Statistics of quantum transport with orthogonal or …...IntroductionResultsMethodConclusion...

Introduction Results Method Conclusion

Statistics of quantum transport with orthogonalor symplectic symmetry

Nick SimmIn collaboration with Francesco Mezzadri

School of Mathematics, Bristol University

Brunel-Bielefeld workshop on RMTDecember 16, 2011

The Quantum Transport Problem

Two lead scattering problem - S relates amplitudes ofincoming and outgoing states.

Boundary conditions lead to n propagating modes in eachlead.

|cin| = |cout | ⇒ S is Unitary (flux conservation).

Chaotic classical dynamics inside dot.

Structure of the S-matrix. Physical Observables.

Division of states into left and right cLin, cR

in , cLout , cR

out gives blockdecomposition

[r t ′

t r ′

], (2n × 2n)

r and t : reflection and transmission matrices.

Landauer formalism: Scattering observables cast in terms ofeigenvalues T1, . . . ,Tn of tt†.

Flow of current characterized by Conductance G = Tr[tt†]

Current fluctuations ⇒ Shot noise: P = Tr[tt†(1− tt†)]

Chaotic classical dynamics inside cavity. Search for universalproperties of G and P.

in , cLout , cR

[r t ′

t r ′

], (2n × 2n)

r and t : reflection and transmission matrices.Landauer formalism: Scattering observables cast in terms ofeigenvalues T1, . . . ,Tn of tt†.

in , cLout , cR

[r t ′

t r ′

], (2n × 2n)

in , cLout , cR

[r t ′

t r ′

], (2n × 2n)

in , cLout , cR

[r t ′

t r ′

], (2n × 2n)

Symmetry and Statistics

Chaotic classical dynamics → Random Matrix Hypothesis

ChooseS ∈ U(2n) with Haar measure constrained by

TRI - S is unitary and symmetric (COE : β = 1)

TRI and spin - S = σ2STσ2 is unitary and self-dual (CSE :β = 4)

No other symmetries (CUE : β = 2)

The distribution of the Tj ’s is (Jalabert et al. ’94) (0 ≤ Tj ≤ 1)

P(T1, . . . ,Tn) =1

n∏j=1

Tβ/2−1j |∆(T)|β

where ∆(T ) :=∏

1≤p<q≤n(Tq − Tp) - Vandermonde Determinant

Chaotic classical dynamics → Random Matrix Hypothesis ChooseS ∈ U(2n) with Haar measure constrained by

P(T1, . . . ,Tn) =1

n∏j=1

Tβ/2−1j |∆(T)|β

where ∆(T ) :=∏

P(T1, . . . ,Tn) =1

n∏j=1

Tβ/2−1j |∆(T)|β

where ∆(T ) :=∏

P(T1, . . . ,Tn) =1

n∏j=1

Tβ/2−1j |∆(T)|β

where ∆(T ) :=∏

The distribution of the Tj ’s

is (Jalabert et al. ’94) (0 ≤ Tj ≤ 1)

P(T1, . . . ,Tn) =1

n∏j=1

Tβ/2−1j |∆(T)|β

where ∆(T ) :=∏

P(T1, . . . ,Tn) =1

n∏j=1

Tβ/2−1j |∆(T)|β

where ∆(T ) :=∏

Moments of Conductance and Shot Noise

Using the j.p.d.f. it can be shown that (Beenakker et al. early 90s):

E(G ) =n2

2n − 1 + 2β

2n→∞

Var(G ) =2

n2(n − 1 + 2/β)2

(2n − 2 + 2/β)(2n − 1 + 2β)2(2n − 1 + 4/β)→ 1

Semiclassical method of chaotic quantum transportin agreement with RMT to all orders! (November 2011 preprint ofKuipers & Berkolaiko, see also Novaes 2011).But besides this agreement, how can we actually calculatehigher order moments/cumulants?

