Resource Theories

A quantum resource theory asks which transformations become possible when a particular capability is scarce. Alice and Bob may lack quantum communication, a laboratory may lack a phase reference, or a computer may have only stabilizer operations. States that supply the missing capability become resources. A useful answer can be an explicit conversion protocol, an impossibility theorem, or a quantitative bound; these are different kinds of result.

We develop finite-dimensional state resources here. The same state can be useful in one theory and free in another. For example, a state diagonal in the energy basis has no coherence in that basis but can still have populations different from thermal equilibrium.

Helpful background. Density operators describes states and conditional measurement outputs; quantum channels supplies completely positive trace-preserving maps; quantum entropy explains the relative entropy used below.

Free states and allowed quantum transformations

Let $\mathcal D(\mathcal H_S)$ be the density operators on a finite-dimensional system $S$, and let

$$\mathcal F_S\subseteq\mathcal D(\mathcal H_S)$$

be the states that can be prepared without consuming the resource. Write $\mathfrak O(S\!\to\!T)$ for the free transformations from $S$ to $T$. A free channel must at least preserve the free set:

$$\Lambda(\mathcal F_S)\subseteq\mathcal F_T.$$

Preserving free states does not uniquely determine the allowed operations. LOCC is smaller than the maximal class of channels that preserve separability. Similarly, operations implemented with a thermal bath and energy-conserving unitary need not exhaust all Gibbs-preserving channels. One must specify how the resource may be manipulated, as well as which states are free.

Several closure properties have direct physical meanings. Identity and composition are needed for the exact conversion relation below to be a preorder. Tensor-product closure determines whether independent free procedures remain free. Free ancillas, classical records, conditioning, and discarding control what selective protocols can do. A phase reference, work battery, or communication channel can itself supply the resource; include it in the physical model when it is used.

Convexity is also an assumption, not a default. If free randomization is allowed and its record may be forgotten, the free set and transformation class are often convex. If the random choice itself is costly or unavailable, convex mixtures need not be free. Topological closure matters when an optimum is approached but not attained.

Channel and measurement resources have different inputs and outputs from state resources. They require transformations of processes or simulation procedures; an instrument also includes its conditional output state. Generalized measurements and instruments explains this distinction.

Exact and Approximate Conversion Preorders

For exact deterministic state conversion, define

$$\rho\succeq_{\mathfrak O}\sigma \quad\Longleftrightarrow\quad \exists\,\Lambda\in\mathfrak O(S\!\to\!T) \ \text{such that}\ \Lambda(\rho)=\sigma.$$

If identity maps are free, then $\rho\succeq_{\mathfrak O}\rho$. If free maps are closed under composition, then

$$\rho\succeq_{\mathfrak O}\sigma, \quad \sigma\succeq_{\mathfrak O}\tau \quad\Longrightarrow\quad \rho\succeq_{\mathfrak O}\tau.$$

The relation is therefore reflexive and transitive, but generally not antisymmetric or total. Two objects can be incomparable, and two distinct objects can be mutually interconvertible.

Declare the trace norm by

$$\lVert X\rVert_1 = \operatorname{Tr}\sqrt{X^\dagger X}.$$

For normalized state outputs, a deterministic approximate conversion error is

$$\begin{aligned} \varepsilon_{\mathfrak O}(\rho\!\to\!\sigma) &= \inf_{\Lambda\in\mathfrak O} T\!\left(\Lambda(\rho),\sigma\right),\\ T(\omega,\sigma) &= \frac12\lVert\omega-\sigma\rVert_1. \end{aligned}$$

The factor $1/2$ fixes the conventional state trace distance. This formula must not be copied unchanged to a subnormalized success branch, channel, measurement, or instrument. A selective branch requires its success probability and an explicit convention for conditional versus subnormalized error. Channel and measurement resources require a declared free supermap or simulability relation and an operationally appropriate process distance.

Approximate convertibility also depends on the order of limits. Allowing an auxiliary dimension to diverge before sending the output error to zero can produce a different claim from fixing the auxiliary system first. This issue is central for approximate catalysts and embezzling constructions.

Monotones, Witnesses, and Complete Criteria

A deterministic resource monotone $M$ for the declared free operations satisfies

$$M(\Lambda(\rho))\leq M(\rho) \qquad \text{for every }\Lambda\in\mathfrak O.$$

Thus $M(\rho)<M(\sigma)$ rules out $\rho\to\sigma$. The converse generally fails: $M(\rho)\geq M(\sigma)$ does not construct a free protocol, and equality of one monotone does not establish interconvertibility. A complete family of monotones or a task-specific theorem is stronger than one scalar diagnostic. A witness can certify that an object lies outside the free set without quantifying a conversion rate.

