# Von Neumann’s information engine without the spectral theorem

Shintaro Minagawa<EMAIL_ADDRESS>Graduate School of
Informatics, Nagoya University, Chikusa-Ku, Nagoya 464-8601, Japan  Hayato
Arai<EMAIL_ADDRESS>Graduate School of Mathematics, Nagoya
University, Chikusa-Ku, Nagoya 464-8602, Japan  Francesco Buscemi
<EMAIL_ADDRESS>Graduate School of Informatics, Nagoya University,
Chikusa-Ku, Nagoya 464-8601, Japan

###### Abstract

Von Neumann obtained the formula for the entropy of a quantum state by
assuming the validity of the second law of thermodynamics in a thought
experiment involving semipermeable membranes and an ideal gas of quantum-
labeled particles. Despite being operational in the most part, von Neumann’s
argument departs from an operational narrative in its use of the spectral
theorem. In this work we show that the role of the spectral theorem in von
Neumann’s argument can be taken over by the operational assumptions of
repeatability and reversibility, and using these we are able to explore the
consequences of the second law also in theories that do not possess a unique
spectral decomposition. As a byproduct, we obtain the Groenewold–Ozawa
information gain as a natural monotone for a suitable ordering of instruments,
providing it with an operational interpretation valid in quantum theory and
beyond.

##  I Introduction

Ever since Maxwell summoned his demon [1], the mutual influence between
physics (i.e., the representation of a system’s physical properties),
information (i.e., the representation of an agent’s knowledge about a physical
system), and the measurement process (i.e., the interaction between system and
agent) has emerged as one of the main themes of debate in theoretical physics.
Von Neumann is surely among the most influential names to have contributed to
this discussion. In his mathematical formulation of quantum theory [2], much
space is devoted to a careful analysis of the interplay between quantum
measurement theory and thermodynamics. Therein, von Neumann approaches the
problem using the artifact—common, as he explicitly remarks, in
phenomenological thermodynamics and used before him also by Einstein [3] (see
also Ref. [4]) and Szilard [5, 6]—of an ideal gas of particles, whose
mechanical degrees of freedom obey the laws of classical mechanics, while
their states (which should really be thought of, in this context, as mere
labels) are described according to quantum theory, but are otherwise
irrelevant from an energetic viewpoint. In this way, it is possible to
separate, on the one side, the mechanical and thermal properties of the gas,
and, on the other side, the information that an agent acting on the gas has
about its particles. The “missing link” between physics and information is
provided by two assumptions: the first is the existence of particular devices
called semipermeable membranes; we will discuss them extensively in what
follows. The second assumption is about the validity of the second law of
thermodynamics, which is posited by von Neumann ab initio. Following this line
of thought, von Neumann was able to explore the consequences of the second law
in quantum theory and obtain his famous formula for the entropy of quantum
states.

The argument constructed by von Neumann, although operational in the most part
(i.e., the quantum entropy is defined using a thermodynamic protocol, which in
principle also provides a way to measure the quantum entropy), still relies on
the structure of Hilbert spaces. In particular, a crucial role is played by
the spectral decomposition of self-adjoint operators [2]. A natural question
is then to see how far von Neumann’s discussion can be reconstructed from
purely operational assumptions. The common way to approach this kind of
problems utilizes the framework of general probabilistic theories (GPTs; see,
e.g., Ref. [7, 8, 9, 10, 11, 12, 13, 14, 15, 16]). These provide a modern take
on the operational approach to quantum theory, which can be traced back to
works by Ludwig [17, 18], Davies and Lewis [19], Gudder [20], and Ozawa [21].

Without Hilbert spaces, in GPTs there exist various ways to introduce the
entropy functional in an operational way, which turn out to be all equivalent
in conventional quantum theory, but are not so in general [22, 12, 23, 24]. A
possible approach is to add assumptions that are strong enough to conclude
that a unique spectral decomposition exists [25, 26]. However, in more general
setups, the uniqueness of the spectral decomposition is not guaranteed [15,
27, 28]: in all such cases, von Neumann’s argument seems to be a nonstarter.

In this paper, we clarify the importance of two implicit assumptions in von
Neumann’s argument: the existence of repeatable measurement processes, on the
one hand, and of states that are the fixed point of some non-trivial
repeatable measurement, on the other. These two operational assumptions—we
argue—can take over the role played, in von Neumann’s argument, by the
spectral theorem, which instead is not operational. In this way, we can
provide von Neumann’s thought experiment with a fully operational narrative,
and to explore the consequences of the second law of thermodynamics also in
GPTs that do not have a unique spectral decomposition. However, to achieve
this, some modifications to von Neumann’s argument are needed: in particular,
the thermodynamic process must be modified into a cycle. As a byproduct, our
argument allows us to obtain also an analogue of the Groenewold–Ozawa
information gain [29, 30] in a wide range of GPTs, and to equip it with the
operational meaning of monotone, with respect to a suitable preorder of the
measurement processes.

##  II Basic definitions

In order to discuss von Neumann’s thought experiment in GPTs, we begin by
briefly reviewing the basics of a simple single-system theory (see, e.g.,
Refs. [8, 9] and references therein). A single-system GPT is determined by
providing all possible events, thought of as “black boxes” with an input and
an output, which can be either the system of the theory or the trivial system
(i.e., a system with only one possible state). The input-output arrangement
determines the type of the event. Events with trivial input are interpreted as
preparation events, whereas events with trivial output are interpreted as
observation events, or effects. Families of events of the same type form a
test: one has, therefore, preparation tests and observation tests. The latter
are usually called measurements. Tests describe what can happen in an
experiment: among the events it contains, one and only one will occur in any
given repetition of the same experiment. Therefore, tests with only one
element describe deterministic events. In particular, a common assumption
(corresponding to a “no-signaling from the future” principle [31, 32]) is that
only one deterministic measurement exists. Instead, a theory typically
provides many possible deterministic preparations: these are the (normalized)
states of the theory and form its state space, denoted by $\Omega$. Normalized
states naturally form a convex set: its extremal points are called pure
states, otherwise they are mixed.

