sums may be performed using geometric series and yield our main result. Along with (13a) and (13b) it contains all rhe diagrams, and their semiclassical contributions, generated recursively. V. CONDUCTANCE WITH Note that if the numbers of channels of the supercon- SUPERCONDUCTING ISLANDS ducting leads are equal the symmetry of P towards the phase implies that Aα and Bα are symmetric in φ yield- l l ing Ae = Ah and Be = Bh and thus Tee = Thh and We now evaluatethe conductanceofAndreev billiards l l l l ij ij The =Teh. withtwo normalleads. We firstconsidera chaoticquan- ij ij tum dot coupled to two normal conducting leads and Therefore we now have all the necessary utilities to one or two isolated superconductors with equal number calculate the conductance of Andreev billiards with two of channels as shown in Fig. 1(c). The chemical poten- superconducting islands. When the incoming and out- tial of the superconducting lead is then adjusted by the going lead are the same, i = j, and the first encounter dotsuchthatthe netcurrentinthe superconductorvan- is an 2-encounter this encounter may enter the lead. In ishes. The dimensionless conductance g = π~I/(e2V) this case however the encounter simply contributes N i and the diagrams consist of one ζ-side tree and one ζ′- with I the current and V the voltage drop between the two normal leads, in this case is given at zero tempera- sidetree. Thecontributionofthesediagramsistherefore ture by3 simply δijNi|Pe|2 if the dot is entered by an electron, (18a) TheThe−TheThe g =Tee+The+2 11 22 21 12 . (19) δijNi|Ph|2 if the dot is entered by a hole. (18b) 21 21 T1h1e+T2h2e+T2h1e+T1h2e