E(G ) =n2

2n − 1 + 2β

2n→∞

Var(G ) =2

n2(n − 1 + 2/β)2

(2n − 2 + 2/β)(2n − 1 + 2β)2(2n − 1 + 4/β)→ 1

E(G ) =n2

2n − 1 + 2β

2n→∞

Var(G ) =2

n2(n − 1 + 2/β)2

(2n − 2 + 2/β)(2n − 1 + 2β)2(2n − 1 + 4/β)→ 1

E(G ) =n2

2n − 1 + 2β

2n→∞

Var(G ) =2

n2(n − 1 + 2/β)2

(2n − 2 + 2/β)(2n − 1 + 2β)2(2n − 1 + 4/β)→ 1

E(G ) =n2

2n − 1 + 2β

2n→∞

Var(G ) =2

n2(n − 1 + 2/β)2

(2n − 2 + 2/β)(2n − 1 + 2β)2(2n − 1 + 4/β)

E(G ) =n2

2n − 1 + 2β

2n→∞

Var(G ) =2

n2(n − 1 + 2/β)2

(2n − 2 + 2/β)(2n − 1 + 2β)2(2n − 1 + 4/β)→ 1

E(G ) =n2

2n − 1 + 2β

2n→∞

Var(G ) =2

n2(n − 1 + 2/β)2

(2n − 2 + 2/β)(2n − 1 + 2β)2(2n − 1 + 4/β)→ 1

Semiclassical method of chaotic quantum transportin agreement with RMT to all orders! (November 2011 preprint ofKuipers & Berkolaiko, see also Novaes 2011).

But besides this agreement, how can we actually calculatehigher order moments/cumulants?

E(G ) =n2

2n − 1 + 2β

2n→∞

Var(G ) =2

n2(n − 1 + 2/β)2

(2n − 2 + 2/β)(2n − 1 + 2β)2(2n − 1 + 4/β)→ 1

Approaches to higher moments

Two key approaches to calculation of higher cumulants/momentscurrently exist.

1 The Selberg Integral Approach:

Based on the mathematicaltheory surrounding Selberg’s integral. Works for all β, buthard to go beyond 4th moment. Khoruzhenko, Savin,Sommers, Wieczorek (2006-2009).

2 The Integrable Systems Approach: Gives recursion relationsfor higher cumulants. Main problem is that it was onlyapplicable for β = 2. V. Osipov and Kanzieper (2008/2009).

We extend approach 2 considerably:

We apply it to other transport statistics, and discover newrelations to Painleve equations.Crucially, we describe how to adapt it for the troublesomeβ = 1, 4 symmetry classes.We resolve some conjectures which were guessed usingapproach 1

1 The Selberg Integral Approach: Based on the mathematicaltheory surrounding Selberg’s integral. Works for all β, buthard to go beyond 4th moment. Khoruzhenko, Savin,Sommers, Wieczorek (2006-2009).

2 The Integrable Systems Approach:

Gives recursion relationsfor higher cumulants. Main problem is that it was onlyapplicable for β = 2. V. Osipov and Kanzieper (2008/2009).

We apply it to other transport statistics, and discover newrelations to Painleve equations.

Crucially, we describe how to adapt it for the troublesomeβ = 1, 4 symmetry classes.We resolve some conjectures which were guessed usingapproach 1

We apply it to other transport statistics, and discover newrelations to Painleve equations.Crucially, we describe how to adapt it for the troublesomeβ = 1, 4 symmetry classes.

We resolve some conjectures which were guessed usingapproach 1

Conjectures of Khoruzhenko, Savin, Sommers 2009

This work was originally motivated by the following conjectures forlimiting statistics of conductance and shot noise:

Conjecture

Cumulants of conductance, l > 2, β = 1

limn→∞

nl−1κl,0 = − (l − 2)!

), l even.

limn→∞

nlκl,0 =(l − 1)!

4(2)l, l odd.

Cumulants of shot noise, k > 2, β = 1

limn→∞

nk−1κ0,k = − (k − 2)!

), k even.

limn→∞

nkκ0,k =(k − 1)!

8(4)k, k odd.

Conjecture

limn→∞

nl−1κl,0 = − (l − 2)!

), l even.

limn→∞

nlκl,0 =(l − 1)!

4(2)l, l odd.

limn→∞

nk−1κ0,k = − (k − 2)!

), k even.

limn→∞

nkκ0,k =(k − 1)!

8(4)k, k odd.

Conjecture

limn→∞

nl−1κl,0 = − (l − 2)!

), l even.

limn→∞

nlκl,0 =(l − 1)!

