Introduction to Quantum Information Hermann Kampermann Heinrich-Heine-Universität Düsseldorf Theoretische Physik III Summer school Bleubeuren July 014
Contents 1 Quantum Mechanics........................... Qubits................................... 3 3 Composite systems and entanglement................. 6 4 Entanglement as a resource: Quantum teleportation........ 9 5 Superdense coding............................ 11 1
1 Quantum Mechanics 1 Quantum Mechanics 1.1 Schrödinger equation i ψ(t) = Ĥ ψ(t) (1) t ψ(t) State vector; element of Hilbert space H (= vector space over complex numbers, with scalar product) Ĥ Hamilton operator (self-adjoint (hermitean) operator on a Hilbert space H) Formal solution for Ĥ time-independent ψ(t) = U(t) ψ(0) with U(t) = e i Ĥt () U(t) Time-evolution operator. U(t) is unitary: U(t)U(t) = e i Ĥt e i Ĥt hermiticity = 1 = U(t) U(t) (3) 1. Measurements: Positive Operator Valued Measure (POVM) Def. (POVM): A set of positive operators {ʵ}, i.e. Ê µ 0 with µ ʵ = 1 is a POVM. Ê µ is called POVM element. The state of ϱ after the outcome i occurs becomes ϱ i = U i Êi ϱ Êi U i ). (4) tr(êi ϱ ) Probability to find outcome µ is p µ = tr(êµ ϱ. POVM can be seen as a v. Neumann measurement on a higher-dim. Hilbert space Example: Unambiguous state discrimination Consider two non-orthogonal pure states u = cos α 0 + sin α 1 v = sin α 0 + cos α 1, where u v = sin(α) (5) Task: distinguish u and v without making mistake: 1 Ê u = v v 1 + sin(α) 1 Ê v = u u 1 + sin(α) detects u with certainty detects v with certainty Ê? = 1 Êu Êv inconclusive result (6)
CONTENTS 3 Qubits classical information processing with binary alphabet: variable can take values 0 or 1, bit quantum information processing: information carriers are quantum states, basis states 0 and 1, but the sate ψ can be in any superposition. Realization: spin- 1 particle, linearly polarized photon, atom in ground/exited state, etc. General normalized pure state of a qubit: ψ = cos θ 0 + eiφ sin θ 1, with i j = δ ij (7) and 0 θ π and 0 φ π. Basis states: ( ) 1 0, 0 ( ) 0 1. 1 (8).1 The density matrix For pure states: ψ = i c i i with i j = δ ij, i c ic i = 1 (normalization) Representation of pure states as density matrix: ϱ pure = ψ ψ (9) Mixed quantum states, i.e. mixture of projectors onto pure states: ϱ mixed := j p j ψj ψj with pi 0, p i = 1 (10) Interpretation: A source emits state ψ i with probability p i, or ϱ describes a subsystem of a larger Hilbert space: ϱ = ϱ tr(ϱ) = 1 (normalization) ϱ 0 (ϱ is positive semidefinite, i.e. it has nonnegative eigenvalues) Only for pure states: ϱ = ϱ (here ϱ is a projector) i
4 Qubits Density operator for a pure qubit: ϱ pure = ( ) ψ ψ cos θ 1 = e iφ sin θ 1 eiφ sin θ sin θ General mixed qubit state: ( ) ϱ00 ϱ ϱ mixed = 01 ϱ, (1) 01 ϱ 11 with ϱ 00, ϱ 11 real. 3 real parameters describe ϱ fully (remember: tr(ϱ) = ϱ 00 +ϱ 11 = 1).. The Bloch sphere Decompose ϱ into 1, σ x, σ y, σ z as (11) ϱ = 1 (1 + s σ), (13) with the (real) Bloch vector s x s = s y s z (14) and the vector of Pauli matrices σ x σ = σ y. (15) σ z Pauli matrices: ( ) 0 1 σ x = 1 0 ( ) 0 i σ y = i 0 ( ) 1 0 σ z = 0 1 (16) It follows ( 1 ϱ = + 1s 1 z (s ) x is y ) 1 (s 1 x + is y ) 1s z s x = Reϱ 01 s y = Imϱ 01 s z = ϱ 00 ϱ 11 (17)
