Solutions Problems of Chap. 2 2.1 Mixture of pure states. It is easy to see that ρ(x) = ρ(0)δ x0 + ρ(1)δ x1 which suggests that the distribution of the mixing weights must coincide with the dis- tribution ρ. Accordingly, in the case of continuous phase space, we can choose w(¯ x)= ρ(¯ x) to mix the pure states δ(x − ¯ x): ρ(x)= w(¯ x)δ(x − ¯ x)d¯ x. 2.2 Probabilistic or non-probabilistic mixing? Mixing n zeros and n ones is realized by their random permutation that amounts to the following 2n-partite com- posite state: ρ(x 1 ,..., x 2n )= δ x10 ...δ xn0 δ xn+11 ...δ x2n1 + permutations of x 1 ,... x 2n # of permutations . In case of the probabilistic mixing, however, we make 2n random decisions as to choose a zero or a one and then we mix the 2n elements. The probabilistic mixing amounts to ρ(x 1 ) ...ρ(x 2n ) which is obviously different from the above composite state. 2.3 Classical separability. If we choose w(¯ x A , ¯ x B )= ρ AB (¯ x A , ¯ x B ) and replace summation over the weights by integration, it works: ρ AB (x A ,x B )= w(¯ x A , ¯ x B )δ(x A − ¯ x A )δ(x B − ¯ x B )d¯ x A d¯ x B . Thus all ρ AB are mixtures of the uncorrelated pure states δ(x A − ¯ x A )δ(x B − ¯ x B ). 2.4 Decorrelating a single state? The map ρ AB → ρ A ρ B is nonlinear: ρ AB (x A ,x B ) −→ ρ AB (x A ,x B )dx B ρ AB (x A ,x B )dx A . Hence the map is not a real operation. 2.5 Decorrelating an ensemble. The collective state ρ ×2n AB is granted to start with. We interchange the first n and the second n subsystems A. Then we trace over
Transcript
Solutions
Problems of Chap. 2
2.1 Mixture of pure states. It is easy to see that ρ(x) = ρ(0)δx0 + ρ(1)δx1 whichsuggests that the distribution of the mixing weights must coincide with the dis-tribution ρ. Accordingly, in the case of continuous phase space, we can choosew(x) = ρ(x) to mix the pure states δ(x − x):
ρ(x) =∫
w(x)δ(x − x)dx .
2.2 Probabilistic or non-probabilistic mixing? Mixing n zeros and n ones isrealized by their random permutation that amounts to the following 2n-partite com-posite state:
In case of the probabilistic mixing, however, we make 2n random decisions as tochoose a zero or a one and then we mix the 2n elements. The probabilistic mixingamounts to ρ(x1) . . . ρ(x2n) which is obviously different from the above compositestate.
2.3 Classical separability. If we choose w(xA, xB) = ρAB(xA, xB) and replacesummation over the weights by integration, it works:
ρAB(xA, xB) =∫
w(xA, xB)δ(xA − xA)δ(xB − xB)dxAdxB .
Thus all ρAB are mixtures of the uncorrelated pure states δ(xA − xA)δ(xB − xB).
2.4 Decorrelating a single state? The map ρAB → ρAρB is nonlinear:
ρAB(xA, xB) −→∫
ρAB(xA, x′B)dx′
B
∫
ρAB(x′A, xB)dx′
A .
Hence the map is not a real operation.
2.5 Decorrelating an ensemble. The collective state ρ×2nAB is granted to start with.
We interchange the first n and the second n subsystems A. Then we trace over
110 Solutions
the second n composite systems AB. We get (ρAρB)×n. This can be verified if,e.g., we calculate the expectation values of observables like f(xA1)g(xB1), andf1(xA1)f2(xA2)g1(xB1)g2(xB2), etc.