Throughout this page logarithms are base $2$. For a declared free-state set, the relative-entropy resource is

$$\begin{aligned} D_{\mathcal F}(\rho) &= \inf_{\tau\in\mathcal F}D(\rho\Vert\tau),\\ D(\rho\Vert\tau) &= \operatorname{Tr}\!\left[ \rho\left(\log_2\rho-\log_2\tau\right) \right]. \end{aligned}$$

Set $D(\rho\Vert\tau)=+\infty$ unless $\operatorname{supp}\rho\subseteq\operatorname{supp}\tau$. If $\Lambda$ preserves the free set, then data processing gives

$$\begin{aligned} D_{\mathcal F}(\Lambda\rho) &= \inf_{\omega\in\mathcal F} D(\Lambda\rho\Vert\omega)\\ &\leq \inf_{\tau\in\mathcal F} D(\Lambda\rho\Vert\Lambda\tau)\\ &\leq D_{\mathcal F}(\rho). \end{aligned}$$

The middle line restricts the output optimization to images of free inputs. This proof uses data processing and free-set preservation; an arbitrary geometric distance to $\mathcal F$ is not automatically monotone.

Another common quantity is the generalized robustness

$$R_{\mathcal F}^{\mathrm g}(\rho) = \inf\left\{ s\geq0: \exists\tau\in\mathcal D(\mathcal H),\ \frac{\rho+s\tau}{1+s}\in\mathcal F \right\}.$$

The mixing state $\tau$ is arbitrary here. Standard or free robustness restricts the mixing state and is a different quantity. Witness duals and operational discrimination interpretations require their own finite-dimensional, closedness, and convexity hypotheses.

Vanishing, faithfulness, convexity, deterministic monotonicity, strong monotonicity, additivity, asymptotic continuity, and computability are separate properties. A theorem establishing one of these properties does not automatically establish the others.

Selective Operations, Success Probability, and Strong Monotonicity

Let $\{\Lambda_x\}_x$ be a free selective instrument whose sum is trace preserving. For input $\rho$, define the subnormalized branch $\Lambda_x(\rho)$, its probability, and its normalized conditional state by

$$\begin{aligned} p_x&=\operatorname{Tr}\Lambda_x(\rho),\\ \rho_x&=\frac{\Lambda_x(\rho)}{p_x} \qquad (p_x>0). \end{aligned}$$

When $M$ is strongly monotone for this instrument class,

$$M(\rho)\geq\sum_x p_xM(\rho_x).$$

Deterministic monotonicity of the averaged channel $\sum_x\Lambda_x$ does not by itself prove this branch-averaged inequality. Convexity also does not replace the needed selective-operation statement. The transformation class and monotone must support the exact claim being made.

A selected branch can have more resource than the input without violating strong monotonicity. For example, local filtering can yield a Bell pair from a partially entangled pair on a successful branch. To compare input and output, include the branch probability and the failure state, as in the worked example below.

Normalization conventions are equally important. If $\widetilde\rho_x=\Lambda_x(\rho)$ is subnormalized, then $\operatorname{Tr}\widetilde\rho_x=p_x$. A monotone defined only on normalized states is evaluated on $\rho_x$, not silently on $\widetilde\rho_x$. Conversely, an optimization written for subnormalized operators should not be read as an ordinary state-distance statement.

One-Shot, Asymptotic, and Catalytic Regimes

A one-shot conversion uses one specified input and returns one specified output, perhaps with a declared error or success probability. A finite-block statement fixes a finite number of copies. An iid asymptotic statement concerns a sequence of transformations and an order of limits.

For vanishing trace-distance error, define the iid asymptotic rate by the existence of free maps:

$$\begin{aligned} r_{\mathfrak O}(\rho\!\to\!\sigma) = \sup\bigg\{r\geq0: &\ \exists\{\Lambda_n\},\ \Lambda_n\in\mathfrak O_n,\\ &\lim_{n\to\infty} \frac12\left\lVert \Lambda_n(\rho^{\otimes n}) -\sigma^{\otimes\lfloor rn\rfloor} \right\rVert_1=0 \bigg\}. \end{aligned}$$

This rate counts target copies per input copy while the global output error tends to zero. Define

$$M^\infty(\rho) = \lim_{n\to\infty} \frac{M(\rho^{\otimes n})}{n}$$

only when the limit exists. A converse for vanishing-error conversion needs both a declared tensor-power scaling condition—additivity or a suitable regularization—and a continuity bound strong enough to make the output approximation contribute only $o(n)$. Neither ingredient replaces the other. Such a converse does not establish an achievable protocol or general asymptotic reversibility.

