Any pure quantum state ψ (qubit) of this system can be written, up to a phase, as a superposition (linear combination of the states)

Transcription

1 Chapter Qubits A single qubit is a two-state system, such as a two-level atom The states (kets) h and v of the horizontaland vertical polarization of a photon can also be considered as a two-state system Another example is the relative phase and intensity of a single photon in two arms of an interferometer The underlying Hilbert space for the qubit is C An arbitraryorthonormal basis for C is denoted by {, }, where (scalar product) = =, = = Any pure quantum state ψ (qubit) of this system can be written, up to a phase, as a superposition (linear combination of the states) ψ = α +β, α + β =, α,β C The classical boolean states, and, can be represented by a fixed pair of orthonormal states of the qubit If the qubit represents a mixed state one uses a two-dimensional density matrix for its representation We therefore express one qubit as ρ = (I +n σ) (I +n σ +n σ +n 3 σ 3 ) where n R 3, n n n +n +n 3, and σ = (σ,σ,σ 3 ) denote the Pauli spin matrices For pure states we have n n = and ρ = ψ ψ 3

2 4 Problems and Solutions Problem Any state (qubit) in C can be written as α, α,β C, α + β = β Find a parameter representation (i) if the underlying field is the set of real numbers (ii) if the underlying field is the set of complex numbers Solution (i)usingα = cosθ, β = sinθ andtheidentitycos θ+sin θ for all θ R we have cosθ sinθ (ii) We have as a representation e iφ cosθ sinθ where θ,φ R and e iφ e iφ = Problem Consider the normalized states (θ,θ [,π)) cosθ cosθ, sinθ sinθ Find the condition on θ and θ such that cosθ cosθ + sinθ sinθ is normalized Solution From the condition that the vector cosθ +cosθ is normalized it follows that sinθ +sinθ (sinθ +sinθ ) +(cosθ +cosθ ) = Thus we have It follows that sinθ sinθ +cosθ cosθ = cos(θ θ ) =

3 Qubits 5 Therefore, θ θ = π/3 or θ θ = 4π/3 Problem 3 Let {, } be an orthonormal basis in the Hilbert space R Let A := + Consider the three cases (i) := (ii) := ( (iii) :=, := cosθ, := sinθ ), := sinθ cosθ Find the matrix representation of A in these bases Solution 3 We find (i) A= + = (ii) A= + = ( cos (iii) A= θ cosθsinθ sin θ cosθsinθ sin + θ cosθsinθ = ) cosθsinθ cos θ For all three cases A = I, where I is the unit matrix Obviously, the third case contains the first two as special cases This is the completeness relation Problem 4 Let {, } be an orthonormal basis in the Hilbert space C The NOT operation (unitary operator) is defined as, (i) Find the unitary operator U NOT which implements the NOT operation with respect to the basis {, } (ii) Let =, = Find the matrix representation of U NOT for this basis (iii) Let =, =

4 6 Problems and Solutions be the Hadamard basis Find the matrix representation of U NOT for this basis Solution 4 (i) Obviously, U NOT = + since = = and = = (ii) For the standard basis we find U NOT = (iii) For the Hadamard basis we find U NOT = Thus we see that the respective matrix representations for the two bases are different Problem 5 The Walsh-Hadamard transform is a -qubit operation, denoted by H, and performs the following linear transform ( + ), ( ) (i) Find the unitary operator U H which implements H with respect to the basis {, } (ii) Find the inverse of this operator (iii) Let =, = be the standard basis in C Find the matrix representation of U H for this basis (iv) Let =, = be the Hadamard basis in C Find the matrix representation of U H for this basis Solution 5 (i) Obviously, U H = ( + ) + ( ) = ( + )+ ( )

5 Qubits 7 (ii)theoperatoru H isunitaryandtheinverseisgivenbyu H = U H = U H, where denotes the adjoint (iii) For the standard basis we find (iv) For the Hadamard basis we find U H = U H = We see that the matrix representations for each of the two bases are the same Problem 6 The Hadamard operator on one qubit can be written as U H = (( + ) +( ) ) (i) Calculate the states U H and U H (ii) Calculate U H U H Solution 6 (i) We obtain the normalized states U H = ( + ), U H = ( ) (ii) Since = = and = = we obtain U H U H = + = I where I is the identity operator ( unit matrix) Problem 7 Consider the Hilbert space C and the linear operator ( matrix) Π(n) := 3 I + n j σ j where n := (n,n,n 3 ) (n j R) is a unit vector, ie, n + n + n 3 = Here σ, σ, σ 3 are the Pauli matrices σ =, σ = j= i, σ i 3 =