2.6 Classical indirect measurement. Imagine a detector of discrete state spacen. Let the composite state of the system and the detector be ρ(x, n) = ρ(x)Πn(x)where ρ(x) stands for the reduced state of the system. Observe that Πn(x) becomesthe conditional state ρ(n|x) of the detector provided the system is in the pure statex. Now we perform a projective measurement on the detector quantity n. Formally,the partition Pm of the detector state space must be defined as Pm(n) = δmn forall m. According to the rules of projective measurement, the state change will be:
ρ(x, n) → ρm(x, n) ≡ 1pm
δmnρ(x, n) ,
with probability pm =∫
ρ(x,m)dx. For the reduced state ρ(x) =∑
m ρ(x,m),the above projective measurement induces the desired non-projective measurementof the effects Πn(x).
Problems of Chap. 3
3.1 Bohr quantization of the harmonic oscillator. The sum of the kinetic andpotential energies yields the total energy E = 1
2p2+ 1
2ω2q2 which is constant during
the motion. Therefore the phase space point (q, p) moves on the ellipse of surface2πE/ω. The surface plays a role in the Bohr-Sommerfeld q-condition because thecontour-integral of pdq for one period is equal to the surface of the enclosed ellipse.Hence the canonical action takes the form E/ω and we get:
E/ω = (n + 12 ) .
3.2 The role of adiabatic invariants. The canonical actions Ik are adiabatic in-variants of the classical motion. This means that they remain approximately con-stant against whatever large variations of the external parameters of the Hamilton-function provided the variations are slow with respect to the motion. So, the canoni-cal action I of the oscillator will be invariant against the variation of ω in the Hamil-tonian 1
2p2 + 1
2ω2(t)q2 provided |ω| ω2. The q-condition remains satisfied with
the same q-number n.
3.3 Classical-like or q-like motion. The Bohr-Sommerfeld q-condition restricts thecontinuum of classical motions to a discrete infinite sequence. For small q-numbersthis restriction is relevant since the allowed phase space trajectories are well sepa-rated. For large q-numbers, typically, the allowed trajectories become quite densein phase space and might fairly approximate any classical trajectory which doesotherwise not satisfy the q-conditions.
Solutions 111
Problems of Chap. 4
4.1 Decoherence-free projective measurement. Let us construct the spectral ex-pansion A =
∑
λ AλPλ and the post-measurement state ρ′ =∑
λ PλρPλ. If[A, ρ] = 0 then ρ′ = ρ since [A, ρ] = 0 is equivalent with [Pλ, ρ] = 0 for allλ. To prove the inverse statement, we consider the following identity:
[ρ′ − ρ, Pλ] = [Pλ, ρ] .
If ρ′ = ρ then the l.h.s. is zero for all λ which implies that the r.h.s. is zero for all λwhich implies [A, ρ] = 0.
4.2 Mixing the eigenstates. Let us consider the spectral expansion of the matrix ρ:
ρ =∑
λ
ρλPλ .
If ρ is non-degenerate then the Pλ’s correspond to the pure eigenstates of ρ andtheir mixture yields the state ρ if the corresponding eigenvalues make the mixingweights: wλ = ρλ. In the general case, the spectral expansion implies the mixtureρ =
∑
λ wλρλ with wλ = dλρλ and ρλ = Pλ/dλ where dλ is the dimension of Pλ.
4.3 Separability of pure states. If |ψAB〉 = |ψA〉 ⊗ |ψB〉 then the compositedensity matrix ρAB is a single tensor product and it is trivially separable. The otherway around, when the pure state satisfies the separability condition (4.47):
|ψAB〉 〈ψAB | =∑
λ
wλρAλ ⊗ ρBλ ,
then it follows that the matrices on both sides have rank 1. Accordingly, the r.h.s.must be equivalent to the tensor product of rank-one (i.e.: pure state) density matri-ces:
|ψAB〉 〈ψAB | = |ψA〉 〈ψA| ⊗ |ψB〉 〈ψB | ,
which implies the form |ψAB〉 = |ψA〉 ⊗ |ψB〉.4.4 Unitary cloning? Let us suppose that we have duplicated two states |ψ〉 and|ψ′〉:
The inner product of the two initial composite states is 〈ψ|ψ′〉 while the inner prod-uct of the two final composite states is 〈ψ|ψ′〉2. Therefore the above process of stateduplication can not be unitary.
