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OPTI647 Fall 2019 Lecture 3

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OPTI647 Fall 2019 Lecture 3 Photonic Quantum Information Processing OPTI 647: Lecture 3 Saikat Guha September 03, 2019 Associate Professor College of Optical Sciences Meinel 523 Outline for today • Coherent states and linear optics • Quantizing the field • Distinguishing pure states General pure state of a single mode Mode φ ( t ) , a quantum system, is excited in a coherent state ↵ , ↵ C | i 2 If we do ideal direct detection of mode φ ( t ) , the total number of photons K is a Poisson random variable of mean N Mode φ ( t ) , a quantum system, is excited in a number state n ,n 0, 1,..., | i 2 { 1} If we do ideal direct detection of mode φ ( t ) , the total number of This orthogonality photons K = n (exactly so; K is not a random variable). (in the Hilbert A mode of ideal laser light is in a coherent state. state) is different Number (Fock) state of a given mode is very hard to produce from that of experimentally modes (in L2 norm space) There are infinitely many other types of “states” of the mode φ ( t ) . Coherent state and Fock state are just two example class of states n ,n 0, 1,..., Fock states of a mode are special: they form an orthonormal | i 2 { 1} basis that spans any general quantum state of that mode 1 1 | i m n = δ and = c n , c 2 =1 mn | i n| i | n| h | i n=0 n=0 X X Coherent state as a quantum state ↵ 2 Fock states can | | n 1 e− 2 ↵ be thought of as ↵ = n infinite-length | i pn! | i unit-length n=0 ! column vectors X cn corresponding to 1 0 0 the orthogonal 0 1 0 axes of an 0 = 2 3 1 = 2 3 2 = 2 3 ... | i 0 | i 0 | i 1 infinite- 6 7 6 7 6 7 6 7 6 7 6 7 dimensional 6 7 6 7 6 7 4 5 4 5 4 5 vector space Ideal photon detection is a von Neumann quantum measurement described by projectors, n n ,n=0, 1,..., {| ih |} 1 Ideal direct detection on a coherent state ↵ produces outcome “n” | i N n 2 2 e− N (i.e., n “clicks”) with probability, p = n ↵ = c = n |h | i| | n| n! Poisson detection statistics in a laser pulse is a result of the projection of the quantum state of the laser pulse—a coherent state–on to one of the Fock states Coherent states and “linear-optical transformations” (beam-splitters) β2 | i⌘ (0, 1) Transmissivity, ↵ 2 β | 1i | 1i ↵ | 2i Complex Unitary matrix, Mach Zehnder Interferometer (MZI): An arbitrary 2-mode linear optical unitary can be realized with a MZI – two 50-50 beam-splitters and two phases Arbitrary N-mode linear optical unitary • Any N-by-N unitary U can be realized with M = N(N-1)/2 Mach Zehnder Interferometers, . So, we need N(N-1) 50-50 beam-splitters and N(N-1) tunable phases to realize any N-mode linear optical unitary U 1 2 ... ... U Reck et al., PRL 73, 1 (1994) N Clements et al., Optica 3 (12), 1460-1465 (2016) A coherent state is always single mode • By an appropriate choice of modal basis, any “multi-mode” coherent state can always be expressed as a single mode coherent state ↵ ↵ ... ↵ β 0 ... 