Tests can be composed, following the idea that any experiment can be seen as a
chain preparation–process–measurement. Since states can be convexly mixed,
tests are naturally assumed to satisfy linearity on such convex combinations.
This observation allows for a more concrete definition of single-system tests
as families $\\{s_{j}\\}_{j\in J}$ of affine maps acting on $\Omega$ 111Notice
that we are not assuming that any family of affine maps constitutes a
legitimate test of the theory: further conditions, most notably that of
complete positivity, may restrict the set of possible tests. However, this
point is irrelevant for the present discussion.. Denoting by $u$ the unique
deterministic measurement, normalization of probability requires that
$\sum_{j}(u\circ s_{j})\rho=1$, for all $\rho\in\Omega$. In conventional
quantum theory, it is easy to recognize that tests are quantum instruments
[34], whereas the composition $\\{u\circ s_{j}\\}_{j\in J}$ provides the
generalization of positive operator-valued measures (POVMs). For this reason,
in what follows we will call instruments those tests that have both input and
output non-trivial, while the term measurement will only be used to denote the
analogue of POVMs. Summarizing, in what follows we will work with normalized
states, denoted by $\rho$, $\sigma$ etc., instruments, denoted by
$\\{s_{j}\\}_{j\in J}$, $\\{t_{k}\\}_{k\in K}$ etc., and measurements, denoted
by $\\{e_{j}\\}_{j\in J}$, $\\{f_{k}\\}_{k\in K}$ etc.

Finally, an important notion is that of perfect distinguishability: a family
of normalized states $\\{\rho_{j}\\}_{j\in J}$, $\rho_{j}\in\Omega$ is said to
be _perfectly distinguishable_ if there exists a measurement
$\\{e_{j}\\}_{j\in J}$ such that

$e_{j}(\rho_{j^{\prime}})=\delta_{jj^{\prime}}\;,$

for all $j,j^{\prime}\in J$.

The measurement stage.—As anticipated in the introduction, von Neumann’s
thermodynamic thought experiment makes use of semipermeable membranes (SPMs).
These are devices that are can separate, reversibly and without any
thermodynamic cost, the particles of a gas, as long as their states are
distinguishable—at least in principle, of course. Von Neumann goes to a great
length to justify the use of such idealized devices, which, he argues,
represent “the thermodynamic definition of difference” [2]. In what follows,
we characterize SPMs from an operational viewpoint.

The first property of SPMs, implicit in their definition, is their
repeatability: the first time a particle collides with the membrane, a
measurement occurs determining whether the arriving particle is a “pass” or a
“bounce,” and it will remain so until the end of the experiment, even if the
same particle collides with the membrane multiple times. Hence, our first
assumption is that the theory contains repeatable instruments. Formally, an
instrument $\\{s_{j}\\}_{j\in J}$ is repeatable if and only if

$s_{j^{\prime}}\circ s_{j}=\delta_{j^{\prime}j}s_{j}\;,$

holds, for all $j,j^{\prime}\in J$.

The second requirement needed to go along with von Neumann’s discussion is the
existence of states that are the fixed points of some repeatable instrument.
This is necessary if we want to speak, as von Neumann does, of thermodynamic
reversibility with respect to the system’s initial state. Since even in
quantum theory there exist repeatable instruments without invariant states
[35], we need to treat this requirement as a further assumption. In
thermodynamic terms, the assumption of state invariance makes it reasonable to
assume that there are no hidden thermodynamic costs (besides the macroscopic
mechanical ones) incurred when using SPMs that preserve the mixture being
separated. Formally, we say that an instrument $\\{s_{j}\\}_{j\in J}$ is
$\rho$-preserving whenever $\sum_{j}s_{j}(\rho)=\rho$.

In the assumption of state invariance, however, we need to exclude the trivial
case of the deterministic identical instrument, which is of course repeatable
and for which all states are fixed points. We thus strengthen our requirements
as follows: first, we define a set of “maximal” instruments with respect to a
suitable preorder, and then require the existence of states that are fixed
points of some instrument that is maximal and repeatable. As a straightforward
extension of the post-processing preorder of POVMs [36] to the case of
instruments, we introduce the following preorder (see also Fig. 1):

Figure 1: A schematic representation of the notion of Groenewold majorization
$t\succ_{\rho}s$.

######  Definition 1 (Groenewold majorization).

Given two instruments $s=\\{s_{j}\\}_{j\in J}$ and $t=\\{t_{k}\\}_{k\in K}$,
and a state $\rho\in\Omega$, we say that $t$ _Groenewold-majorizes_ $s$ given
$\rho$, in formula,

$t\succ_{\rho}s\;,$

if and only if there exists a conditional probability distribution $p(j|k)$
such that

$s_{j}(\rho)=\sum_{k\in K}p(j|k)t_{k}(\rho)\;,$ (1)

for all $j\in J$.

In the above definition, we could have in fact chosen a more general scenario,
including also some suitable transformations before and after the
randomization. However, since for our purposes we are only interested in the
maximal points of $\succ_{\rho}$ and since these are the same with or without
the extra transformations, for the sake of simplicity, we choose to work with
the above definition.

######  Definition 2 (Fine-grained instruments).

An instrument $s=\\{s_{j}\\}_{j\in J}$ is said to be _fine-grained_ if and
only if, for all preorders $\succ_{\rho}$ (i.e., for all states $\rho$), the
condition $t\succ_{\rho}s$ implies that also $s\succ_{\rho}t$ holds.

######  Definition 3 (MPP instruments).

An instrument $s=\\{s_{j}\\}_{j\in J}$ is said to be of the _measure-and-
prepare-pure (MPP)_ form if and only if it is completely characterized by one
measurement $\\{e_{j}\\}_{j\in J}$ and one family of normalized pure states
$\\{\sigma_{j}\\}_{j\in J}$, such that the $j$-th state $\sigma_{j}$ is
prepared whenever the $j$-th effect $e_{j}$ occurs.

The following lemma holds (for the proof, see Appendix A) .

######  Lemma 1.

Let $s=\\{s_{j}\\}_{j\in J}$ be a fine-grained instrument. Then, $s_{j}(\rho)$
is (up to normalization) a pure state, for all states $\rho\in\Omega$ and all
$j\in J$. In other words, $s$ is an _MPP_ instrument.