6 8 Problems and Solutions and I is the unit matrix (i) Describe the properties of Π(n), ie, find Π (n), tr(π(n)) and Π (n) (ii) Find the vector (φ,θ R) e Π(n) iφ cosθ sinθ Discuss Solution 7 (i) Forthe Paulimatriceswehaveσ = σ, σ = σ, σ 3 = σ 3 Thus Π(n) = Π (n) Since trσ = trσ = trσ 3 =, tri =, and the trace operation is linear, we obtain tr(π(n)) = Since and σ = σ = σ 3 = I [σ,σ ] + =, [σ,σ 3 ] + =, [σ 3,σ ] + = where [A,B] + := AB +BA denotes the anticommutator, the expression Π (n) = 3 I + n j σ j = 4 4 I + 3 n j σ j n j n k σ j σ k 4 simplifies to j= Π (n) = 4 I + j= 3 n j σ j + 4 j= 3 n ji j= j= k= Using n +n +n 3 = we obtain Π (n) = Π(n) Thus Π(n) is a projection matrix (ii) We find Π(n) e iφ cosθ = sinθ (+n3 )e iφ cosθ +(n in )sinθ (n +in )e iφ cosθ +( n 3 )sinθ Problem 8 The qubit trine is defined by the following states ψ =, ψ = 3, ψ = + where {, } is an orthonormal basis Find the probabilities ψ ψ, ψ ψ, ψ ψ 3 Solution 8 Using =, = and = we find ψ ψ = 4, ψ ψ = 4, ψ ψ = 4

7 Qubits 9 Problem 9 The kets h and v are states of horizontal and vertical polarization, respectively Consider the normalized states ψ = ( h + 3 v ) ψ = ( h 3 v ) ψ 3 = h φ = ( h + e πi/3 v ) 3 φ = ( h + e +πi/3 v ) 3 φ 3 = ( h + v ) 3 Give an interpretation of these states Solution 9 Since h h = v v = and v h = h v = we find ψ ψ =, ψ ψ 3 =, ψ ψ 3 = Since the solution to cos(α) = / is given by α = o or α = 4 o we find that that the first three states ψ, ψ, ψ 3 correspond to states of linear polarization separated by o We find φ φ = i 3 The states φ and φ correspond to elliptic polarization and the third state φ 3 corresponds to linear polarization Problem Let ψ = e iφ cosθ sinθ where φ,θ R (i) Find the density matrix ρ := ψ ψ (ii) Find trρ (iii) Find ρ Solution (i) Since ψ = (e iφ cosθ,sinθ)

8 Problems and Solutions we obtain the density matrix ( cos ρ = ψ ψ = θ e iφ sinθcosθ ) e iφ sinθcosθ sin θ (ii) Since cos θ +sin θ = we obtain from (i) that trρ = (iii) Since ψ ψ = we have ρ = ( ψ ψ ) = ψ ψ ψ ψ = ψ ψ = ρ Problem Given the Hamilton operator (i) Find the solution of the Schrödinger equation Ĥ = ωσ x ψ(t) = e iĥt/ ψ(t = ) i d ψ = Ĥ ψ dt with the initial conditions ψ(t = ) = (ii) Find the probability ψ(t = ) ψ(t) (iii) The solution of the Heisenberg equation of motion is given by i dσ z dt = [σ z,ĥ](t) σ z (t) = e iĥt/ σ z e iĥt/ where σ z (t = ) = σ z Calculate σ z (t) (iv) Show that ψ(t = ) σ z (t) ψ(t = ) = ψ(t) σ z ψ(t) Solution (i) The solution of the Schrödinger equation is given by ψ(t) = exp( iĥt/ ) ψ(t = )

9 Qubits Since σ x = I we find the unitary matrix ( cos(ωt) exp( iĥt/ ) U(t) = i sin(ωt) ) isin(ωt) cos(ωt) Thus the normalized state at time t is ψ(t) = U(t) = (ii) We find the probability cos(ωt) i sin(ωt) ψ(t = ) ψ(t) = cos (ωt) (iii) Since the commutators are given by [σ z,ĥ] = ω[σ z,σ x ] = i ωσ y, [σ y,ĥ] = ω[σ y,σ x ] = i ωσ z we obtain the linear system of matrix-valued differential equations dσ z dt = ωσ y (t), dσ y dt = ωσ z (t) with the initial conditionsσ z (t = ) = σ z andσ y (t = ) = σ y Hereweused the Heisenberg equation of motion for σ y to obtain the second differential equation The solution of this system of matrix-valued linear differential equations is given by σ z (t) = σ z cos(ωt)+σ y sin(ωt) (iv) We find and σ y (t) = σ y cos(ωt) σ z sin(ωt) ψ(t = ) σ z (t) ψ(t = ) = cos(ωt) ψ(t) σ z ψ(t) = cos (ωt) sin (ωt) = cos(ωt) Problem Consider a Mach-Zehnder interferometer in which the beam pair spans a two-dimensional Hilbert space with orthonormal basis {, } The state vectors and can be considered as orthonormal wave packets that move in two given directions defined by the geometry of the interferometer We may represent mirrors, beam splitters and relative U P phase shifts by the unitary matrices U M =, U B = e iχ, U P =

