Thermodynamics and the structure of quantum theory

In this section we introduce the two postulates that express key operational concepts from thermodynamics. The first postulate is motivated by the universality of thermodynamics and the distinction between microscopic and macroscopic behaviour. At first we consider the universality of thermodynamics, in the sense that thermodynamics is a very general theory whose basic principles can be applied to many possible implementations, as already noticed by N Carnot [44]:

'In order to consider in the most general way the principle of the production of motion by heat, it must be considered independently of any mechanism or any particular agent. It is necessary to establish principles applicable not only to steam engines but to all imaginable heat-engines, whatever the working substance and whatever the method by which it is operated.'

Recalling von Neumann's thought experiment in the case of quantum theory, we can think of thermodynamical protocols (which will ultimately also include heat engines) as acting on a given ensemble, defined as a probabilistic mixture of pure states chosen from a fixed basis. If we interpret ensembles with different choices of basis as different 'working substances', then Carnot's principle should apply: protocols that can be implemented on one ensemble (say, ensemble 1) can also be implemented on the other (say, ensemble 2)¹³ . In quantum theory, this universality is ensured by the existence of unitary transformations: all orthonormal bases can be translated into each other by a unitary and therefore reversible map. In this sense, the state of ensemble 1 can in principle be transferred to ensemble 2, then the thermodynamic protocol of ensemble 2 can be performed (if we have also transformed the projectors describing the membranes accordingly), and then one can transform back. Even if this cannot always be achieved in practice, the corresponding unitary symmetry of the quantum state space (considered as passive transformations between different descriptions) enforces the aforementioned universality¹⁴ .

This universality of implementation, as well as independence of the choice of labels and descriptions, should continue to hold in all generalised theories that we consider. An orthonormal basis from quantum theory is nothing else than a set of perfectly distinguishable pure states, i.e. an n-frame. Therefore, in our generalised theories, we expect that this universality of implementation is achieved by the existence of reversible transformations that, in analogy to unitary maps, transform any given n-frame into any other:

Postulate 1. For each $n\in {\mathbb{N}}$, all sets of n perfectly distinguishable pure states are equivalent. That is, if $\{{\omega }_{1},\ldots ,{\omega }_{n}\}$ and $\{{\varphi }_{1},\ldots ,{\varphi }_{n}\}$ are two such sets, then there exists a reversible transformation T with $T{\omega }_{j}={\varphi }_{j}$ for all j.

Furthermore, postulate 1 expresses a physical property that is crucial for thermodynamics: that of microscopic reversibility. Many characteristic properties of thermodynamics arise from limited experimental access to the microscopic degrees of freedom, which by themselves undergo reversible time evolution. This reversibility, for example, forbids evolving two microstates into one, which is at the heart of the non-decrease of entropy. If the experimenter had full access to the microscopic degrees of freedom, then he or she could convert any state of maximal knowledge to any other one as long as he or she preserved distinguishability. Postulate 1 formalises this microscopic basis of thermodynamics by demanding the existence of 'enough' distinguishability-preserving, microscopic transformations T, which can be understood as reversible time evolutions.

Postulate 1 has substantial information-theoretical justifications and consequences. The basic concepts of both thermodynamics and information processing are independent of the choice of implementation. For information processing this is formalised by the Turing machine which admits a multitude of physical realisations. Perfectly distinguishable pure states can be taken as bits, and postulate 1 expresses that all bits (or their higher-dimensional analogues) are equivalent. It is for this reason that postulate 1 was called generalised bit symmetry in [34], and its restriction to pairs of distinguishable states was called bit symmetry in [64]. Starting with Landauer's principle, 'thermodynamics of computation' [92] has become a fruitful paradigm that relates the two apparently disjoint fields. The two complementary interpretations of postulate 1 are one instance of this.

Now we turn to our second postulate. We are looking for theories very similar to the thermodynamics we are used to; thus it is essential that we can adopt basic notions of standard thermodynamics unchanged or with only very small alterations. Two such notions of great importance are (Shannon) entropy $S=-{k}_{B}{\sum }_{j}{p}_{j}\mathrm{log}{p}_{j}$ and majorization theory. In classical and quantum thermodynamics, these notions operate on the coefficients in a decomposition of a state into perfectly distinguishable pure states (in quantum theory, the eigenvalues). In order to not change thermodynamic theory too much, we would also like this to be possible in our more general state spaces. Thus, we demand that every state has a convex decomposition into perfectly distinguishable pure states.

Note that this was indeed one of our assumptions in von Neumann's thought experiment in section 3. There, it allowed us to realise any state ω as a 'quasiclassical ensemble', i.e. as an ensemble of states that behave like classical labels. This gives us a further justification of our second postulate: thermodynamic (thought) experiments require that states have an ensemble interpretation. An unambiguous notion of 'counting of microstates' demands that the ensembles consist of perfectly distinguishable, pure states. Without this, obtaining a phenomenological thermodynamics for which the theory is the underlying microscopic theory seems problematic. Thus, our second postulate is

Postulate 2. Every state $\omega \in {{\rm{\Omega }}}_{A}$ has a convex decomposition $\omega ={\sum }_{j}{p}_{j}{\omega }_{j}$ into perfectly distinguishable pure states ${\omega }_{j}$.

It is tempting to interpret the two postulates as reflecting the microscopic and the macroscopic aspects of thermodynamics, respectively: while postulate 1 describes microscopic reversibility of the pure states that may describe single particles in thermodynamics, postulate 2 ensures that mixed states can be interpreted macroscopically as descriptions of quasiclassical ensembles, composed of a large number of particles that are separately in unknown but distinguishable microstates.

We will not introduce any further postulates. In particular, we will not make any assumptions on the composition of systems. All our results are therefore independent from notions like tomographic locality [51] (which is arguably dispensable in many important situations [81]) or purification [56] (which is a rather strong assumption); we do not assume either of the two.

Postulates 1 and 2 have been analysed in [43], but in a different context: instead of investigating thermodynamics, the goal in [43] was to obtain a reconstruction of quantum theory, by supplementing postulates 1 and 2 with further postulates. Some of the insights from [43] will be important here, and are therefore briefly discussed below. Starting with section 5.4, we will also obtain new results that are interesting in a thermodynamic context.

In contrast to Hilbert space, there is no apriori notion of inner product for GPTs. However, as shown in [64], we get a natural inner product $\langle \cdot ,\,\cdot \rangle$ as a consequence of postulates 1 and 2: it satisfies $\langle T\omega ,T\varphi \rangle =\langle \omega ,\varphi \rangle$ for all reversible transformations T, and $0\leqslant \langle \omega ,\varphi \rangle \leqslant 1$ for all states $\omega ,\varphi \in {{\rm{\Omega }}}_{A}$. Furthermore, $\langle \omega ,\omega \rangle =1$ for all pure $\omega \in {{\rm{\Omega }}}_{A}$ and $\langle \varphi ,\varphi \rangle \lt 1$ for all mixed $\varphi \in {{\rm{\Omega }}}_{A}$, and $\langle \omega ,\varphi \rangle =0$ if $\omega ,\varphi \in {{\rm{\Omega }}}_{A}$ are perfectly distinguishable. Thus, all perfectly distinguishable states are orthogonal, as in quantum theory.

Moreover, the cone of unnormalized states becomes self-dual with this choice of inner product. In particular, every effect e can be taken as a vector in ${A}_{+}$, such that $e(\omega )=\langle e,\omega \rangle$. In standard quantum theory, this is the Hilbert Schmidt inner product on the real vector space of Hermitian matrices: $\langle X,Y\rangle =\mathrm{tr}({XY})$ for $X={X}^{\dagger }$, $Y={Y}^{\dagger }$.

Quantum theory has more structure: the convex set of density matrices ${{\rm{\Omega }}}_{A}$ has faces¹⁵ , and these faces are in one-to-one correspondence to subspaces of Hilbert space (namely, a face F contains all density matrices that have support on the corresponding Hilbert subspace). To every face F, we can associate a number $| F|$ which is the dimension of the corresponding Hilbert subspace, and $F\subsetneq G$ implies $| F| \lt | G|$. Every face F is generated by $| F|$ pure and perfectly distinguishable states in F (an $| F|$-frame in F), and every (smaller) frame that is a subset of F can be completed, or extended, to a frame which has $| F|$ elements and thus generates F.

In all theories that satisfy postulates 1 and 2, all these properties hold in complete analogy [43]. However, since faces do not any more correspond to Hilbert spaces, the numbers $| F|$ do not have an interpretation as the dimension of a subspace. Instead, we call $| F|$ the rank of F. If von Neumann's thought experiment is supposed to make sense for these theories, we need a way to formalise the working of a semipermeable membrane, which in quantum theory is done via projective measurements.

Since we are dealing with unnormalized states, the corresponding analogue in GPTs will be formulated in terms of the set of unnormalized states ${A}_{+}$. As one can see in the case of the gbit, it is not automatic that we have any notion of 'projective measurements' for any given state space. However, postulates 1 and 2 turn out to ensure that projective measurements exist. For any face F of ${A}_{+}$ (the non-negative multiples of the corresponding face of ${{\rm{\Omega }}}_{A}$), consider the orthogonal projector P_F onto the span of F. One can show that P_F is positive, i.e. maps (unnormalized) states to (unnormalized) states [43]. Moreover, P_F does not disturb the states in the face F.

Thus, to a given set of mutually orthogonal faces ${F}_{1},\ldots ,{F}_{m}$ such that $| {F}_{1}| \,+\text{}\ldots \,+\,| {F}_{m}| ={N}_{A}$, we can associate an instrument with transformations ${T}_{i}:= {P}_{{F}_{i}}$, which describes a projective measurement, as in a semipermeable membrane. Transformation T_i leaves the states in face F_i unperturbed, but fully blocks out states in the other faces, i.e. ${T}_{i}\omega =0$ for $\omega \in {F}_{j}$, $i\ne j$. In standard quantum theory, these transformations are ${P}_{{F}_{i}}\rho ={\pi }_{i}\rho {\pi }_{i}$, where ${\pi }_{i}$ is the orthogonal Hilbert space projector onto the ith Hilbert subspace. The rank condition becomes $\mathrm{tr}({\pi }_{1})+\,\ldots \,+\,\mathrm{tr}({\pi }_{m})={N}_{A}$ (the total Hilbert space dimension), and mutual orthogonality is ${\pi }_{i}{\pi }_{j}={\delta }_{{ij}}{\pi }_{i}$. We will show in section 5.4 that the mutually orthogonal faces replace the eigenspaces from quantum theory and that the projective measurement described here can be interpreted as measuring an observable.

The Hilbert space projector ${\pi }_{i}$ therefore also has an interpretation as an effect in standard quantum theory: it yields the probability of outcome i in the projective measurement on a state ρ, namely $\mathrm{tr}({\pi }_{i}\rho )$. The analogous effect in a GPT that satisfies postulates 1 and 2, corresponding to a face F, is

$$\begin{eqnarray*}&&{u}_{F}:= {P}_{F}{u}_{A}\end{eqnarray*}$$

(identifying the effect u_A with a vector via the inner product). The effect u_F is sometimes called the 'projective unit' of F. In quantum theory, we can write ${\pi }_{i}={\sum }_{j}| {\psi }_{j}\rangle \langle {\psi }_{j}|$, where the $| {\psi }_{j}\rangle$ are an orthonormal basis of the corresponding Hilbert subspace. The same turns out to be true in our GPTs: we have

$$\begin{eqnarray}&&{u}_{F}=\displaystyle \sum _{j=1}^{| F| }{\omega }_{i},\end{eqnarray} \tag{ 4 }$$

where ${\omega }_{1},\ldots ,{\omega }_{| F| }$ is any frame that generates F. Therefore, the probability to obtain outcome i in the projective measurement above on state ω is $\langle {u}_{{F}_{i}},\omega \rangle =\langle {u}_{A},{P}_{{F}_{i}}\omega \rangle$.

