The Statistical Signature of BosonSampling Mattia Walschaers, Jack - - PowerPoint PPT Presentation

The Statistical Signature

• f BosonSampling

Mattia Walschaers, Jack Kuipers, Juan-Diego Urbina, Klaus Mayer, Malte Christopher Tichy, Klaus Richter and Andreas Buchleitner

Luchon, March 2015

Bosons are Remarkable

50/50 Beamsplitter Bunching

BosonSampling

~ i ~

• U

~ i,~

• =

      U3,1 U3,2 U3,3 U3,4 U3,12 U6,1 U6,2 U6,3 U6,4 U6,12 U10,1 U10,2 U10,3 U10,4 U10,12 U11,1 U11,2 U11,3 U11,4 U11,12 U12,1 U12,2 U12,3 U12,4 U12,12      

p

~ i→~

• = perm
• U

~ i,~

• 2

comp

Distinguishable p

~ i→~

• =
• det U

~ i,~

• 2

Fermions Computationally Complex Bosons p

~ i→~

• =
• perm U

~ i,~

• 2

Aaronson and Arkhipov, Theory of Computing 4, 143 (2013)

U Random

W i t h w h a t p r

• b

a b i l i t y d

• e

s t h i s e v e n t

• c

c u r ?

Certification

Does the machine work? Let’s calculate the result! The reason why we like BosonSampling is also the reason why we cannot directly certify it So-called 
 BosonSampler

Obtaining the Statistical Signature

Cij = hˆ niˆ nji hˆ niihˆ nji

Calculate this quantity for all modes i and j

U

C-dataset

C12

C13 C14 Cm-1m

…

Bosons Distinguishable Fermions

• 0.004 -0.003 -0.002 -0.001 0.000

0.001 500 1000 1500 C = <NiNj> - <Ni><Nj> PHCL

Histogram

Benchmarking the Statistical Signature

Random Matrix Theory Averaging over the Unitary

group allows to analytically estimate the first moments

• f the C-dataset

EU(Ci,j) EU(C2

i,j)

EU(C3

i,j) 2 m(m − 1) X

i>j

Ci,j 2 m(m − 1) X

i>j

Ci,j

2

2 m(m − 1) X

i>j

Ci,j

3

≈

• 1.2
• 1.1
• 1.0
• 0.9
• 0.8
• 0.7
• 0.6
• 3.5
• 3.0
• 2.5
• 2.0
• 1.5
• 1.0
• 0.5

0.0 CV S Bosons Distinguishable Fermions Simulated Bosons Analytical Predictions Numerical Mean

Statistical Certification

Different data points obtained by either changing the circuit or varying the input state

6 particles in 120 modes

2nd and 3rd moment of C-dataset for one U

Normalised 3rd Moment Normalised 2nd Moment

Take Home Message

Two-point correlation functions contain a significant amount of information on many-body interference Doing statistics on all possible two-point correlation functions -“C-dataset”- allows us to certify that the sampled particles are bosons Interested? arXiv:1410.8547 Complex systems require a statistical treatment

Extra Slide: Bosons Bunch…

Clouding B u n c h i n g

Carolan et al, Nat. Photon. 8, 621 (2014) Tichy, J. Phys. B: At. Mol. Opt. Phys. 47, 103001 (2014)

Idea: Bosonic quantum statistics enhances the probability of events with multiple particles per output mode.

…but they are not alone

Idea: Mean-field theories have been effectively applied to mimic many-particle behaviour. “Simulated Bosons”

U

1 √n

eiθ1 eiθ2 eiθ3

eiθ4

eiθ12

Single-particle state

Repeat n times for

• ne sampling event

Add random phase to each sampling event

BUNCHING & CLOUDING

Tichy et al, PRL 113, 020502 (2014)