Problems of Chap. 5
5.1 Pure state fidelity from density matrices. Observe that 〈m|n〉2 equals thetrace of the product |n〉 〈n| times |m〉 〈m|. Let us invoke the Pauli-representationof these two density matrices and evaluate the trace of their product:
112 Solutions
| 〈m|n〉 |2 = tr
(
I + nσ
2I + mσ
2
)
=1 + nm
2,
which yields cos2(ϑ/2).
5.2 Unitary rotation for |↑〉 −→ |↓〉 . Since |↑〉corresponds to the north pole and|↓〉corresponds to the south pole on the Bloch-sphere, we need a π-rotation around,e.g., the x-axis. The rotation vector is α = (π, 0, 0) and the corresponding unitarytransformation becomes:
U(α) ≡ exp(
− i
2ασ
)
= −iσx .
We can check the result directly:
−iσx |↑〉= −i[0 11 0
] [10
]
= −i[01
]
= −i |↓〉 .
5.3 Density matrix eigenvalues and -states in terms of polarization. Considerthe density matrix 1
2 (I +sσ) and find the spectral expansion of sσ. We learned thatif s is a unit vector then sσ |↑s〉 = |↑s〉 and sσ |↓s〉 = − |↓s〉. If s ≤ 1, the twoeigenstates remain the same and we keep the simple notations |↑s〉 , |↓s〉 to denotequbits polarized along or, respectively, opposite to the direction s. The eigenvalueswill change trivially and we have sσ |↑s〉 = s |↑s〉 and sσ |↓s〉 = −s |↓s〉. Then wecan summarize the eigenvalues and eigenstates of the density matrix in the followingway:
I + sσ
2|↑s〉 =
1 + s
2|↑s〉 ,
I + sσ
2|↓s〉 =
1 − s
2|↓s〉 .
5.4 Magnetic rotation for |↑〉 −→ |↓〉 . We must implement a π-rotation of thepolarization vector and we can choose the rotation vector (π, 0, 0) which means π-rotation around the x-axis. In magnetic field ω, the polarization vector s satisfiesthe classical equation of motion s = ω × s meaning that s will rotate around thedirection ω of the field at angular velocity ω. Accordingly, we can choose the fieldto point along the x-axis: ω = (ω, 0, 0). The rotation angle π is achieved if weswitch on the field for a period t = π/ω.
5.5 Interrelated qubit physical quantities.
Pn + P−n = I
2Pn − σn = 2P−n + σn = I
5.6 Mixing non-orthogonal polarizations. Since the qubit density matrix is a lin-ear function of the polarization vector, mixing the density matrices means averagingtheir polarization vectors with the mixing weights. Therefore our mixture has thefollowing polarization vector:
s =13× (0, 0, 1) +
23× (1, 0, 0) = (1/3, 0, 2/3) .
Solutions 113
Problems of Chap. 6
6.1 Universality of Hadamard and phase operations. The unitary rotations U ofthe qubit space are, apart from an irrelevant phase, equivalent to the spatial rotationsof the corresponding Bloch sphere. This time the three Euler angles ψ, θ, φ are thenatural parameters. We can write the unitary rotations, corresponding to the spatialones, into this form:
U(ψ, θ, φ) = exp(
− i
2ψσz
)
exp(
− i
2θσx
)
exp(
− i
2φσz
)
.
The middle factor, too, becomes rotation around the z-axis if we sandwich it be-tween two Hadamard operations because HσxH = σz . We can thus express ther.h.s. in the desired form:
U(ψ, θ, φ) = T (ψ)HT (θ)HT (φ) .