0 | 1i| 2i | K i⌘| i| i | i – In other words… if we have a deterministic field in any spatio-temporal shape (of any given polarization), we can always represent that as a single-mode coherent state of an appropriate normalized mode – We will see later, this is not true for other quantum states in general. For example, a multimode thermal state or a multimode squeezed state, etc. Slicing a coherent state pulse (in time) (t) E 1 1/⌧ t t 0 T| p T| (1) Single-mode coherent state of this mode: φ(t) 2(t) ↵ , ↵ = pN, N = E2T 1/⌧ | i | t φ(t) p ⌧ T 1/T t 1/⌧ p 0 T| t p T| (2) M-mode coherent state of the modes: k(t),k =1,...,M T Orthogonal M = N ⌧ temporal modes β β ... β , β = | i| i | i M M (t) r 1/⌧ t p T| Slicing a coherent state pulse (in space) E E t pM t | 0 T ↵ β 0 T| | i | i E 0 β pM t | i | i 0 T| 0 β E | i pM t | i 0 T| ... ... E 0 β pM t | i | i 0 T| Examples of optical qubits • Single-rail qubit 0 = 0 , 1 = 1 | i | i | i | i • Dual-rail qubit 0 = 0, 1 , 1 = 1, 0 | i | i | i | i • Cat-basis qubit 0 = N ( ↵ + ↵ ), | i + | i | − i 1 = N ( ↵ ↵ ) | i − | i| − i Prove that the cat-basis qubit states are mutually orthogonal, and find the normalization constants N+ and N- in terms of ↵ Problem 4 Binary pure-state discrimination 1 ( H1) 2 ( H2) Inner product between | i | i 1 the two states – Assume equal priors: p1 = p2 = 2 1 2 = σ Consider a von Neumann h | i 1 1 projective measurement: w1 = 0 | i p2 1 ⇧1 = w1 w1 1 = | ih | | i 1 1 p1+σ ⇧ = w w ✓ ◆ 1 = 2 2 2 | i p2 p1 σ | ih | − 1 0 = | i 0 1 p1+σ ✓ ◆ = 2 p2 p1 σ | i − − 1 1 w2 = | i p2 1 − P = P (H )P (H H )+P (H )P (H H ) e 1 2| 1 2 1| 2 1 1 = w 2 + w 2 2 |h 2| 1i| 2 |h 1| 2i| 1 2 Show that: P e = 1 1 σ and find the expression 2 − − | | Problem 5 for minimum average error probability for P (H1)=p, P (H2)=1 p h p i − Coherent states are not orthogonal • Distinguishability of two coherent states ↵ 2 | | n 1 e− 2 ↵ – Recall: ↵ = n | i p | i n=0 n! ! X – Inner product between two coherent states: Binary phase shift keying (BPSK) coherent-state modulation E 2↵ σ = ↵ ↵ = e− 0 T h | − i 1 = ( ↵ ↵ ) | i N− | i| − i w1 0 T | i E 1 = ↵ − | i | i 0 = ( ↵ + ↵ ) | i N+ | i | − i 2 = ↵ | i w | − i | 2i • Optimal measurement operators are cat states • Minimum probability of error 1 P = 1 1 e 4 ↵ 2 e 2 − − − | | h p i Kennedy receiver: a suboptimal receiver γ γ + β Im | i | i ⌘ 1 β ⇡ Re p1 ⌘ β ↵ | i | i − • Displace the BPSK states, then use direct detection β2 X =1 ↵ e− Y =1 | − i 2 β (2↵+β)2 1 e− e − − X =2 ↵ (2↵+β)2 Y =2 | i 1 e− − 1 (2↵+β)2 1 β2 Pe(N)=min e− + 1 e− β 2 2 − ⇣ ⌘ 1 4N = e− , β =0 (exact nulling case) 2 BPSK error probability 1 4N How to design a e− 2 structured receiver (Dolinar receiver) that achieves the minimum error probability. See OPTI 495B/595B 1 4N 1 1 e− [Dolinar, MIT Ph.D. Probabilityerror of 2 − − h p i Thesis, 1976] N Optimize (minimize) the probability of error of the optimal-nulling Kennedy receiver (find optimal β ) and plot the probabilities of error as function of N, as above Problem 6 Density operator – pure and mixed states • State of a quantum system – Complete knowledge is a pure state | i – Incomplete knowledge is a statistical (classically- random) mixture of pure states • density operator: positive and unit trace, ⇢ = | ih | ⇢ = p (x) = λ λ λ X X | xih x| i| iih i| x i X X Mixture of pure states, p X ( x )=1 Spectral decomposition x the states don’t have toX be orthogonal λi =1 i Take the statistical mixture intuition with a X pinch of salt -- same density operator can λi λj = δij be expressed as different mixtures h | i 16 Projective measurement on mixed state • Measure state, with projective measurement Probability of outcome j Conditional