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

The Statistical Signature of BosonSampling Mattia Walschaers, Jack Kuipers, Juan-Diego Urbina, Klaus Mayer, Malte Christopher Tichy, Klaus Richter and Andreas Buchleitner Luchon, March 2015

Bosons are Remarkable Bunching 50/50 Beamsplitter

BosonSampling Distinguishable 2 � � p � U Random � � o = perm ~ ~ i → ~ i, ~ o � comp U Fermions 2 y t � � i l i b a ? o = � det U p b r � � u o r c ~ ~ p c i → ~ i, ~ o o t � a t h n w e v h e t i s W i h t s e o d ~ ~ i o Bosons 2 � � p � perm U � � o = ~ ~ i → ~ i, ~ o �   U 3 , 1 U 3 , 2 U 3 , 3 U 3 , 4 U 3 , 12 U 6 , 1 U 6 , 2 U 6 , 3 U 6 , 4 U 6 , 12   Computationally Complex   U U 10 , 1 U 10 , 2 U 10 , 3 U 10 , 4 U 10 , 12 ~ o =   i, ~   U 11 , 1 U 11 , 2 U 11 , 3 U 11 , 4 U 11 , 12   U 12 , 1 U 12 , 2 U 12 , 3 U 12 , 4 U 12 , 12 Aaronson and Arkhipov, Theory of Computing 4 , 143 (2013)

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

Obtaining the Statistical Signature C ij = h ˆ n j i � h ˆ n i ih ˆ n j i n i ˆ U Calculate this quantity for all modes i and j Histogram C-dataset Bosons 1500 Distinguishable … C 12 C 13 C 14 C m-1m Fermions P H C L 1000 500 0 - 0.004 - 0.003 - 0.002 - 0.001 0.000 0.001 C = < N i N j > - < N i >< N j >

Benchmarking the Statistical Signature Random Matrix Theory Averaging over the Unitary group allows to analytically estimate the first moments of the C-dataset 2 X C i,j E U ( C i,j ) m ( m − 1) i>j 2 E U ( C 2 X 2 i,j ) ≈ C i,j m ( m − 1) i>j E U ( C 3 i,j ) 2 X 3 C i,j m ( m − 1) i>j

Statistical Certification 2nd and 3rd moment of C-dataset for one U 6 particles in 120 0.0 modes Normalised 3rd Moment - 0.5 - 1.0 Bosons Distinguishable - 1.5 S Fermions - 2.0 Simulated Bosons - 2.5 Analytical Predictions Numerical Mean - 3.0 - 3.5 - 1.2 - 1.1 - 1.0 - 0.9 - 0.8 - 0.7 - 0.6 CV Normalised 2nd Moment Different data points obtained by either changing the circuit or varying the input state

Take Home Message Complex systems require a statistical treatment 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

Extra Slide: Bosons Bunch… Idea: Bosonic quantum statistics enhances the probability of events with multiple particles per output mode . 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)

…but they are not alone Idea: Mean-field theories have been effectively applied to mimic many-particle behaviour. “Simulated Bosons” e i θ 1 Repeat n times for e i θ 2 e i θ 3 one sampling event e i θ 4 1 Add random phase to √ n U each sampling event BUNCHING e i θ 12 & CLOUDING Single-particle state Tichy et al , PRL 113 , 020502 (2014)

Extra Slide: Correlation Functions C ij = h φ | ˆ n j | φ i � h φ | ˆ n i | φ i h φ | ˆ n j | φ i n i ˆ m X | φ i = i n | Ω i U q 1 ,i 1 a ∗ i 1 . . . U q n ,i n a ∗ i 1 ,...,i n =1 input mode q 1 , . . . q n output mode i, j = 1 , . . . m n n X X C B U q k ,i U q k ,j U ⇤ q k ,i U ⇤ U q k ,i U q l ,j U ⇤ q l i U ⇤ q k ,j , ij = − q k ,j + k =1 k 6 = l =1 n n X X C F U q k ,i U q k ,j U ⇤ q k ,i U ⇤ U q k ,i U q l ,j U ⇤ q l ,i U ⇤ ij = − q k ,j − q k ,j k =1 k 6 = l =1 n X C D U q k ,i U q k ,j U ⇤ q k ,i U ⇤ ij = − q k ,j k =1 n n ✓ ◆ 1 − 1 q s ,j − 1 X X C S U q s ,i U q r ,j U ⇤ q r ,i U ⇤ U q r ,i U q s ,j U ⇤ q r ,i U ⇤ ij = q s ,j n n r,s =1 r 6 = s =1