The above lemma guarantees that fine-grained instruments are all physically
admissible, simply because they can be physically realized as measurements
followed by the preparation of pure states that only depend on the outcome. As
such, they do not require any notion of composition (i.e., complete
positivity) to be discussed. We denote fine-grained instruments as pairs
$\\{e_{j},\sigma_{j}\\}_{j\in J}$, where $\\{e_{j}\\}$ is a measurement and
$\\{\sigma_{j}\\}$ is a family of normalized pure states.

Hence, the following two assumptions, that is,

  1. 1.

the existence of fine-grained and repeatable instruments; and,

  2. 2.

the existence of states that are fixed points of some fine-grained and
repeatable instrument,

together with Lemma 1, imply that the theory contains states that can be
decomposed on a set of perfectly distinguishable pure (PDP) states. Following
[15, 26], we call such states weakly spectral. For simplicity, we summarize
the previous discussion into one assumption as follows:

Assumption 1 (Weak spectrality). We assume that the theory contains weakly
spectral states $\rho\in\Omega$, that is, states that admit a (possibly non-
unique) convex decomposition into PDP states $\rho=\sum_{i}p_{i}\rho_{i}$.

Notice that we are not assuming that all states of the theory are weakly
spectral. In what follows, for ease of notation, for a weakly spectral $\rho$,
we denote by $\mathcal{D}(\rho)$ the set of all possible probability
distributions $\\{p_{i}\\}$ that appear in at least one of its PDP
decompositions.

Assumption 1 (A1) above is “weak” because in the literature its “strong”
version is often encountered, and the separation between the two is strict.
More precisely, instead of A1, the property of (unique or strong) spectrality
assumes that all PDP decompositions of the same state correspond to
distributions $\\{p_{i}\\}$ which differ at most in a permutation of the
indices [27, 28]. Strong spectrality is hence akin to assuming the spectral
theorem from the onset. Instead, the existence of weakly spectral states is
guaranteed as soon as there exist at least two perfectly distinguishable pure
states. Examples of theories that satisfy weak spectrality but not strong
spectrality are given in Ref. [28]. For what follows, it is convenient to
introduce the following definition:

######  Definition 4 ($\rho$-separating SPMs).

A set of SPMs is said $\rho$-_separating_ if it corresponds to a repeatable
_MPP_ instrument $\\{e_{j},\sigma_{j}\\}_{j\in J}$, which is in particular
$\rho$-preserving, that is

$\rho=\sum_{j\in J}e_{j}(\rho)\sigma_{j}\;.$

From the above discussion, it is clear that $\rho$-separating SPMs exist if
only if $\rho$ is weakly spectral.

Figure 2: The process of separating particles in a state
$\rho=q_{1}\sigma_{1}+q_{2}\sigma_{2}$ using suitable $\rho$-separating SPMs.
The dashed line is a SPM transparent to $\sigma_{1}$ but perfectly elastic for
$\sigma_{2}$; the dashed-dotted line vice versa.

##  III The feedback stage

Von Neumann’s argument also involves a feedback control stage, during which a
suitable transformation is applied to the system, depending on the measurement
outcome. Again, to follow von Neumann’s narrative, we need an assumption, that
we identify in the following:

Assumption 2 (Free pure-state transformations [13, 37, 26]). For any pair of
pure states, $\rho_{\textrm{in}}$ and $\rho_{\textrm{out}}$, the theory
contains a deterministic event $F:\Omega\to\Omega$ which is reversible, that
is, there exists another deterministic event $G$ such that $G\circ
F=\operatorname{id}$, and satisfies
$F(\rho_{\textrm{in}})=\rho_{\textrm{out}}$.

According to the thermodynamic narrative, a reversible operation is one that
does not cause any change to the entropy of the thermodynamic universe. Notice
that we do not put any constraints on what the operation does on states other
than $\rho_{\textrm{in}}$. In this sense, Assumption 2 (A2), like A1 before,
is “weak”: instead of A2, in the literature it is common to find the
assumption of strong symmetry, which assumes that any two collections of PDP
states are connected by one simultaneous reversible process [15, 28, 26]. A
counterexample of a theory that satisfies A2 but is not strongly symmetric is
given in Appendix B, where we explicitly construct a GPT that contains two
pairs of PDP states that cannot be simultaneously and reversibly converted.

We conclude this section by noticing that conventional quantum theory
satisfies both strong spectrality and strong symmetry. In fact, it is known
that any GPT satisfying both strong spectrality and strong symmetry becomes to
a large extent akin to quantum theory [38].

##  IV Entropy from thermodynamic considerations

We are now ready to formulate our version of von Neumann’s thought experiment
in the language of GPTs. As already noticed, we follow von Neumann’s argument,
in that the particles’ mechanical degrees of freedom obey the classical laws
of ideal gases, whereas the non-classical degrees of freedom, i.e., the
generalized states labeling the different “isomers,” are thought of as
internal degrees of freedom with a completely degenerate Hamiltonian so that
they do not directly enter in the energetic balance of the process.

We begin with the calculation of the work needed to separate the particles by
means of SPMs. The separation process is depicted in Fig. 2. An ideal,
thermostatted gas contains $N$ particles in the mixture state $\rho$.
According to the above discussion, $\rho$ is assumed to satisfy the property
of weak spectrality. For simplicity, we assume that $\rho$ contains only two
pure components, that is, $\rho=q_{1}\sigma_{1}+q_{2}\sigma_{2}$, where
$\\{q_{j}\\}_{j=1,2}$ is a probability distribution and
$\\{\sigma_{j}\\}_{j=1,2}$ are two PDP states. The generalization to a larger
number of components is straightforward.