4(2)l, l odd.

limn→∞

nk−1κ0,k = − (k − 2)!

), k even.

limn→∞

nkκ0,k =(k − 1)!

8(4)k, k odd.

Conjecture

limn→∞

nl−1κl,0 = − (l − 2)!

), l even.

limn→∞

nlκl,0 =(l − 1)!

4(2)l, l odd.

limn→∞

nk−1κ0,k = − (k − 2)!

), k even.

limn→∞

nkκ0,k =(k − 1)!

8(4)k, k odd.

Conjecture

limn→∞

nl−1κl,0 = − (l − 2)!

), l even.

limn→∞

nlκl,0 =(l − 1)!

4(2)l, l odd.

limn→∞

nk−1κ0,k = − (k − 2)!

), k even.

limn→∞

nkκ0,k =(k − 1)!

8(4)k, k odd.

Conjecture

limn→∞

nl−1κl,0 = − (l − 2)!

), l even.

limn→∞

nlκl,0 =(l − 1)!

4(2)l, l odd.

limn→∞

nk−1κ0,k = − (k − 2)!

), k even.

limn→∞

nkκ0,k =(k − 1)!

8(4)k, k odd.

Conjecture

limn→∞

nl−1κl,0 = − (l − 2)!

), l even.

limn→∞

nlκl,0 =(l − 1)!

4(2)l, l odd.

limn→∞

nk−1κ0,k = − (k − 2)!

), k even.

limn→∞

nkκ0,k =(k − 1)!

8(4)k, k odd.

Proof of Conjectures

Our main contribution is the following Theorem, interpolatingbetween the previous conjectures:

Theorem

Let β ∈ {1, 2, 4}. We have,

limn→∞

nl+kκl,k =

2− 1

)2(l + k − 1)!

β(2β)l(4β)k, l odd

limn→∞

nl+k−1κl,k =

2− 1

)(l + k − 2)!

(4β)l(−4β)k

k∑j=0

(2j + l

j + l/2

)(−2)−j , l even

Proves three of the conjectures in a more general setting.

Limiting distribution: G and P → independent Gaussians.

Qualitatively: Convergence slower for β = 1, 4.

Theorem

Let β ∈ {1, 2, 4}. We have,

limn→∞

nl+kκl,k =

2− 1

)2(l + k − 1)!

limn→∞

nl+k−1κl,k =

2− 1

)(l + k − 2)!

(4β)l(−4β)k

k∑j=0

(2j + l

j + l/2

)(−2)−j , l even

Theorem

Let β ∈ {1, 2, 4}. We have,

limn→∞

nl+kκl,k =

2− 1

)2(l + k − 1)!

limn→∞

nl+k−1κl,k =

2− 1

)(l + k − 2)!

(4β)l(−4β)k

k∑j=0

(2j + l

j + l/2

)(−2)−j , l even

Theorem

Let β ∈ {1, 2, 4}. We have,

limn→∞

nl+kκl,k =

2− 1

)2(l + k − 1)!

limn→∞

nl+k−1κl,k =

2− 1

)(l + k − 2)!

(4β)l(−4β)k

k∑j=0

(2j + l

j + l/2

)(−2)−j , l even

Theorem

Let β ∈ {1, 2, 4}. We have,

limn→∞

nl+kκl,k =

2− 1

)2(l + k − 1)!

limn→∞

nl+k−1κl,k =

2− 1

)(l + k − 2)!

(4β)l(−4β)k

k∑j=0

(2j + l

j + l/2

)(−2)−j , l even

Random matrix approach to time delay

Time delay in quantum mechanics: Smith (1960) introducedthe Hermitian matrix Q = −i~S† dS

Eigenvalues λ1, . . . , λn of Q - proper delay times.

Their average τW = 1n

∑nj=1 λj - Wigner time delay.

What if S ∼ CβE ? Brouwer, Frahm, Beenakker (1997) showedthat

P(λ1, . . . , λn) ∝n∏

λ−bj exp

(− βn

)|∆(λ)|β

Here, b = −3βn/2− 2 + β. It is an inverted Wishart or Laguerretype distribution.

