1. Quick check §
Name the free states in computational-basis coherence.
Hint 1
Which states have no off-diagonal entries?
Worked solution
All density matrices diagonal in that declared basis, including their probabilistic mixtures.
Quantum Information Theory · Chapter 09
Quantum Resource Theories
How do operational restrictions turn coherence, entanglement, or asymmetry into useful resources?
In this chapter · 3 sections
Reading guide · preparation and approach
Starting point: Chapters 1–8, especially data processing and entanglement transformations.
Anchor example: Coherence of a qubit relative to a fixed computational basis.
Method: Define free operations first, then construct monotones and conversion bounds.
Watch for: A function decreasing under free operations need not completely characterize convertibility.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
9.1 Choosing a resource theory
9.1.1 Free states alone are not enough
A resource theory specifies a set of free states \(\mathcal F\) and a collection of allowed channels that preserve this set. Identity and composition should be allowed; tensor-product rules must also be stated for multi-system tasks. A state outside \(\mathcal F\) is a resource only relative to these restrictions. Unrestricted quantum operations could simply prepare the desired state, making the conversion problem trivial.
For entanglement, free states are separable and a physically motivated operation class is LOCC. For basis coherence, free states are diagonal in a declared basis. Different operation classes can preserve those same diagonal states while permitting different transformations. Calling an operation “free” is therefore a substantive modeling choice, not a consequence of a preferred numerical measure. The broad framework is surveyed by Chitambar and Gour.
Take a qubit whose free basis is |0⟩,|1⟩. The state |+⟩ has coherence, but |0⟩ does not. In the X eigenbasis, these roles change. This is not an inconsistency: the laboratory’s reference frame and permitted controls determine which basis has physical significance. For asymmetry under a symmetry group, free states are invariant and free channels are typically chosen covariant with that group; arbitrary basis dephasing is not a substitute for specifying the symmetry.
Worked check 9.1. Is the Hadamard gate free in a resource theory where all diagonal states must remain diagonal?
Reasoned solution
No. It sends the free state |0⟩⟨0| to the coherent state |+⟩⟨+|. The preservation requirement already excludes it.
9.2 Relative-entropy monotones
9.2.1 A general proof and a qubit calculation
Define \(R(\rho)=\inf_{\sigma\in\mathcal F}D(\rho\|\sigma)\). If a channel \(\Lambda\) maps every free state to a free state, then \(R(\Lambda\rho)\le D(\Lambda\rho\|\Lambda\sigma)\le D(\rho\|\sigma)\) for any free \(\sigma\). Taking the infimum proves monotonicity. The first inequality uses free-state preservation; the second is data processing. This short proof shows exactly which physical assumption supports the bound.
For coherence, write \(\Delta\) for complete dephasing in the chosen basis. If \(\delta\) is diagonal, then \(\tr\rho\log\delta=\tr\Delta(\rho)\log\delta\). Adding and subtracting \(S(\Delta\rho)\) gives \(D(\rho\|\delta)=S(\Delta\rho)-S(\rho)+D(\Delta\rho\|\delta)\). Nonnegativity makes the optimum occur at \(\delta=\Delta\rho\). Thus \(C_r(\rho)=S(\Delta\rho)-S(\rho)\).
For \(\rho=\begin{pmatrix}1/2&r/2\\r/2&1/2\end{pmatrix}\), with real \(|r|\le1\), the eigenvalues are \((1\pm r)/2\). Therefore \(C_r=1-h_2((1+r)/2)\). At r=0 the state is free and the value is zero; at r=1 it is |+⟩ and the value is one. The formula was derived for a fixed basis and should not be presented as basis independent.
Worked check 9.2. Compute Cᵣ for r=0.6.
Reasoned solution
The eigenvalues are (0.8,0.2), so Cᵣ=1−h₂(0.2)≈0.278072 bits of relative-entropy coherence.