For a fixed declared catalyst $\omega_C$, exact uncorrelated catalytic conversion requires a free map on the joint systems:

$$\begin{aligned} \rho\succeq_{\mathfrak O}^{\,\omega_C}\sigma \quad\Longleftrightarrow\quad &\exists\,\Lambda\in \mathfrak O(S\!\otimes\!C\to T\!\otimes\!C)\ \text{such that}\\ &\Lambda(\rho\otimes\omega_C) =\sigma\otimes\omega_C. \end{aligned}$$

A claim that some catalyst exists must also quantify over the admissible catalyst system $C$ and state $\omega_C$; the catalyst can itself be resourceful even though it is returned. Exact product return is different from returning only the catalyst marginal while allowing correlations, and both differ from approximate return. For approximate return, dimension and correlations matter as well as the distance between the initial and final catalyst. A growing catalyst can permit conversions that a fixed finite catalyst cannot; a small return error on each use need not remain small after many uses. Exact return of a pure catalyst marginal forces the joint output to factor, since a pure marginal cannot be correlated with another system.

Examples of different resources

Resource	Free states and operations	Physical capability
Entanglement	Separable states and a specified LOCC class	Sharing quantum correlations across a named split; see entanglement measures.
Coherence	States diagonal in a fixed basis; the allowed channels require a separate choice	Using phase relations in that basis; see coherence and preferred bases.
Athermality	A Gibbs state at specified Hamiltonian and temperature; for example, thermal operations	Departing from equilibrium, with work storage included when relevant; see thermal operations.
Magic	Stabilizer mixtures and specified stabilizer operations	Supplying operations beyond the stabilizer model; see magic-state distillation.
Asymmetry	States invariant under a chosen group representation, and covariant channels	Supplying a reference for an otherwise unavailable orientation or phase; see Gour and Spekkens.

These theories are not interchangeable. Coherence in an arbitrary basis is not automatically useful for work extraction, and a resource monotone need not equal a gate count, success probability, or conversion rate.

Worked example: coherence under dephasing

Take $\rho_+=|+\rangle\langle+|$, with $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$. Let $\Delta$ remove off-diagonal matrix elements in this computational basis. A dephasing-covariant incoherent operation, abbreviated DIO, is a channel satisfying $\Delta\Lambda=\Lambda\Delta$. The channel

$$\Phi_\eta(\rho)=\begin{pmatrix} \rho_{00}&\eta\rho_{01}\\ \eta\rho_{10}&\rho_{11} \end{pmatrix},\qquad -1\le\eta\le1,$$

belongs to this class. One implementation applies $I$ with probability $(1+\eta)/2$ and $Z$ with probability $(1-\eta)/2$, which proves complete positivity and trace preservation. It also makes clear why negative $\eta$ is allowed: a phase reversal can accompany the decay of coherence magnitude.

The relative entropy of coherence is

$$C_r(\rho)=\inf_{\tau\text{ diagonal}}D(\rho\Vert\tau) =S(\Delta\rho)-S(\rho).$$

The last identity follows by writing $D(\rho\Vert\tau)=S(\Delta\rho)-S(\rho)+D(\Delta\rho\Vert\tau)$ and minimizing at $\tau=\Delta\rho$. Because a DIO channel preserves diagonal states, the earlier data-processing argument proves that $C_r$ cannot increase.

For $\eta=0.6$, the output is

$$\Phi_{0.6}(\rho_+)=\begin{pmatrix}0.5&0.3\\0.3&0.5\end{pmatrix}, \qquad \operatorname{spec}\Phi_{0.6}(\rho_+)=(0.8,0.2).$$

Its diagonal state is $I/2$, so $C_r=1-h_2(0.8)\simeq0.278072$ bits, reduced from the input value of one bit. The separate diagnostic $C_{l_1}=\sum_{i\ne j}|\rho_{ij}|$ equals $0.6$. The two values are different functions of the state, not two estimates of one amount. We use $C_r$ for the general DIO monotonicity argument; the $l_1$ calculation alone supplies no general conversion criterion.