P(λ1, . . . , λn) ∝n∏

λ−bj exp

(− βn

)|∆(λ)|β

P(λ1, . . . , λn) ∝n∏

λ−bj exp

(− βn

)|∆(λ)|β

P(λ1, . . . , λn) ∝n∏

λ−bj exp

(− βn

)|∆(λ)|β

P(λ1, . . . , λn) ∝n∏

λ−bj exp

(− βn

)|∆(λ)|β

P(λ1, . . . , λn) ∝n∏

λ−bj exp

(− βn

)|∆(λ)|β

Our Results: Statistics of τW

Exact formulae for first few cumulants:

κ1 = 1, κ2 =4

(n + 1)(nβ − 2), κ3 =

(n + 1)(n + 2)(nβ − 2)(nβ − 4)

Further:

New inductive differential equations for log E(ezτW ) ifβ = 1, 4. If β = 2 we get a non-linear ODE related toPainleve III (Chen and Its 2009).

Efficient recursion relations describing all higher cumulants forfinite n and β ∈ {1, 2, 4}e.g. when β = 2, we get (pl = limn→∞ n2l−2κl/(l − 1)!)

(l + 1)pl+1 = 2(2l − 1)pl + 2l−1∑i=0

(3i + 1)(l − i)pi+1pl−i

with p1 = 1. Solutions are all integers. Why?

κ1 = 1, κ2 =4

(n + 1)(nβ − 2), κ3 =

(n + 1)(n + 2)(nβ − 2)(nβ − 4)

Further:

(l + 1)pl+1 = 2(2l − 1)pl + 2l−1∑i=0

(3i + 1)(l − i)pi+1pl−i

κ1 = 1, κ2 =4

(n + 1)(nβ − 2), κ3 =

(n + 1)(n + 2)(nβ − 2)(nβ − 4)

Further:

Efficient recursion relations describing all higher cumulants forfinite n and β ∈ {1, 2, 4}

e.g. when β = 2, we get (pl = limn→∞ n2l−2κl/(l − 1)!)

(l + 1)pl+1 = 2(2l − 1)pl + 2l−1∑i=0

(3i + 1)(l − i)pi+1pl−i

κ1 = 1, κ2 =4

(n + 1)(nβ − 2), κ3 =

(n + 1)(n + 2)(nβ − 2)(nβ − 4)

Further:

(l + 1)pl+1 = 2(2l − 1)pl + 2l−1∑i=0

(3i + 1)(l − i)pi+1pl−i

κ1 = 1, κ2 =4

(n + 1)(nβ − 2), κ3 =

(n + 1)(n + 2)(nβ − 2)(nβ − 4)

Further:

(l + 1)pl+1 = 2(2l − 1)pl + 2l−1∑i=0

(3i + 1)(l − i)pi+1pl−i

κ1 = 1, κ2 =4

(n + 1)(nβ − 2), κ3 =

(n + 1)(n + 2)(nβ − 2)(nβ − 4)

Further:

(l + 1)pl+1 = 2(2l − 1)pl + 2l−1∑i=0

(3i + 1)(l − i)pi+1pl−i

Outline of Method - Integrable Systems (β = 1)

Based on work of Adler et al. (90s) and Morozov et al. (late 80s).

1 Integral representation for MGF, MX (z), of transport quantity.

2 Define τ -function τn(t) where t = (t1, t2, . . . , ) andτn(0) = MX (z).

3 τn(t) satisfies universal PDEs of ‘Pfaffian’ type.

4 τn(t) obeys non-universal Virasoro constraints.

5 Combine (3) & (4) at t = 0 to give differential-difference eqnfor MX (z).

6 Taylor expanding MX (z) finally yields desired recurrence oncumulants of X .

Integral Representation (Conductance, β = 1)

The moment generating function of conductance is:

MG (z) := E(

∫[0,1]n

n∏j=1

ρz(Tj)|∆(T)|dT

The weight or measure is ρz(T ) = T−1/2ezT .

How can we calculate the cumulants from this multipleintegral?

The basic idea: Find differential equations for log MG (z).

Then, expanding log MG (z) =∑∞

l=1 κlzl/l! we get recursion

relations for cumulants.

