Starting point: Channels, Holevo information, teleportation, dense coding, conditional entropy and error correction; capacity theorems are introductory and their asymptotic assumptions are stated.
Keep in view: Compare tasks performed using repeated uses of a noisy qubit channel.
We now know how to protect a qubit against a known set of errors, at the price of using several physical qubits for one logical qubit. Classically, Shannon's capacity told us the best rate any code can achieve over a noisy channel, and the Holevo bound limits the bits that one encoding can deliver.
A quantum channel raises a sharper version of the capacity question, because it can be asked to carry different things. It may carry classical bits or unknown quantum states, with or without shared entanglement as help. What is the best rate for each task? A simple channel that sometimes loses its qubit, and says so, lets us compute every answer. We first separate the tasks and the resources they may use. Then we define what a capacity claim asserts. The Holevo information gives the classical rate, with a subtlety when inputs are entangled across uses. A Bell pair sent through the channel gives the entanglement-assisted rate. The entanglement that survives with a reference gives the quantum rate.
15.1 Communication Tasks
What does it mean to send information through a quantum channel? We introduce a channel that either delivers a qubit intact or replaces it with a flag. Comparing what a classical message and a quantum message need from it shows that they are different tasks. Teleportation and dense coding then show that the answer also depends on what other resources are free.
A qubit erasure channel delivers its input unchanged with probability \(1-e\). With probability \(e\) it replaces the input with a flag state \(\ket\perp\), which is orthogonal to both \(\ket0\) and \(\ket1\). The output therefore lives in three dimensions, and its density matrix has two blocks:
a \(2\times2\) block for the delivered qubit and a single entry for the flag. This is often written \((1-e)\rho\oplus e\ket\perp\bra\perp\). The blocks carry classical probabilities; the output is not a superposition of a qubit and a flag.
Send \(\ket+\) with \(e=1/4\). Three quarters of uses deliver plus exactly, and one quarter deliver the flag. Bob can ask "erased or not?" with the projective measurement onto the flag and onto the qubit subspace. Because the flag is orthogonal to the whole qubit subspace, this question does not disturb a delivered qubit. He learns where the losses are, but not what was lost.
What should count as success? For a classical message, Bob must identify Alice's label. For a quantum message, an unknown state must arrive intact, including any entanglement it has with a reference system that never enters the channel. The two can differ sharply. A device that measures Z and resends the result is a perfect channel for the labels 0 and 1, yet it destroys every superposition. So the success criterion must match the kind of information requested.
Does teleportation make sending qubits the same as sending classical bits? With one ebit in hand, two classical bits deliver one qubit. Dense coding runs the other way: with an ebit, one qubit delivers two classical bits. So the rate a channel supports depends on what else Alice and Bob may use for free.
We therefore always name the resource model. An unassisted code uses only the channel. An entanglement-assisted code may also use any number of ebits shared in advance. Some codes may also use classical communication, one way or in both directions. Each model has its own capacity, and rates from different models cannot be compared without counting the extra resources.
Problems
Problem 15.1. Why does a known erasure differ from an unflagged bit flip?
Worked solution
The receiver knows the failed positions and can adapt decoding to that side information. The channel output alphabet includes a flag distinct from the qubit subspace.
15.2 Coding and Rates
For each task and resource model we want the best rate. What does a claim that a rate is "best" actually assert? We state the two halves of a capacity theorem. Then we find the classical rate of the erasure channel, first with a simple code and then with the Holevo information, and ask what changes when inputs may be entangled across uses. Adding shared entanglement doubles the rate, as dense coding suggests.
Classically, a capacity was a rate below which long codes work and above which none do. The same two halves define every quantum capacity. A rate \(R\) is achievable if there are codes using the channel \(n\) times that carry about \(nR\) bits or qubits with error tending to zero as \(n\) grows. A converse shows that no rate above the capacity is achievable under the same error criterion.
A rate counts message size divided by all channel uses, including the erased ones, since the code is chosen before anyone knows which uses will fail. And a finite demonstration with a small error establishes neither half of a capacity theorem. Throughout, we assume a memoryless channel: independent uses of the same channel, \(\mathcal N^{\otimes n}\), on finite-dimensional systems.