Extra Slide: Correlation Functions

Cij = hφ| ˆ niˆ nj |φi hφ| ˆ ni |φi hφ| ˆ nj |φi

|φi =

m

X

i1,...,in=1

Uq1,i1a∗

i1 . . . Uqn,ina∗ in |Ωi

CB

ij = − n

X

k=1

Uqk,iUqk,jU ⇤

qk,iU ⇤ qk,j + n

X

k6=l=1

Uqk,iUql,jU ⇤

qliU ⇤ qk,j,

CF

ij = − n

X

k=1

Uqk,iUqk,jU ⇤

qk,iU ⇤ qk,j − n

X

k6=l=1

Uqk,iUql,jU ⇤

ql,iU ⇤ qk,j

CD

ij = − n

X

k=1

Uqk,iUqk,jU ⇤

qk,iU ⇤ qk,j

CS

ij =

✓ 1 − 1 n ◆

n

X

r6=s=1

Uqs,iUqr,jU ⇤

qr,iU ⇤ qs,j − 1

n

n

X

r,s=1

Uqr,iUqs,jU ⇤

qr,iU ⇤ qs,j

input mode q1, . . . qn

• utput mode i, j = 1, . . . m

Extra Slide: RMT averaging

EU(Ua1,b1 . . . Uan,bnU ∗

α1,β1 . . . U ∗ αn,βn)

= X

σ,π∈Sn

VN(σ−1π)

n

Y

k=1

δ(ak − ασ(k))δ(bk − βπ(k)), Can be obtained recursively In practice you look them up in tables

Extra Slide: Results written out
 Fermions

EU (CF ) = n(n − m) m (m2 − 1), EU

• CF

2

= 2n(n + 1)(m − n)(m − n + 1) m2(m + 2)(m + 3) (m2 − 1) , EU

• CF

3

= − 6n(n + 1)(n + 2)(m − n)(m − n + 1)(m − n + 2) m2(m + 1)(m + 2)(m + 3)(m + 4)(m + 5) (m2 − 1),

Extra Slide: Results written out
 Distinguishable Particles

EU (CD) = − n m(m + 1), (1) EU

• CD

2

= n

• m2n + 3m2 + mn − 5m + 2n − 2
• m2(m + 2)(m + 3) (m2 − 1)

, (2) EU

• CD

3

= −n

• m2n2 + 9m2n + 26m2 + 5mn2 + 21mn − 62m + 12n2 + 60n − 72
• m2(m + 2)(m + 3)(m + 4)(m + 5) (m2 − 1)

, (3)

EU (CB) = n(−m − n + 2) m (m2 − 1) , (1) EU

• CB

2

= 2n

• m2n + m2 + 9mn − 11m + n3 − 2n2 + 5n − 4
• m2(m + 2)(m + 3) (m2 − 1)

, (2) EU

• CB

3

= −2n ✓m3n2 + 15m3n + 2m3 + 3m2n3 + 6m2n2 + 213m2n − 222m2 − 3mn4 m2(m + 1)(m + 2)(m + 3)(m + 4)(m + 5) (m2 − 1) +45mn3 + 32mn2 + 372mn − 464m + 3n5 − 6n4 + 45n3 + 78n2 + 168n − 288 m2(m + 1)(m + 2)(m + 3)(m + 4)(m + 5) (m2 − 1) ◆ , (3)

Extra Slide: Results written out
 Bosons

Extra Slide: Results written out
 Simulated Bosons

EU (CS) = −n(m + n − 2) m (m2 − 1) , (1) EU

• CS

2

= 4mn − m − 14n2 + 8n − 2 m2(m + 2)(m + 3) (m2 − 1) n + 2m2n3 − m2n2 + 4m2n − m2 + 18mn3 − 25mn2 + 2n5 − 4n4 + 10n3 m2(m + 2)(m + 3) (m2 − 1) n , (2) EU

• CS

3

= ✓−2m3n5 − 21m3n4 + 30m3n3 − 41m3n2 − 10m3n + 8m3 − 6m2n6 − 3m2n5 (m − 1)m2(m + 1)2(m + 2)(m + 3)(m + 4)(m + 5)n2 +−285m2n4 + 261m2n3 + 75m2n2 − 66m2n + 24m2 + 6mn7 − 90mn6 − 55mn5 (m − 1)m2(m + 1)2(m + 2)(m + 3)(m + 4)(m + 5)n2 + −360mn4 + 591mn3 + 8mn2 − 128mn + 64m (m − 1)m2(m + 1)2(m + 2)(m + 3)(m + 4)(m + 5)n2 +−6n8 + 12n7 − 90n6 − 120n5 − 24n4 + 396n3 − 168n2 − 48(n − 1) (m − 1)m2(m + 1)2(m + 2)(m + 3)(m + 4)(m + 5)n2 ◆ . (3)