It is easy to see that both quantum and classical state spaces satisfy postulates 1 and 2. By a 'classical state space', we mean a state space that consists of discrete probability distributions. Concretely, for any number $N\in {\mathbb{N}}$ of mutually exclusive alternatives, consider the state space

$$\begin{eqnarray*}&&{\rm{\Omega }}:= \{({p}_{1},\ldots ,{p}_{N})| \,{p}_{i}\geqslant 0,\,\displaystyle \sum _{i}{p}_{i}=1\}.\end{eqnarray*}$$

Any pure state is given by a deterministic probability vector, i.e. ${\omega }_{i}=(0,\ldots ,0,1,0,\ldots ,0)$ (where 1 is on the ith place). If we have two equally sized sets of such vectors (as in postulate 1), then there is always a permutation that maps one set to the other. In fact, the reversible transformations correspond to the permutations of the entries. Postulate 2 is then simply the statement that

$$\begin{eqnarray*}&&({p}_{1},\ldots ,{p}_{N})={p}_{1}{\omega }_{1}+{p}_{2}{\omega }_{2}\,+\,\ldots \,+\,{p}_{N}{\omega }_{N}.\end{eqnarray*}$$

Which state spaces are there, in addition to standard complex quantum theory and classical probability theory, that satisfy postulates 1 and 2? We think that this question is very difficult to answer. Thus, we formulate the following

Open problem 1. Classify all state spaces that satisfy postulates 1 and 2.

From the results in [43], we know which state spaces satisfy postulates 1 and 2 and one additional property: the absence of third-order interference. The notion of higher-order interference has been introduced by Sorkin [76], and has since been the subject of intense theoretical [93, 95, 96] and experimental [97–102] interest.

For the main idea, think of three mutually exclusive alternatives in quantum theory (such as three slits in a triple-slit experiment), described by orthogonal projectors ${\pi }_{1},{\pi }_{2},{\pi }_{3}$. The event that alternative 1 or alternative 2 takes place is described by the projector ${\pi }_{12}={\pi }_{1}+{\pi }_{2};$ similarly, we have ${\pi }_{13}$, ${\pi }_{23}$ and ${\pi }_{123}$. Their actions on density matrices are described by superoperators

$$\begin{eqnarray*}&&\rho \mapsto {P}_{12}(\rho ):= {\pi }_{12}\rho {\pi }_{12}\end{eqnarray*}$$

(and similarly for the other projectors). As a consequence, we obtain that ${P}_{12}\ne {P}_{1}+{P}_{2}$, which expresses the phenomenon of interference. However, it is easy to check that

$$\begin{eqnarray}&&{P}_{123}={P}_{12}+{P}_{13}+{P}_{23}-{P}_{1}-{P}_{2}-{P}_{3},\end{eqnarray} \tag{ 5 }$$

which means that interference over three alternatives can be reduced to contributions from interferences of pairs of alternatives. Similar identities hold for an arbitrary number $n\geqslant 4$ of alternatives: quantum theory admits only pairwise interference, and no 'third-order interference' which would be characterised by a violation of this equality.

In the context of postulates 1 and 2, we have an analogous notion of orthogonal projectors, and thus we can consider (5) and its generalisation to $n\geqslant 4$ alternatives on a state space with $N\geqslant n$ perfectly distinguishable states. Postulating this 'absence of third-order interference' in addition to postulates 1 and 2 gives us the following:

Theorem 2 The possible state spaces which satisfy postulates 1 and 2 and which do not admit third-order interference, in addition to classical state spaces, are the following. First, for $N\geqslant 4$ perfectly distinguishable states, there are only three possibilities:

• Standard complex quantum theory.
• Quantum theory over the real numbers. That is, only real entries are allowed in the $N\times N$ density matrices.
• Quantum theory over the quaternions. The state spaces are the self-adjoint $N\times N$ quaternionic matrices of unit trace.

For $N=3$ perfectly distinguishable states, all of the above and one exceptional solution are possible, namely quantum theory over the octonions (but only for the case of $3\times 3$ unit trace density matrices).

For $N=2$ (the 'bit' case), we have the $d$-dimensional Bloch ball state spaces ${{\rm{\Omega }}}_{d}:= \{{(1,r)}^{T}| r\in {{\mathbb{R}}}^{d},\parallel r\parallel \leqslant 1\}$ with $d\geqslant 2$. They are analogous to the standard Bloch ball ${{\rm{\Omega }}}_{3}$ of quantum theory, with very similar descriptions of effects etc. Their group of reversible transformations may either be $\mathrm{SO}(d)$ (which corresponds to $\mathrm{PU}(2)$ for $d=3$), or some subgroup of ${\rm{O}}(d)$ which is transitive on the sphere (such as $\mathrm{SU}(2)$ for $d=4$).

Mathematically, these examples correspond to the state spaces of the finite-dimensional irreducible formally real Jordan algebras [24, 43]. We do not know whether there are theories that satisfy postulates 1 and 2 but admit higher-order interference and therefore do not appear on this list. In theorem 12, we will show that the question whether a theory has third-order interference is related to the properties of its Rényi entropies.

A central part of physics are observables and how they can be measured. In standard quantum theory, we can introduce observables in two different ways, which both equivalently lead to the prescription that observables are described by Hermitian operators/matrices.

First, in finite dimensions, we can characterise observables as those objects that linearly assign real expectation values to states. In the case of quantum theory it follows that observables are represented by matrices X, and Hermiticity $X={X}^{\dagger }$ implies that expectation values $\mathrm{tr}(\rho X)$ are always real. Linearity is enforced by the statistical interpretation of states, for the same reason that effects in GPTs are linear.

Second, we can introduce observables by saying that there is a projective measurement ${\pi }_{1},\ldots ,{\pi }_{n}$ that measures this observable, and which has outcomes ${x}_{1},\ldots ,{x}_{n}\in {\mathbb{R}}$. This leads to the Hermitian operator $X={\sum }_{i=1}^{n}{x}_{i}{\pi }_{i}$. Since every Hermitian operator can be diagonalized, these two definitions are equivalent.

Our two postulates provide the structure to introduce observables in a completely analogous way. First, using the inner product, we can define observables as linear maps of the form

$$\begin{eqnarray*}&&\omega \mapsto \langle x,\omega \rangle \end{eqnarray*}$$

and thus identify them with elements $x\in A$ of the vector space that carries the states (as in quantum theory, where this vector space is the space of Hermitian matrices). As noticed in [62], every such vector has a representation of the form

$$\begin{eqnarray}&&x=\displaystyle \sum _{i}{x}_{i}{u}_{i},\end{eqnarray} \tag{ 6 }$$

where the u_i are projective units corresponding to mutually orthogonal faces F_i, ${x}_{i}\in {\mathbb{R}}$, and ${x}_{i}\ne {x}_{j}$ for $i\ne j$. The analogy with quantum theory goes even further: due to (4), we have $x={\sum }_{i}{x}_{i}{\sum }_{j}{\omega }_{i}^{(j)}$, whenever ${\omega }_{i}^{(1)},\ldots ,{\omega }_{i}^{(| {F}_{i}| )}$ is a frame on F_i. This corresponds to the identity $X={\sum }_{i}{x}_{i}{\sum }_{j}| {\psi }_{i}^{(j)}\rangle \langle {\psi }_{i}^{(j)}|$ in standard quantum theory. In analogy to quantum theory we will call the F_i eigenfaces and the x_i eigenvalues. To further justify this terminology, note that the x_i are eigenvalues of the map ${\sum }_{i}{x}_{i}{P}_{i}$, where P_i are the orthogonal projectors onto the spans of the faces F_i.

Theorem 3. If postulates 1 and 2 hold, then every element $x\in A$ has a representation of the form $x={\sum }_{j=1}^{n}{x}_{j}{u}_{j}$ where ${x}_{j}\in {\mathbb{R}}$ are pairwise different and the ${u}_{j}$ are the projective units of pairwise orthogonal faces ${F}_{j}$ such that ${\sum }_{j}{u}_{j}={u}_{A}$. This decomposition $x={\sum }_{j=1}^{n}{x}_{j}{u}_{j}$ is unique up to relabelling. In analogy to quantum theory, we will call the ${x}_{j}$ eigenvalues and the ${F}_{j}$ eigenfaces.

Furthermore, for every real function $f$ with suitable domain of definition, we can define

$$\begin{eqnarray}&&f(x):= \displaystyle \sum _{j=1}^{n}f({x}_{j}){u}_{j}\end{eqnarray} \tag{ 7 }$$

as in spectral calculus.

If ${P}_{j}$ is the orthogonal projector onto the span of ${F}_{j}$, then $({P}_{1},\ldots ,{P}_{n})$ is a well-defined instrument with induced measurement $({u}_{1},\ldots ,{u}_{n})$ which leaves the elements of $\mathrm{span}({F}_{j})$ invariant:

$$\begin{eqnarray*}&&{P}_{j}(\omega )={\delta }_{{jk}}\cdot \omega \quad {\rm{for}}\,{\rm{all}}\,{\rm{}}\omega \in {F}_{k}.\end{eqnarray*}$$

In analogy to quantum theory, we will call this instrument the projective measurement of the observable $x$.

We will give a proof in the appendix ¹⁶ . Equation (7) allows us to define a notion of entropy, in full analogy to quantum mechanics.

Definition 4 If A is a state space that satisfies postulates 1 and 2, we define the ${\rm{spectral}}\ {\rm{entropy}}$ for any state $\omega \in {{\rm{\Omega }}}_{A}$ as

$$\begin{eqnarray*}&&S(\omega ):= -\displaystyle \sum _{i}{p}_{i}\mathrm{log}{p}_{i},\end{eqnarray*}$$

where $\omega ={\sum }_{i}{p}_{i}{\omega }_{i}$ is any convex decomposition of ω into pure and perfectly distinguishable states ${\omega }_{i}$, and $0\mathrm{log}0:= 0$.

Theorem 3 tells us that this definition is independent of the choice of decomposition: it is easy to check that

$$\begin{eqnarray*}&&S(\omega )=-\langle \omega ,\mathrm{log}\omega \rangle ,\end{eqnarray*}$$

where $\mathrm{log}\omega$ is understood in the sense of spectral calculus as in (7). The right-hand side is manifestly independent of the decomposition. It can also be written $S(\omega )={u}_{A}(\eta (\omega ))$, where $\eta (x)=-x\mathrm{log}x$ for $x\gt 0$ and $\eta (0)=0$. In particular,

$$\begin{eqnarray}&&\omega \,{\rm{is}}\,{\rm{a}}\,{\rm{pure}}\,{\rm{state}}\iff S(\omega )=0.\end{eqnarray} \tag{ 8 }$$

To see this, note that any pure state ${\omega }_{1}=\omega$ can be extended to a set of perfectly distinguishable pure states ${\omega }_{1},{\omega }_{2},\ldots ,{\omega }_{{N}_{A}}$ such that $\omega =1\cdot {\omega }_{1}+0\cdot {\omega }_{2}\,+\,\ldots \,+\,0\cdot {\omega }_{{N}_{A}}$. Conversely, if $S(\omega )=0$, then any decomposition of ω must have coefficients $(1,0,\ldots ,0)$.