15.2.2 Classical and entanglement-assisted capacities§
How many classical bits per use can the erasure channel carry? Send \(\ket0\) or \(\ket1\) with equal probability. When the qubit arrives, Bob reads the bit exactly. When it is erased, the bit stays fair. So \(H(X\mid Y)=e\), and the mutual information is \(1-e\) bits per use.
Could cleverer inputs beat \(1-e\)? For any encoding, Bob's information is limited by the Holevo information of the output states, \(\chi\). The output for input \(\rho_x\) is \((1-e)\rho_x\oplus e\ket\perp\bra\perp\). Its entropy has a term \(h_2(e)\) for which block occurred, plus \((1-e)S(\rho_x)\) from the qubit block. The flag block is the same for every \(x\), so it cancels in \(\chi\), leaving \(\chi=(1-e)\bigl[S(\bar\rho)-\sum_xp_xS(\rho_x)\bigr]\leq1-e\). For one use, orthogonal inputs are already the best.
Two refinements turn this into a capacity. Write \(\chi^*(\mathcal N)\) for the largest Holevo information of the outputs over all input ensembles. We quote the theorem of Holevo (1998) and of Schumacher and Westmoreland (1997): unentangled inputs, with Bob measuring long blocks of outputs together, achieve rate \(\chi^*\). But Alice may also feed the channel inputs entangled across several uses. To allow that, apply the same quantity to \(n\) uses at once and divide by \(n\):
Taking the \(n\)-use quantity, dividing by \(n\) and letting \(n\) grow is called regularization. It cannot be dropped in general: Hastings (2009) showed that for some channels entangled inputs do better than \(\chi^*\). For the erasure channel they do not, and \(C=1-e\) is a proved channel-specific result (Bennett, DiVincenzo and Smolin, 1997).
With unlimited shared entanglement the answer changes. We quote the theorem of Bennett, Shor, Smolin and Thapliyal (2002): the entanglement-assisted classical capacity is the largest \(I(R:B)\) obtained by sending the system \(A\) of a pure state \(\ket\psi_{RA}\) through the channel, with \(R\) kept as a reference. No regularization is needed. For the erasure channel send one half of a Bell pair. On delivery, \(RB\) is a Bell pair; on erasure, \(R\) is \(I/2\) and \(B\) holds the flag. The joint state has eigenvalues \(1-e\), \(e/2\) and \(e/2\), so \(S(RB)=h_2(e)+e\). Bob's output has eigenvalues \((1-e)/2\), \((1-e)/2\) and \(e\), so \(S(B)=h_2(e)+(1-e)\). With \(S(R)=1\),
At \(e=0\) this is dense coding's two bits per qubit.
Problems
Problem 15.2. For a fair classical bit through an erasure channel with e=1/2, compute the mutual information.
Worked solution
The input entropy is one bit. On half the trials the output reveals it completely; on the other half it reveals only an erasure flag, leaving one bit of uncertainty. Thus H(X|Y)=1/2 and I(X:Y)=1/2 bit per use.
15.3 Quantum Transmission
For a quantum message, what must survive is the entanglement between the transmitted system and a reference. How do we measure how much survives one use of the channel? Negative conditional entropy signalled entanglement, so we use its negative, the coherent information. A cited theorem turns it into the quantum capacity, and the erasure channel shows all three capacities side by side, together with a reason why its quantum capacity vanishes at \(e=1/2\).
Send half of a Bell pair through the erasure channel, keeping the reference \(R\). How much of the entanglement with \(R\) is still in Bob's hands? A negative conditional entropy \(S(R\mid B)\) meant that \(B\) holds part of a purification of \(R\). So we measure the surviving entanglement by its negative,
\[I_c=-S(R\mid B)=S(B)-S(RB).\]
This is the coherent information of the input through the channel. When the state matters we write \(S(B)_\omega\) for the entropy of \(B\) in the output state \(\omega_{RB}\). For the erasure channel, the entropies we just computed give \(I_c=\bigl[h_2(e)+1-e\bigr]-\bigl[h_2(e)+e\bigr]=1-2e\). The term \(h_2(e)\), which records whether an erasure happened, cancels.
Is coherent information the quantum capacity? We quote the theorem of Lloyd, Shor and Devetak: the unassisted quantum capacity is
maximized over inputs to \(n\) uses at once. The regularization matters here too: for some channels, inputs entangled across uses give more than any single-use value. A positive coherent information for one input shows an achievable rate. It does not by itself give the capacity.
