# Does von Neumann Entropy Correspond to Thermodynamic Entropy?

Eugene Chua
University of California San Diego
Department of Philosophy
<EMAIL_ADDRESS>I would like to thank Craig Callender, Eddy Keming Chen, Erik
Curiel, Tim Maudlin, John Norton, Sai Ying Ng, Adrian K. Yee, and participants
at the 2019 summer school The Nature of Entropy I for their feedback,
discussion and comments - they have contributed to this paper in one way or
another. I would also like to thank two anonymous reviewers for pushing me to
improve on various aspects of this paper.

###### Abstract

Conventional wisdom holds that the von Neumann entropy corresponds to
thermodynamic entropy, but Hemmo and Shenker (2006) have recently argued
against this view by attacking von Neumann’s (1955) argument. I argue that
Hemmo and Shenker’s arguments fail due to several misunderstandings: about
statistical-mechanical and thermodynamic domains of applicability, about the
nature of mixed states, and about the role of approximations in physics. As a
result, their arguments fail in all cases: in the single-particle case, the
finite particles case, and the infinite particles case.

##  1 Introduction

According to conventional wisdom in physics, von Neumann entropy corresponds
to phenomenological thermodynamic entropy. The origin of this claim is von
Neumann’s (1955) argument that his proposed entropy corresponds to the
thermodynamic entropy, which appears to be the only explicit argument for the
equivalence of the two entropies. However, Hemmo and Shenker (H$\&$S) (2006) –
and earlier, Shenker (1999) – have argued that this correspondence fails,
contrary to von Neumann. If so, this leaves conventional wisdom without
explicit justification.

Correspondence can be understood, at the very least, as a numerical
consistency check: in this context, this means that the von Neumann entropy
has to be included in calculating thermodynamic entropy to ensure consistent
accounting in contexts where both thermodynamic and von Neumann entropy are
physically relevant. Successful correspondence provides strong evidence of
equivalence. While it does not guarantee equivalence, it seems to be at least
a necessary condition for equivalence. If thermodynamic entropy and von
Neumann entropy correspond, then we have reason to think that von Neumann
entropy is rightfully thermodynamic in nature, since proper accounting of
thermodynamic entropy would demand von Neumann entropy. By contrast, a failure
of correspondence seems to entail that the von Neumann entropy is not
thermodynamical in nature, since it is irrelevant to thermodynamic
calculations in contexts where both entropies are physically significant (e.g.
when a system has both quantum degrees of freedom and is sufficiently large to
warrant thermodynamical considerations).

Although Henderson (2003), in my view, has successfully criticized Shenker’s
earlier argument, little has been done in the philosophical literature to
evaluate H$\&$S’s more recent arguments.111It is only slightly better in the
physics literature: Deville and Deville (2013) appears to be the only paper to
critique H$\&$S. On the philosophical side, one (very recent) exception is
Prunkl (ms), though she restricts discussion to the single-particle case and
appears to conflate information entropy with thermodynamic entropy. See
$\S$4.1/$\S$4.2 for why this is not obviously right. This lacuna is striking
because, as I mentioned, von Neumann’s argument appears to be the only
explicit argument for correspondence for the two entropies.

My goal in this paper is to fill this lacuna by providing a novel set of
criticisms to H$\&$S. Here’s the plan: I introduce key terms ($\S$2) and then
present von Neumann’s thought-experiment which aims to establish the
correspondence between thermodynamic entropy and von Neumann entropy; along
the way, a novel counterpart to the usual argument for correspondence is
discussed ($\S$3). I then present and criticize H$\&$S’s arguments for the
single-particle case in the context of thermodynamics ($\S$4.1) and in the
context of statistical mechanics ($\S$4.2), the N-particles case ($\S$4.3),
and the infinite-particles case ($\S$4.4). I conclude that their argument
fails in all cases – in turn, we have good reasons to reject their claim that
the von Neumann entropy fails to correspond to thermodynamic entropy, and
hence the claim that von Neumann entropy is not thermodynamic in nature.

##  2 Key Terms

Let me first define the notions of thermodynamic entropy and von Neumann
entropy. Following H$\&$S, I define the change in thermodynamic entropy
S${}_{\text{TD}}$ between two thermodynamic states in an isothermal quasi-
static process,222There is no change in temperature in an isothermal quasi-
static process, which is why $T$ is taken to be constant. As a matter of
historical note, von Neumann uses an isothermal set-up in his argument, with a
box containing a quantum ideal gas coupled to a (much larger) heat sink
ensuring constant temperature over time (von Neumann 1955, 361/371). as:

$\Delta S_{TD}=\frac{1}{T}\int P\text{ }dV$ (1)

We will restrict our discussion to ideal gases in equilibrium (i.e. systems
where pressure $P$, volume $V$, and temperature $T$ remain constant).