MG (z) := E(

∫[0,1]n

n∏j=1

ρz(Tj)|∆(T)|dT

MG (z) := E(

∫[0,1]n

n∏j=1

ρz(Tj)|∆(T)|dT

MG (z) := E(

∫[0,1]n

n∏j=1

ρz(Tj)|∆(T)|dT

MG (z) := E(

∫[0,1]n

n∏j=1

ρz(Tj)|∆(T)|dT

Tau-Functions of Quantum Transport

The first step: We introduce a τ -function: t = (t1, t2, t3, . . .).

τn(t) =1

∫[0,1]n

n∏j=1

ρz(Tj) exp

( ∞∑i=1

)|∆(T)|dT

The parameters tj turn our moment generating function into anintegrable system. ASvM (1999,2000) discovered its law ofevolution:(

∂t41

+ 3∂2

∂t22

− 4∂2

∂t1∂t3

)log τn(t) + 6

∂t21

log τn(t)

= 12τn−2(t)τn+2(t)

(τn(t))2.

We need the projection of this equation at t = 0. This isperformed with help of Virasoro constraints.

τn(t) =1

∫[0,1]n

n∏j=1

ρz(Tj) exp

( ∞∑i=1

)|∆(T)|dT

∂t41

+ 3∂2

∂t22

− 4∂2

∂t1∂t3

)log τn(t) + 6

∂t21

log τn(t)

= 12τn−2(t)τn+2(t)

(τn(t))2.

τn(t) =1

∫[0,1]n

n∏j=1

ρz(Tj) exp

( ∞∑i=1

)|∆(T)|dT

The parameters tj turn our moment generating function into anintegrable system.

ASvM (1999,2000) discovered its law ofevolution:(

∂t41

+ 3∂2

∂t22

− 4∂2

∂t1∂t3

)log τn(t) + 6

∂t21

log τn(t)

= 12τn−2(t)τn+2(t)

(τn(t))2.

τn(t) =1

∫[0,1]n

n∏j=1

ρz(Tj) exp

( ∞∑i=1

)|∆(T)|dT

∂t41

+ 3∂2

∂t22

− 4∂2

∂t1∂t3

)log τn(t) + 6

∂t21

log τn(t)

= 12τn−2(t)τn+2(t)

(τn(t))2.

τn(t) =1

∫[0,1]n

n∏j=1

ρz(Tj) exp

( ∞∑i=1

)|∆(T)|dT

∂t41

+ 3∂2

∂t22

− 4∂2

∂t1∂t3

)log τn(t) + 6

∂t21

log τn(t)

= 12τn−2(t)τn+2(t)

(τn(t))2.

Virasoro Constraints

They are a sequence of linear differential operators Vp with theimportant property:

Vpτn(t) ≡ 0 p = 1, 2, 3, . . .

They arise due to certain symmetries of the τ -function τn(t)

1 Change variables Tj → εf (Tj)T p+1j where

ddT log ρ(T ) = −g(T )

f (T ) (rational log-derivative)

2 Compute first order variation of the integral in ε (zero)

3 Recognise the appearance of the Virasoro operators

Lp =∞∑

ktk∂

∂tk+p+

p−1∑k=1

∂tk∂tp−k

satisfying the Virasoro Algebra [Lp,Lq] = (p − q)Lp+q.

Vpτn(t) ≡ 0 p = 1, 2, 3, . . .

Lp =∞∑

ktk∂

∂tk+p+

p−1∑k=1

∂tk∂tp−k

Vpτn(t) ≡ 0 p = 1, 2, 3, . . .

Lp =∞∑

ktk∂

∂tk+p+

p−1∑k=1

∂tk∂tp−k

Vpτn(t) ≡ 0 p = 1, 2, 3, . . .

Lp =∞∑

ktk∂

∂tk+p+

p−1∑k=1

∂tk∂tp−k

What do these operators look like?