If a state space satisfies postulates 1 and 2, then it also satisfies all the assumptions that we have made in von Neumann's thought experiment. It is easy to check all items in assumptions 1: (a) is simply postulate 2, and (c) is a consequence of postulate 1. As we have seen in the previous section, our two postulates imply that we have orthogonal projectors sharing important properties with those of standard quantum theory. If we make the physical assumption that we can actually implement them by means of semipermeable membranes (as in quantum theory), we obtain (b). Item (e) is the same as (8). Note that assumption (d) is not a mathematical assumption about the state space, but a physical assumption about thermodynamic entropy. This shows part of the following (the full proof will be given in the appendix):

Observation 5. von Neumann's thought experiment, as explained in section 3, can be run for every state space that satisfies postulates 1 and 2. The notion of thermodynamic entropy $H$ that one obtains from that thought experiment turns out to equal spectral entropy $S$ as given in definition 4,

$$\begin{eqnarray*}&&H(\omega )=S(\omega )\qquad {\rm{for}}\,{\rm{all}}\,{\rm{states}}\,{\rm{}}\omega .\end{eqnarray*}$$

This is consistent with assumptions 1. Furthermore, it is also consistent with Petz' version of the thought experiment, because spectral entropy satisfies

$$\begin{eqnarray}&&S(\omega )=\displaystyle \sum _{j}{p}_{j}S({\omega }_{j})-\displaystyle \sum _{j}{p}_{j}\mathrm{log}{p}_{j}\end{eqnarray} \tag{ 9 }$$

for every convex decomposition $\omega ={\sum }_{j}{p}_{j}{\omega }_{j}$ of ω into perfectly distinguishable, not necessarily pure states ${\omega }_{j}$.

Thus, spectral entropy S gives meaningful and consistent physical predictions in situations like von Neumann's and Petz' thought experiments. However, we clearly do not know whether S is a consistent notion of physical entropy in all thermodynamical situations.

It turns out that there are further properties of S that encourage its physical interpretation as a thermodynamical entropy. In particular, we will now show that the second law holds in two important situations. We start by considering projective measurements ${P}_{1},\ldots ,{P}_{n}$. Projective measurements can model semipermeable membranes as in von Neumann's thought experiment, or they describe the measurement of an observable as explained in section 5.4. Consider the action of this measurement on a given state ω. With probabilities $({u}_{A}\circ {P}_{j})(\omega )$, this measurement yields the outcome j with post-measurement state ${\omega }_{j}:= {P}_{j}\omega /({u}_{A}\circ {P}_{j}(\omega ))$. Performing this measurement on every particle of an ensemble (without learning the outcomes) yields a new ensemble, described by the post-measurement state

$$\begin{eqnarray*}&&\omega ^{\prime} =\displaystyle \sum _{j:\,{u}_{A}\circ {P}_{j}(\omega )\ne 0}({u}_{A}\circ {P}_{j})(\omega )\cdot {\omega }_{j}=\displaystyle \sum _{j}{P}_{j}\omega .\end{eqnarray*}$$

Projective measurements do not decrease the entropy of the ensemble:

Theorem 6. Suppose postulates 1 and 2 are satisfied. Let ${P}_{1},\ldots ,{P}_{n}$ be orthogonal projectors which form a valid instrument. Then the induced measurement with post-measurement ensemble state $\omega ^{\prime} ={\sum }_{j}{P}_{j}w$ does not decrease entropy: $S(\omega ^{\prime} )\geqslant S(\omega )$.

The proof will be given in the appendix. As in standard quantum theory, projectors P_j form a valid instrument if and only if they are mutually orthogonal, i.e. ${P}_{i}{P}_{j}={\delta }_{{ij}}{P}_{i}$, and complete: ${\sum }_{i}{u}_{A}\circ {P}_{i}={u}_{A}$.

Another important manifestation of the second law is in mixing procedures as in figure 4. Consider tanks that are separated by walls. Similarly to von Neumann's thought experiment, let the jth tank contain an N_j -particle gas that represents an ${\omega }_{j}$-ensemble. Furthermore, assume that all the gases are at the same pressure and density. Identifying thermodynamic entropy H with spectral entropy S (as suggested by observation 5), the entropy of the GPT-ensemble in tank j is ${N}_{j}S({\omega }_{j})$, where S is the entropy per system. Thus the total GPT-entropy is ${\sum }_{j}{N}_{j}S({\omega }_{j})$. We remove the walls and let the gases mix. Then we put the walls back in. Now all the tanks contain gases hosting ${\sum }_{j}\tfrac{{N}_{j}}{N}{\omega }_{j}$ ensembles at the same conditions as before, where $N={\sum }_{j}{N}_{j}$. The total GPT-entropy in the end is given by ${\sum }_{j}{N}_{j}S\left({\sum }_{k}\tfrac{{N}_{k}}{N}{\omega }_{k}\right)={NS}\left({\sum }_{k}\tfrac{{N}_{k}}{N}{\omega }_{k}\right)$. As the gases in the tanks have the same density, volume, temperature and pressure as before, the only difference in entropy is due to the GPT-ensembles. The second law requires that the entropy does not decrease in this process, i.e. that ${\sum }_{j}{N}_{j}S({\omega }_{j})\leqslant {NS}\left({\sum }_{j}\tfrac{{N}_{j}}{N}{\omega }_{j}\right)$ and thus ${\sum }_{j}\tfrac{{N}_{j}}{N}S({\omega }_{j})\leqslant S\left({\sum }_{j}\tfrac{{N}_{j}}{N}{\omega }_{j}\right)$. The following theorem shows that our two postulates guarantee that this is true:

Zoom In Zoom Out Reset image size

Theorem 7. Assume postulates 1 and 2. Then entropy is concave, i.e. for ${\omega }_{1},\ldots ,{\omega }_{n}\in {{\rm{\Omega }}}_{A}$ and ${p}_{1},\ldots ,{p}_{n}$ a probability distribution, we have

$$\begin{eqnarray}&&S\left(\displaystyle \sum _{j}{p}_{j}{w}_{j}\right)\geqslant \displaystyle \sum _{j}{p}_{j}S({w}_{j}).\end{eqnarray} \tag{ 10 }$$

Thus, the second law automatically holds for mixing processes. One way to prove (10) is to see that S equals 'measurement entropy' as we will show in section 5.6, proven to be concave in [66, 67]. However, there is a simpler proof that uses a notion of relative entropy, which is an important notion in its own right.

Definition 8. For state spaces A that satisfy postulates 1 and 2, we define the ${relative}\ {entropy}$ of two states $\omega ,\varphi \in {{\rm{\Omega }}}_{A}$ as

$$\begin{eqnarray*}&&S(\omega \parallel \varphi ):= -S(\omega )-\langle \omega ,\mathrm{log}\varphi \rangle .\end{eqnarray*}$$

Here, for $\varphi ={\sum }_{j}{q}_{j}{\varphi }_{j}$ any decomposition into a maximal frame, $\mathrm{log}\varphi ={\sum }_{j}\mathrm{log}({q}_{j}){\varphi }_{j}$ according to theorem 3. (As in quantum theory, this can be infinite if there are q_j = 0 such that $\langle \omega ,{\varphi }_{j}\rangle \ne 0$).

A notion of relative entropy in GPTs has also been defined in Scandolo's Master Thesis [48], but under different assumptions, as discussed in the introduction. Relative entropy continues to satisfy Klein's inequality, a fact that is useful in proving theorem 7. The proof is similar to that within standard quantum theory and deferred to the appendix.

Theorem 9 For all $\omega ,\varphi \in {{\rm{\Omega }}}_{A}$,

$$\begin{eqnarray*}&&S(\omega \parallel \varphi )\geqslant 0.\end{eqnarray*}$$

Klein's inequality can be used to give a simple proof of theorem 7:

$$\begin{eqnarray*}&&0\leqslant \displaystyle \sum _{j}{p}_{j}S\left({\omega }_{j}\parallel \displaystyle \sum _{k}{p}_{k}{\omega }_{k}\right)=-\displaystyle \sum _{j}{p}_{j}S({\omega }_{j})-\langle \displaystyle \sum _{j}{p}_{j}{\omega }_{j},\mathrm{log}\left(\displaystyle \sum _{k}{p}_{k}{\omega }_{k}\right)\rangle =-\displaystyle \sum _{j}{p}_{j}S({\omega }_{j})+S\left(\displaystyle \sum _{k}{p}_{k}{\omega }_{k}\right).\end{eqnarray*}$$

Given all the calculations in this subsection in terms of orthogonal projections, it may seem at first sight as if every statement or calculation in quantum theory can be analogously made in the more general state spaces that satisfy postulates 1 and 2. However, this may not quite be true, as the fact that the following is an open problem shows:

Open problem 2. For state spaces satisfying postulates 1 and 2, if ω is a pure state, and $P$ an orthogonal projection, then is $P\omega$ also (up to normalisation) a pure state?

In classical and quantum state spaces, the answer is 'yes', but we do not know if a positive answer follows from postulates 1 and 2 alone. We will return to this problem in theorem 12.

Note that Chiribella and Scandolo have applied similar techniques and found beautiful results, including some which are comparable to some of ours, in [45, section 7] (see also [48]). They derive diagonalizability of states from a very different set of postulates.

So far we have considered entropy from a thermodynamic perspective. But entropies also arise in information theory, and as the GPT framework is mostly studied in quantum information theory, indeed there have been many results on entropy from a information-theoretic perspective. Our exposition will mainly follow [66], but has also been given in a slightly different formalism in [67].

Let $e=({e}_{1},\ldots ,{e}_{n})$ and $f=({f}_{1},\ldots ,{f}_{m})$ be two measurements such that there exists a map $M\,:\{1,\ldots ,n\}\to \{1,\ldots ,m\}$ with

$$\begin{eqnarray*}&&\displaystyle \sum _{\{j| M(j)=k\}}{e}_{j}={f}_{k}\qquad (k=1,\ldots ,m).\end{eqnarray*}$$

If M is bijective, then the measurement f is simply a re-labelling of e. If there exists a k with $M(j)\ne k\ \ \forall j$, then because of the normalisation of the e-measurement, f_k = 0, i.e. f_k corresponds to a trivial outcome that never happens. If M is not injective, then f is a coarse-graining of e (or vice versa, e a refinement of f) in the sense that f is obtained from e by collecting several outcomes of e and giving them a common new outcome label (and by possibly adding the 0-effect a few times), see figure 5. In this sense, we do not care about which of the e_j triggered the new effect.

Zoom In Zoom Out Reset image size

However, there exist trivial refinements/coarse-grainings: for those, ${e}_{j}\propto {f}_{M(j)}\ \ \forall j$. We write ${e}_{j}={p}_{j}{f}_{M(j)}$. Then such a measurement can be obtained by performing f, and if outcome k is triggered, we activate a classical random number generator which generates the final outcome j among those j with $M(j)=k$ with probability

$$\begin{eqnarray*}&&\displaystyle \frac{{p}_{j}}{{\sum }_{\{a| M(a)=k\}}{p}_{a}}.\end{eqnarray*}$$

Thus, a trivial refinement does not yield any additional information about the GPT-system. We call a measurement fine-grained if it does not have any non-trivial refinements. The set of fine-grained measurements on any state space A is denoted ${{ \mathcal E }}_{A}^{* }$.

Now we consider the Rényi entropies [65], which are defined for probability distributions ${\bf{p}}=({p}_{1},\ldots ,{p}_{n})$ as

$$\begin{eqnarray*}&&{H}_{\alpha }({\bf{p}})=\displaystyle \frac{1}{1-\alpha }\mathrm{log}\left(\displaystyle \sum _{j}{p}_{j}^{\alpha }\right),\end{eqnarray*}$$

where $\alpha \in (0,\infty )\setminus \{1\}$. Furthermore,

$$\begin{eqnarray*}&&{H}_{0}({\bf{p}}):= \mathop{\mathrm{lim}}\limits_{\alpha \to 0}{H}_{\alpha }({\bf{p}})=\mathrm{log}| \mathrm{supp}({\bf{p}})| \end{eqnarray*}$$

with $\mathrm{supp}({\bf{p}})=\{{p}_{j}\ | \ {p}_{j}\gt 0\}$ is called the max-entropy, and

$$\begin{eqnarray*}&&{H}_{\infty }({\bf{p}}):= \mathop{\mathrm{lim}}\limits_{\alpha \to \infty }{H}_{\alpha }({\bf{p}})=-\mathrm{log}\mathop{\max }\limits_{j}{p}_{j}\end{eqnarray*}$$

is called the min-entropy. Also,

$$\begin{eqnarray*}&&{H}_{1}({\bf{p}}):= \mathop{\mathrm{lim}}\limits_{\alpha \to 1}{H}_{\alpha }({\bf{p}})=-\displaystyle \sum _{j}{p}_{j}\mathrm{log}{p}_{j}=H({\bf{p}})\end{eqnarray*}$$

is just the regular Shannon entropy H.