Worked example: converting a partially entangled pair

Alice and Bob share

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

and want a Bell pair $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$ using LOCC, with no additional entanglement. Three different answers apply to three different tasks.

Exact deterministic conversion of one copy is impossible. Nielsen’s theorem requires the sorted input Schmidt probabilities to be majorized by the target probabilities. For vectors $p,q$, this means $\sum_{i=1}^kp_i\le\sum_{i=1}^kq_i$ for every $k$, with equal total sums. Here the first inequality would be $0.9\le0.5$, which fails.

Exact probabilistic conversion succeeds with probability at most $0.2$. Vidal’s pure-state formula minimizes the ratios of Schmidt tail sums. Here the two ratios are $1$ and $0.1/0.5$, giving $p_{\max}=0.2$. A local two-outcome filter on Alice actually attains this value:

$$K_s=\begin{pmatrix}1/3&0\\0&1\end{pmatrix}, \qquad K_f=\begin{pmatrix}\sqrt8/3&0\\0&0\end{pmatrix}.$$

They satisfy $K_s^\dagger K_s+K_f^\dagger K_f=I$. The successful vector is $\sqrt{0.1}(|00\rangle+|11\rangle)$, with norm squared $0.2$, while failure produces $|00\rangle$ with probability $0.8$. The mean output entanglement is $0.2$ ebits, smaller than the input’s $h_2(0.1)\simeq0.4690$ ebits.

Many identical copies yield $h_2(0.1)$ Bell pairs per input copy asymptotically. Collective pure-state concentration achieves this larger rate with error tending to zero. Repeating the single-pair filter is a valid but suboptimal many-copy strategy. Neither the asymptotic rate nor the single-pair success probability is a guarantee of a particular output count from a finite block. Entanglement distillation develops concrete protocols.

Exercises

1. Changing the preferred basis

For $|+\rangle$, compute relative-entropy coherence in the computational basis and in the $X$ eigenbasis. Explain why the answers differ without any physical change to the state.

Solution

The pure state has entropy zero. Computational-basis dephasing gives $I/2$, so $C_r=1$. Dephasing in the $X$ eigenbasis leaves $|+\rangle\langle+|$ unchanged, so $C_r=0$. Changing the preferred basis changes which states and controls count as free. It changes the resource theory, rather than revealing an inconsistency in the measure.

2. Test the preorder assumptions

Prove reflexivity and transitivity of $\succeq_{\mathfrak O}$ from identity and composition closure. Then give a declared transformation set without composition closure for which calling the relation a preorder is invalid.

Solution

If $\operatorname{id}_S\in\mathfrak O(S\!\to\!S)$, then $\operatorname{id}_S(\rho)=\rho$, so $\rho\succeq_{\mathfrak O}\rho$. If $\Lambda(\rho)=\sigma$ and $\Gamma(\sigma)=\tau$, composition closure gives $\Gamma\circ\Lambda\in\mathfrak O$ and

$$(\Gamma\circ\Lambda)(\rho)=\tau,$$

which proves transitivity.

For a counterexample, consider three perfectly distinguishable classical states $0,1,2$ embedded as diagonal quantum states. Let $f$ be a measure-and-prepare channel with $f(0)=1$, $f(1)=2$, and $f(2)=2$, and declare $\mathfrak O=\{\operatorname{id},f\}$ while excluding $f^2$. Then $0\succeq1$ and $1\succeq2$, but neither admitted map sends $0$ to $2$. The relation is not transitive, exactly because the declared set is not closed under composition.

3. Propagate uncertainty in the coherence parameter

Suppose the dephasing parameter has central value $0.600$ and standard uncertainty $0.010$. Use first-order uncertainty propagation to estimate the uncertainties of the output matrix, its eigenvalues, $C_r$, and $C_{l_1}$. The two off-diagonal entries depend on the same uncertain parameter; do not treat their errors as independent.