post- measurement state Projective measurement on mixed state • Consider ensemble of pure states, = pX (x), x E { | i} ⇢ = p (x) – Density operator, X | xih x| x • Measurement projectors,X • Assume the state in the ensemble was | xi – Post-measurement states: – • If we get outcome j, we have conditional ensemble – with Projective measurement on mixed state • Density operator of this post- measurement ensemble of states Ej Projective measurement on mixed state • Probability of outcome j Statistical mixture of coherent states • Classical state: P function representation ⇢ = P (↵) ↵ ↵ d2↵ | ih | What is the ZC P function of a coherent state β ? • Statistical field in classical EM theory | i • Multimode classical state ⇢ = P (↵) ↵ ↵ d2n↵ n | ih | ZC where ↵ = ↵ ... ↵ | i | 1i | ni – cannot in general be written as a single mode state unlike a multimode coherent state Single-mode thermal state • Gaussian mixture of coherent states 2 e ↵ /N ⇢ = P (↵) ↵ ↵ d2↵ P (↵)= −| | | ih | ⇡N ZC – Probability distribution for photon counting, ⇧ = n n { n | ih |} P (n)=Tr(n n ⇢)= n ⇢ n | ih | N nh | | i – Show that P (n)= ; n =0, 1,... (1 + N)1+n – Show that, n ⇢ m =0, m = n h | | i 6 1 – Hence, ⇢ = P (n) n n | ih | n=0 X Problem 7 Phase scrambled coherent state • Application of a phase, U ↵ = ↵ei✓ ✓| i | i • Consider the state after application of a random phase to a coherent state: 2⇡ ⇢ = U ↵ ↵ U †d✓ ✓| ih | ✓ Z0 e N N n – Show that: P (n)= n ⇢ n = − ; n =0, 1,... h | | i n! – and that n ⇢ m =0, m = n h | | i 6 1 – So, ⇢ is diagonal in the number basis, ⇢ = P (n) n n | ih | n=0 Problem 8 X • Circularly-symmetric states are diagonal in the number basis (we will revisit this later) Annihilation operator of a mode • Recall field quantization: K K E(t)= aiφi(t) Eˆ(t)= aˆiφi(t) kX=1 k=1 Coherent state of modes φ i (t) Field operatorX for the set of with complex field modes φ i ( t ) , with modal amplitudes, ai annihilation operators aˆi • Annihilation operator aˆ of a single mode – Eigenstate is a coherent state aˆ ↵ = ↵ ↵ | i | i – “Annihilates” photon number, – Number operator, Nˆ =ˆa†aˆ Nˆ n = n n | i | i – Show that, [ˆa, aˆ†]=ˆaaˆ† aˆ†aˆ = Iˆ Problem 9 − The phase operator • Application of a phase, U ↵ = ↵ei✓ ✓| i | i • The phase operator, i✓Nˆ i✓aˆ†a U✓ = e = e • Show that random phase scrambling of any pure state leads to a number diagonal state 2⇡ 1 = P (n) n n ⇢ = U✓ U✓†d✓ | ih | | ih | n=0 Z0 X – With P (n)= n 2 |h | i| Problem 10 Coherent states resolve the identity 1 • Recall: Iˆ = n n | ih | n=0 • Prove that: X 1 ↵ ↵ d2↵ = Iˆ Problem 11 ⇡ | ih | ZC • Coherent states are an “over-complete” basis • They are not orthogonal (and hence cannot be distinguished perfectly) • They form a positive operator valued measure (POVM) – the most general description of a quantum measurement Quantization of the field: summary • Classical (deterministic) field (coherent state) K E(t)= aiφi(t) K kX=1 • Quantum description of the field: Eˆ(t)= aˆiφi(t) – Field becomes an operator kX=1 – Field described by a quantum state of constituent modes – Modal annihilation operator: aˆi – Classical field is a special case: each mode i is excited in a coherent state ↵ , ↵i = ai | ii – Classical statistical field is a mixture of coherent states, density operator ⇢ = P ( ↵ ) ↵ ↵ d ↵ , ↵ = ↵ ↵ ..
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