10 Problems and Solutions respectively Consider the density matrix ρ in = where {, } denotes the standard basis Using this basis find ρ out = U B U M U P U B ρ in U B U P U M U B Give an interpretation of the result Solution Since ρ in = = ( ) = and we obtain U B U M U P U B = ρ out = e iχ + e iχ e iχ + e iχ +cos(χ) isin(χ) i sin(χ) cos(χ) This yields the intensity along as I +cos(χ) Thus the relative U P phase χ could be observed in the output signal of the interferometer Problem 3 Let {, } be an orthonormal basis in C (i) Find the commutator [ ], (ii) Find the operator exp(t ) (iii) Find the operator exp(t ) (iv) Find the operator exp(t ) exp(t ) (v) Find the operator exp(t( + )) (vi) Is exp(t( + )) = exp(t ) exp(t )? Solution 3 (i) We have [ ], =

11 Qubits 3 since = = and = = We see that the commutator is nonzero (ii) Since = = we find (iii) Analogously exp(t ) = exp(t ) = j= j= t j j! ( )j = I +t t j j! ( )j = I +t (iv) Multiplying the results found above we obtain (v) Since we obtain exp(t )exp(t ) = I +t( + )+t exp(t + t ) = ( + ) = I j= (vi) Clearly when t we have t j (j)! I + j= t j+ (j +)! ( + ) =cosh(t)i +sinh(t)( + ) exp(t( + )) exp(t ) exp(t ) Problem 4 Consider the unitary matrix for the NOT gate U NOT = Show that we can find a unitary matrix V such that V = U NOT Thus V would be the square root NOT gate What are the eigenvalues of V? Solution 4 We find the unitary matrix V = +i i i +i Obviously V is also a square root The eigenvalues of V are and i The eigenvalues of V are and i Note that the eigenvalues of U NOT are and

12 4 Problems and Solutions Problem 5 Let σ,σ,σ 3 be the Pauli spin matrices Let n be a unit vector in R 3 We define the operator Σ := n σ n σ +n σ +n 3 σ 3 (i) Calculate Σ From this result and the fact that Σ is hermitian show that Σ is unitary (ii) Find the eigenvalues of Σ (iii) Let ψ = Calculate the state Σ ψ and the probability ψ Σ ψ Solution 5 (i) Using n +n +n 3 =, σ = σ = σ 3 = I and we obtain σ σ +σ σ =, σ σ 3 +σ 3 σ =, σ σ 3 +σ 3 σ = Σ =(n σ +n σ +n 3 σ 3 ) =(n +n +n 3)I +n n (σ σ +σ σ )+n n 3 (σ 3 σ +σ σ 3 )+n n 3 (σ σ 3 +σ 3 σ ) =I Since Σ is hermitian, ie Σ = Σ and Σ = I we find that Σ is a unitary matrix with Σ = Σ (ii) Since Σ is hermitian and unitary the eigenvalues λ, λ can only be ± Since trσ = = λ +λ we obtain that the eigenvalues are + and (iii) We find Σ ψ = n +n +n i 3 It follows that ψ Σ ψ = n 3 Problem 6 Let n be a unit vector in R 3, σ = (σ,σ,σ 3 ) and n σ := n σ +n σ +n 3 σ 3 (i) Find the unitary matrix exp(iθn σ), where θ R (ii) Find the state exp(iθn σ)

13 Qubits 5 Solution 6 (i) Since σ j σ k = δ jk I +i 3 ǫ jkl σ l where ǫ 3 = ǫ 3 = ǫ 3 =, ǫ 3 = ǫ 3 = ǫ 3 = and otherwise, we obtain l= exp(iθn σ)=i cosθ+i(n σ)sinθ cosθ +in3 sinθ i(n = in )sinθ i(n +in )sinθ cosθ in 3 sinθ Note that we could also use (n σ) = I to find the result (ii) Using (i) we find cosθ+in3 sinθ exp(iθn σ) = i(n +in )sinθ Problem 7 Consider the Pauli spin matrix σ z and the state in C Calculate the variance ψ = and discuss the dependence on θ cos(θ) sin(θ) V σz (ψ) := ψ σ z ψ ( ψ σ z ψ ) Solution 7 Using that σ z = I we have V σz (ψ)= ψ I ψ ( ψ σ z ψ ) sin(θ) = (cos(θ) sin(θ)) cos(θ) = cos(θ)sin(θ) For θ = we have V σz (ψ) = The minimum value is, for example at θ = π/4 The maximum value is, for example at θ = 3π/4 Problem 8 Consider the Hamilton operator α Ĥ = ω α