6.2 Statistical error of qubit determination. Out of N , we allocate Nx, Ny, Nz
qubits to estimate sx, sy, sz , respectively. We learned that the estimated value of sx
takes this form:N↑x − N↓x
N↑x + N↓x=
2N↑x
Nx− 1 ,
because on a large statistics Nx = N↑x + N↓x the ratio N↑x/Nx converges to theq-theoretical prediction p↑x = 〈↑x| ρ |↑x〉 ≡ 1
2 (1 + sx). The statistical error ofthe estimation takes the form 2∆N↑x/Nx and we are going to determine the meanfluctuation ∆N↑x. The statistical distribution of the count N↑x is binomial:
p(N↑x) =(
Nx
N↑x
)
pN↑x
↑x pN↓x
↓x ,
hence the mean squared fluctuation of the count N↑x takes the form (∆N↑x)2 =Nxp↑xp↓x = Nx(1 − s2
x)/4. This yields the ultimate form of the estimation error:
∆sx =
√
1 − s2x
Nx,
and we could get similar results for ∆sy and ∆sz .
6.3 Fidelity of qubit determination. If the state |n〉 sent by Alice and the polar-ization σm chosen by Bob were fixed then the structure of the expected fidelity ofBob’s guess would be this:
| 〈n|m〉 |2p↑m + | 〈n|−m〉 |2p↓m .
Here we have understood that Bob’s optimum guess must always be the post-measurement state |±m〉 based on the measurement outcome σm = ±1, respec-tively. Now we recall that | 〈n|m〉 |2 = p↑m = cos2(ϑ/2) where cos ϑ = nm, and
114 Solutions
| 〈n|−m〉 |2 = p↓m = sin2(ϑ/2). Hence the above fidelity takes the simple formcos4(ϑ/2) + sin4(ϑ/2) which we rewrite into the equivalent form 1
2 + 12 cos2(ϑ).
The average of cos2(ϑ) = (nm)2 over the random independent n and m yields1/3 therefore the expected fidelity of Bob’s guess becomes 2/3.
6.4 Post-measurement depolarization. Let σn denote the polarization chosen byBob. The non-selective measurement induces the change ρ → PnρPn + P−nρP−n
of the state. Inserting the Pauli-representation of ρ and the projectors P±n yields:
I + sσ
2→ I + σn
2I + sσ
2I + σn
2+
I − σn
2I + sσ
2I − σn
2=
=I + sσ
4+
I + sσnσσn
4.
Since Bob’s choice is random regarding n we shall average n over the solid angle.Averaging the structure σnσσn yields −σ/3 hence the average influence of Bob’snon-selective measurements can be summarized as:
I + sσ
2→ I + sσ/3
2.
6.5 Anti-linearity of polarization reflection. Let us calculate the influence of theanti-unitary transformation T on a pure state qubit:
T |n〉 = T
(
cosθ
2|↑〉+ eiϕ sin
θ
2|↓〉)
= − cosθ
2|↓〉+e−iϕ sin
θ
2|↑〉= e−iϕ |−n〉 .
6.6 General qubit effects. We can suppose that the weights wn are non-vanishing.First, we have to impose the conditions |an| ≤ 1 since otherwise the matrices wouldbe indefinite. Second, the request Πn ≥ 0 implies the conditions wn > 0. And third,the request
∑
n Πn = I implies the conditions∑
n wn = 1 and∑
n wnan = 0.
Problems of Chap. 7
7.1 Schmidt orthogonalization theorem. Let r be the rank of c and let us considerthe non-negative matrices cc† and c†c of rank r both. Their spectrum is non-negativeand identical. Indeed, if c†c |R〉 = w |R〉, i.e., w and |R〉 are an eigenvalue and a(normalized) eigenvector of c†c then |L〉 = c |R〉 /
√w will be a (normalized) eigen-
vector of cc† with the same eigenvalue w. This can be seen by direct inspection. Nowwe determine the r positive eigenvalues wλ for λ = 1, 2, . . . , r and the correspond-ing orthonormal eigenstates |λ;R〉 of c†c. Then, by |λ;L〉 = c |λ;R〉 /
√wλ, we
define the r orthonormal eigenstates of cc† which belong to the common positiveeigenvalues wλ, for λ = 1, 2, . . . , r. Now we can see that
c |λ;R〉 =√
wλ |λ;L〉 ,
Solutions 115
for all λ = 1, 2, . . . , r. We have thus proven that there exists the following Schmidtdecomposition of the matrix c:
c =r∑
λ=1
√wλ |λ;L〉 〈λ;R| .