Closely following von Neumann, we apply a set of $\rho$-separating SPMs,
corresponding to the repeatable MPP instrument
$\\{e_{j},\sigma_{j}\\}_{j\in\\{1,2\\}}$ with effects such that
$e_{j}(\sigma_{j^{\prime}})=\delta_{jj^{\prime}}$. This is physically modeled
by two SPMs with opposite mechanical behaviors: if one SPM is transparent for,
say, $\sigma_{1}$, the other is transparent for $\sigma_{2}$. After the
separation, which is done isothermally, the two species ($\sigma_{1}$ and
$\sigma_{2}$) are contained in two separate chambers, both of the same size as
the initial chamber. The number of particles in a state $\sigma_{j}$ is
$e_{j}(\rho)N=q_{j}N$. In agreement with our preceding discussion and previous
analyses [2, 37, 26], we can assume that this first step of the separation is
basically a solid translation of coordinates, so that the work worth of this
step is zero.

The two SPMs are then replaced by impermeable walls and we isothermally
compress the chamber with $\sigma_{1}$ (resp., $\sigma_{2}$) until its volume
becomes $q_{1}V$ (resp., $q_{2}V$), so that after the compression both
chambers have the same (initial) pressure. From the ideal gas law, the amount
of work needed for the isothermal compression c$\to$d is

$\displaystyle-\int_{V}^{q_{1}V}\frac{q_{1}NkT}{V^{\prime}}dV^{\prime}-\int_{V}^{q_{2}V}\frac{q_{2}NkT}{V^{\prime}}dV^{\prime}$
$\displaystyle=H(\\{q_{j}\\})NkT\ln 2\;,$ (2)

where $T$ is the temperature of the environment, $k$ is the Boltzmann
constant, and $H(\\{q_{j}\\}):=-\sum_{j}q_{j}\ln q_{j}$. This ends the
separation protocol. If the decomposition of $\rho$ has more than two PDP
states, we repeat this protocol for each perfectly distinguishable state and
obtain the same result.

Next, we consider the mixing process. Here we deviate from von Neumann’s
process, in that we want to go back to the initial mixed state, whereas von
Neumann’s final state is pure. After the separation process has been
completed, we are in the situation in which the two (pure, distinguishable)
species $\sigma_{1}$ and $\sigma_{2}$ are in two separate chambers with equal
pressure (i.e., the initial pressure). The total volume of the two chambers
together equals the initial volume, but now we know which species is present
in each chamber. If we were to replace the wall between the two chambers again
with the same SPMs used during the separation step, by letting the two species
slowly and independently expand, we would be able to gain back the work
invested in the separation stage with the gas restored to its initial state.
We would then have achieved a trivial, reversible cycle.

Figure 3: The cycle used in the proof of Theorem 1. [$\alpha\to\beta$]: work
is invested to achieve the separation as in Fig. 2. [$\beta\to\gamma$]:
additional partition plates are inserted at proper positions (zero work).
[$\gamma\to\delta$]: to each chamber, a suitable reversible transformation,
with zero work-cost, is applied and the states therein contained are
transformed into $\rho_{i}$. [$\delta\to\alpha$]: work is extracted, by
running the separation process for the decomposition
$\rho=\sum_{i}p_{i}\rho_{i}$ backwards.

Instead, we consider a different decomposition of $\rho$ in pure
distinguishable states, as assumption A1 does not exclude such a possibility.
Let us denote the alternative decomposition of $\rho$ by
$\sum_{i}p_{i}\rho_{i}$. As depicted in Fig. 3, we can freely insert
additional partitions in a suitable way, and transform the pure states (which
are known) in each partition into $\rho_{1}$ or $\rho_{2}$, as needed. In this
way (by removing additional walls as necessary), we have transformed the
arrangement corresponding to the decomposition $\sum_{j}q_{j}\sigma_{j}$ into
the arrangement corresponding to $\sum_{i}p_{i}\rho_{i}$ without the need to
account for any new term in the thermodynamic balance, thanks to assumption
A2.

Then, by using two new SPMs tuned to match the new PDP decomposition
$\sum_{i}p_{i}\rho_{i}$, and by following the separation steps backward, we
can bring the system back to its original state having gained in the process
(isothermal expansion) an amount of work equal to:

$H(\\{p_{i}\\})NkT\ln 2.$ (3)

As a whole, therefore, the amount of work we can extract from this cycle is
calculated from (2) and (3) as follows:

$\displaystyle\Delta W=[H(\\{p_{i}\\})-H(\\{q_{j}\\})]\;NkT\ln 2\;.$ (4)

We now invoke, following von Neumann, the second law of thermodynamics, which
implies that $\Delta W$ cannot be strictly positive; otherwise, we would have
constructed a perpetuum mobile of the second kind. Moreover, since the same
must hold also if we exchange the role of the two decompositions, we conclude
that $\Delta W$ must be exactly zero. But this must hold for any PDP
decomposition of $\rho$. Hence we obtain the following:

######  Theorem 1.

Under assumptions A1 and A2, the second law of thermodynamics implies that for
all weakly spectral $\rho\in\Omega$ we have

$H(\\{p_{i}\\})=H(\\{q_{i}\\})\;,$ (5)

for all $\\{p_{i}\\},\\{q_{i}\\}\in\mathcal{D}(\rho)$. In other words, a
necessary condition for the validity of the second law is that any weakly
spectral state has a unique spectral entropy.

Indeed, we provide an explicit example of a GPT where the spectral entropy is
not uniquely defined in Appendix C.

Here we derive concavity of the spectral entropy under the assumption that all
the states of the theory are weakly spectral and that the second law is valid.

######  Theorem 2 (Concavity).

Let $\rho_{1},\rho_{2}\in\Omega$ be states. Then we have the following
inequality:

$H\left(p\rho_{1}+(1-p)\rho_{2}\right)\geq pH(\rho_{1})+(1-p)H(\rho_{2})\;,$
(6)

for any value $0\leq p\leq 1$.