The sequence of operators Vp depends on the details of themeasure ρz(T ). The first couple:

V1 =∞∑

∂tk+1− ∂

)+ (z + n + 1/2)

∂t1− z

∂t2− n2/2

V2 =∞∑

∂tk+2− ∂

∂tk+1

∂t21

− (n − 1/2)∂

+ (z + n + 1)∂

∂t2− z

which both satisfy Vpτn(t) ≡ 0. Setting t = 0 gives a system ofsimultaneous equations for the partial derivatives (F = log τn(t))(

∂t2,∂F

∂t3,∂2F

∂t1∂t2,∂2F

∂t1∂t3,∂2F

∂t22

) ∣∣∣∣t=0

V1 =∞∑

∂tk+1− ∂

)+ (z + n + 1/2)

∂t1− z

∂t2− n2/2

V2 =∞∑

∂tk+2− ∂

∂tk+1

∂t21

− (n − 1/2)∂

+ (z + n + 1)∂

∂t2− z

∂t2,∂F

∂t3,∂2F

∂t1∂t2,∂2F

∂t1∂t3,∂2F

∂t22

) ∣∣∣∣t=0

V1 =∞∑

∂tk+1− ∂

)+ (z + n + 1/2)

∂t1− z

∂t2− n2/2

V2 =∞∑

∂tk+2− ∂

∂tk+1

∂t21

− (n − 1/2)∂

+ (z + n + 1)∂

∂t2− z

which both satisfy Vpτn(t) ≡ 0.

Setting t = 0 gives a system ofsimultaneous equations for the partial derivatives (F = log τn(t))(

∂t2,∂F

∂t3,∂2F

∂t1∂t2,∂2F

∂t1∂t3,∂2F

∂t22

) ∣∣∣∣t=0

V1 =∞∑

∂tk+1− ∂

)+ (z + n + 1/2)

∂t1− z

∂t2− n2/2

V2 =∞∑

∂tk+2− ∂

∂tk+1

∂t21

− (n − 1/2)∂

+ (z + n + 1)∂

∂t2− z

∂t2,∂F

∂t3,∂2F

∂t1∂t2,∂2F

∂t1∂t3,∂2F

∂t22

) ∣∣∣∣t=0

Cumulant Recursion Formula

Finally, we get an exact recurrence relation for conductancecumulants with β = 1

A(l , n)κl+1 +l−1∑i=0

κi+1κl−iB(i , l) + C(l , n)κl + D(l , n)κl−1

= E(l , n)µl−3

µl =∑

∏B∈π

(κn→n−2|B| + κn→n+2

|B| − 2κ|B|

)over all

partitions π of {1, 2, . . . , l}.The RHS appears due to the non-zero RHS in the Pfaff-KPequation, and vanishes in the simpler case β = 2.

Large-n limit selects only dominant partition π∗ - results inlinear, homogeneous recurrence!

A(l , n)κl+1 +l−1∑i=0

= E(l , n)µl−3

µl =∑

∏B∈π

(κn→n−2|B| + κn→n+2

|B| − 2κ|B|

)over all

partitions π of {1, 2, . . . , l}.

The RHS appears due to the non-zero RHS in the Pfaff-KPequation, and vanishes in the simpler case β = 2.

A(l , n)κl+1 +l−1∑i=0

= E(l , n)µl−3

µl =∑

∏B∈π

(κn→n−2|B| + κn→n+2

|B| − 2κ|B|

)over all

A(l , n)κl+1 +l−1∑i=0

= E(l , n)µl−3

µl =∑

∏B∈π

(κn→n−2|B| + κn→n+2

|B| − 2κ|B|

)over all

A(l , n)κl+1 +l−1∑i=0

= E(l , n)µl−3

µl =∑

∏B∈π

(κn→n−2|B| + κn→n+2

|B| − 2κ|B|

)over all

Conclusion

We have also done the same calculation for asymmetric cavities,cavities coupled to superconducting regions (Andreev billiards),and statistics of Wigner time delay.

In conclusion, we provided:

Recursive procedures valid for finite n and β ∈ {1, 2, 4},allowing a non-perturbative calculation of various transportstatistics.

Pointed out new connections to Painleve equations for theshot noise (V) and Wigner time delay (III).

Solved and generalized some conjectures of Khoruzhenko etal. Is our formula true for any β > 0?

Thank you for listening.

Conclusion

We have also done the same calculation for asymmetric cavities,cavities coupled to superconducting regions (Andreev billiards),and statistics of Wigner time delay. In conclusion, we provided:

Conclusion

Statistics of quantum transport with orthogonal or …...IntroductionResultsMethodConclusion...

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