Extra Slide: RMT averaging E U ( U a 1 ,b 1 . . . U a n ,b n U ∗ α 1 , β 1 . . . U ∗ α n , β n ) n X Y V N ( σ − 1 π ) = δ ( a k − α σ ( k ) ) δ ( b k − β π ( k ) ) , k =1 σ , π ∈ S n Can be obtained recursively In practice you look them up in tables

Extra Slide: Results written out 
 Fermions E U ( C F ) = n ( n − m ) m ( m 2 − 1) , = 2 n ( n + 1)( m − n )( m − n + 1) 2 � � C F m 2 ( m + 2)( m + 3) ( m 2 − 1) , E U = − 6 n ( n + 1)( n + 2)( m − n )( m − n + 1)( m − n + 2) 3 � � E U C F m 2 ( m + 1)( m + 2)( m + 3)( m + 4)( m + 5) ( m 2 − 1) ,

Extra Slide: Results written out 
 Distinguishable Particles n E U ( C D ) = − (1) m ( m + 1) , m 2 n + 3 m 2 + mn − 5 m + 2 n − 2 � � = n 2 � � (2) E U C D , m 2 ( m + 2)( m + 3) ( m 2 − 1) m 2 n 2 + 9 m 2 n + 26 m 2 + 5 mn 2 + 21 mn − 62 m + 12 n 2 + 60 n − 72 � � = − n 3 � � E U C D , m 2 ( m + 2)( m + 3)( m + 4)( m + 5) ( m 2 − 1) (3)

Extra Slide: Results written out 
 Bosons E U ( C B ) = n ( − m − n + 2) , (1) m ( m 2 − 1) m 2 n + m 2 + 9 mn − 11 m + n 3 − 2 n 2 + 5 n − 4 � � = 2 n 2 � � (2) E U C B , m 2 ( m + 2)( m + 3) ( m 2 − 1) ✓ m 3 n 2 + 15 m 3 n + 2 m 3 + 3 m 2 n 3 + 6 m 2 n 2 + 213 m 2 n − 222 m 2 − 3 mn 4 3 � � = − 2 n E U C B m 2 ( m + 1)( m + 2)( m + 3)( m + 4)( m + 5) ( m 2 − 1) +45 mn 3 + 32 mn 2 + 372 mn − 464 m + 3 n 5 − 6 n 4 + 45 n 3 + 78 n 2 + 168 n − 288 ◆ , m 2 ( m + 1)( m + 2)( m + 3)( m + 4)( m + 5) ( m 2 − 1) (3)

Extra Slide: Results written out 
 Simulated Bosons E U ( C S ) = − n ( m + n − 2) (1) m ( m 2 − 1) , 4 mn − m − 14 n 2 + 8 n − 2 2 � � = E U C S m 2 ( m + 2)( m + 3) ( m 2 − 1) n + 2 m 2 n 3 − m 2 n 2 + 4 m 2 n − m 2 + 18 mn 3 − 25 mn 2 + 2 n 5 − 4 n 4 + 10 n 3 , m 2 ( m + 2)( m + 3) ( m 2 − 1) n (2) ✓ − 2 m 3 n 5 − 21 m 3 n 4 + 30 m 3 n 3 − 41 m 3 n 2 − 10 m 3 n + 8 m 3 − 6 m 2 n 6 − 3 m 2 n 5 3 � � = E U C S ( m − 1) m 2 ( m + 1) 2 ( m + 2)( m + 3)( m + 4)( m + 5) n 2 + − 285 m 2 n 4 + 261 m 2 n 3 + 75 m 2 n 2 − 66 m 2 n + 24 m 2 + 6 mn 7 − 90 mn 6 − 55 mn 5 ( m − 1) m 2 ( m + 1) 2 ( m + 2)( m + 3)( m + 4)( m + 5) n 2 − 360 mn 4 + 591 mn 3 + 8 mn 2 − 128 mn + 64 m + ( m − 1) m 2 ( m + 1) 2 ( m + 2)( m + 3)( m + 4)( m + 5) n 2 + − 6 n 8 + 12 n 7 − 90 n 6 − 120 n 5 − 24 n 4 + 396 n 3 − 168 n 2 − 48( n − 1) ◆ . ( m − 1) m 2 ( m + 1) 2 ( m + 2)( m + 3)( m + 4)( m + 5) n 2 (3)