7.2 Swap operation. For convenience, we introduce σ(±) = (σx ± iσy)/2 insteadof σx, σy . Now we express the Pauli matrices in the up-down basis:
7.3 Singlet density matrix. The singlet state ρ(singlet) is invariant under rotationsof the Bloch sphere. Therefore ρ(singlet) must be of the form:
ρ(singlet) =I ⊗ I + λσ ⊗ σ
4
because there are no further rotational invariant mathematical structures. We coulddetermine the parameter λ from the pure state condition [ρ(singlet)]2 = ρ(singlet),yielding λ = −1. However, we can spare these calculations if we recall the swapS. It is Hermitian, rotation invariant and idempotent: S2 = I ⊗ I . Hence we get thesinglet state directly in the form:
ρ(singlet) =I ⊗ I − S
2=
I ⊗ I − σ ⊗ σ
4.
7.4 Local measurement of expectation values. Alice and Bob will determine〈A ⊗ B〉 and 〈A′ ⊗ B′〉 separately on two independent sub-ensembles and willfinally add them since the expectation value is additive. Still we have to show thatthe expectation value of a tensor product, like 〈A ⊗ B〉, can be determined in localmeasurements. We introduce the local spectral expansions A =
∑
λ AλPλ and B =∑
µ BµQµ. Alice and Bob perform local measurements of A and B in coincidence,yielding the measurement outcomes A1, B1, A2, B2, . . . , Ar, Br, . . . AN , BN whereAr is always an eigenvalue Aλ and the case is similar for the Br’s. Then Alice andBob can calculate the q-expectation value asymptotically:
〈A ⊗ B〉 = limN→∞
1N
∑
r
ArBr .
116 Solutions
To see that this is indeed the right expression of 〈A ⊗ B〉 we have to rethink thenonlocal measurement of A ⊗ B itself. Its spectral expansion is
A ⊗ B =∑
(λ,µ)
(AλBµ)(Pλ ⊗ Qµ) ,
and the corresponding q-measurement will obviously yield the same statistics of theoutcomes ArBr like in case of the local-measurements.
7.5 Local measurement of certain nonlocal quantities. If we measure σz ⊗ σz ona singlet state we always get −1 and the singlet state remains the post-measurementstate. In the attempted local measurement, the entanglement is always destroyed andwe get either |↑↓〉 or |↓↑〉 for the post-measurement state. Obviously the degeneratespectrum of σz⊗σz plays a role in the nonlocality. If, in the general case, we supposeA ⊗ B has a non-degenerate spectrum then the post-measurement states will be thesame pure states in both the nonlocal measurement of A ⊗ B and the simultaneouslocal measurements of A and B.
7.6 Nonlocal hidden parameters. Let the further hidden parameter ν take values1, 2, 3, 4 marking whether Alice and Bob measures A⊗B, A′⊗B, A⊗B′ or A′⊗B′,respectively. Then, according to the hidden variable concept, the assignment of allfour polarization values will uniquely depend on the composite hidden variable rν:
Contrary to the local assignment (7.36), the above assignments are called nonlo-cal since the hidden variable rν is nonlocal: it depends on both Alice’s and Bob’smeasurement setup. The statistical relationships, cf. (7.37), become modified:
〈A ⊗ B〉 = limN1→∞
1N1
∑
r∈Ω1
Ar1Br1; N1 = |Ω1|,
〈A′ ⊗ B〉 = limN2→∞
1N2
∑
r∈Ω2
Ar2Br2 ; N2 = |Ω2| ,
etc. for the other two cases ν = 3, 4. The assignments are independent for the fourdifferent values of ν. There is no constraint combining the Arν’s with different ν’s.Hence it has become straightforward to reproduce the above said four q-theoreticalpredictions including of course correlations 〈C〉 that are higher than 2.
7.7 Does teleportation clone the qubit? The selective post-measurement state ofthe two qubits on Alice’s side is one of the four maximally entangled Bell-states.Therefore the reduced state of the qubit that she had teleported is left in the totallymixed state independently of its original form as well as of the four outcomes ofAlice’s measurement. Note that the form (7.47) of the three-qubit pre-measurementstate shows that Alice’s measurement outcome is always random. The four outcomeshave probability 1/4 each.