###### Proof.

Consider a state $\rho$ whose decomposition into PDP states is
$\rho=\sum_{i}q_{i}\sigma_{i}$. First, as shown in Fig. 4, we separate it into
its pure components using a suitable set of $\rho$-separating SPMs. The
resulting arrangement is shown as (b) in Fig. 4 and costs an amount of work
proportional to $H(\rho)$. Then, by adding partitions, transforming pure
states, and mixing, we can arrive at the arrangement shown in (c) of Fig. 4.
In this step we can gain an amount of work proportional to
$\sum_{j}p_{j}H(\rho_{j})$, where $p_{1}\rho_{1}+p_{2}\rho_{2}=\rho$. Finally,
since $\rho$ is the convex combination of $\rho_{1}$ and $\rho_{2}$, we can
accomplish the process (c)$\to$(a) by only removing the partion, without
requiring any work. Therefore, the work we can extract from the isothermal
cycle (a)$\to$(b)$\to$(c)$\to$(a) is $\Delta
W=\left[-H(\rho)+\sum_{j}p_{j}H(\rho_{j})\right]NkT\ln 2$. The second law
implies that $\Delta W\leq 0$, which completes the proof.

Figure 4: The proof of the concavity of spectral entropy. [(a)$\to$(b)]:
separate $\rho$ into PDP states; this process needs work $H(\rho)NkT\ln 2$.
[(b)$\to$(c)]: create $\rho_{1}$ and $\rho_{2}$ from pure states; the amount
of work $\sum_{i}p_{i}H(\rho_{i})NkT\ln 2$ can be extracted from this process.
[(c)$\to$(a)]: since $\rho$ is the convex combination of $\rho_{1}$ and
$\rho_{2}$, we can accomplish this process by just removing the wall with no
work.

∎

##  V Groenewold–Ozawa information gain

The uniqueness of the spectral entropy functional can be used to extend the
definition of the Groenewold–Ozawa information gain [29, 30] to GPTs that
satisfy weak spectrality as follow:

######  Definition 5.

For any state $\rho\in\Omega$ and instrument $s$, we define the
Groenewold–Ozawa information gain as follows:

$I_{\mathrm{G}}(\rho,s):=H(\rho)-\sum_{j}e_{j}(\rho)H\left(\frac{s_{j}(\rho)}{e_{j}(\rho)}\right)\;,$
(7)

where $e_{j}(\rho):=(u\circ s_{j})(\rho)$ is the probability of the $j$-th
outcome.

The Groenewold–Ozawa information gain earns an operational interpretation in
terms of the preorder introduced in Definition 1 as a consequence of the
following fact, proved in Appendix D.

######  Theorem 3.

For any state $\rho\in\Omega$ and instruments $t$ and $s$, $t\succ_{\rho}s$
implies the following inequality:

$I_{\mathrm{G}}(\rho,t)\geq I_{\mathrm{G}}(\rho,s)\;.$ (8)

Recently, Ref. [39] proved a relation between the Groenewold–Ozawa information
gain and the heat absorbed by the system during a measurement process. This
result together with our inequality (8) suggests a link between the
thermodynamic and the resource-theoretic characterization of quantum
measurements. We leave this point open for future studies.

##  VI Conclusions

In this work, we have shown how von Neumann’s thought experiment can be
formulated in a purely operational language, without resorting to the
structure of Hilbert spaces or the spectral theorem. An advantage of our
reformulation is that we can now appreciate how important it is, in von
Neumann’s argument, to assume the validity of the second law of thermodynamics
from the beginning. It is the second law, and not the spectral theorem, to
force the entropy to be unique, lest we build a perpetuum mobile of the second
kind. In this sense, the second law is used by von Neumann as a consistency
check, a first principle of logic rather than a law of physics [40, 41, 42,
43].

A problem left open is that of finding relations between the spectral entropy
considered in this work and other entropies that can be considered in GPTs
[22, 12, 23, 44]. Moreover, since we can regard von Neumann’s device as a
process for extracting work from measurements, there may be a close
relationship between the present discussion and previous research on work
extraction from quantum measurement processes [45]. We leave these questions
for future works.

##  VII Acknowledgments

The authors are very grateful to Michele Dall’Arno, Masanao Ozawa, and Mark M.
Wilde for their useful comments. Support from MEXT Quantum Leap Flagship
Program (MEXT QLEAP) Grant No. JPMXS0120319794 is acknowledged. S.M. would
like to take this opportunity to thank the “Nagoya University
Interdisciplinary Frontier Fellowship” supported by JST and Nagoya University.
H.A. is supported by a JSPS Grant-in-Aids for JSPS Fellows No. JP22J14947, a
JSPS Grant-in-Aids for Scientific Research (B) Grant No. JP20H04139, and a
Grant-in-Aid for JST SPRING No. JPMJSP2125. F.B. acknowledges support from
MEXT Quantum Leap Flagship Program (MEXT QLEAP) Grant No. JPMXS0120319794;
from MEXT-JSPS Grant-in-Aid for Transformative Research Areas (A) “Extreme
Universe”, No. 21H05183; from JSPS KAKENHI Grants No. 19H04066 and 20K03746.

Author contributions.—S. M. and H. A. contributed to this work equally.

##  Appendix A Proof of Lemma 1

Firstly we prove that the output states of fine-grained instruments are pure.
Consider a state space $\Omega$ and a state $\rho\in\Omega$. Let
$s=\\{s_{j}\\}$ be a fine-grained instrument and $e_{j}(\rho)$ denotes
$(u\circ s_{j})(\rho)$ for simplicity. Without loss of generality, suppose
that a post-measurement state corresponding to the outcome $j=|J|$,

$\sigma_{|J|}(\rho):=\frac{1}{e_{|J|}(\rho)}s_{|J|}(\rho)$ (9)

is not pure, that is, there is a convex decomposition like

$\sigma_{|J|}(\rho)=\sum_{l\in L}q_{l}\sigma_{l},$ (10)

where $\\{q_{l}\\}_{l\in L}$ is a probability distribution and $|L|>1$. Note
that both $\\{q_{l}\\}_{l\in L}$ and $\sigma_{l}$ depend on $\rho$.

If we multiply both side of this equation by $e_{|j|}(\rho)$, we obtain

$s_{|J|}(\rho)=\sum_{l\in L}e_{|J|}(\rho)q_{l}\sigma_{l}.$ (11)

Let us now introduce events $\tilde{s}_{l,|J|}$ such that

$\tilde{s}_{l,|J|}(\rho)=e_{|J|}(\rho)q_{l}\sigma_{l}.$ (12)

Let $K^{J}$ and $K^{L}$ be sets such that

$\displaystyle K_{1}$ $\displaystyle:=\\{1,\dots,j,\dots,|J|-1\\}$ (13)
$\displaystyle K_{2}$
$\displaystyle:=\\{(1,|J|),\dots,(l,|J|),\dots,(|L|,|J|)\\},$ (14)

where $j\in J\setminus\\{|J|\\}$ and $l\in L$.