Solution

Multiplying each off-diagonal entry of $\rho_+$ by $\eta=0.600\pm0.010$ gives

$$\Phi_\eta(\rho_+) = \begin{pmatrix} 0.5&0.300\pm0.005\\ 0.300\pm0.005&0.5 \end{pmatrix}.$$

Its eigenvalues are $(1\pm\eta)/2$, hence $0.800\pm0.005$ and $0.200\mp0.005$. Since dephasing leaves the maximally mixed diagonal state,

$$C_r=1-h_2(0.8)=0.278071905\ldots\ \text{bits}.$$

At $\eta=0.6$, $dC_r/d\eta=1$, so first-order propagation gives standard uncertainty $0.010$ bit. The $l_1$ sum contains two entries of magnitude $\eta/2$, giving $C_{l_1}=\eta=0.600\pm0.010$. Relative-entropy coherence is measured in bits and has an information-theoretic definition; $l_1$ coherence is dimensionless and has a different operational meaning. Their numerical values cannot be compared as though they were two estimates of one quantity.

4. Prove relative-entropy monotonicity

Prove $D_{\mathcal F}(\Lambda\rho)\leq D_{\mathcal F}(\rho)$ from data processing and free-set preservation, retaining the support condition.

Solution

For every free $\tau\in\mathcal F$, preservation gives $\Lambda\tau\in\mathcal F$. Therefore

$$\begin{aligned} D_{\mathcal F}(\Lambda\rho) &=\inf_{\omega\in\mathcal F}D(\Lambda\rho\Vert\omega)\\ &\leq D(\Lambda\rho\Vert\Lambda\tau)\\ &\leq D(\rho\Vert\tau), \end{aligned}$$

where the last line is quantum relative-entropy data processing. Taking the infimum over free $\tau$ proves the claim. The definition uses $D(\rho\Vert\tau)=+\infty$ when $\operatorname{supp}\rho\nsubseteq\operatorname{supp}\tau$; CPTP maps can restore support overlap, so the extended-real data-processing inequality remains the correct statement.

5. Include the unsuccessful branch

A local filter returns a Bell pair with probability $0.20$ and a product failure state otherwise. Compute the average output entanglement and compare it with the $0.468996$-ebit input.

Solution

The accepted Bell branch contains one ebit and the product failure branch contains zero. The branch-averaged output is

$$0.20(1\ \text{ebit})+0.80(0\ \text{ebits}) =0.20\ \text{ebits}.$$

Strong monotonicity requires $0.468996\geq0.20$, which holds. The one-ebit accepted branch is therefore compatible with monotonicity; reporting it without the $20\%$ probability and failure state would be the error.

6. Separate three entanglement regimes

For an input with Schmidt probabilities $(0.9,0.1)$ and a Bell-pair target, verify the majorization test, tail-ratio bound and entropy rate. Explain why passing an entropy comparison alone would not prove an exact one-copy conversion.

Solution

For deterministic exact one-copy LOCC, Nielsen’s criterion requires $\lambda^\psi\prec\lambda^\Phi$. The first partial sum fails:

$$0.9\nleq0.5.$$

Thus deterministic one-copy conversion is impossible. For probabilistic exact one-copy LOCC, Vidal’s tail ratios are $1$ for the full sum and $0.1/0.5=0.2$ for the second tail, so $p_{\max}=0.20$. For iid asymptotic pure-state distillation with vanishing trace-distance error,

$$h_2(0.1)=0.468995593\ldots$$

Bell pairs are obtained per input copy. These are respectively an impossibility result, an optimal success probability, and an asymptotic rate. None is a finite-block yield.

7. Derive a monotone converse bound

Given $M^\infty(\rho)=0.4$ and $M^\infty(\sigma)=0.8$, derive the converse bound $r\leq0.5$ under explicitly supplied assumptions. Explain why it does not prove achievability.

Solution

Assume $M$ is monotone for the admissible asymptotic maps, has the normalization used in the supplied values, and obeys the tensor-power scalings

$$\begin{aligned} M(\rho^{\otimes n}) &=nM^\infty(\rho)+o(n),\\ M(\sigma^{\otimes m_n}) &=m_nM^\infty(\sigma)+o(n). \end{aligned}$$

Additivity can supply these scalings, but it does not control approximation error. Separately assume an asymptotic-continuity bound that turns the vanishing trace-distance error into an $o(n)$ correction. Monotonicity and continuity then give

$$\begin{aligned} M(\rho^{\otimes n}) &\geq M\!\left(\Lambda_n(\rho^{\otimes n})\right)\\ &\geq M(\sigma^{\otimes m_n})-o(n), \end{aligned}$$

and hence

$$nM^\infty(\rho)+o(n) \geq m_nM^\infty(\sigma)-o(n).$$

Dividing by $nM^\infty(\sigma)$ and taking the limit superior yields

$$r =\limsup_{n\to\infty}\frac{m_n}{n} \leq\frac{0.4}{0.8}=0.5.$$

This is a converse: rates above $0.5$ are excluded. It supplies no sequence of free maps attaining $0.5$ or any positive rate, so it does not prove achievability or reversibility.