Solutions 117
Problems of Chap. 8
8.1 All q-operations are reductions of unitary dynamics. Given the trace-preserving q-operation Mρ =
∑
n MnρM†n, we have to construct the unitary in-
teraction matrix U acting on the composite state of the system and environment.Let us introduce the composite basis |λ〉 ⊗ |n;E〉 where λ = 1, 2, . . . , d andn = 1, 2, . . . , dE . Let us define the influence of U on a subset of the compositebasis:
U (|λ〉 ⊗ |1;E〉) =dE∑
n=1
Mn |λ〉 ⊗ |n;E〉 , λ = 1, 2, . . . , d .
This definition is possible because the above map generates orthonormal vectors:
dE∑
m=1
〈µ| M†m ⊗ 〈m;E|
dE∑
n=1
Mn |λ〉 ⊗ |n;E〉 =dE∑
n=1
〈µ| M†nMn |λ〉 = δλµ .
The further matrix elements of U , i.e. those not defined by our first equation above,can be chosen in such a way that U is unitary on the whole composite state. Usingthis definition of U in the equation (8.3) of reduced dynamics we can directly inspectthat the resulting operation is Mρ =
∑
n MnρM†n, as expected.
8.2 Non-projective effect as averaged projection. Let us substitute the proposedform of the effects Πn into the equation pn = tr(Πnρ) introduced for non-projectivemeasurement in Sect. 4.4.2:
tr(
Πnρ)
= tr(
trE PnρE ρ)
= tr(
PnρE ρ)
.
In this formalism, i.e., without the ⊗’s, the matrices of different subsystems com-mute hence ρE ρ = ρρE . Thus we obtain the following result: tr(Πnρ) = tr(PnρρE).We recognize the coincidence of the r.h.s. with the r.h.s. of (8.18). Since this coin-cidence is valid for all possible ρ, it verifies the proposed form of Πn.
8.3 Q-operation as supermatrix. We start from the Kraus representation Mρ =∑
n MnρM†n. We take the matrix elements of both sides and we also sandwich the
ρ between the identities∑
λ′ |λ′〉 〈λ′| and∑
µ′ |µ′〉 〈µ′| on the r.h.s.:
〈λ|Mρ |µ〉 = 〈λ|∑
n
Mn
∑
λ′
|λ′〉 〈λ′| ρ∑
µ′
|µ′〉 〈µ′| M†n |µ〉 .
Comparing the r.h.s. with∑
λ′µ′ Mλµλ′µ′ρλ′µ′ , we read out the components of the
supermatrix: Mλµλ′µ′ =∑
n 〈λ| Mn |λ′〉 〈µ′| M†n |µ〉.
8.4 Environmental decoherence, time-continuous depolarization. The equationtakes the Lindblad form with H = 0 and with three hermitian Lindblad matricesidentified by the Cartesian components of (σ/2
√τ). For convenience, we shall use
the Einstein convention to sum over repeated indices, e.g.: sσ = saσa. We write
118 Solutions
the r.h.s. of the master equation into the equivalent form −(1/8τ)[σb, [σb, ρ] ] andinsert ρ = 1
2 (I + saσa) into it. The master equation reduces to:
saσa = − 18τ
[σb, [σb, saσa] ] = −1τ
saσa ,
which means the simple equation s = −s/τ for the polarization vector. Its solu-tion is s(t) = e−t/τs(0). Therefore the τ may be called depolarization time, ordecoherence time as well.
8.5 Kraus representation of depolarization. The map should be of the formMρ = (1 − 3λ)ρ + λσρσ with 0 ≤ λ ≤ 1/3 since there exist no other isotropicKraus structures for a qubit. The depolarization channel decreases the polarizationvector s by a factor 1 − w and we have to find the parameter λ as function of w.Inserting ρ = 1
2 (I + sσ) we get
M I + sσ
2=
I + (1 − 4λ)sσ
2,
which means that λ = w/4. Four Kraus matrices make the depolarization channel:√
1 − 3w/4I and the three components of√
w/4σ.
