Ph219C/CS219C
Due: Thursday 5 May 2022
2.1 A quantum version of Fano’s inequality
a) In a d-dimensional system, suppose a density operator ρ approxi-
mates the pure state |ψ⟩ with fidelity
F =⟨ψ|ρ|ψ⟩=1−ε. (1)
H(ρ) ≤ H2(ε) + ε log2(d − 1). (2)
Hint: Recall that if a complete orthogonal measurement per- formed on the state ρ has distribution of outcomes X, then H(ρ) ≤ H(X), where H(X) is the Shannon entropy of X.
b) As in §10.7.2, suppose that the noisy channel NA→B acts on the pure state ψRA, and is followed by the decoding map DB→C. Show that
H(R)ρ − Ic(R ⟩B)ρ ≤ 2H(RC)σ, (3) ρRB = N(ψRA), σRC = D ◦ N(ψRA). (4)
Therefore, if the decoder’s output (the state of RC) is almost pure, then the coherent information of the channel N comes close to matching its input entropy. Hint: Use the data processing inequality Ic(R ⟩C)σ ≤ Ic(R ⟩B)ρ and the subadditivity of von Neumann entropy. It is convenient to consider the joint pure state of the reference system, the output, and environments of the dilations of N and D.
c) Suppose that the decoding map recovers the channel input with high fidelity,
F (D ◦ N (ψRA), ψRC ) = 1 − ε. (5)
Programming Help
H(R)ρ − Ic(R ⟩B)ρ ≤ 2H2(ε) + 2ε log2(d2 − 1), (6)
assuming that R and C are d-dimensional. This is a quantum version of Fano’s inequality, which we may use to derive an upper bound on the quantum channel capacity of N.
2.2 Noisy superdense coding and teleportation.
a) By converting the entanglement achieved by the mother protocol into classical communication, prove the noisy superdense coding resource inequality:
NoisySD: ⟨φABE⟩+H(A)[q→q]≥I(A;B)[c→c]. (7) Verify that this matches the standard noiseless superdense coding
resource inequality when φ is a maximally entangled state of AB.
b) By converting the entanglement achieved by the mother protocol into quantum communication, prove the noisy teleportation re- source inequality:
NoisyTP: ⟨φABE⟩+I(A;B)[c→c]≥Ic(A⟩B)[q→q]. (8) Verify that this matches the standard noiseless teleportation re-
source inequality when φ is a maximally entangled state of AB. 2.3 Degradability of amplitude damping and erasure
The qubit amplitude damping channel NA→B(p) discussed in §3.4.3
has the dilation UA→BE such that U :|0⟩A → |0⟩B ⊗ |0⟩E ,
|1⟩A →1−p|1⟩B ⊗|0⟩E +√p|0⟩B ⊗|1⟩E;
a qubit in its “ground state” |0⟩A is unaffected by the channel, while a qubit in the “excited state” |1⟩A decays to the ground state with probability p, and the decay process excites the environment. Note that U is invariant under interchange of systems B and E accompanied by transformation p ↔ (1 − p). Thus the channel complementary to
NA→B(p) is NA→E(1 − p). a.d. a.d.
Computer Science Tutoring
3 a) Show that NA→B(p) is degradable for p ≤ 1/2. Therefore, the
quantum capacity of the amplitude damping channel is its opti-
mized one-shot coherent information. Hint: It suffices to show that
NA→E(1 − p) = NB→E(q) ◦ NA→B(p), (9) a.d. a.d. a.d.
where 0 ≤ q ≤ 1.
The erasure channel NA→B(p) has the dilation UA→BE such that
U : |ψ⟩A → 1 − p |ψ⟩B ⊗ |e⟩E + √p |e⟩B ⊗ |ψ⟩E; (10)
Alice’s system passes either to Bob (with probability 1 − p) or to
Eve (with probability p), while the other party receives the “erasure
symbol” |e⟩, which is orthogonal to Alice’s Hilbert space. Because U
is invariant under interchange of systems B and E accompanied by
transformation p ↔ (1 − p), the channel complementary to N A→B (p)
is NA→E(1 − p). erase
b) Show that NA→B(p) is degradable for p ≤ 1/2. Therefore, the erase
quantum capacity of the erasure channel is its optimized one-shot coherent information. Hint: It suffices to show that
NA→E(1 − p) = NB→E(q) ◦ NA→B(p), (11) erase erase erase
where 0 ≤ q ≤ 1.
c) Show that for p ≤ 1/2 the quantum capacity of the erasure channel
Q(NA→B(p))=(1−2p)log d, (12) erase 2
where A is d-dimensional, and that the capacity vanishes for 1/2 ≤ p ≤ 1.
2.4 Proof of the decoupling inequality
In this problem we complete the derivation of the decoupling inequality sketched in §10.9.1. Equation numbers of the form (10.xxx) refer to Chapter 10 of the lecture notes.
a) Verify eq.(10.336).
To derive the expression for EU [MAA′(U)] in eq.(10.340), we first note that the invariance property eq.(10.325) implies that EU [M AA′ (U )] commutes with V ⊗V for any unitary V . Therefore, by Schur’s lemma, EU [MAA′(U)]isaweightedsumofprojectionsontoirreduciblerepre- sentations of the unitary group. The tensor product of two fundamen- tal representations of U(d) contains two irreducible representations — the symmetric and antisymmetric tensor representations. Therefore we may write
E [M ′(U)]=c Π(sym) +c Π(anti); (13) U AA sym AA′ anti AA′
here Π(sym) is the orthogonal projector onto the subspace of AA′ sym- AA′
metric under the interchange of A and A′, Π(anti) is the projector onto AA′
the antisymmetric subspace, and csym, canti are suitable constants. Note that
Π(sym)=1(I ′+S ′), AA′ 2AA AA
Π(anti)=1(I ′−S ′), (14) AA′ 2AA AA
where SAA′ is the swap operator, and that the symmetric and an- tisymmetric subspaces have dimension 12|A|(|A|+1) and dimension
12 |A| (|A| − 1) respectively.
Even if you are not familiar with group representation theory, you might regard eq.(13) as obvious. We may write MAA′(U) as a sum of two terms, one symmetric and the other antisymmetric under the in- terchange of A and A′. The expectation of the symmetric part must be symmetric, and the expectation value of the antisymmetric part must be antisymmetric. Furthermore, averaging over the unitary group en- sures that no symmetric state is preferred over any other.
b) To evaluate the constant csym, multiply both sides of eq.(13) by Π(sym) and take the trace of both sides, thus finding
csym = |A1| + |A2|. (15) |A|+1
c) To evaluate the constant canti, multiply both sides of eq.(13)) by
Π(anti) and take the trace of both sides, thus finding AA′
canti = |A1| − |A2|. (16) |A|−1
cI = 21 (csym + canti) , cS = 12 (csym − canti) (17) prove eq.(10.341).
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