For $\alpha \in [0,\infty ]$ and GPTs satisfying postulates 1 and 2, we generalise the classical Rényi entropies:

$$\begin{eqnarray*}&&{S}_{\alpha }(\omega ):= {H}_{\alpha }({\bf{p}}),\end{eqnarray*}$$

where $\omega ={\sum }_{j}{p}_{j}{w}_{j}$ is any decomposition into perfectly distinguishable pure states. According to theorem 3, the result is independent of the choice of decomposition. We have ${S}_{1}=S$, the spectral entropy of definition 4.

Following [66], for every $\alpha \in [0,\infty ]$ and $\omega \in {{\rm{\Omega }}}_{A}$, we define the order-α Rényi measurement entropy as

$$\begin{eqnarray*}&&{\widehat{S}}_{\alpha }(\omega )=\mathop{\inf }\limits_{e\in {{ \mathcal E }}_{A}^{* }}{H}_{\alpha }({e}_{1}(\omega ),{e}_{2}(\omega ),\ldots ),\end{eqnarray*}$$

where H_α on the right-hand side denotes the classical Rényi entropy. The order-α Rényi decomposition entropy is defined as

$$\begin{eqnarray}&&{\check{S}}_{\alpha }(\omega ):= \mathop{\inf }\limits_{\omega ={\displaystyle \sum }_{j}{q}_{j}{\varphi }_{j}}{H}_{\alpha }({\bf{q}}),\end{eqnarray} \tag{ 11 }$$

where the infimum is over all convex decompositions of ω into pure states ${\varphi }_{j}\in {{\rm{\Omega }}}_{A}$.

The idea of measurement entropy is to characterise the state before a measurement. For example, in quantum theory, particles prepared in a state $| \psi \rangle$ which all give the same result in energy measurements would be said to be in an energy eigenstate. If instead we performed a position measurement, the resulting distribution of positions would have non-zero entropy. However, this entropy would arguably not come from the initial state, but from the measurement process itself due to the uncertainty principle.

Suppose we would like to prepare a state ω by using states of maximal knowledge (i.e. pure states) ${\varphi }_{j}$, and a random number generator which gives output j with probability p_j. Then the decomposition entropy quantifies the smallest information content (entropy) of a random number generator that would be necessary to build such a device. For more detailed operational interpretations of measurement and decomposition entropy, in particular for $\alpha =1$, see [66, 67] Note that in quantum theory, measurement, decomposition and spectral Rényi entropies all coincide, with the $\alpha =1$ case giving von Neumann entropy, $S(\omega )=-\mathrm{tr}(\omega \mathrm{log}\omega )$.

Our first result is that the spectral and measurement definitions of the entropies agree:

Theorem 10. Consider any state space $A$ which satisfies postulates 1 and 2. Then the Rényi entropies ${S}_{\alpha }$ and the Rényi measurement entropies ${\widehat{S}}_{\alpha }$ coincide, and upper-bound the Rényi decomposition entropy ${\check{S}}_{\alpha }$, i.e.

$$\begin{eqnarray*}&&{\check{S}}_{\alpha }(\omega )\leqslant {S}_{\alpha }(\omega )={\widehat{S}}_{\alpha }(\omega )\,{\rm{for}}\,{\rm{all}}\ \omega \in {{\rm{\Omega }}}_{A},\alpha \in [0,\infty ].\end{eqnarray*}$$

In particular, for $\alpha =1$, the measurement entropy $\widehat{S}$ is the same as the spectral entropy S from definition 4, which we have identified with thermodynamical entropy $H$ in observation 5.

The inequality ${\check{S}}_{\alpha }\leqslant {S}_{\alpha }$ is easy to see: for a decomposition $\omega ={\sum }_{i}{p}_{i}{\omega }_{i}$ into perfectly distinguishable pure states ${\omega }_{i}$, the states ${\omega }_{i}$ can also be seen as a fine-grained measurement, yielding outcome probabilities P_i. So taking the infimum over all decompositions gives at most ${H}_{\alpha }({\bf{p}})={S}_{\alpha }(\omega )$. The equality between S_α and ${\widehat{S}}_{\alpha }$ is shown in the appendix.

We do not know in general whether postulates 1 and 2 imply that ${\check{S}}_{\alpha }={S}_{\alpha }$ for all α. Interestingly, we know it for $\alpha =2$ and $\alpha =\infty$:

Theorem 11. If a state space satisfies postulates 1 and 2, then ${\check{S}}_{2}(\omega )={S}_{2}(\omega )$ and ${\check{S}}_{\infty }(\omega )={S}_{\infty }(\omega )$ for all states ω.

Proof. To give the reader an idea of the kind of arguments involved, we present the proof for S₂, but defer the proof for ${S}_{\infty }$ to the appendix. If $\omega ={\sum }_{j}{p}_{j}{\omega }_{j}$ is any convex decomposition into a maximal set of perfectly distinguishable pure states (without loss of generality ${p}_{1}\geqslant {p}_{2}\geqslant \ldots$), and $\omega ={\sum }_{j}{q}_{j}{\varphi }_{j}$ any (other) convex decomposition into pure states ${\varphi }_{j}$ (also with ${q}_{1}\geqslant {q}_{2}\geqslant \ldots$,) then ${\sum }_{j}{p}_{j}^{2}=\langle \omega ,\omega \rangle ={\sum }_{j}{q}_{j}^{2}+{\sum }_{j\ne k}{q}_{j}{q}_{k}\langle {\varphi }_{j},{\varphi }_{k}\rangle \geqslant {\sum }_{j}{q}_{j}^{2}$ since $\langle {\varphi }_{j},{\varphi }_{k}\rangle \geqslant 0$. Thus, we have

$$\begin{eqnarray*}&&{S}_{2}(\omega )=-\mathrm{log}\displaystyle \sum _{j}{p}_{j}^{2}\leqslant -\mathrm{log}\displaystyle \sum _{j}{q}_{j}^{2}={H}_{2}({\bf{q}}),\end{eqnarray*}$$

and since ${\check{S}}_{2}(\omega )$ is defined as the infimum over the right-hand side, we obtain that ${\check{S}}_{2}(\omega )\geqslant {S}_{2}(\omega );$ we find the converse inequality in theorem 10.□

We do not know whether the same identity holds for the most interesting case $\alpha =1$, the case of standard thermodynamic entropy $S={S}_{1}$. In the max-entropy case $\alpha =0$, however, we have a surprising relation to higher-order interference:

Theorem 12. Consider a state space satisfying postulates 1 and 2. Then the following statements are all equivalent:

• (i)  
  The state space does not have third-order interference.
• (ii)  
  The measurement and decomposition versions of max-entropy coincide, i.e. ${\check{S}}_{0}(\omega )={S}_{0}(\omega )$ for all states ω.
• (iii)  
  The state space is either classical, or one on the list of theorem 2.
• (iv)  
  If ω is a pure state and ${P}_{F}$ any orthogonal projection onto any face $F$, then ${P}_{F}\omega$ is a multiple of a pure state.
• (v)  
  The 'atomic covering property' of quantum logic holds.

The equivalences $({\rm{i}})\iff ({\rm{iii}})\iff ({\rm{iv}})\iff ({\rm{v}})$ are shown in [43]; our new result is the equivalence to (ii), which is shown in the appendix.

Absence of third-order interference is meant in the sense of equation (5), as introduced originally by Sorkin [76]: only pairs of mutually exclusive alternatives can possibly interfere. It is interesting that this is related to an information-theoretic property of max-entropy S₀, as given in (ii). We do not currently know whether S₀ (or, in particular, the identity of ${\check{S}}_{0}$ and S₀) has any thermodynamic relevance in the class of theories that we are considering, but it certainly does within quantum theory, where it attains operational meaning in single-shot thermodynamics [28, 29].

As (iii) shows, this theorem is closely related to open problem 1: it gives properties of conceivable state spaces that satisfy postulates 1 and 2, but are not on the list of known examples (namely, they do not satisfy any of $({\rm{i}})\mbox{--}({\rm{v}})$). Similarly, (iv) shows the relation of higher-order interference to open problem 2, and (v) relates all these items to quantum logic. In fact, one can show that postulates 1 and 2 imply that the set of faces of the state space has the structure of an orthomodular lattice, which is often seen as the definition of quantum logic. For readers who are familiar with the terminology of quantum logic, we give some additional remarks in section A.3 in the appendix.

A.1.1. Proof that observables are well-defined

In this appendix, a decomposition of a state into perfectly distinguishable pure states (which always exists due to postulate 2) will be called a 'classical decomposition'.

Lemma 13. Assume postulates 1 and 2. Let $F\ne \{0\}$ be a face of ${A}_{+}$ and $\omega \in {{\rm{\Omega }}}_{A}\cap F$. Then there exists a classical decomposition $\omega ={\sum }_{j}{p}_{j}{\omega }_{j}$ with ${\omega }_{j}\in F$ for all j.

Proof. Let $\omega ={\sum }_{j}{p}_{j}{\omega }_{j}$ be a classical decomposition with ${p}_{j}\ne 0$. As $\omega \in F$ and F a face, ${\omega }_{j}\in F$ for all j.□

Proof of theorem 3. Let $x\in A$ be arbitrary. By lemma 5.46 from [62] there exists a frame $\{{\omega }_{j}\}$ and ${x}_{j}^{\prime }\in {\mathbb{R}}$ such that $x={\sum }_{j}{x}_{j}^{\prime }\,{\omega }_{j}$. We extend $\{{\omega }_{j}\}$ to a maximal frame by adding ${x}_{j}^{\prime }:= 0$ for the new indices j. Now we group together the j with the same ${x}_{j}^{\prime }$ value, and by relabelling we find that $x={\sum }_{k=1}^{n}{x}_{k}{\sum }_{i}{\omega }_{k;i}$ where the x_k are pairwise different values of the ${x}_{j}^{\prime }$ and the ${\omega }_{k;i}$ are the ${\omega }_{j}$ that belong to this ${x}_{j}^{\prime }$ value. For any given k, the ${\omega }_{k;i}$ generate a face F_k with projective unit ${u}_{k}={\sum }_{i}{\omega }_{k;i}$.

Therefore we find a decomposition $x={\sum }_{k=1}^{n}{x}_{k}{u}_{k}$ with x_k pairwise different real numbers and u_k order units of faces F_k and ${\sum }_{k=1}^{n}{u}_{k}={u}_{A}$.

Now we show that the faces F_k are mutually orthogonal:

Let $\omega \in {F}_{k}$ be an arbitrary normalised state. By lemma 13 it has a classical decomposition $\omega ={\sum }_{j}{p}_{j}{\omega }_{j}^{(k)}$ which uses only pure states ${\omega }_{j}^{(k)}\in {F}_{k}$. W.l.o.g. we assume that these pure states form a generating frame of F_k, by extending the frame and adding p_j = 0 to the decomposition. Consider another face F_m, i.e. $m\ne k$. Likewise to ω, let $\omega ^{\prime} \in {F}_{m}$ be an arbitrary normalised state and $\omega ^{\prime} ={\sum }_{j}{q}_{j}{\omega }_{j}^{(m)}$ be a classical decomposition with ${\omega }_{j}^{(m)}$ a generating frame for F_m. For the other faces define ${\omega }_{j}^{(i)}:= {\omega }_{i;j}$. Then ${u}_{i}={\sum }_{j}{\omega }_{j}^{(i)}$ and in total ${u}_{A}={\sum }_{i}{u}_{i}={\sum }_{i}{\sum }_{j}{\omega }_{j}^{(i)}$. As $\langle \nu ,\nu \rangle =1$ for all pure states $\nu \in {{\rm{\Omega }}}_{A}$, this implies that the ${\omega }_{j}^{(i)}$ are mutually orthogonal:

$$\begin{eqnarray*}&&1={u}_{A}({\omega }_{h}^{(g)})=\displaystyle \sum _{i}\displaystyle \sum _{j}\langle {\omega }_{j}^{(i)},{\omega }_{h}^{(g)}\rangle =1+\displaystyle \sum _{(i,j)\ne (g,h)}\langle {\omega }_{j}^{(i)},{\omega }_{h}^{(g)}\rangle \end{eqnarray*}$$

and therefore $\langle {\omega }_{j}^{(i)},{\omega }_{h}^{(g)}\rangle \geqslant 0$ implies $\langle {\omega }_{j}^{(i)},{\omega }_{h}^{(g)}\rangle =0$ for all $(i,j)\ne (g,h)$. Thus we find $\langle \omega ,\omega ^{\prime} \rangle ={\sum }_{j}{\sum }_{b}{p}_{j}{q}_{b}\langle {\omega }_{j}^{(k)},{\omega }_{b}^{(m)}\rangle =0$ because $m\ne k$. As $\omega \in {F}_{k}$ and $\omega ^{\prime} \in {F}_{m}$ were arbitrary (normalised) states, this implies that F_k and F_m are orthogonal. As $k\ne m$ were arbitrary, all the faces are mutually orthogonal.