Problems of Chap. 9
9.1 Positivity of relative entropy. We can write:
S(ρ′‖ρ) =∑
x
ρ(x) logρ(x)ρ′(x)
.
We invoke the inequality ln λ > 1 − λ−1 valid for λ = 1 and apply it to λ = ρ/ρ′.This yields:
S(ρ′‖ρ) >1
ln 2
∑
x
ρ(x)[
1 − ρ′(x)ρ(x)
]
= 0 ,
which always holds if ρ′ = ρ.
9.2 Concavity of entropy. Suppose we have a long message x(1)1 x(1)
2 . . . x(1)n where
ρ1(x) is the apriori distribution of one letter. Let S1 stand for the single-letter en-tropy S(ρ1). The number of the typical messages is 2nS1 so that their shortest codeis nS1 bits. Consider a second message from the same alphabet and suppose thesingle-letter distribution ρ2(x) is different from ρ1(x). Let us concatenate the twomessages:
x(1)1 x(1)
2 x(1)3 . . . x(1)
n x(2)1 x(2)
2 x(2)3 . . . x(2)
m ,
where the two lengths n and m may be different. Obviously, the number of thetypical ones among such composite messages is 2nS1×2mS2 and their shortest codeis nS1+mS2 bits. Now imagine that we permute the n+m letters randomly. On one
Solutions 119
hand, the composite messages become usual (n + m)-letter-long messages wherethe single letter distribution is always the same, i.e., the mixture ρ = w1ρ1 + w2ρ2
with weights w1 = n/(n + m) and w2 = m/(n + m). Therefore the number ofthe typical messages is 2(n+m)S(ρ) and the shortest code is (n + m)S(ρ) bits. Onthe other hand, we can inspect that the number of the typical messages 2(n+m)S(ρ)
is greater than 2nS1 × 2mS2 because the number of inequivalent permutations hasincreased: the first n letters have become permutable with the last m letters. Thismeans that (n + m)S(ρ) > nS1 + mS2 which is just the concavity of the entropy:S(w1ρ1 + w2ρ2) > w1S(ρ1) + w2S(ρ2), for ρ1 = ρ2.
9.3 Subadditivity of entropy. We can make the choice ρ′AB(x, y) = ρA(x)ρB(y).To calculate S(ρ′AB‖ρAB) = −S(ρAB) −
∑
x,y ρAB(x, y) log ρ′AB(x, y), we notethat the second term is −
∑
x,y ρAB(x, y) log[ρA(x)ρB(y)] = S(ρA) + S(ρB).Hence the positivity of the relative entropy S(ρ′AB‖ρAB) ≥ 0 proves subadditivity:S(ρAB) ≤ S(ρA) + S(ρB).
9.4 Coarse graining increases entropy. Let us identify our system by the k-partite composite system of the k bits x1, x2, . . . , xk. Then the coarse grained sys-tem corresponds to the (k − 1)-partite sub-system consisting of the first k − 1 bitsx1, x2, . . . , xk−1. The coarse grained state ρ is just a reduced reduced state w.r.t. ρ.Hence we see that coarse graining increases the entropy because reduction does it.
Problems of Chap. 10
10.1 Subadditivity of q-entropy. Let us calculate
S(ρA ⊗ ρB‖ρAB) = −S(ρAB) − tr[ρAB log(ρA ⊗ ρB)]
and note that the second term is S(ρA) + S(ρB). Hence the Klein inequalityS(ρ′‖ρ) ≥ 0 proves the subadditivity: S(ρAB) ≤ S(ρA) + S(ρB).
10.2 Concavity of q-entropy, Holevo entropy. We assume a certain environmentalsystem E and a basis |n;E〉 for it. Let us construct a composite state:
ρbig =∑
n
wnρn ⊗ |n;E〉 〈n;E| .