9.3 Conversion bounds and their limits
9.3.1 Monotones constrain protocols
If \(\rho\) converts deterministically to \(\tau\) by a free channel, every monotone obeys \(R(\rho)\ge R(\tau)\). A violated inequality rules out the conversion. A satisfied inequality alone usually does not prove a protocol exists. Entanglement majorization provided a complete family of inequalities for a special pure-state problem; a single entropy loses part of that ordering information.
If a monotone is additive, an exact conversion from n copies of \(\rho\) to m copies of \(\tau\) implies \(nR(\rho)\ge mR(\tau)\). When the target value is positive, this yields \(m/n\le R(\rho)/R(\tau)\). Relative-entropy coherence is additive because both entropy and dephasing factor on product inputs. Extending an exact bound to approximate asymptotic conversion requires an appropriate continuity argument; it does not follow just by deleting the word “exact.”
Catalysts add another resource question: an auxiliary state participates and must be returned according to a stated criterion. Exact return as an uncorrelated product, approximate return, and returning only the same marginal are different requirements. A resource accounting that hides correlations with the catalyst can produce misleading conclusions. A research calculation should name its free operations, error tolerance, and return condition before optimizing a conversion rate.
Worked check 9.3. What exact conversion bound follows from an input with Cᵣ≈0.278072 per copy to |+⟩ copies?
Reasoned solution
For free operations preserving diagonality, additivity and monotonicity require m/n≤0.278072. This is only a necessary condition; the calculation alone does not construct an exact distillation protocol.
Summary
- A resource is defined relative to states and operations declared free.
- Data processing constructs relative-entropy monotones; for coherence, \(C_r=S(\Delta\rho)-S(\rho)\).
- A conversion bound is not a protocol, and catalyst return conditions must be explicit.
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 is Cᵣ(|+⟩)?
Hint 1
The state is pure and its dephasing is I/2.
Worked solution
Cᵣ=1−0=1 bit.
3. Quick check §
What is Cᵣ(I/2)?
Hint 1
Dephasing leaves it unchanged.
Worked solution
Cᵣ=1−1=0.
4. Worked problem §
Compute Cᵣ for √0.9|0⟩+√0.1|1⟩.
Hint 1
Its entropy is zero; dephasing gives probabilities 0.9 and 0.1.
Worked solution
Cᵣ=h₂(0.1)≈0.468996 bits.
5. Worked problem §
Show that dephasing is idempotent.
Hint 1
Apply it to a matrix already diagonal.
Worked solution
\(\Delta(\Delta\rho)=\Delta\rho\), because the first application removes every off-diagonal element and the second has nothing further to remove.
6. Worked problem §
Prove additivity of Cᵣ on product states.
Hint 1
Use product dephasing and entropy additivity.
Worked solution
\(\Delta(\rho\otimes\tau)=\Delta\rho\otimes\Delta\tau\); subtracting the additive input entropy from the additive dephased entropy gives \(C_r(\rho\otimes\tau)=C_r(\rho)+C_r(\tau)\).
7. Worked problem §
Bound the number of exact |+⟩ outputs from ten copies of the r=0.6 state.
Hint 1
Apply the additive monotone and integer rounding.
Worked solution
m≤10(0.278072), so m≤2. This upper bound does not guarantee that two are achievable.
8. Misconception §
Does one monotone inequality prove a state conversion exists?
Hint 1
Compare an obstruction with a construction.
Worked solution
No. It gives a necessary condition. Sufficiency requires a theorem or an explicit allowed protocol.
9. Misconception §
Do identical free-state sets force identical resource theories?
Hint 1
Consider which channels are allowed.
Worked solution
No. Different operation classes can preserve the same free states while supporting different conversions and achievable rates.
10. Extension §
Explain why a catalyst’s unchanged marginal does not mean it was returned uncorrelated.
Hint 1
Construct a correlated classical pair with fixed local distributions.
Worked solution
Two copies of one fair classical bit each remain uniformly distributed, but their joint state is correlated rather than a product. Returning a catalyst marginal alone can hide correlations and differs from returning the original independent resource.