Now we define an index $k\in K$ as a direct sum of $K^{J}$ and $K^{L}$:

$K:=K_{1}\cup K_{2}$ (15)

There is a family of pure states $\\{t_{k}(\rho)\\}_{k\in K}$ such that

$t_{k}(\rho):=\begin{cases}&s_{k}(\rho)\quad\mathrm{if}\>k\in K_{1}\\\
&\tilde{s}_{k}(\rho)\quad\mathrm{if}\>k\in K_{2}\end{cases}$ (16)

Since we have $s_{|J|}(\rho)=\sum_{k\in K_{2}}\tilde{s}_{k}(\rho)$, we have
$t\succ_{\rho}s$. However, since the $|j|$-th output state of $s$ is mixed,
and it is not possible to make a mixed state pure by further convex mixtures,
it is clear that $s\nsucc_{\rho}t$. This contradicts the assumption that $s$
is fine-grained, thus proving the first part, that is, the states
$\frac{1}{e_{j}(\rho)}s_{j}(\rho)$ must be pure for all outcomes $j$ and all
initial states $\rho$.

Now we show that fine-grained instruments are MPP instruments. Suppose that a
state $\rho\in\Omega$ has a convex decomposition
$\rho=p\rho_{1}+(1-p)\rho_{2}$, where $\rho_{1},\rho_{2}$ are two different
state on $\Omega$. From the affinity of $s_{j}$,
$s_{j}(\rho)=ps_{j}(\rho_{1})+(1-p)s_{j}(\rho_{2})$ holds. For what we said
before, $s_{j}(\rho)$ is proportional to a pure state. Therefore, it must be
that both $s_{j}(\rho_{1})$ and $s_{j}(\rho_{2})$ are proportional to
$s_{j}(\rho)$. This means that the post-measurement state does not depend on
the initial state, if not from the outcome $j$, which implies that $s$ is MPP.

##  Appendix B A theory satisfying A2 but not strong symmetry

Here we give an explicit example of a theory that satisfies property A2 above,
that is, the existence of free pure-state transformations, but does not
satisfy strong symmetry.

Let $\mathcal{H}$ be a finite-dimensional Hilbert space and let $\mathcal{S}$
be a state space on $\mathcal{H}$, that is, a convex set of positive
semidefinite matrices on $\mathcal{H}$ with unit trace. Also, let
$\mathcal{P}\subseteq\mathcal{S}$ be a set of rank one matrices, which
corresponds to the set of pure states. Consider two systems $\mathcal{H}_{A}$
and $\mathcal{H}_{B}$ and let
$\mathcal{P}_{\mathrm{A}}\otimes\mathcal{P}_{\mathrm{B}}$ be the set of
product pure states. The state space $\Omega$ of the theory is the convex hull
$\mathrm{SEP(A;B)}$ of
$\mathcal{P}_{\mathrm{A}}\otimes\mathcal{P}_{\mathrm{B}}$.

For a state space $\Omega$, define a class of transformation
$\mathcal{F}(\Omega)$ as the set of all linear maps $F$ satisfying
$F(\Omega)=\Omega$. Any reversible transformation clearly belongs to
$\mathcal{F}(\Omega)$.

######  Definition 6 ($k$-symmetry).

We say that a state space $\Omega$ is $k$-symmetric if there exists a map
$F\in\mathcal{F}(\Omega)$ such that $\rho_{i}=F(\sigma_{i})$, for
$i=1,\cdots,k$, for any pair of $k$-tuples of perfectly distinguishable pure
states $\\{\rho_{i}\\}_{i=1}^{k}$ and $\\{\sigma_{i}\\}_{i=1}^{k}$.

Now we show the difference between strong symmetry and weak symmetry by giving
the following counterexample.

######  Theorem 4.

$\mathrm{SEP(A;B)}$ is $1$-symmetric but not $2$-symmetric.

We invoke the following two results. The first result is about the necessary
and sufficient condition for two states in SEP to be perfectly distinguished
[46, Theorem 2.4]. Notice that non-orthogonal states can be perfectly
distinguished in SEP because the set of all measurements in SEP is larger than
the set of bipartite POVMs.

######  Lemma 2.

In $\mathrm{SEP(A;B)}$, two pure states
$\rho_{1}=\rho^{\mathrm{A}}_{1}\otimes\rho^{\mathrm{B}}_{1}$ and
$\rho_{2}=\rho^{\mathrm{A}}_{2}\otimes\rho^{\mathrm{B}}_{2}$ are perfectly
distinguishable if and only if they satisfy

$\operatorname{Tr}\rho^{\mathrm{A}}_{1}\rho^{\mathrm{A}}_{2}+\operatorname{Tr}\rho^{\mathrm{B}}_{1}\rho^{\mathrm{B}}_{2}\leq
1.$ (17)

The second result gives the form of the transformation maps in
$\mathcal{F}(\mathrm{SEP(A;B)})$ concretely [47, Theorem 3].

######  Lemma 3.

For the linear map $F$ from
$\mathcal{T}(\mathcal{H}_{\mathrm{A}}\otimes\mathcal{H}_{\mathrm{B}})\to\mathcal{T}(\mathcal{H}_{\mathrm{A}}\otimes\mathcal{H}_{\mathrm{B}})$,
the following are equivalent:

  1. (i)

$F\in\mathcal{F}(\mathrm{SEP(A;B)})$.

  2. (ii)

$F(\mathcal{P}_{\mathrm{A}}\otimes\mathcal{P}_{\mathrm{B}})=\mathcal{P}_{\mathrm{A}}\otimes\mathcal{P}_{\mathrm{B}}$.