Now we will show that the decomposition $x={\sum }_{j}{x}_{j}{u}_{j}$ is unique. So assume there are two decompositions $x={\sum }_{j=1}^{{n}_{a}}{a}_{j}{u}_{j}^{(a)}={\sum }_{j=1}^{{n}_{b}}{b}_{j}{u}_{j}^{(b)}$ with ${a}_{j}\in {\mathbb{R}}$ pairwise different and projective units ${u}_{j}^{(a)}$ that add up to the order unit (analogously for b) and belong to pairwise orthogonal faces ${F}_{j}^{(a)}$. W.l.o.g. we assume that the a_j and b_j are ordered by size, i.e. ${a}_{1}\lt {a}_{2}\lt ...\,\lt \,{a}_{{n}_{a}}$. We want to show ${a}_{1}={b}_{1}$. The ${u}_{j}^{(a)}$ generate the faces ${F}_{j}^{(a)}$. Let ${\omega }_{j;i}^{(a)}$ be a generating frame for the face ${F}_{j}^{(a)}$, especially ${\sum }_{i}{\omega }_{j;i}^{(a)}={u}_{j}^{(a)}$. As the faces are mutually orthogonal and the projective units add up to u_A, the ${\omega }_{j;i}^{(a)}$ form a maximal frame; in particular they add up to u_A (likewise for b). Therefore:

$$\begin{eqnarray*}&&{a}_{1}=\langle {\omega }_{1;j}^{(a)},x\rangle =\displaystyle \sum _{k,i}{b}_{k}\langle {\omega }_{1;j}^{(a)},{\omega }_{k;i}^{(b)}\rangle \geqslant \displaystyle \sum _{k,i}{b}_{1}\langle {\omega }_{1;j}^{(a)},{\omega }_{k;i}^{(b)}\rangle ={b}_{1}{u}_{A}({\omega }_{1;j}^{(a)})={b}_{1}.\end{eqnarray*}$$

Analogously show ${b}_{1}\geqslant {a}_{1}$, i.e. ${b}_{1}={a}_{1}$ in total.

Now suppose there was a $k\gt 1$ and an i with $\langle {\omega }_{1;j}^{(a)},{\omega }_{k;i}^{(b)}\rangle \ne 0$, i.e. $\langle {\omega }_{1;j}^{(a)},{\omega }_{k;i}^{(b)}\rangle \gt 0$. Then

$$\begin{eqnarray*}{a}_{1} & = & \langle {\omega }_{1;j}^{(a)},x\rangle =\displaystyle \sum _{k,i}{b}_{k}\langle {\omega }_{1;j}^{(a)},{\omega }_{k;i}^{(b)}\rangle ={a}_{1}\displaystyle \sum _{i}\langle {\omega }_{1;j}^{(a)},{\omega }_{1;i}^{(b)}\rangle +\displaystyle \sum _{k\gt 1,i}{b}_{k}\langle {\omega }_{1;j}^{(a)},{\omega }_{k;i}^{(b)}\rangle \gt {a}_{1}\displaystyle \sum _{i}\langle {\omega }_{1;j}^{(a)},{\omega }_{1;i}^{(b)}\rangle \\ & & +\displaystyle \sum _{k\gt 1,i}{a}_{1}\langle {\omega }_{1;j}^{(a)},{\omega }_{k;i}^{(b)}\rangle ={a}_{1}\displaystyle \sum _{k,i}\langle {\omega }_{1;j}^{(a)},{\omega }_{k;i}^{(b)}\rangle ={a}_{1}{u}_{A}({\omega }_{1;j}^{(a)})={a}_{1}.\end{eqnarray*}$$

This is a contradiction. Thus $\langle {\omega }_{1;j}^{(a)},{\omega }_{k;i}^{(b)}\rangle =0$ for all $k\gt 1$ and i. Therefore we find ${u}_{1}^{(b)}({\omega }_{1;i}^{(a)})={\sum }_{j}\langle {\omega }_{1;j}^{(b)}\,,$ ${\omega }_{1;i}^{(a)}\rangle ={\sum }_{j,k}\langle {\omega }_{k,j}^{(b)},\,$ ${\omega }_{1;i}^{(a)}\rangle ={u}_{A}({\omega }_{1;i}^{(a)})=1$ and analogously ${u}_{1}^{(a)}({\omega }_{1;i}^{(b)})=1$. By proposition 5.29 from [62], we have ${{\rm{\Omega }}}_{A}\cap F=\{\omega \in {{\rm{\Omega }}}_{A}\ | \ {u}_{F}(\omega )=1\}$. Therefore a generating frame of ${F}_{1}^{(a)}$ is contained in ${F}_{1}^{(b)}$ and vice versa. Thus we find ${F}_{1}^{(a)}={F}_{1}^{(b)}$ and ${u}_{1}^{(a)}={u}_{1}^{(b)}$.

For the remaining indices, we construct an inductive proof: choose $L\in {\mathbb{R}}$ large enough such that ${a}_{1}+L\gt \max \{{a}_{{n}_{a}},{b}_{{n}_{b}}\}$, and define $x^{\prime} := x+L\cdot {u}_{1}^{(a)}$, i.e. $x^{\prime} ={\sum }_{j=1}^{{n}_{a}}({a}_{j}+{\delta }_{j,1}\cdot L)\,$ ${u}_{j}^{(a)}={\sum }_{j=1}^{{n}_{b}}({b}_{j}+{\delta }_{j,1}\cdot L){u}_{j}^{(b)}$. Furthermore defining ${a}_{1}^{\prime }:= {a}_{2}$, ${a}_{2}^{\prime }:= {a}_{3}$ ,..., ${a}_{{n}_{a}}^{\prime }:= {a}_{1}+L$, ${u}_{1}^{(a^{\prime} )}:= {u}_{2}^{(a)}$, ${u}_{2}^{(a^{\prime} )}:= {u}_{3}^{(a)}$,..., ${u}_{{n}_{a}}^{(a^{\prime} )}:= {u}_{1}^{(a)}$ and likewise for ${b}_{j}^{\prime }$, we find $x^{\prime} ={\sum }_{j=1}^{{n}_{a}}{a}_{j}^{\prime }{u}_{j}^{(a^{\prime} )}={\sum }_{j=1}^{{n}_{b}}{b}_{j}^{\prime }{u}_{j}^{(b^{\prime} )}$ with ${a}_{1}^{\prime }\lt {a}_{2}^{\prime }\lt ...\,\,\lt \text{}{a}_{{n}_{a}}^{\prime }$ and ${b}_{1}^{\prime }\lt {b}_{2}^{\prime }\lt ...\,\lt \,{b}_{{n}_{b}}^{\prime }$. Repeating the exact same procedure as before, we obtain ${a}_{1}^{\prime }={b}_{1}^{\prime }$ and ${u}_{1}^{(a^{\prime} )}={u}_{1}^{(b^{\prime} )}$, i.e. ${a}_{2}={b}_{2}$ and ${u}_{2}^{(a)}={u}_{2}^{(b)}$. We iterate to find a_j = b_j and ${u}_{j}^{(a)}={u}_{j}^{(b)}$ for all j. Note that as all maximal frames have the same size and as the projective units add up to u_A, necessarily n_a = n_b.

At last we construct the projective measurement that corresponds to measuring the observable x: for F_k, let P_k be the orthogonal projector onto the span of F_k (in particular, ${P}_{k}\,:A\to \mathrm{span}({F}_{k})$ surjective). We know that these projectors are positive and linear and satisfy ${u}_{A}\circ {P}_{k}={u}_{k}$. Furthermore $0\leqslant {u}_{k}={u}_{A}\circ {P}_{k}\leqslant {u}_{A}$ and ${\sum }_{k}{u}_{A}\circ {P}_{k}={\sum }_{k}{u}_{k}={u}_{A}$, i.e. we obtain a well-defined measurement; therefore the P_k form a well-defined instrument. As they are projectors, the P_k leave the elements of F_k unchanged.□

A.1.2. Proof of observation 5

In order to show that $H(\omega )=S(\omega )$ is consistent with assumptions 1, we only have to show that $\omega \mapsto S(\omega )$ is continuous, to comply with assumption (d). According to theorem 10 (which we will prove below), the spectral entropy $S(\omega )$ equals measurement entropy $\widehat{S}(\omega )$. But it is well-known [67] and easy to see from its definition that $\widehat{S}$ is continuous.

It remains to show equation (9). So let $\omega ={\sum }_{j}{p}_{j}{\omega }_{j}$ be any decomposition of ω into perfectly distinguishable, not necessarily pure states ${\omega }_{i}$. Decompose all the ${\omega }_{i}$ into perfectly distinguishable pure states ${\omega }_{j}^{(i)}$, i.e. ${\omega }_{j}={\sum }_{i}{q}_{j}^{(i)}{\omega }_{j}^{(i)}$. Perfectly distinguishable states live in orthogonal faces, thus $\langle {\omega }_{i},{\omega }_{j}\rangle =0$ for $i\ne j$ (note that this is a conclusion that follows from postulates 1 and 2, but could not be drawn from bit symmetry alone in [64]). Thus, we also have $\langle {\omega }_{i}^{(j)},{\omega }_{k}^{(l)}\rangle =0$ for $i\ne k$ or $j\ne l$, and so $\omega ={\sum }_{{ij}}{p}_{j}{q}_{j}^{(i)}{\omega }_{j}^{(i)}$ is a decomposition of ω into perfectly distinguishable pure states. Define the real function $\eta :[0,1]\to {\mathbb{R}}$ via $\eta (x):= -x\mathrm{log}x$ for $x\gt 0$ and $\eta (0)=0$. Due to theorem 3 and $\eta ({xy})=-{xy}\mathrm{log}x-{xy}\mathrm{log}y$, we have

$$\begin{eqnarray*}&&\eta (\omega )=\displaystyle \sum _{{ij}}\eta ({p}_{j}{q}_{j}^{(i)}){\omega }_{j}^{(i)}=-\displaystyle \sum _{{ij}}{p}_{j}{q}_{j}^{(i)}\mathrm{log}{p}_{j}{\omega }_{j}^{(i)}-\displaystyle \sum _{{ij}}{p}_{j}{q}_{j}^{(i)}(\mathrm{log}{q}_{j}^{(i)}){\omega }_{j}^{(i)},\end{eqnarray*}$$

and therefore

$$\begin{eqnarray*}&&S(\omega )={u}_{A}(\eta (\omega ))=-\displaystyle \sum _{{ij}}{p}_{j}{q}_{j}^{(i)}\mathrm{log}{p}_{j}-\displaystyle \sum _{{ij}}{p}_{j}{q}_{j}^{(i)}\mathrm{log}{q}_{j}^{(i)}=-\displaystyle \sum _{j}{p}_{j}\mathrm{log}{p}_{j}+\displaystyle \sum _{j}{p}_{j}\mathop{\underbrace{\left(-\displaystyle \sum _{i}{q}_{j}^{(i)}\mathrm{log}{q}_{j}^{(i)}\right)}}\limits_{S({\omega }_{j})}\end{eqnarray*}$$