CONTENTS 5 Bloch ball for qubits s z 0 S s x s y 1 Figure 1: Bloch sphere representation a qubit in a pure (green) and a mixed (blue) state. The vector for pure states can take any point on the surface of the sphere and is represented by two parameters (ϑ and ϕ). For mixed states the Bloch vector s can take any point inside the sphere. Bloch vector: sin ϑ cos φ s = s sin ϑ sin φ, (18) cos ϑ with s = 1 for pure states, s < 1 for mixed states. 3 parameters of s: ϑ, φ, s determine ϱ pure states on surface, mixed states in interior of Bloch ball orthogonal states have relative angle of 180 every Bloch vector corresponds to physical state
6 3 Composite systems and entanglement 3 Composite systems and entanglement Total Hilbert space: H A H B The tensor product is bilinear, i.e. ψ A, φ A H A and ψ B H B and c C: c ( ψ A + φ A ) ψ B (19) = ψ A (c ψ B ) + φ A (c ψ B ). Example for tensor product: ψ A ψ B = ( ) α β A ( ) γ = δ B αγ αδ βγ βδ Notation: 0 A 0 B 0 A 0 B 00 AB 00 General pure state of qubits: AB Ψ = A 00 00 + A 01 01 + A 10 10 + A 11 11, (0) A ij = 1 (1) ij 3.1 Entanglement Def. (separability): A pure state is called separable iff it is a product state, i.e. Ψ sep = ψ A ψ B. () A state which is not separable is called entangled. Example for separable state: Ψ = 1 ( 00 + 01 ) = 0 1 ( 0 + 1 ) Examples for entangled states: Φ ± := 1 ( 00 ± 11 ) Ψ ± := 1 ( 01 ± 10 ), (3) which are the Bell states (orthogonal entangled bases of dim Hilbert space). Given ψ, is it separable or entangled? Use Schmidt decomposition : dim H A = d A, dim H B = d B, H AB = H A H B, any state ψ H AB can be written as ψ = r a k e k A f k B, (4) k=1 with Schmidt coefficients a k real, positive, k a k = 1 and e i e j A = δ ij = f i f j B, with Schmidt rank r, fulfills 1 r min(d A, d B ).
CONTENTS 7 ψ is separable iff r = 1. Proof: ( Singular) value decomposition of A (see Eq. 1): A00 A A = 01 = USV A 10 A. Here U, V are unitary matrices, where the columns 11 are the Schmidt vectors (U = ( e 1, e ), V = ( f 1, f ), and S is a diagonal matrix ( ( containing )) the singular values which represent the Schmidt coefficients a1 0 S =. 0 a Partial trace and the Schmidt rank Partial trace: Given a density matrix ϱ AB acting on a state of H AB = H A H B. The partial trace over subsystem B leads to ϱ A = tr B (ϱ AB ) = i (1 A i B ) ϱ AB (1 A i B ), (5) where ϱ A is the reduced density matrix of subsystem A. Partial trace of ψ AB in Schmidt decomposition: ϱ A = tr B (ϱ AB ) = ( r ) ( r ) (1 A f k ) a i e i f i a j e j f j (1 A f k ) k i=1 i=1 = a k ek ek, (6) k the eigenvalues of ϱ A are the squared Schmidt coeff. of ψ AB, i.e. Schmidt-rank (ϱ AB ) = rank (ϱ A ) = rank (ϱ B ). Example: A two qubit state: ξ = 1 ( 00 + 01 + 10 11 ) with corresponding density operator ϱ AB = ξ ξ = 1 4 1 1 1 1 1 1 1 1 1 1 1 1. (7) 1 1 1 1
8 3 Composite systems and entanglement This leads to ϱ A =tr B (ϱ AB ) = i (1 A i B ) ϱ AB (1 A i B ) = (1 A 0 ) ϱ AB (1 A 0 ) + (1 A 1 ) ϱ AB (1 A 1 ) = ( ) 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 0 0 4 1 1 1 1 1 1 1 1 ( ) 0 0 1 0 1 0 0 0 1 4 = 1 ( ) 1 1 + 1 4 1 1 4 = 1 ( ) 1 0 0 1 1 0 0 1 0 0 + 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 1 0 0 1 0 1 1 1 1 0 1 ) ( 1 1 1 1 = 1 1. (8) The reduced state of subsystem A is the maximally mixed state, i.e. the state ξ has Schmidt rank two and it is maximally entangled (see next section). Quantifying pure state entanglement Def. (von Neumann