  3. (iii)

$F(A\otimes B)=F_{\mathrm{A}}(A)\otimes F_{\mathrm{B}}(B)$, or
$\dim\mathcal{H}_{\mathrm{A}}=\dim\mathcal{H}_{\mathrm{B}}$ and $F(A\otimes
B)=F_{\mathrm{B}}(B)\otimes F_{\mathrm{A}}(A)$, where
$F_{\mathrm{A}}(A)=U_{\mathrm{A}}AU_{\mathrm{A}}^{\dagger}$ or
$U_{\mathrm{A}}A^{\mathsf{T}}U_{\mathrm{A}}^{\dagger}$ and
$F_{\mathrm{B}}(B)=V_{\mathrm{B}}BV_{\mathrm{B}}^{\dagger}$ or
$V_{\mathrm{B}}B^{\mathsf{T}}V_{\mathrm{B}}^{\dagger}$.

###### Proof of Theorem 4.

Since $\mathcal{F}(\mathrm{SEP(A;B)})$ contains all local unitary maps,
$\mathrm{SEP(A;B)}$ clearly satisfies $1$-symmetry.

Now, we show that $\mathrm{SEP(A;B)}$ is not $2$-symmetric by giving a
counterexample. Take the following four separable pure states:

$\displaystyle\rho_{1}$
$\displaystyle=\rho_{1}^{\mathrm{A}}\otimes\rho_{1}^{\mathrm{B}}=\begin{bmatrix}1&0\\\
0&0\end{bmatrix}\otimes\begin{bmatrix}1&0\\\ 0&0\end{bmatrix},$ (18)
$\displaystyle\rho_{2}$ $\displaystyle=\frac{1}{2}\begin{bmatrix}1&1\\\
1&1\end{bmatrix}\otimes\frac{1}{2}\begin{bmatrix}1&1\\\ 1&1\end{bmatrix},$
(19) $\displaystyle\sigma_{1}$ $\displaystyle=\begin{bmatrix}1&0\\\
0&0\end{bmatrix}\otimes\begin{bmatrix}1&0\\\ 0&0\end{bmatrix},$ (20)
$\displaystyle\sigma_{2}$ $\displaystyle={}\begin{bmatrix}0&0\\\
0&1\end{bmatrix}\otimes\begin{bmatrix}0&0\\\ 0&1\end{bmatrix}.$ (21)

By direct inspection, we can verify that the two dichotomies
$\\{\rho_{1},\rho_{2}\\}$ and $\\{\sigma_{1},\sigma_{2}\\}$ both satisfy
condition (17) and thus, by Lemma 2, both contain perfectly distinguishable
pure states in $\mathrm{SEP(A;B)}$.

Assume that there is a map $F\in\mathcal{F}(\mathrm{SEP}(A;B))$ where
$\sigma_{1}=F(\rho_{1})$ and $\sigma_{2}=F(\rho_{2})$. From Lemma 3, the
following equality should holds:

$\begin{split}\operatorname{Tr}\\{\sigma_{1}\sigma_{2}\\}&=\operatorname{Tr}\\{F_{\mathrm{A}}(\rho_{1}^{\mathrm{A}})F_{\mathrm{A}}(\rho_{2}^{\mathrm{A}})\otimes
F_{\mathrm{B}}(\rho_{1}^{\mathrm{B}})F_{\mathrm{B}}(\rho_{2}^{\mathrm{B}})\\}\\\
&=\operatorname{Tr}\\{\rho_{1}^{\mathrm{A}}\rho_{2}^{\mathrm{A}}\otimes\rho_{1}^{\mathrm{B}}\rho_{2}^{\mathrm{B}}\\}\\\
&=\operatorname{Tr}\\{\rho_{1}\rho_{2}\\}.\end{split}$ (22)

However, now we have

$\operatorname{Tr}\\{\rho_{1}\rho_{2}\\}=\frac{1}{4},\quad\operatorname{Tr}\\{\sigma_{1}\sigma_{2}\\}=0.$
(23)

This contradicts (22). Thus $\mathrm{SEP}(A;B)$ is $1$-symmetric but not
$2$-symmetric. ∎

##  Appendix C Non-uniqueness of spectral entropy

Theorem 1 says that the entropies corresponding to different PDP
decompositions must all be equal if the second law is valid. However, in
general, there exist theories where the same state can have decompositions
with different entropies. As a consequence of Theorem 1, all such theories are
intrinsically incompatible with the second law.

We consider again the state space SEP. Lemma 2 implies that the following two
separable states are perfectly distinguishable:

$\displaystyle\rho_{1}$ $\displaystyle=\begin{bmatrix}1&0\\\
0&0\end{bmatrix}\otimes\begin{bmatrix}1&0\\\ 0&0\end{bmatrix},$ (24)
$\displaystyle\rho_{2}$ $\displaystyle=\cfrac{1}{2}\begin{bmatrix}1&1\\\
1&1\end{bmatrix}\otimes\cfrac{1}{2}\begin{bmatrix}1&1\\\ 1&1\end{bmatrix}.$
(25)

Besides, the reference [46] gives the following measurement
$\\{e_{1},e_{2}\\}$ that discriminate $\\{\rho_{1},\rho_{2}\\}$ perfectly:

$\displaystyle e_{1}(\rho)$
$\displaystyle=\operatorname{Tr}\left\\{\cfrac{1}{2}\begin{bmatrix}2&0&0&-1\\\
0&0&-1&0\\\ 0&-1&0&0\\\ -1&0&0&2\end{bmatrix}\rho\right\\},$ (26)
$\displaystyle e_{2}(\rho)$
$\displaystyle=\operatorname{Tr}\left\\{\cfrac{1}{2}\begin{bmatrix}0&0&0&1\\\
0&2&1&0\\\ 0&1&2&0\\\ 1&0&0&0\end{bmatrix}\rho\right\\}.$ (27)

The two matrices appearing above have negative eigenvalues: this is because
SEP does not contain any entangled state.