This completes the proof of observation 5.□

A.1.3. Proof of the second half of theorem 11

Use the notation of the first half of the proof. We claim that ${\max }_{\varphi \in {\rm{\Omega }}}\langle \omega ,\varphi \rangle ={p}_{1}$. The inequality '$\geqslant$' is trivial (consider the special case $\varphi ={\omega }_{1}$). To see the inequality '$\leqslant$', note that $\langle \omega ,\varphi \rangle ={\sum }_{j}{p}_{j}{\lambda }_{j}$, where ${\lambda }_{j}:= \langle {\omega }_{j},\varphi \rangle \in [0,1]$ satisfies ${\sum }_{j}{\lambda }_{j}=\langle {\sum }_{j}{\omega }_{j},\varphi \rangle =\langle u,\varphi \rangle =1$, and so $\langle \omega ,\varphi \rangle \leqslant {p}_{1}$ for all φ. Thus

$$\begin{eqnarray*}&&{S}_{\infty }(\omega )=-\mathrm{log}{p}_{1}=-\mathrm{log}\mathop{\max }\limits_{\varphi \in {\rm{\Omega }}}\langle \omega ,\varphi \rangle \leqslant -\mathrm{log}\langle \omega ,{\varphi }_{1}\rangle =-\mathrm{log}\left({q}_{1}+\displaystyle \sum _{j\geqslant 2}{q}_{j}\langle {\varphi }_{j},{\varphi }_{1}\rangle \right)\leqslant -\mathrm{log}{q}_{1}={H}_{\infty }({\bf{q}}).\end{eqnarray*}$$

Similarly as in the first part of the proof, we obtain ${\check{S}}_{\infty }(\omega )\geqslant {S}_{\infty }(\omega )$. The converse inequality from theorem 10 for $\alpha =\infty$ concludes the proof. □

A.1.4. Proof of Klein's inequality and the second law for projective measurements

We consider an ensemble of systems described by an arbitrary state $\omega \in {{\rm{\Omega }}}_{A}$. To all systems of this ensemble we apply a projective measurement described by orthogonal projectors P_a which form an instrument, resulting in a new ensemble state $\omega ^{\prime}$. The P_a project onto the linear span of faces F_a that replace the eigenspaces from quantum theory. We want to show that the measurement cannot decrease the entropy of the ensemble, i.e.

$$\begin{eqnarray*}&&S(\omega ^{\prime} )\geqslant S(\omega ).\end{eqnarray*}$$

We decompose the proof into several steps. Our basic idea follows the proof of a similar statement for quantum theory in [50]: we reduce the proof of the second law to Klein's inequality. But as we do not have access to an underlying pure state Hilbert space, we will need to use a different argument for why Klein's inequality implies the second law for projective measurements.

So at first we prove Klein's inequality, adapting the proof of [50]. We note that a similar proof has also been found by Scandolo [48], albeit under different assumptions.

Proof of theorem 9. We consider two arbitrary states $\omega ,\nu$ with classical decompositions $\omega ={\sum }_{j}{p}_{j}{\omega }_{j}$, $\nu ={\sum }_{k}{q}_{k}{\nu }_{k}$, where w.l.o.g. the ${\omega }_{j}$ and the ${\nu }_{k}$ form maximal frames. We define the matrix ${P}_{{jk}}:= \langle {\omega }_{j},{\nu }_{k}\rangle$. All its components are non-negative, i.e. ${P}_{{jk}}\geqslant 0$, because the scalar product itself is non-negative for all states. As all maximal frames have the same size, the matrix is a square matrix; as maximal frames sum to u_A, the rows and columns sum to one: ${\sum }_{j}{P}_{{jk}}={\sum }_{k}{P}_{{jk}}=1$. Thus, we get

$$\begin{eqnarray*}&&S(\omega \parallel \nu )=-S(\omega )-\langle \omega ,\mathrm{log}\nu \rangle =\displaystyle \sum _{j}{p}_{j}\mathrm{log}{p}_{j}-\displaystyle \sum _{{jk}}{p}_{j}\mathrm{log}{q}_{k}\langle {\omega }_{j},{\nu }_{k}\rangle =\displaystyle \sum _{j}{p}_{j}\left(\mathrm{log}{p}_{j}-\displaystyle \sum _{k}{P}_{{jk}}\mathrm{log}{q}_{k}\right).\end{eqnarray*}$$

We define ${r}_{j}:= {\sum }_{k}{P}_{{jk}}{q}_{k}$. Note that the r_j form a probability distribution: ${r}_{j}\geqslant 0$ and ${\sum }_{j}{r}_{j}={\sum }_{k}{\sum }_{j}{P}_{{jk}}{q}_{k}={\sum }_{k}{q}_{k}=1$. Using the strict concavity of the logarithm, we find:

$$\begin{eqnarray*}&&\mathrm{log}{r}_{j}=\mathrm{log}\left(\displaystyle \sum _{k}{P}_{{jk}}{q}_{k}\right)\geqslant \displaystyle \sum _{k}{P}_{{jk}}\mathrm{log}{q}_{k}.\end{eqnarray*}$$

Therefore we get

$$\begin{eqnarray*}&&S(\omega \parallel \nu )=\displaystyle \sum _{j}{p}_{j}\left(\mathrm{log}{p}_{j}-\displaystyle \sum _{k}{P}_{{jk}}\mathrm{log}{q}_{k}\right)\geqslant \displaystyle \sum _{j}{p}_{j}(\mathrm{log}{p}_{j}-\mathrm{log}{r}_{j})=\displaystyle \sum _{j}{p}_{j}\mathrm{log}\left(\displaystyle \frac{{p}_{j}}{{r}_{j}}\right).\end{eqnarray*}$$

We recognise the last expression as the classical relative entropy of the probability distributions p_j and r_j. This classical relative entropy has the important property that it is never negative. Thus:

$$\begin{eqnarray*}&&S(\omega \parallel \nu )\geqslant 0.\end{eqnarray*}$$

□

In order to get the main proof less convoluted, we will state some technical parts as lemmas.

Lemma 14. Assume postulate 1 and 2. Consider orthogonal projectors ${P}_{j}$ which form an instrument. Then the ${P}_{j}$ are mutually orthogonal:

$$\begin{eqnarray*}&&{P}_{k}{P}_{j}={\delta }_{{jk}}{P}_{j}.\end{eqnarray*}$$

Proof. We prove ${P}_{k}{P}_{j}\omega =0$ for all $\omega \in A$, $j\ne k$. If ${P}_{j}\omega =0$ this is trivial, so from now on assume ${P}_{j}\omega \ne 0$. As the cone is generating (i.e. $\mathrm{Span}({A}_{+})=A$) and the projectors linear, it is sufficient to show ${P}_{k}{P}_{j}\omega =0$ for all $w\in {A}_{+}$. As P_j is positive, ${P}_{j}\omega \ne 0$ implies that $({u}_{A}\circ {P}_{j})(\omega )\gt 0$ because only the zero-state is normalised to 0. Using ${u}_{A}={u}_{A}\circ ({\sum }_{j}{P}_{j})={\sum }_{j}{u}_{A}\circ {P}_{j}$ and ${P}_{j}{P}_{j}={P}_{j}$:

$$\begin{eqnarray*}1 & = & {u}_{A}\left(\displaystyle \frac{{P}_{j}\omega }{({u}_{A}\circ {P}_{j})(\omega )}\right)=\left({u}_{A}\circ \displaystyle \,\sum _{k}{P}_{k}\right)\left(\displaystyle \frac{{P}_{j}\omega }{({u}_{A}\circ {P}_{j})(\omega )}\right)={u}_{A}\left(\displaystyle \frac{{P}_{j}{P}_{j}\omega }{({u}_{A}\circ {P}_{j})(\omega )}\right)+{u}_{A}\left(\displaystyle \frac{{\displaystyle \sum }_{k| k\ne j}{P}_{k}{P}_{j}\omega }{({u}_{A}\circ {P}_{j})(\omega )}\right)\\ & = & {u}_{A}\left(\displaystyle \frac{{P}_{j}\omega }{({u}_{A}\circ {P}_{j})(\omega )}\right)+{u}_{A}\left(\displaystyle \frac{{\displaystyle \sum }_{k| k\ne j}{P}_{k}{P}_{j}\omega }{({u}_{A}\circ {P}_{j})(\omega )}\right)\qquad \Rightarrow \qquad 0={u}_{A}\left(\displaystyle \frac{{\displaystyle \sum }_{k| k\ne j}{P}_{k}{P}_{j}\omega }{({u}_{A}\circ {P}_{j})(\omega )}\right)=\displaystyle \sum _{k| k\ne j}{u}_{A}({P}_{k}{P}_{j}\omega ).\end{eqnarray*}$$

As the projectors are positive and only the zero-state is normalised to 0, this shows ${P}_{k}{P}_{j}\omega =0$ for $k\ne j$.□

Lemma 15. Assume postulates 1 and 2. Consider an orthogonal projector $P$ which projects onto the linear span of a face $F$ of ${A}_{+}$. Then for all states $\omega \in {A}_{+}$ we find $P\omega \in F$.

Proof. From basic convex geometry (see e.g. proposition 2.10 in [63]), we know that $F=\mathrm{span}(F)\cap {A}_{+}$. Since P is positive, we have $P\omega \in {A}_{+};$ furthermore, since P projects onto F, we have $P\omega \in \mathrm{span}(F)$, thus $P\omega \in F$.□

Proof of theorem 6. We know that $S(\omega \parallel \omega ^{\prime} )=-S(\omega )-\langle \omega ,\mathrm{log}\omega ^{\prime} \rangle \geqslant 0$. As in theorem 11.9 from [50] , we claim $-\langle \omega ,\mathrm{log}\omega ^{\prime} \rangle =S(\omega ^{\prime} )$ and therefore $-S(\omega )+S(\omega ^{\prime} )\geqslant 0$. Thus we only have to prove $-\langle \omega ,\mathrm{log}\omega ^{\prime} \rangle =S(\omega ^{\prime} )$. But as we do not have access to an underlying pure state Hilbert space, our proof is different from [50].

By lemma 14, the P_a are mutually orthogonal, i.e. ${P}_{a}{P}_{b}={\delta }_{{ab}}{P}_{b}$. By symmetry of the P_a also the ${P}_{a}\omega$ are mutually orthogonal: $\langle {P}_{a}\omega ,{P}_{b}\omega \rangle =\langle \omega ,{P}_{a}{P}_{b}\omega \rangle =0$ for $a\ne b$. This also shows that the F_a are mutually orthogonal. If ${P}_{a}\omega =0$ we use the decomposition ${P}_{a}\omega ={u}_{A}({P}_{a}\omega ){\sum }_{k}{r}_{{ak}}{w}_{{ak}}$ with ${r}_{{ak}}={\delta }_{{ak}}$ and w_ak an arbitrary generating frame of F_a. If ${P}_{a}\omega \ne 0$, then $\tfrac{{P}_{a}\omega }{{u}_{A}({P}_{a}\omega )}\in {F}_{a}\cap {{\rm{\Omega }}}_{A}$ and by lemma 13, there is a classical decomposition $\tfrac{{P}_{a}\omega }{{u}_{A}({P}_{a}\omega )}={\sum }_{k}{r}_{{ak}}{\omega }_{{ak}}$ with ${\omega }_{{ak}}\in {F}_{a}$. We complete the ${\omega }_{{ak}}$ to generating frames of the F_a by adding terms with ${r}_{{ak}}=0$. As we are using classical decompositions/frames, we know $\langle {\omega }_{{aj}},{\omega }_{{ak}}\rangle ={\delta }_{{jk}}$. Furthermore, as the F_a are mutually orthogonal, we know $\langle {\omega }_{{aj}},{\omega }_{{bk}}\rangle =0$ for $a\ne b$.