15.3.2 Regularization and finite-resource limitations§
What are the three capacities of the erasure channel? For this channel they are known exactly, by channel-specific theorems (Bennett, DiVincenzo and Smolin, 1997, and the entanglement-assisted theorem above):
At \(e=1/4\) they are \(3/4\) bit, \(3/2\) bits and \(1/2\) qubit per use.
Why does \(Q\) reach zero at \(e=1/2\)? Picture the environment receiving the qubit whenever Bob receives the flag. At \(e=1/2\) Bob and the environment are in identical positions. If Bob could decode an unknown state, the environment could decode the same state by the same method, and the state would have been cloned. So no quantum message gets through.
At \(e=3/4\), the Bell-half input has coherent information \(1-2e=-1/2\). A capacity cannot be negative, because sending nothing achieves rate zero. That is why the formula has the maximum with zero. It is a theorem about this channel, not permission to replace the regularized formula by one example in general.
The model matters to the end. Two-way classical communication would raise the quantum rate, so it must not slip into an unassisted claim. And finite block lengths, decoding effort and experimental noise remain practical limits on any real code.
These rates describe reliable delivery. Can the same physics, no-cloning and the disturbance caused by measurement, also keep a message secret from an eavesdropper?
Problem 15.3. Why is a single-use coherent information not generally “the quantum capacity”?
Worked solution
Block encodings can use correlations across channel inputs and improve the per-use value. The regularized maximization is essential without a theorem proving additivity for the channel.
Transfer problem. Compute the three stated erasure capacities at e=3/4. Why does the positive entanglement-assisted classical rate not contradict zero unassisted quantum capacity?
Worked solution
The classical capacity is 1/4 bit per use, entanglement-assisted classical capacity is 1/2 bit per use, and unassisted quantum capacity is zero. The positive rate concerns classical labels with prior shared entanglement; the zero rate concerns unknown quantum states without that resource. Teleportation would consume additional ebits and classical communication.
Summary
The Holevo bound \(\chi = S(\sum_x p_x\rho_x) - \sum_x p_x S(\rho_x)\) caps the accessible classical mutual information extractable from an ensemble, and the Holevo–Schumacher–Westmoreland (HSW) theorem achieves this rate using collective measurements.
Entanglement-assisted classical capacity simplifies additively to a single-letter quantum mutual information formula.
The quantum capacity \(Q(\mathcal{N})\) for transmitting intact qubits and entanglement is governed by the regularized coherent information \(I_c(\rho, \mathcal{N}) = S(\mathcal{N}(\rho)) - S((\mathcal{N}\otimes I)(|\Psi_
ho\rangle))\).
Questions and worked problems
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
Using \(C=1-e\), \(C_E=2(1-e)\), and \(Q=\max(0,1-2e)\) for the qubit erasure channel, evaluate all three at \(e=0.1\) and \(e=\frac12\). What is special about \(e=\frac12\)?
Hint 1
Substitute directly.
Hint 2
Compare \(Q\) with \(C\) at each value.
Worked solution
At \(e=0.1\): \(C=0.9\) bits, \(C_E=1.8\) bits, \(Q=0.8\) qubits per use. At \(e=\frac12\): \(C=0.5\), \(C_E=1\), \(Q=0\). At \(e=\frac12\) the channel can still carry classical information, and even more with shared entanglement, yet it cannot carry quantum information at any positive rate without assistance. Quantum transmission is the most demanding of the three tasks.
Which kind of transmission does each task require? (a) Sending the result of a measurement; (b) sending half of a Bell pair so that Alice and Bob share entanglement; (c) transferring an unknown qubit using a pre-shared Bell pair. Name the resources each consumes.
Hint 1
Ask whether the receiver must identify a label or preserve coherence with a reference.
Hint 2
Teleportation needs classical messages and an ebit.
Worked solution
(a) Classical: the receiver must identify a label, so a measure-and-resend channel would suffice. (b) Quantum: the transmitted qubit must stay coherent with Alice’s half, so the channel must preserve entanglement, which is exactly what quantum capacity measures. (c) Classical, with entanglement assistance: two classical bits per qubit and one ebit, with no quantum channel used during the protocol. The same physical outcome, a qubit at Bob, can be achieved with very different resource accounts.