8. An entanglement catalyst

Consider pure bipartite states with sorted Schmidt probabilities

$$p=(0.4,0.4,0.1,0.1),\qquad q=(0.5,0.25,0.25,0).$$

Show that direct deterministic LOCC conversion is impossible, but becomes possible with a pure catalyst of Schmidt probabilities $c=(0.6,0.4)$, returned exactly and uncorrelated. Must exact return of this pure catalyst’s marginal also imply product return?

Solution

Direct majorization fails at the second partial sum, $0.8>0.75$. After tensoring with the catalyst, the sorted probabilities are

$$\begin{aligned} (p\otimes c)^\downarrow &=(0.24,0.24,0.16,0.16, 0.06,0.06,0.04,0.04),\\ (q\otimes c)^\downarrow &=(0.30,0.20,0.15,0.15, 0.10,0.10,0,0). \end{aligned}$$

The cumulative sums are

$$\begin{aligned} P&=(0.24,0.48,0.64,0.80, 0.86,0.92,0.96,1),\\ Q&=(0.30,0.50,0.65,0.80, 0.90,1,1,1). \end{aligned}$$

Every $P_k\le Q_k$, so Nielsen’s theorem guarantees the joint conversion to the exact product target. The catalyst changes the ordering of joint Schmidt probabilities without being consumed. It is entangled and was not available for free.

A pure catalyst marginal cannot be correlated with the rest of the output, so exact marginal return of this catalyst forces product return. This implication need not hold for a mixed catalyst. Approximate return is also a different condition; a small marginal error alone does not specify the joint target error or control repeated use.

References

• T. Baumgratz, M. Cramer, and M. B. Plenio, “Quantifying Coherence,” Physical Review Letters 113, 140401 (2014), doi:10.1103/PhysRevLett.113.140401.
• F. G. S. L. Brandão and G. Gour, “Reversible Framework for Quantum Resource Theories,” Physical Review Letters 115, 070503 (2015), with erratum 115, 199901 (2015), doi:10.1103/PhysRevLett.115.070503.
• E. Chitambar and G. Gour, “Quantum Resource Theories,” Reviews of Modern Physics 91, 025001 (2019), doi:10.1103/RevModPhys.91.025001.
• B. Coecke, T. Fritz, and R. W. Spekkens, “A Mathematical Theory of Resources,” Information and Computation 250, 59–86 (2016), doi:10.1016/j.ic.2016.02.008.
• G. Gour and R. W. Spekkens, “The Resource Theory of Quantum Reference Frames: Manipulations and Monotones,” New Journal of Physics 10, 033023 (2008), doi:10.1088/1367-2630/10/3/033023.
• D. Jonathan and M. B. Plenio, “Entanglement-Assisted Local Manipulation of Pure Quantum States,” Physical Review Letters 83, 3566–3569 (1999), doi:10.1103/PhysRevLett.83.3566.
• M. Lostaglio, “An Introductory Review of the Resource Theory Approach to Thermodynamics,” Reports on Progress in Physics 82, 114001 (2019), doi:10.1088/1361-6633/ab46e5.
• M. A. Nielsen, “Conditions for a Class of Entanglement Transformations,” Physical Review Letters 83, 436–439 (1999), doi:10.1103/PhysRevLett.83.436.
• R. Takagi and B. Regula, “General Resource Theories in Quantum Mechanics and Beyond: Operational Characterization via Discrimination Tasks,” Physical Review X 9, 031053 (2019), doi:10.1103/PhysRevX.9.031053.
• V. Veitch, S. A. H. Mousavian, D. Gottesman, and J. Emerson, “The Resource Theory of Stabilizer Quantum Computation,” New Journal of Physics 16, 013009 (2014), doi:10.1088/1367-2630/16/1/013009.
• G. Vidal, “Entanglement of Pure States for a Single Copy,” Physical Review Letters 83, 1046–1049 (1999), doi:10.1103/PhysRevLett.83.1046.
• G. Vidal, “Entanglement Monotones,” Journal of Modern Optics 47, 355–376 (2000), doi:10.1080/09500340008244048.