Note that the reduced state of the system is invariably ρ =∑
n wnρn and the re-duced state of the environment is ρE =
∑
n wn |n;E〉 〈n;E|. Subadditivity guar-antees that S(ρbig) ≤ S(ρ) + S(ρE). Let us calculate and insert the entropiesS(ρbig) = S(w) +
∑
n wnS(ρn) and S(ρE) = S(w) which results in the desiredinequality:
∑
n wnS(ρn) ≤ S(ρ).
10.3 Data compression of the non-orthogonal code. The density matrix of thecorresponding 1-letter q-message reads:
ρ =|↑z〉 〈↑z| + |↑x〉 〈↑x|
2=
I + σn/√
22
,
120 Solutions
which is a partially polarized state along the skew direction n = (1, 0, 1)/√
2, cf.(6.29). The eigenvalues of this density matrix are the following:
p+ =1 + 1/
√2
2, p− =
1 − 1/√
22
,
hence its von Neumann entropy amounts to:
S(ρ) = −p+ log p+ − p− log p− ≈ 0.60 .
According to the q-data compression theorem, we can compress one qubit of theq-message into 0.6 qubit on average, and this is the best maximum faithful com-pression.
10.4 Distinguishing two non-orthogonal qubits: various aspects. In fact, wemeasure the polarization component orthogonal to the polarization of the single-letter density matrix. The measurement outcomes ±1 on both q-states |↑z〉 , |↑x〉will appear with probabilities p+ and p− (cf. Prob. 10.3), in alternating order ofcourse:
10.5 Simple optimum q-code. The q-data compression theory says that a pure stateq-code is not compressible faithfully (i.e.: allowing the same accessible information)if and only if the single-letter average state has the maximum von Neumann entropy.In our case, we must assure the following:
|R〉 〈R| + |G〉 〈G| + |B〉 〈B|3
=I
2,
which is possible if we chose three points on a main circle of the Bloch-sphere, atequal distances from each other.
Problems of Chap. 11
11.1 Creating the totally symmetric state.
|S〉 ≡ 12n/2
2n−1∑
x=1
|xnxn−1 . . . x1〉 =1∑
xn=0
1∑
xn−1=0
· · ·1∑
x1=0
|xn〉⊗|xn−1〉⊗ . . .⊗|x1〉
= H |0〉 ⊗ H |0〉 ⊗ . . . ⊗ H |0〉 ≡ H⊗n |0〉⊗n.
Solutions 121
11.2 Constructing Z-gate from X-gate.
HXH =1√2
[1 11 −1
] [0 11 0
]1√2
[1 11 −1
]
=[
1 00 −1
]
= Z .
The inverse relationship follows from H2 = I .
11.3 Constructing controlled Z-gate from cNOT-gate.
H H
Z
11.4 Q-circuit to produce Bell states. Let us calculate the successive actions ofthe H-gate and the cNOT-gate:
|00〉 −→ |00〉 + |10〉 −→ |00〉 + |11〉 −→∣∣Φ(+)
⟩
|01〉 −→ |01〉 + |11〉 −→ |01〉 + |10〉 −→∣∣Ψ (+)
⟩
|10〉 −→ |00〉 − |10〉 −→ |00〉 − |11〉 −→∣∣Φ(−)
⟩
|11〉 −→ |01〉 − |11〉 −→ |01〉 − |10〉 −→∣∣Ψ (−)
⟩
The trivial factors 1/√
2 in front of the intermediate states have not been denoted.
11.5 Q-circuit to measure Bell states. The task is the inverse task of preparing theBell states. Since both the H-gate and the cNOT-gate are the inverses of themselves,respectively, we can simply use them in the reversed order w.r.t. the circuit thatprepared the Bell states (cf. Prob. 11.4):
M
H M
The boxes M stand for projective measurement of the computational basis.
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Index
Alice, Bob 36, 44Alice, Bob, Eve 48
Bellbasis, states 58inequality 63nonlocality 62
bit 11
channel capacity 83code 80
optimum 83q- 88superdense 64
completely positive map 69contrary to classical 20, 22, 23, 29
data compression 80, 89decoherence 23density matrix 19, 34