A team reports sending \(900\) message bits reliably over \(1000\) uses of an erasure channel with \(e=0.25\). Another reports \(700\). Which claim is consistent with the capacity, and what must the reported rate count?
Hint 1
\(C=1-e=0.75\).
Hint 2
Rate is message bits divided by all channel uses.
Worked solution
The capacity is \(0.75\) bits per use, so \(900\) bits in \(1000\) uses (rate \(0.9\)) cannot be sent reliably without some other resource, while \(700\) (rate \(0.7\)) is below capacity and plausible for a good code. The rate must count all channel uses, erased ones included; dividing by the \(750\) or so successful positions would overstate it. A claim above capacity signals an error or an unaccounted resource, such as feedback or shared entanglement.
The dephasing channel applies \(Z\) with probability \(p\). (a) Show that its classical capacity is \(1\) bit per use, using inputs \(\ket0,\ket1\). (b) Send half of \(\ket{\Phi^+}\) through it and compute \(S(B)\), \(S(RB)\), and the coherent information. (c) Evaluate at \(p=0.1\). (The quantum capacity of this channel equals the single-use value found in (b).)
Hint 1
\(Z\) leaves \(\ket0\) and \(\ket1\) unchanged.
Hint 2
The output joint state is \((1-p)\ket{\Phi^+}\bra{\Phi^+}+p\ket{\Phi^-}\bra{\Phi^-}\).
Worked solution
(a) \(\ket0\) and \(\ket1\) pass unchanged, so one bit per use is sent perfectly, and a qubit cannot carry more than one bit. (b) Bob’s output is \(I/2\), so \(S(B)=1\). The joint output is \((1-p)\ket{\Phi^+}\bra{\Phi^+}+p\ket{\Phi^-}\bra{\Phi^-}\), with eigenvalues \(1-p,p\), so \(S(RB)=h_2(p)\). The coherent information is \(1-h_2(p)\). (c) At \(p=0.1\), about \(0.531\) qubits per use. Phase noise is harmless for classical bits in the computational basis but costs quantum capacity, because it damages coherence.
Send a classical bit over the erasure channel by repetition: transmit it \(r\) times and decode from any unerased copy. (a) Find the failure probability and rate. (b) Evaluate for \(r=3\) and \(e=\frac14\). (c) Compare the rate with the capacity \(1-e\), and describe a better strategy that uses the receiver’s knowledge of erasures.
Hint 1
Failure means all \(r\) copies are erased.
Hint 2
Long block codes for erasures can approach rate \(1-e\).
Worked solution
(a) Failure requires every copy to be erased: probability \(e^r\), at rate \(1/r\). (b) For \(r=3\), \(e=\frac14\): failure \(\frac{1}{64}\approx0.016\) at rate \(\frac13\). (c) The capacity is \(0.75\), more than twice that rate. Because Bob knows which positions were erased, a long code needs only enough redundancy to fill in the missing fraction, about \(e\) of the positions, approaching rate \(1-e\) with vanishing error. Repetition wastes the information already carried by the unerased uses.
Show that dense coding achieves the entanglement-assisted rate \(2(1-e)\) on the erasure channel. (a) What happens to each dense-coded qubit when it is erased and when it is not? (b) Combine with an outer classical erasure code to find the rate. (c) Evaluate at \(e=\frac14\) and compare with \(C=\frac34\).
Hint 1
A successful use delivers both bits; an erasure delivers a flag.
Hint 2
The receiver knows which uses were erased.
Worked solution
(a) If the qubit arrives, Bob holds both halves of the encoded Bell state and decodes two bits exactly; if it is erased, he gets the flag and learns nothing, but knows the two bits are missing. (b) Each use is a two-bit symbol erasure channel with erasure probability \(e\), and an outer erasure code over these symbols approaches rate \(2(1-e)\) bits per use. (c) At \(e=\frac14\), \(1.5\) bits per use, twice the unassisted \(0.75\). The extra bit is paid for by one ebit per use, distributed in advance.
With free two-way classical communication, Alice can send halves of Bell pairs through an erasure channel, learn from Bob which arrived, keep only those pairs, and then teleport her qubits. (a) What rate of qubits per channel use does this give? (b) Compare with \(Q=\max(0,1-2e)\) at \(e=\frac14\) and \(e=\frac34\). (c) Why is this not a contradiction?