We note that the the w_aj form a maximal frame:

$$\begin{eqnarray*}&&{u}_{A}=\displaystyle \sum _{a}{u}_{A}\circ {P}_{a}=\displaystyle \sum _{a}{u}_{{F}_{a}}=\displaystyle \sum _{a}\displaystyle \sum _{j}{\omega }_{{aj}}.\end{eqnarray*}$$

For $a\ne b$ we have ${P}_{b}{\omega }_{{aj}}={P}_{b}{P}_{a}{\omega }_{{aj}}=0$, so we have a classical decomposition

$$\begin{eqnarray*}&&\omega ^{\prime} =\displaystyle \sum _{a}{P}_{a}\omega =\displaystyle \sum _{a}\displaystyle \sum _{j}{u}_{A}({P}_{a}\omega ){r}_{{aj}}{\omega }_{{aj}}\end{eqnarray*}$$

with ${\omega }_{{aj}}$ a maximal frame that satisfies ${P}_{a}{\omega }_{{bj}}={\delta }_{{ab}}{\omega }_{{bj}}$. Note that we do not need to normalise $\omega ^{\prime}$ as the measurement itself is required to be normalised. Using

$$\begin{eqnarray*}&&\displaystyle \sum _{a}{P}_{a}\mathrm{log}\omega ^{\prime} =\displaystyle \sum _{{bj}}\mathrm{log}({u}_{A}({P}_{b}\omega ){r}_{{bj}})\displaystyle \sum _{a}{P}_{a}{\omega }_{{bj}}=\displaystyle \sum _{{bj}}\mathrm{log}({u}_{A}({P}_{b}\omega ){r}_{{bj}}){\omega }_{{bj}}=\mathrm{log}\omega ^{\prime} \end{eqnarray*}$$

and

$$\begin{eqnarray}&&-S(\omega ^{\prime} )=-\displaystyle \sum _{{bj}}({u}_{A}({P}_{b}\omega ){r}_{{bj}})\mathrm{log}({u}_{A}({P}_{b}\omega ){r}_{{bj}})=\langle \omega ^{\prime} ,\mathrm{log}\omega ^{\prime} \rangle \end{eqnarray} \tag{ 12 }$$

as well as the symmetry of the P_a we finally find:

$$\begin{eqnarray}&&-S(\omega ^{\prime} )=\langle \omega ^{\prime} ,\mathrm{log}\omega ^{\prime} \rangle =\left\langle\displaystyle \sum _{a}{P}_{a}\omega ,\mathrm{log}\omega ^{\prime} \right\rangle = \left\langle\omega ,\displaystyle \sum _{a}{P}_{a}\mathrm{log}\omega ^{\prime}\right\rangle =\langle \omega ,\mathrm{log}\omega ^{\prime} \rangle .\end{eqnarray} \tag{ 13 }$$

□

A.1.5. Proof that measurement and spectral entropies are identical

In the main text we encountered different ways to define the entropy. One of them is to adapt classical entropy definitions by using the coefficients of a classical decomposition. Another is to adapt classical entropy definitions by using measurement probabilities and minimising over all fine-grained measurements. Here we will show that in the context of postulates 1 and 2, these two concepts yield the same Rényi entropies.

To prove this, we will first analyse fine-grained measurements in further detail. The results will allow us to reproduce the quantum proof found in [66] for our GPTs.

Lemma 16. Assume postulates 1 and 2. Consider an arbitrary fine-grained measurement $({e}_{1},\ldots ,{e}_{n})$. Then for all $j$ there exist some ${c}_{j}\in [0,1]$ and a pure state ${\omega }_{j}\in {{\rm{\Omega }}}_{A}$ such that ${e}_{j}={c}_{j}\langle {\omega }_{j},\cdot \rangle$.

Proof. If e_j = 0, we can just take c_j = 0 and any pure state ${\omega }_{j}$. So from now on assume ${e}_{j}\ne 0$.

Because of self-duality there exists some $\omega ^{\prime} \in {A}_{+}$ such that $\langle \omega ^{\prime} ,\cdot \rangle ={e}_{j}$. As ${e}_{j}\ne 0$ also $\omega ^{\prime} \ne 0$ and therefore ${u}_{A}(\omega ^{\prime} )\ne 0$. With ${A}_{+}={{\mathbb{R}}}_{\geqslant 0}\cdot {{\rm{\Omega }}}_{A}$ and ${c}_{j}:= {u}_{A}(\omega ^{\prime} )\gt 0$ there exists an $\omega \in {{\rm{\Omega }}}_{A}$ such that $\omega ^{\prime} ={c}_{j}\cdot \omega$. We want to prove that ω is pure, so assume it was not pure. Then it has a classical decomposition $\omega ={\sum }_{k=0}^{N}{p}_{k}{\omega }_{k}$ with ${p}_{k}\gt 0$ and $N\geqslant 1$. By relabelling we can assume j = n, i.e. we consider ${e}_{n}={c}_{j}{\sum }_{k=0}^{N}{p}_{k}\langle {\omega }_{k},\cdot \rangle$. Define a measurement $({e}_{1}^{\prime },\ldots ,{e}_{n+N}^{\prime })$ by ${e}_{k}^{\prime }:= {e}_{k}$ for all $k=1,2,\ldots ,n-1$ and ${e}_{n+i}^{\prime }:= {c}_{j}{p}_{i}\langle {\omega }_{i},\cdot \rangle$ for all $i=0,1,\ldots ,N$. Because of $0\leqslant {c}_{j}{p}_{i}\langle {\omega }_{i},\cdot \rangle ={e}_{n+i}^{\prime }$ and ${\sum }_{k=1}^{n+N}{e}_{k}^{\prime }={\sum }_{k=1}^{n-1}{e}_{k}+{\sum }_{i=0}^{N}{c}_{j}{p}_{i}\langle {\omega }_{i},\cdot \rangle ={\sum }_{k=1}^{n}{e}_{k}={u}_{A}$ this is a well-defined measurement.

Now define $M:\{1,\ldots ,n+N\}\to \{1,\ldots ,n\}$ by $M(i):= i$ for all $i=1,\ldots ,n-1$ and $M(i):= n$ for all $i=n,\ldots ,n+N$. Then we get

$$\begin{eqnarray*}&&\displaystyle \sum _{\{a| M(a)=i\}}{e}_{a}^{\prime }={e}_{i}\qquad \ \mathrm{for}\ i\lt n\displaystyle \sum _{\{a| M(a)=i\}}{e}_{a}^{\prime }=\displaystyle \sum _{a=n}^{n+N}{e}_{i}^{\prime }={e}_{n}\qquad \ \mathrm{for}\ i=n.\end{eqnarray*}$$

Thus the measurement $({e}_{1}^{\prime },\ldots ,{e}_{n+N}^{\prime })$ is a refinement of $({e}_{1},\ldots ,{e}_{n})$. With ${e}_{n}^{\prime }({\omega }_{0})={c}_{j}{p}_{0}={e}_{n}({\omega }_{0})$ and ${e}_{n}^{\prime }({\omega }_{1})=0\ne {e}_{n}({\omega }_{1})$ we find that ${e}_{n}^{\prime }$ is not proportional to e_n, thus the fine-graining is non-trivial. This is in contradiction to our assumptions. Thus ω has to be pure. Furthermore $1={u}_{A}(\omega )\geqslant {e}_{j}(\omega )={c}_{j}\langle \omega ,\omega \rangle ={c}_{j}$.

So in total we have found that ${e}_{j}={c}_{j}\langle \omega ,\cdot \rangle$ with $\omega \in {{\rm{\Omega }}}_{A}$ pure and ${c}_{j}\in [0,1]$.□

Lemma 17. Assume postulates 1 and 2. Let $\omega \in {{\rm{\Omega }}}_{A}$ and $\omega ={\sum }_{j=1}^{d}{p}_{j}{\omega }_{j}$ be a decomposition into a maximal frame. Then the measurement that perfectly distinguishes the ${\omega }_{j}$ (i.e. ${e}_{k}({\omega }_{j})={\delta }_{{jk}}$) can be chosen to be fine-grained.

Proof. Define ${e}_{j}:= \langle {\omega }_{j},\cdot \rangle$. As maximal frames add up to the order unit, this is a well-defined measurement and it satisfies ${e}_{j}({\omega }_{k})={\delta }_{{jk}}$. It remains to show that this measurement is fine-grained.

Consider a fine-graining ${e}_{k}^{\prime }$ with ${e}_{i}={\sum }_{\{j| M(j)=i\}}{e}_{j}^{\prime }$. By self-duality, there exist ${c}_{j}\geqslant 0$ and ${\omega }_{j}^{\prime }\in {{\rm{\Omega }}}_{A}$ such that ${e}_{j}^{\prime }={c}_{j}\langle {\omega }_{j}^{\prime },\cdot \rangle$ and therefore ${\sum }_{\{j| M(j)=k\}}{c}_{j}{\omega }_{j}^{\prime }={\omega }_{k}$. As $1={u}_{A}({\omega }_{k})={\sum }_{\{j| M(j)=k\}}{c}_{j}{u}_{A}({\omega }_{j}^{\prime })={\sum }_{\{j| M(j)=k\}}{c}_{j}$ we find that ${\sum }_{\{j| M(j)=k\}}{c}_{j}{\omega }_{j}^{\prime }={\omega }_{k}$ is a convex decomposition of a pure state. This requires c_j = 0 or ${\omega }_{j}^{\prime }={\omega }_{k}$. In both cases ${e}_{j}^{\prime }={c}_{j}\langle {\omega }_{k},\cdot \rangle ={c}_{j}{e}_{k}$ holds true for all j with $M(j)=k$. Therefore, the fine-graining is trivial.□

Lemma 18. Assume postulates 1 and 2. Consider a fine-grained measurement ${\bf{e}}=({e}_{1},\ldots ,{e}_{N})\in {{ \mathcal E }}^{* }$. Then the maximal number of perfectly distinguishable states $d$ (often denoted as ${N}_{A}$) satisfies $d\leqslant N$.

Furthermore, consider a state $\omega \in {{\rm{\Omega }}}_{A}$ with classical decomposition $\omega ={\sum }_{j=1}^{d}{p}_{j}{\omega }_{j}$ into a maximal frame. Define the vector ${\bf{q}}:= {({e}_{j}(\omega ))}_{1\leqslant j\leqslant N}$ of outcome probabilities and the $N$-component vector ${\bf{p}}=({p}_{1},\ldots ,{p}_{d},0,\ldots ,0)\in {{\mathbb{R}}}^{N}$. Then ${\bf{q}}\,\prec \,{\bf{p}}$, i.e. there exists a bistochastic $N\times N$-matrix $M$ with ${\bf{q}}=M{\bf{p}}$.

Proof. By lemma 16 there exist ${c}_{j}\in [0,1]$ and pure ${\omega }_{j}^{\prime }\in {{\rm{\Omega }}}_{A}$ such that ${e}_{j}={c}_{j}\langle {\omega }_{j}^{\prime },\cdot \rangle$. Define ${q}_{l}:= {e}_{l}(\omega )={c}_{l}\langle {\omega }_{l}^{\prime },\omega \rangle$. Using ${\sum }_{j=1}^{N}{e}_{j}={u}_{A}$ and ${\sum }_{j=1}^{d}{\nu }_{j}={u}_{A}$ for an arbitrary maximal frame $({\nu }_{1},\ldots ,{\nu }_{d})$ we find:

$$\begin{eqnarray*}&&\displaystyle \sum _{j=1}^{N}{c}_{j}=\displaystyle \sum _{j=1}^{N}{c}_{j}{u}_{A}({\omega }_{j}^{\prime })={u}_{A}\left(\displaystyle \sum _{j=1}^{N}{c}_{j}{\omega }_{j}^{\prime }\right)={u}_{A}({u}_{A})={u}_{A}\left(\displaystyle \sum _{j=1}^{d}{\nu }_{j}\right)=\displaystyle \sum _{j=1}^{d}{u}_{A}({\nu }_{j})=\displaystyle \sum _{j=1}^{d}1=d.\end{eqnarray*}$$

As ${c}_{j}\in [0,1]$, ${\sum }_{j=1}^{N}{c}_{j}=d$ shows $d\leqslant N$.

