1. Quick check §
Compute χ when every signal state is identical.
Hint 1
The average equals each signal.
Worked solution
χ=0. No label information reaches the receiver through the signal.
Quantum Information Theory · Chapter 06
Classical Communication over Quantum Channels
How many classical bits can quantum signals carry through a noisy channel?
In this chapter · 3 sections
Reading guide · preparation and approach
Starting point: Chapters 1–5; mutual information, channels, and coding rates.
Anchor example: Two equally likely nonorthogonal pure signals, and a flagged erasure channel.
Method: Use a classical label register to derive a measurement bound by data processing.
Watch for: The Holevo quantity is not automatically attained by a single-signal measurement.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
6.1 An ensemble is a communication model
6.1.1 The Holevo quantity
A sender samples a classical label \(X=x\) with probability \(p_x\), prepares \(\rho_x\), and transmits it. The joint description is \(\rho_{XB}=\sum_xp_x|x\rangle\langle x|\otimes\rho_x\). Because the label blocks are orthogonal, its entropy is \(H(p)+\sum_xp_xS(\rho_x)\). Consequently, \(I(X:B)=S(\bar\rho)-\sum_xp_xS(\rho_x)\), where \(\bar\rho=\sum_xp_x\rho_x\). This expression is the Holevo quantity \(\chi\).
The receiver measures B and obtains a classical record Y. Measurement is a channel, so data processing gives \(I(X:Y)\le I(X:B)=\chi\). This is the Holevo bound. Maximizing the left-hand side over all POVMs defines accessible information. It need not equal \(\chi\). The bound separates information present in a quantum ensemble from information accessible through a chosen single-copy measurement.
For two equally likely pure states with overlap magnitude \(c\), the average state has eigenvalues \((1+c)/2\) and \((1-c)/2\). Each signal has zero entropy, so \(\chi=h_2((1-c)/2)\). For |0⟩ and |+⟩ this is approximately 0.600876 bits. Continuously many possible amplitudes have not created continuously many perfectly distinguishable messages.
Worked check 6.1. What is χ for equally likely orthogonal qubit signals?
Reasoned solution
The average state is I/2 and the individual signals are pure, so χ=1−0=1 bit. Measuring in their common orthogonal basis attains the bound.
6.2 Coding over independent uses
6.2.1 Why a capacity may need regularization
A block code encodes one of \(M_n\) messages into \(n\) uses of a memoryless channel \(\mathcal N\), followed by a joint decoding measurement. Its rate is \(\log_2M_n/n\). We define unassisted classical capacity using uniformly distributed messages and average decoding error tending to zero, with no preshared entanglement or free side communication. Arbitrary input states across a block are allowed.
For one use, define \(\chi^*(\mathcal N)=\max_{\{p_x,\rho_x\}}\chi(\{p_x,\mathcal N(\rho_x)\})\). Product-state coding and collective decoding attain rates approaching this quantity. With arbitrary entangled block inputs the full expression is \(C(\mathcal N)=\sup_{n\ge1}\chi^*(\mathcal N^{\otimes n})/n\). This coding theorem is stated with its resource assumptions; it is not proved by applying the single-copy Holevo bound. See Watrous, Chapter 8.
Regularization matters: optimizing one-use ensembles does not generally settle the unrestricted capacity. Conversely, a particular ensemble’s \(\chi\) is not even the optimized one-use value. These three quantities—chosen ensemble information, optimized one-use information, and regularized capacity—must be named separately. Joint decoding may help even when the sender uses product codewords, so collective measurements and entangled encodings should not be conflated.
Worked check 6.2. What does a code with 1024 messages and 100 channel uses transmit per use?
Reasoned solution
Its rate is log₂1024/100=0.1 bits per use. This is an achieved code rate only after its decoding error is specified; it is not by itself a capacity.