Hint 1
A fraction \(1-e\) of Bell halves arrive intact.
Hint 2
Teleportation needs only classical messages once the pairs are shared.
Worked solution
(a) On average \(1-e\) pairs per use survive, and each teleports one qubit using classical messages, so the rate is \(1-e\) qubits per use. (b) At \(e=\frac14\): \(0.75\) against \(Q=0.5\); at \(e=\frac34\): \(0.25\) against \(Q=0\). (c) The unassisted capacity forbids classical feedback, and without it Alice does not know which uses failed, so she must protect every qubit in advance. Feedback lets her discard failures after the fact. Capacities depend on the assistance allowed and must be compared under the same resource model.
A student argues: “On an erasure channel with \(e=\frac14\), three quarters of the qubits arrive perfectly, so the quantum capacity is \(\frac34\).” Explain why the unassisted quantum capacity is only \(\frac12\), identifying what Alice does not know when she encodes.
Hint 1
Without feedback, Alice must choose the encoding before knowing which positions will be erased.
Hint 2
Compare with the classical capacity, which is \(\frac34\).
Worked solution
The \(\frac34\) counts successful positions after the fact. Alice must encode an unknown quantum state before transmission, without knowing which positions will be erased, and the encoding cannot copy her state onto several positions (no-cloning). It must spread the state so that any typical subset of \(\frac34\) of the positions suffices, while the erased \(\frac14\) reveal nothing. The coding theorem shows this works only up to \(1-2e=\frac12\). Classical bits can be repeated freely, so the classical capacity reaches \(\frac34\); quantum information cannot.
Use no-cloning to show that the erasure channel with \(e=\frac12\) has zero quantum capacity. Model the channel as an isometry that sends the input to Bob with probability \(\frac12\) and to the environment otherwise, each time leaving the other party a flag. Argue that a code allowing Bob to recover the quantum message would also let the environment recover it.
Hint 1
At \(e=\frac12\) the roles of Bob and the environment are symmetric.
Hint 2
Two parties each holding a decodable copy of an unknown state would amount to cloning.
Worked solution
At \(e=\frac12\), each use delivers the input to Bob or to the environment with equal probability, and whoever does not receive it gets a flag: the channel to Bob and the channel to the environment are identical. If some code let Bob reconstruct an unknown quantum message with fidelity near \(1\), the same decoder applied by the environment to its outputs would also reconstruct it. Together they would produce two copies of an unknown state, which no-cloning forbids. So no positive rate is possible at \(e=\frac12\), and by degrading the channel further, none for \(e>\frac12\) either, consistent with \(Q=\max(0,1-2e)\).
For amplitude damping with parameter \(\gamma\) and the maximally mixed input (half of \(\ket{\Phi^+}\)), (a) find Bob’s output and \(S(B)\); (b) argue that \(S(RB)\) equals the entropy of the environment, which receives the excitation with probability \(\gamma/2\); (c) compute the coherent information \(h_2\big(\frac{1-\gamma}{2}\big)-h_2\big(\frac\gamma2\big)\) for \(\gamma=0.2\) and \(\gamma=0.5\).
Hint 1
The output of \(I/2\) is \(\mathrm{diag}\big(\frac{1+\gamma}{2},\frac{1-\gamma}{2}\big)\).
Hint 2
The total state of \(R\), \(B\), and the environment is pure, so \(S(RB)=S(E)\).
Worked solution
(a) Amplitude damping maps \(I/2\) to \(\mathrm{diag}\big(\frac{1+\gamma}{2},\frac{1-\gamma}{2}\big)\), so \(S(B)=h_2\big(\frac{1-\gamma}{2}\big)\). (b) Modelling the channel as a unitary with an environment in \(\ket0\), the joint state of \(R\), \(B\), and \(E\) is pure, so \(S(RB)=S(E)\). The environment ends in \(\ket1\) exactly when a decay occurs, which happens with probability \(\gamma\cdot\frac12\), so \(S(E)=h_2(\gamma/2)\). (c) For \(\gamma=0.2\): \(h_2(0.4)-h_2(0.1)\approx0.971-0.469=0.502\). For \(\gamma=0.5\): \(h_2(0.25)-h_2(0.25)=0\). With this input, coherent information vanishes once the environment learns as much as Bob; the quantum capacity of amplitude damping is in fact zero for \(\gamma\ge\frac12\).