Set ${q}_{l| j}:= {e}_{l}({\omega }_{j})$, introduce the N-component vector ${\bf{p}}:= ({p}_{1},\ldots ,{p}_{d},0,\ldots ,0)$ and use that measurement effects and states of maximal frames add up to the order unit:

$$\begin{eqnarray*}\displaystyle \sum _{j=1}^{d}{q}_{l| j}{p}_{j} & = & \displaystyle \sum _{j=1}^{d}{e}_{l}({p}_{j}{\omega }_{j})={e}_{l}(\omega )={q}_{l},\displaystyle \sum _{j=1}^{d}{q}_{l| j}=\displaystyle \sum _{j=1}^{d}{e}_{l}({\omega }_{j})={c}_{l}\displaystyle \sum _{j=1}^{d}\langle {\omega }_{l}^{\prime },{\omega }_{j}\rangle ={c}_{l}{u}_{A}({\omega }_{l}^{\prime })={c}_{l},\\ \displaystyle \sum _{l=1}^{N}{q}_{l| j} & = & \displaystyle \sum _{l=1}^{N}{e}_{l}({\omega }_{j})={u}_{A}({\omega }_{j})=1.\end{eqnarray*}$$

For $j\leqslant d$ we define ${M}_{l,j}:= {q}_{l| j}$. If $d\lt N$ we also define ${M}_{l,j}:= \tfrac{1-{c}_{l}}{N-d}$ for $N\geqslant j\gt d$. M is an N × N-matrix and it is bistochastic: first of all, ${M}_{l,j}\geqslant 0$ for all $l,j$. Furthermore:

$$\begin{eqnarray*}\displaystyle \sum _{l=1}^{N}{M}_{l,j} & = & \displaystyle \sum _{l=1}^{N}{q}_{l| j}=1\qquad \quad \ \mathrm{for}\ j\leqslant d,\\ \displaystyle \sum _{l=1}^{N}{M}_{l,j} & = & \displaystyle \sum _{l=1}^{N}\displaystyle \frac{1-{c}_{l}}{N-d}=\displaystyle \frac{N-d}{N-d}=1\qquad \ \mathrm{for}\ j\gt d,\\ \displaystyle \sum _{j=1}^{N}{M}_{l,j} & = & \displaystyle \sum _{j=1}^{d}{q}_{l| j}+(N-d)\cdot \displaystyle \frac{1-{c}_{l}}{N-d}={c}_{l}+1-{c}_{l}=1.\end{eqnarray*}$$

This bistochastic matrix maps ${\bf{p}}$ to ${\bf{q}}$, i.e. $M\cdot {\bf{p}}={\bf{q}}$:

$$\begin{eqnarray*}&&\displaystyle \sum _{j=1}^{N}{M}_{l,j}{p}_{j}=\displaystyle \sum _{j=1}^{d}{q}_{l| j}{p}_{j}={q}_{l}.\end{eqnarray*}$$

□

Now we come to the proof of the theorem:

Proof of theorem 10. Consider an arbitrary fine-grained measurement $({e}_{1},\ldots ,{e}_{N})$ and an arbitrary state $\omega \in {{\rm{\Omega }}}_{A}$ with classical decomposition $\omega ={\sum }_{j=1}^{d}{p}_{j}{\omega }_{j}$ into a maximal frame. Define ${q}_{l}:= {e}_{l}(\omega )$ and the N-component vector ${\bf{p}}=({p}_{1},\ldots ,{p}_{d},0,\ldots ,0)$. Let M be the bistochastic matrix from lemma 18 with ${\bf{q}}=M\cdot {\bf{p}}$. By Birkhoff's theorem, it is a convex combination of permutation matrices, i.e. $M={\sum }_{\sigma \in {S}_{N}}{a}_{\sigma }{P}_{\sigma }$ for a probability distribution a_σ and permutation matrices P_σ. W.l.o.g. we only consider the Shannon entropy; the proof for the Rényi entropies works exactly the same way. As the Shannon entropy is Schur-concave and invariant under permutations:

$$\begin{eqnarray*}&&H({\bf{q}})\geqslant \displaystyle \sum _{\sigma \in {S}_{N}}{a}_{\sigma }H({P}_{\sigma }\cdot {\bf{p}})=\displaystyle \sum _{\sigma \in {S}_{N}}{a}_{\sigma }H({\bf{p}})=H({\bf{p}})=S(\omega ).\end{eqnarray*}$$

Furthermore $H({\bf{p}})=-{\sum }_{j=1}^{d}{p}_{j}\mathrm{log}{p}_{j}=S(\omega )$ is the entropy of a measurement that perfectly distinguishes the ${\omega }_{j}$, i.e. ${e}_{j}({\omega }_{k})={\delta }_{{jk}}$. Because of lemma 17, such a measurement can be chosen to be finegrained. Therefore we find:

$$\begin{eqnarray*}&&\widehat{H}(\omega )=\mathop{\inf }\limits_{{\bf{e}}\in {{ \mathcal E }}^{* }}H({\bf{e}}(\omega ))=H({\bf{p}})=S(\omega ).\end{eqnarray*}$$

□

A.1.6. Proof of theorem 12

As mentioned in the main text, the equivalences $({\rm{i}})\iff ({\rm{iii}})\iff ({\rm{iv}})\iff ({\rm{v}})$ are shown in [43]. We will now prove the equivalence $({\rm{ii}})\iff ({\rm{v}})$, which proves theorem 12. Taking into account theorem 10, and formulating the atomic covering property in the context of theories that satisfy postulates 1 and 2, it remains to show the equivalence of the following two statements:

• For all states $\omega \in {{\rm{\Omega }}}_{A}$, we have ${\check{S}}_{0}(\omega )\geqslant {S}_{0}(\omega )$.
• If F is any face of ${A}_{+}$, and ω is any pure state, then the smallest face G that contains both F and ω has rank $| G| \leqslant | F| +1$. (Note that this is trivial if $\omega \perp F$.)

We will first prove that $({\rm{ii}}^{\prime} )\Rightarrow ({\rm{v}}^{\prime} )$, which is equivalent to $\neg ({\rm{v}}^{\prime} )\Rightarrow \neg ({\rm{ii}}^{\prime} )$. So suppose that there exists some face F of ${A}_{+}$ and a pure state ω such that the face G generated by both has rank $| G| \geqslant | F| +2$. Let ${\omega }_{1},\ldots ,{\omega }_{| F| }$ be a frame that generates the face F. Then F is also generated by $\nu := \tfrac{1}{| F| }{\sum }_{j=1}^{| F| }{\omega }_{j}$, i.e. the normalised projective unit of F. This is because every face containing ν also contains all the ${\omega }_{j}$ (and vice versa), and F is the smallest face with this property.

Now consider the state $\tfrac{1}{2}\omega +\tfrac{1}{2}\nu$. The smallest face that contains this state must be G. If this state had a decomposition into $| F| +1$ or fewer perfectly distinguishable pure states, then these would also generate G, and so $| G| \leqslant | F| +1$, in contradiction to our assumption. Thus any decomposition of $\tfrac{1}{2}\omega +\tfrac{1}{2}\nu$ into perfectly distinguishable pure states uses at least $| F| +2$ states with non-zero coefficients, i.e. ${S}_{0}(\tfrac{1}{2}\omega +\tfrac{1}{2}\nu )\geqslant \mathrm{log}(| F| +2)$. But $\tfrac{1}{2}\omega +\tfrac{1}{2}\nu =\tfrac{1}{2| F| }{\sum }_{j=1}^{| F| }{\omega }_{j}+\tfrac{1}{2}\omega$ is a convex decomposition into $| F| +1$ pure states, thus

$$\begin{eqnarray*}&&{\check{S}}_{0}\left(\displaystyle \frac{1}{2}\omega +\displaystyle \frac{1}{2}\nu \right)\leqslant \mathrm{log}(| F| +1)\lt \mathrm{log}(| F| +2)\leqslant {S}_{0}\left(\displaystyle \frac{1}{2}\omega +\displaystyle \frac{1}{2}\nu \right).\end{eqnarray*}$$

It remains to show that $({\rm{v}}^{\prime} )\Rightarrow ({\rm{ii}}^{\prime} )$. So suppose that (v') holds, but that there is a state $\omega \in {{\rm{\Omega }}}_{A}$ with ${\check{S}}_{0}(\omega )\lt {S}_{0}(\omega );$ we will show that this leads to a contradiction. By definition of ${\check{S}}_{0}$, if this is the case, then there exist pure states ${\omega }_{1},\ldots ,{\omega }_{n}$ with $n=\exp ({\check{S}}_{0}(\omega ))$ and ${p}_{1},\ldots ,{p}_{n}\geqslant 0$, ${\sum }_{i}{p}_{i}=1$, such that $\omega ={\sum }_{i=1}^{n}{p}_{i}{\omega }_{i}$. Using property (v'), and recursively looking at the faces generated by ${\omega }_{1}$, generated by ${\omega }_{1},{\omega }_{2}$, generated by ${\omega }_{1},{\omega }_{2},{\omega }_{3}$ and so forth, shows that the rank of the face G generated by ${\omega }_{1},\ldots ,{\omega }_{n}$ can be at most n. Since $\omega \in G$, this shows that ω can be decomposed into n or fewer pure perfectly distinguishable states. Therefore ${S}_{0}(\omega )\leqslant \mathrm{log}n={\check{S}}_{0}(\omega )$.□

The GPT framework has a close relation to quantum logic. This is not surprising, since much of the terminology of GPTs has appeared much earlier, in work on quantum logic and beyond. The approach via convex sets of states and observables can be traced back to Mackey [18] (who immediately made connections to quantum logic), and was developed further through the 1960s and beyond. A partial list of references includes [14, 20, 88], the last two of which offer axiomatic characterisations of quantum theory. Interaction with the quantum logic tradition continued, with the orthomodularity of the lattice of faces of the state and/or effect spaces often providing a point of contact, especially in Ludwig's work [20]. Also closely related to the convex sets approach was the work of Foulis and Randall [10–13] who, for example, studied ways of combining probabilistic systems.

Since postulates 1 and 2 imply that the state cone ${A}_{+}$ is self-dual (so coincides with the effect cone), and that each face of this cone is the intersection of the cone with the image of a filter (equivalently given self-duality, a compresssion in the sense of [24]), we have from these postulates alone, via e.g. [24], theorem 8.10, that the face lattice is orthomodular. The notion of orthomodular partially ordered set, or its special case, the notion of orthomodular lattice, is often taken to define the notion of quantum logic, so we can say that postulates 1 and 2 imply that the face lattice of the cone of states, (equivalently of the cone of effects, or of the set of normalised states) is a quantum logic.

The covering property, in its most common variant the atomic covering property, states that for any element x of the lattice and any atom a not below or equal to x, $a\vee x$ covers x¹⁷ .The equivalence of (iv) and (v), in a setting more general than postulates 1 and 2, is proposition 9.7 of [24] (first appearing in proposition 4.2 in [25] and the discussion preceding it). Along with orthomodularity, the covering law was one of the assumptions of Piron's famous lattice-theoretic characterisation ([94]; also [17] and see the discussion in [27] or for more detail and proofs, [26 pp 18–38, 114–122]) of a class of lattices close to, although larger than, that of real, complex, and quaternionic quantum theory. A generalisation of the covering law was also used in Ludwig's axiomatization (see e.g. [20]¹⁸ ) of quantum theory within the convex sets framework, in which the relevant lattice is a lattice of faces of the state space (equivalent to a lattice of extremal effects in his context), and the result characterised real, complex, and quaternionic quantum theory¹⁹ .