6.3 The erasure channel
6.3.1 A noise model with a readable flag
A d-dimensional erasure channel transmits the input intact with probability \(1-p\) and outputs an orthogonal erasure flag with probability \(p\): \(\mathcal E_p(\rho)=(1-p)\rho\oplus p|e\rangle\langle e|\). The flag lies outside the input’s signal space. It tells the receiver where a loss occurred, without telling them the missing message. This differs from replacing a qubit by I/2 within its ordinary output space.
Use equally likely orthogonal signals. The output reveals the label when the signal survives and reveals none when the flag appears. The mutual information is \((1-p)\log_2d\). The entropy calculation agrees: the average output entropy is \(h_2(p)+(1-p)\log_2d\), while each signal output has entropy \(h_2(p)\). Their difference is the same rate. Classical erasure coding achieves the asymptotic unassisted capacity \(C=(1-p)\log_2d\); the matching converse is a theorem, not established merely by evaluating this ensemble. See Bennett, DiVincenzo, and Smolin.
For a qubit erased with probability 0.2, the classical capacity is 0.8 bits per use. The next chapter asks about preserving an unknown quantum state. The readable flag remains useful, but copying the surviving classical symbols is no longer an available repair strategy.
Worked check 6.3. What is the classical capacity of a four-level erasure channel with p=0.25?
Reasoned solution
It is (1−0.25)log₂4=1.5 bits per use. The erasure flag does not add a message bit because its occurrence is independent of the chosen label.
Summary
- The Holevo bound follows from data processing on a classical–quantum state.
- Accessible information, optimized one-use Holevo information, and unrestricted capacity are distinct.
- Always state encoding restrictions, decoding error, and assistance when reporting a rate.
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.
2. Quick check §
What units does classical channel capacity use?
Hint 1
Count distinguishable classical messages.
Worked solution
Bits per channel use, with the error criterion and assistance model specified.
3. Quick check §
What is the maximum Holevo information of a qubit ensemble?
Hint 1
Use S(average)≤1 and nonnegative signal entropies.
Worked solution
At most one bit without an additional receiver system; orthogonal pure signals attain it.
4. Worked problem §
Evaluate χ for two equally likely pure signals with overlap magnitude 0.6.
Hint 1
The average eigenvalues are 0.8 and 0.2.
Worked solution
χ=h₂(0.2)≈0.721928 bits.
5. Worked problem §
Compute the qubit erasure classical capacity at p=0.6.
Hint 1
Keep the fraction of uses that reveal a signal.
Worked solution
C=0.4 bits per use.
6. Worked problem §
Derive the block-diagonal entropy identity.
Hint 1
The eigenvalues of each block are pₓ times its internal eigenvalues.
Worked solution
Expand \(-\sum_{x,i}p_x\lambda_{xi}\log_2(p_x\lambda_{xi})\). The two terms are \(H(p)\) and \(\sum_xp_xS(\rho_x)\).
7. Worked problem §
Does an ensemble with χ=0.7 establish C≤0.7?
Hint 1
Ask whether the ensemble was optimized.
Worked solution
No. It supplies a rate achievable asymptotically with appropriate coding, hence a lower bound on capacity. An upper bound must cover every allowed code.
8. Misconception §
Does the Holevo bound prohibit superdense coding?
Hint 1
Count the preshared entangled system.
Worked solution
No. Superdense coding includes preshared entanglement; the receiver measures the transmitted system together with their existing half. The unassisted qubit-ensemble model does not describe that resource.
9. Misconception §
Does a collective measurement require entangled transmitted codewords?
Hint 1
Separate encoding from decoding.
Worked solution
No. A decoder can jointly measure a tensor product of individually prepared signals. Entangled encoding is a distinct option.
10. Extension §
Derive χ for a general ensemble through the erasure channel.
Hint 1
Use the block entropy formula for each output and their average.
Worked solution
The common flag term \(h_2(p)\) cancels. The result is \(\chi_{\rm out}=(1-p)\chi_{\rm in}\), because the flag probability is independent of the label.