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Quantum interference
� Aharonov-Bohm effect� ”Aharonov-Bohm” ring� Basic properties of
superconductors� Andreev reflection� Andreev conductance
� Multi-terminal conductors� Scattering matrix� Multi-terminal Landauer formula� Example: Beam splitter� Quantum interference� Example: Double barrier
Multi-terminal conductors
4
5
1I
3I
2I1V
2V
2V
We want to express currentsin terms of voltages
Scattering region
Reservoir
Ideal lead
Gate
Linear regime
4
5
1I
3I
2I1V
2V
2Vk kl l
lI G V=∑
klG - conductance matrix
0 0k klk k
I G= ⇒ =∑ ∑Current conservation:
No current if the potentials at all leads are shifted by an equal amount
0kll
G =∑
Two-terminal conductors
1 1
2 2
I VG GI VG G
− = −
1V2V
1I 2IConductance matrix: Only one independent element
Scattering matrix
;m nsβ α
-amplitude of propagation from the terminal α,transport channel n to the terminal β, transportchannel m....
...
...
1N1 scattering region
s11,12s12,12s13,12
s 21,12
s 22,12
s 23,12
s31,12
s32,12
s33,12
Properties of scattering matrix
;n msα α - reflection back from α.
; ,n msα β α β≠ - transmission from β to α.
Unitarity: �� � 1s s =*
; ;n l n m lmn
s sα γ α β βγα
δ δ=∑Expresses conservation of the
number of particles
Symmetry with respect to the time reversal:
; ;( ) ( )n m m ns B s Bα β β α= −
Landauer formula
( / ) ( )QI G e dE P f Eα αβ ββ
= − ∑∫Relates currents to voltages by means of the scattering matrix
(trace is taken over transport channels)
Current conservation: follows from unitarityNo current at equilibrium (all voltages are the same)-
also follows from unitarity
��P Tr s sαβ αβ αβ αβδ = − Probability of transmission from α to β:
Fool-proof check:
Landauer formula
��QG G Tr s sαβ αβ αβ αβδ = − − Linear regime:
Relation to the two-terminal formula: α,β=L,R
( )� ��LR Q LR LR QG G Tr s s G Tr t t = =
Time-reversal symmetry: ( ) ( )G B G Bαβ βα= −
(In accordance with Onsager symmetry relations)
Example: Beam splitter
3
1
2
a new scattering element
1/ 2 1/ 2 1/ 2
� 1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 0Bs
−
= −
(will use later for quantum interference!)
Example: Beam splitter
3
1
2
Conductance matrix:
3/ 4 1/ 4 1/ 21/ 4 3/ 4 1/ 21/ 2 1/ 2 1
QG Gαβ
− = − −
Example:
1 2 3; 0V V V V= = =QG V
/ 2QG V
/ 2QG V
Quantum interference
Repeat basic quantum mechanics: Double-slit experiment
A B
1
2
Probability of propagation from A to B:
21 2
2 2 *1 2 1 22 Re[ ]
ABP A A
A A A A
= +
= + +
1P 2P
Classical
Interference termQ-mechanical!
Two barriers
L R
Try to guess the result:
Resistances are added - Classically
Is quantum interference important?
1
totr
tott
Need to learn how to combine scattering matrices.
Two barriers
L R
1
totr
totta
b
c
d
'L L
L L
r tt r '
R R
R R
r tt r
00
i
i
ee
ϕ
ϕ
Scattering matrix of the left barrier
Scattering matrix of the right barrier
Propagation between the barriers
kdϕ =
Two barriers
L R
1
totr
totta
ide ϕ
iae ϕ
d
'1ot
iL
L
t L
L
ra
r tr dt e ϕ
=
' 0tot
Ri
R
R R
d aetrt
rt
ϕ =
Two barriers
L R
1
totr
tott
2( )1 2 cos
L Rtot
L R L R
T TT E tR R R R χ
= =+ −
χ � phase accumulated during the round-trip
21 'L R
tot iL R
t ttr r e ϕ=
−
Why does it have anything to do with the double-slit experiment?
Two barriers
21 'L R
tot iL R
t ttr r e ϕ=
−
Process Amplitude Probability
iL Rt t e ϕ
L RT T
3' iL R L Rt t r r e ϕ
L R L RT T R R� � �
Sum of amplitudes:
Sum of probabilities: 1L R
clL R
T TTR R
=−
Two barriers
( )1 2 cos
L R
L R L R
T TT ER R R R χ
=+ −
Quantum result: transmission coefficient depends on energy(Classical result: It does not)
( )T E
E
Take 1L RT T= !
Valley: 2( ) LT E T∝Peak: max 1 (@ 0)T χ= =
Resonant tunneling
ddE
χ τ=
τ � time for the round-trip
Aharonov-Bohm effect
Φ
1
2
Two trajectories enclosing magnetic flux
Phase: ikxWith the vector potential:
( )ek k A xc
→ −" " "
#
1,2 1,21,2
ekL A dlc
θ = − ⋅∫""
#
Aharonov-Bohm effect
Φ
1
2
2 1 2 1( ) ek L L A dlc
θ θ− = − − ⋅∫""
# $
0
2e eA dlc c
πΦ⋅ = Φ =Φ∫
""# #$
02 c
eπΦ = # - flux quantum
All quantities are periodic in Φ, even if there is no magneticfield at the trajectories.
“Aharonov-Bohm” ring
1 tΦ1 2
1b
1d
1a
1c
2b2a
2c2d
0
/ 22 /kLχ
ϕ π== Φ Φ
r
0 1/ 2 1/ 2
1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 1/ 2
− −
1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 0
−
−
/ 2
/ 2
00
i i
i i
ee
χ ϕ
χ ϕ
+
−
/ 2
/ 2
00
i i
i i
ee
χ ϕ
χ ϕ
−
+
Left beam splitter
Lower arm
Right beam splitter
Upper arm
“Aharonov-Bohm” ring
1 tΦ1 2
1b
1d
1a
1c
2b2a
2c2d
1 1
1 1
0 1/ 2 1/ 2 11/ 2 1/ 2 1/ 2
1/ 2 1/ 2 1/ 2
rb ad c
= − −
0
/ 22 /kLχ
ϕ π== Φ Φ
r
“Aharonov-Bohm” ring
1 tΦ
2
2 2
2sin (1 cos )sin 2 [cos2 (1/ 2)(1 cos )]
T χ ϕχ χ ϕ
+=+ − +
02 /ϕ π= Φ Φ
2
(1 cos )2[3 cos ]2
T ϕϕ
πχ +→ =+
= Flux dependent!
“Aharonov-Bohm” ring
02 /ϕ π= Φ Φ2
(1 cos )2[3 cos ]2
T ϕϕ
πχ +→ =+
=
“Aharonov-Bohm” ring
1 t
Φ 02 /ϕ π= Φ Φ
2 20 1 1 2 2
i i i it e e e eϕ ϕ ϕ ϕα α α α α− −− −= + + + + +…
Classical
One turn clockwiseOne turn
counterclockwise
Two turns