Next, we extend the state space of SEP slightly. Consider the following
density matrices with unit rank:

$\displaystyle\sigma_{1}$
$\displaystyle=\cfrac{1}{6}\begin{bmatrix}3&\sqrt{3}&\sqrt{3}&\sqrt{3}\\\
\sqrt{3}&1&1&1\\\ \sqrt{3}&1&1&1\\\ \sqrt{3}&1&1&1\end{bmatrix},$ (28)
$\displaystyle\sigma_{2}$
$\displaystyle=\cfrac{1}{6}\begin{bmatrix}3&-\sqrt{3}&-\sqrt{3}&-\sqrt{3}\\\
-\sqrt{3}&1&1&1\\\ -\sqrt{3}&1&1&1\\\ -\sqrt{3}&1&1&1\end{bmatrix}.$ (29)

Because $\sigma_{1}$ and $\sigma_{2}$ are not separable,
$\sigma_{1},\sigma_{2}\not\in\mathrm{SEP}(A;B)$. Then consider the following
state space $\overline{\Omega}$:

$\displaystyle\overline{\Omega}:=\mathrm{Hul}\left(\mathrm{SEP}(A;B)\cup\\{\sigma_{1},\sigma_{2}\\}\right),$
(30)

where $\mathrm{Hul}(X)$ denotes the convex hull of $X$. We remark that
$\rho_{i}$ and $\sigma_{j}$ are pure because they are rank 1 matrices.

In the model corresponding to this state space, the set of all measurements is
given as the set of all affine functions $\\{e_{j}\\}_{j\in J}$ satisfying
$e_{j}(\rho)\geq 0$ and $\sum_{j}e_{j}(\rho)=1$, for any
$\rho\in\overline{\Omega}$ and $j\in J$. Because the state space
$\overline{\Omega}$ is smaller than the set of all density matrices, but
larger than SEP(A;B), the set of all measurements is larger than the set of
POVMs and smaller than the set of measurements in SEP. In particular, because
$e_{j}(\sigma_{i})\geq 0$ for all $i,j$, the measurement $\\{e_{1},e_{2}\\}$
appearing in Eqs. (26) and (27) is allowed in the model.  Because the two
states $\sigma_{1},\sigma_{2}$ are orthogonal quantum states, they can be
perfectly distinguished in conventional quantum theory and, therefore, are
perfectly distinguishable also in this extended model.

This implies that the state $\rho:=\cfrac{1}{3}\rho_{1}+\cfrac{2}{3}\rho_{2}$
can be decomposed into PDP states in two different ways, as follows:

$\displaystyle\rho$ $\displaystyle=\frac{1}{3}\rho_{1}+\frac{2}{3}\rho_{2},$
(31)
$\displaystyle=\frac{3+\sqrt{3}}{6}\sigma_{1}+\frac{3-\sqrt{3}}{6}\sigma_{2}\;,$
(32)

which clearly possess two different values for the spectral entropy.

##  Appendix D The proof of Theorem 3

Consider two instruments $s=\\{s_{j}\\}_{j\in J}$ and $t=\\{t_{k}\\}_{k\in
K}$, where $e_{j}(\rho)=(u\circ s_{j})(\rho)$ and $f_{k}(\rho)=(u\circ
t_{k})(\rho)$. Suppose that $t\succ_{\rho}s$ holds. Then we have

$\forall j\in J,\quad s_{j}(\rho)=\sum_{k\in K}p(j|k)t_{k}(\rho)$ (33)

where $\sum_{j\in J}p(j|k)=1$ holds for all $k\in K$.

Firstly, we have

$\begin{split}e_{j}(\rho)&=(u\circ s_{j})(\rho)=\left\\{u\circ\left(\sum_{k\in
K}p(j|k)t_{k}\right)\right\\}(\rho)\\\ &=\sum_{k\in K}p(j|k)(u\circ
t_{k})(\rho)=\sum_{k\in K}p(j|k)f_{k}(\rho).\end{split}$ (34)

The third equality is because of the affinity of $u$. Therefore, we have

$\begin{split}\frac{s_{j}(\rho)}{e_{j}(\rho)}&=\frac{\sum_{k\in
K}p(j|k)t_{k}(\rho)}{\sum_{k\in K}p(j|k)f_{k}(\rho)}\\\ &=\sum_{k\in
K}\frac{p(j|k)f_{k}(\rho)}{\sum_{k\in
K}p(j|k)f_{k}(\rho)}\frac{t_{k}(\rho)}{f_{k}(\rho)}.\end{split}$ (35)

This implies that the state $s_{j}(\rho)/e_{j}(\rho)$ is the convex
combination of states $t_{k}(\rho)/f_{k}(\rho)$ with the probability
$p(j|k)f_{k}(\rho)/\sum_{k\in K}p(j|k)f_{k}(\rho)$.

If we use the result of Theorem 3 and (34), we have

$\begin{split}&\sum_{j\in
J}e_{j}(\rho)H\left(\frac{s_{j}(\rho)}{e_{j}(\rho)}\right)\\\ &\geq\sum_{j\in
J}e_{j}(\rho)\sum_{k\in K}\frac{p(j|k)f_{k}(\rho)}{\sum_{k\in
K}p(j|k)f_{k}(\rho)}H\left(\frac{t_{k}(\rho)}{f_{k}(\rho)}\right)\\\
&=\sum_{j\in J}e_{j}(\rho)\sum_{k\in
K}\frac{p(j|k)f_{k}(\rho)}{e_{j}(\rho)}H\left(\frac{t_{k}(\rho)}{f_{k}(\rho)}\right)\\\
&=\sum_{j\in J}\sum_{k\in
K}p(j|k)f_{k}(\rho)H\left(\frac{t_{k}(\rho)}{f_{k}(\rho)}\right)=\sum_{k\in
K}f_{k}(\rho)H\left(\frac{t_{k}(\rho)}{f_{k}(\rho)}\right).\end{split}$ (36)

Therefore, we obtain

$\begin{split}&H(\rho)-\sum_{j\in
J}e_{j}(\rho)H\left(\frac{s_{j}(\rho)}{e_{j}(\rho)}\right)\\\ &\leq
H(\rho)-\sum_{k\in
K}f_{k}(\rho)H\left(\frac{t_{k}(\rho)}{f_{k}(\rho)}\right),\end{split}$ (37)

which is the desired inequality.

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