14 6 Problems and Solutions where α Find α where the energy gap between the two energy levels is the smallest Solution 8 From the eigenvalue equation we find Consequently Thus E ωe = ω α E = ω( +α ), E = ω(+ +α ) E E = ω +α Therefore the shortest energy gap is for α = Problem 9 Consider the Hamilton operator Ĥ = ωσ z + σ x where (i) Find the eigenvalues and the normalized eigenvectors of Ĥ (ii) Use the Cayley-Hamilton theorem to calculate exp( iĥt/ ) Solution 9 (i) The Hamilton operator is given by ω Ĥ = ω From det(ĥ EI ) = we find the two eigenvalues E ± = ± ω + We set E := ω + Then from the eigenvalue equation ω u u = E ω u + u for the eigenvalue E + = E we find u = (E ω)u Thus the eigenvector is given by E ω After normalization we have +(E ω) ( E ω )

15 Qubits 7 Analogously we find for the eigenvalue E = E the normalized eigenvector +(E + ω) E ω (ii) Since E + E and E + = E, E = E we have to solve the system of equations e iet/ = c +c E, e iet/ = c c E for c and c Then e iĥt/ c +c = c I +c Ĥ = ω c c c c ω The solution of the system of equations is given by c =cos(et/ ) c = e iet/ e iet/ E = isin(et/ ) E Thus e iĥt/ cos(et/ ) isin(et/ ) ω/e isin(et/ ) /E = isin(et/ ) /E cos(et/ )+isin(et/ ) ω/e Obviously, exp( iĥt/ ) is a unitary matrix Problem Consider the Pauli spin matrices σ x, σ y and σ z Can one find an α R such that exp(iασ z )σ x exp( iασ z ) = σ y? Solution We have e iα exp(iασ z )σ x exp( iασ z ) = e iα Thus we have to solve For α [,π) we obtain α = 3π/4 exp(iα) = i, exp( iα) = i Problem Consider the unary gates ( unitary matrices) N =, H =,

16 8 Problems and Solutions V = and the normalized state e iπ/, W = e iπ/4 ψ = Calculate the state NHVW ψ and the expectation value ψ NHVW ψ Solution We find the unitary matrix NHVW = ( e i3π/4 e i3π/4 ) Thus we obtain the state NHVW ψ = e i3π/4 +e i3π/4 It follows that ψ NHVW ψ = Problem Let n and m be a unit vectors in R 3, σ = (σ,σ,σ 3 ) and n σ := n σ +n σ +n 3 σ 3 Calculate the commutator [n σ,m σ] Solution We find [n σ,m σ]=i((n m 3 m n 3 )σ +(n 3 m m 3 n )σ +(n m m n )σ 3 ) =i(n m) σ where denotes the vector product Thus the vector n m is perpendicular to the plane spanned by the vectors n and m Problem 3 We define a linear bijection, h, between R 4 and H(), the set of complex hermitian matrices, by t+x y iz (t,x,y,z) y +iz t x We denote the matrix on the right hand side by H

17 Qubits 9 (i) Show that the matrix can be written as a linear combinationof the Pauli spin matrices and the identity matrix I (ii) Find the inverse map (iii) Calculate the determinant of the hermitian matrix H Discuss Solution 3 (i) We have (ii) Consider (a,b R) a c c = b Comparing the entries we obtain (iii) We obtain t = a+b, x = a b H = ti +xσ z +yσ x +zσ y t+x y iz y +iz t x, y = c+c deth = t x y z, z = c c i This is the Lorentz metric Let U be a unitary matrix Then det(uhu ) = det(h) Problem 4 Let ψ and ψ be two normalized states in a Hilbert space H A distance d with d π/ can be defined as cos d = ψ ψ Let H = C and consider the states ψ =, ψ = Find d Solution 4 Since ψ ψ = wehavecos d = andtherefored = π/ Problem 5 Let ρ and ρ be density matrices in the same Hilbert space The Bures distance between the two density matrices is defined as D B (ρ,ρ ) := ( tr((ρ / ρ ρ / ) / )) Consider the density matrices / ρ =, ρ = /

18 Problems and Solutions acting in the Hilbert space C Find the Bures distance Solution 5 Since we obtain Thus ρ / = ρ = ρ / ρ ρ / = D B (ρ,ρ ) = / ( / ) Problem 6 Consider the Hilbert space C Show that Π S = i, Π i A = i i are projection matrices and decompose the Hilbert space into sub-hilbert spaces Solution 6 We have and Consider the state Then Π S ψ = Π S = Π S, Π S = Π S, Π A = Π A, Π A = Π A Π S +Π A = I, Π S Π A = ψ = e iφ sinθ cosθ e iφ sinθ icosθ e ie iφ, Π sinθ +cosθ A ψ = iφ sinθ +icosθ ie iφ sinθ +cosθ with ψ Π A Π S ψ =