Information-Theoretic Foundations

Quantum information asks what a physical state lets us communicate, compute, or infer. To answer, we must distinguish the system that carries information, its preparation, the transformations available, and the measurement that produces a classical result. This chapter develops those distinctions for finite-dimensional systems, with a separate introduction to optical modes.

Helpful background. Quantum states and the Born rule explain how a state predicts probabilities. Tensor products supply the language for dividing a larger system into parts.

From a preparation to a measurement result

Suppose Alice chooses a classical label $x$ with probability $p_x$ and prepares a state $\rho_x$. A channel $\mathcal N$ describes transmission or processing. Bob measures a POVM with effects $M_y$, so

$$p(y|x)=\operatorname{Tr}\!\left[M_y\mathcal N(\rho_x)\right].$$

This equation connects a quantum model to an ordinary conditional probability distribution. A communication problem asks how well Bob can infer $x$ from $y$; a sensing problem asks what $y$ reveals about a parameter in $\mathcal N$.

If Alice’s label is unavailable, Bob describes the input by the average state $\bar\rho=\sum_xp_x\rho_x$. Different preparation ensembles can have the same average state and therefore the same statistics for every measurement on that system. If the label remains available, retain it in a separate classical register. Losing a label can lose correlations even when the density operator of Bob’s system is unchanged.

A POVM specifies outcome probabilities. To predict the state after an outcome, use a quantum instrument: its completely positive map $\mathcal I_y$ produces an unnormalized state, with trace equal to that outcome’s probability. Different instruments can implement the same POVM and disturb the system differently.

Choose a starting question

Question	Read next
How do states, measurements, and resources define an information-processing task?	What Is Quantum Information?
How do classical sources, noisy channels, and codes work?	Classical information review
What distinguishes a bit, a qubit, an encoded qubit, and an oscillator mode?	Bits, qubits, qudits, and modes
How do one-qubit states, gates, and measurements look geometrically?	Bloch sphere
How should I describe uncertainty, a discarded subsystem, or a preparation label?	Density operators for quantum information
Which quantity describes uncertainty, correlation, or distinguishability?	Quantum entropy
How much entanglement does a specified task require or produce?	Entanglement measures
Which transformations become possible when a useful state is supplied?	Resource theories

The number $2^n$ counts the dimension of an $n$-qubit state space. It does not mean that one measurement reveals $2^n$ independently chosen classical numbers. The preparation ensemble, allowed measurements, and available side information determine what can be learned.

A qubit calculation that connects three descriptions

In the ordered basis $|0\rangle,|1\rangle$, consider

$$\rho=\begin{pmatrix}0.8&0.3\\0.3&0.2\end{pmatrix} =\frac{I+\mathbf r\cdot\boldsymbol\sigma}{2}, \qquad \mathbf r=(0.6,0,0.6).$$

The matrix has unit trace and is Hermitian. Its eigenvalues are

$$\lambda_\pm=\frac{1\pm|\mathbf r|}{2} \simeq 0.924264,\ 0.075736.$$

Both are nonnegative, so this is a valid density operator. Its Bloch vector lies inside the unit sphere, which tells us that the state is mixed. The purity is $\operatorname{Tr}\rho^2=0.86$.

Measurements along different axes answer different questions. Here $p(0_z)=\rho_{00}=0.8$, while $p(+_x)=(1+r_x)/2=0.8$ as well. The equality is a feature of this state, not a general identity between incompatible measurements.

Using base-two logarithms, its von Neumann entropy is

$$S(\rho)=-\sum_{s=\pm}\lambda_s\log_2\lambda_s \simeq0.386973\ \text{bits}.$$

By comparison, the entropy of the observed $Z$ outcomes is $h_2(0.2)\simeq0.721928$ bits. A chosen measurement’s outcome entropy is not generally the entropy of the quantum state. Measuring in an eigenbasis gives the latter.

Correlation, entanglement, and useful resources

For a bipartite pure state, the Schmidt decomposition connects entanglement to the eigenvalues of either reduced state. For example,

$$|\psi\rangle=\sqrt{0.9}|00\rangle+\sqrt{0.1}|11\rangle$$

has Schmidt rank two but entanglement entropy only $h_2(0.1)\simeq0.468996$ ebits. Rank detects that this pure state is entangled; the weights determine its entropy. One ebit is the entanglement of a Bell pair. The fractional value has a precise conversion-rate meaning for many identical copies under asymptotic local operations and classical communication, rather than describing a fractional Bell pair inside one copy.

Mixed states require more care: their local entropy can include ordinary preparation uncertainty. Likewise, mutual information counts total correlation and can be positive without entanglement. The exercise below makes these distinctions concrete.

A resource theory specifies both the objects available freely and the allowed free operations. Under local operations and classical communication, entanglement is a resource; under stabilizer operations, magic states supply a different resource. A quantity that cannot increase under the chosen free operations is a monotone. Changing the operations changes the conversion problem, so useful resources need not share a single ranking. Chitambar and Gour give a systematic account of these distinctions.

Exercises

1. Same local uncertainty, different correlations

Compare $\rho_{\mathrm{ind}}=I_4/4$, $\rho_{\mathrm{cl}}=(|00\rangle\langle00|+|11\rangle\langle11|)/2$, and the Bell state $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$. Find their one-qubit states, joint entropies, and mutual informations $I(A:B)=S(A)+S(B)-S(AB)$.

Solution

Each reduced state is $I_2/2$, so $S(A)=S(B)=1$ bit in all three cases. The joint spectra give $S(AB)=2,1,0$ bits respectively. Thus $I(A:B)=0,1,2$ bits. The first state is a product, the second is a mixture of product states with perfectly correlated classical labels, and only the third is entangled. Local entropy alone cannot distinguish them.

2. Changing the measurement

For the qubit above, find $p(+_y)$ and the corresponding outcome entropy. Why can this entropy exceed $S(\rho)$?

Solution

Because $r_y=0$, the two outcomes are equiprobable: $p(+_y)=p(-_y)=1/2$, giving one bit. The $Y$ basis is not an eigenbasis of $\rho$; even a pure state can give random outcomes in an incompatible basis. Measurement randomness and mixedness describe different aspects of a state.

After this chapter, continue to gates and computation models, or to communication protocols to see these quantities at work. For measurement dynamics, use generalized measurements and instruments.

References

• Chitambar, E., and G. Gour. “Quantum resource theories.” Reviews of Modern Physics 91, 025001 (2019). doi:10.1103/RevModPhys.91.025001.
• Nielsen, M. A., and I. L. Chuang. Quantum Computation and Quantum Information, 10th anniversary edition. Cambridge University Press, 2010. doi:10.1017/CBO9780511976667.
• Watrous, J. The Theory of Quantum Information. Cambridge University Press, 2018. Chapters 2, 5, and 6. Author’s manuscript and errata.