Conductance and Thermopower of Ballistic Andreev Cavities

1.2 MB·English

by  Thomas Engl

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