entropy): The von Neumann entropy of a quantum system with density matrix ϱ is S(ϱ) = tr(ϱ log ϱ). (9) (base for log) Note: The logarithm of ϱ is defined via the spectral decomposition: log ϱ = i (log λ i ) i i, (30) where λ i are the eigenvalues and i are the eigenvectors of ϱ. Therefore S(ϱ) = ( ) k λ i log λ i i i k = λ k log λ k. (31) k i k The von Neumann entropy is quantum analogon of the classical Shannon entropy. An entanglement measure for pure states: E( ψ AB = S(ϱ A ) = S(ϱ B ) = a k log a k, (3) k
CONTENTS 9 is the entropy of entanglement, with a k being the Schmidt coefficients. Examples: ψ AB = 00 ϱ A = ( ) 1 0 0 0 φ + = 1 ( 00 + 11 ) ϱ A = 1 S(ϱ A ) = 0 being separable, ( ) 1 0 S(ϱ 0 1 A ) = 1 being maximal entangled. 4 Entanglement as a resource: Quantum teleportation Teleportation (Oxford dictionary): "Teleportation is the apparently instantaneous transportation of persons etc. across space by advanced technological means." [C. Bennett et al., Phys. Rev. Lett. 70, 1895 (1993)] Alice and Bob share ψ AB (a singlet). Alice has an additional qubit in an unknown state ϕ Z = α 0 Z + β 1 Z (33) and wants to transfer it to Bob: Bell measurement ψ AB rotation ϕ Z Alice Bob ϕ send classical bits Write the total state ψ total = ϕ Z ψ AB in the Bell basis for system ZA: 00 = 1 [ φ + + φ ], 01 = 1 [ ψ + + ψ ], 10 = 1 [ ψ + ψ ], 11 = 1 [ φ + φ ]. (34)
10 4 Entanglement as a resource: Quantum teleportation We have ψ total = ϕ Z ψ AB = [α 0 + β 1 ] Z 1 [ 0 A 1 B 1 A 0 B ] The teleportation protocol: = 1 ( α[ φ + + φ ] ZA 1 B α([ ψ + + ψ ] ZA 0 B +β[ ψ + ψ ] ZA 1 B β[ φ + ) φ ] ZA 0 B = 1 ( φ + [α 1 β 0 ] ZA B + φ [α 1 + β 0 ] ZA B + ψ + [ α 0 + β 1 ] ZA B + ψ [ α 0 β 1 ] ZA B). (35) 1. Alice does a Bell measurement on her qubits A and Z, e.g. by unitary operations and single qubit measurements: ( φ +, φ, ψ +, ψ ) H D (0, 1, 0, 1) D (0, 0, 1, 1) The measurement operators are {P µ } = { φ + φ +, φ φ, ψ + ψ +, ψ ψ }. (36) The state after the measurement is given by ψ total µ = P µ 1 B ψ total ψ total P µ 1 B ψ total. (37). Alice tells Bob (classical channel) the outcome of her measurement. 3. Bob rotates his quantum state by a unitary operation (one of the Pauli operators): Alice s outcome φ + φ ψ + ψ Bob s rotation iσ y σ x σ z 1 Remember: σ x = σ y = σ z = ( ) 0 1 = 0 1 + 1 0 1 0 ( ) 0 i = i 0 1 + i 1 0 i 0 ( ) 1 0 = 0 0 1 1. (38) 0 1
CONTENTS 11 Remarks: "Quantum teleportation" transports no matter, only information. The state transfer is not instantaneous, classical information has to be transmitted from Alice to Bob. With classical information, Bob can reconstruct Alice s state. Alice and Bob need ψ AB as a ressource. Successful experiments in Innsbruck [D. Bouwmeester et al., Nature 390, 575 (1997)] and Rome [D. Boschi et al., Phys. Rev. Lett. 80, 111 (1998)]. 5 Superdense coding Transmit two bits of information by transmitting only one qubit [C. Bennett, S. Wiesner, Phys. Rev. Lett. 69, 881 (199)]. Again Alice and Bob share an entangled state ψ AB. The protocol: 1. Alice performs a local rotation, one out of the following four: 1 A 1 B ψ AB = ψ AB (σ x ) A 1 B ψ AB = φ AB (σ y ) A 1 B ψ AB = i φ + AB (σ z ) A 1 B ψ AB = ψ + AB. (39). Alice sends her qubit to Bob. 3. Bob does a Bell measurement on both qubits and finds one of the four outcomes (two bits of information were transmitted).