Next, the von Neumann entropy S${}_{\text{VN}}$, for any pure or mixed quantum
system, is defined as:

$\textit{S}_{\text{VN}}=-kTr(\rho\text{ }log\text{ }\rho)$ (2)

where $k$ is the Boltzmann constant and Tr(.) is the trace function.
Generally, the density matrix $\rho$ is such that:

$\rho=\sum\limits_{n=1}^{i}p_{i}\ket{{\psi}_{i}}\bra{{\psi}_{i}}$ (3)

where ${\psi}_{1},{\psi}_{2},...\text{ }{\psi}_{n}$ correspond to the number
of pure states in a statistical mixture represented by $\rho$, with
$p_{1},p_{2},...\text{ }p_{n}$ being their associated classical probabilities
(which must sum to unity). In the case where there is only one pure state
possible for a system (e.g. when we are absolutely certain about its quantum
state), then $n$ = 1, with probability 1, so the appropriate density matrix is
$\rho=\ket{{\psi}}\bra{{\psi}}$. For such a system in a pure state (i.e.
represented by a single state vector in Hilbert space), $S_{\text{VN}}=0$. For
mixed states (i.e. states which cannot be represented by a single state vector
in Hilbert space, hence mixture of pure states or a mixed state), Tr($\rho$
log $\rho$) $<$ 1 and $S_{\text{VN}}>0$ in general. A mixed state is often
said to represent our ignorance about a system – this will suffice as a first
approximation (more on how to interpret this ignorance in $\S$4.2).

Prima facie, S${}_{\text{VN}}$ and S${}_{\text{TD}}$ appear to share nothing
in common, apart from the word ‘entropy’. However, von Neumann claims that
there are important correlations between the two, which suggests a
correspondence between S${}_{\text{TD}}$ and S${}_{\text{VN}}$.

##  3 Von Neumann’s Thought-Experiment

For the sake of parity, I adopt H$\&$S’s presentation of von Neumann’s
thought-experiment,333It is not clear to me that von Neumann’s original
1932/1955 argument is exactly the same as the argument H$\&$S reproduces.
However, for the sake of argument, I will refer to H$\&$S’s version as von
Neumann’s argument in this paper. which aims to show that changes in
thermodynamic entropy can only be made consistent with the laws of
thermodynamics if we considered the von Neumann entropy as contributing to the
calculation of the thermodynamic entropy. Fig. 1. depicts the stages of the
thought-experiment.

We begin, in stage one, with a box with a partition in the middle. On one side
of the partition there is a gas at volume $V$, constant temperature $T$, and
constant pressure $P$. Each gas particle starts off having the pure state
spin-up along the x-direction $\ket{\psi_{x}^{\uparrow}}$, which is equivalent
to a superposition of spin-up and spin-down pure states along the z-direction,
labelled $\ket{\psi_{z}^{\uparrow}}$ and $\ket{\psi_{z}^{\downarrow}}$
respectively. According to standard quantum mechanics, the state of each
particle is thus
$\frac{1}{\sqrt{2}}(\ket{\psi_{z}^{\uparrow}}+\ket{\psi_{z}^{\downarrow}})$.

In this context, particles with quantum behavior may be taken to be ideal
gases, i.e. sets of particles each of which do not interact with other
particles and take up infinitesimal space. Following von Neumann’s assumptions
(von Neumann 1955, 361),444These assumptions are borrowed from Einstein
(1914). For more, see Peres (Peres 2002, 271). each gas particle is understood
as a quantum particle with a spin degree of freedom contained inside a large
impenetrable box, and each gas particle is put inside an even larger container
isolated from the environment (i.e. the box we began with). This ensures that
each spin degree of freedom is incapable of interacting with other particles.
These boxes’ sizes also ensure that the positions of these boxes (and hence of
the particles) can be approximately classical. Since the container is much
larger than each gas particle, this ensures that the gas particles take up
negligible space relative to the massive container. Accepting these
assumptions, we may then take these quantum particles to behave like an ideal
gas.555I shall follow everyone in this debate in assuming that the above set-
up is physically possible. Following H$\&$S, we further assume that the
position degrees of freedom of the gas particles have no interaction with the
spin degrees of freedom at this point, and “due to the large mass of the
boxes, the position degrees of freedom of the gas may be taken to be classical
and represented by a quantum mechanical mixture”. (Hemmo and Shenker 2006,
155)

Moving on, stage two involves a spin measurement along the z-axis on all the
particles in the container, with a result being an equally weighted
statistical mixture of particles with either $\ket{\psi_{z}^{\uparrow}}$ or
$\ket{\psi_{z}^{\downarrow}}$ states. As a result, the spin state of each
particle is then represented instead by a density matrix $\rho_{spin}$, such
that:

$\rho_{spin}=\frac{1}{2}(\ket{\psi_{z}^{\uparrow}}\bra{\psi_{z}^{\uparrow}}+\ket{\psi_{z}^{\downarrow}}\bra{\psi_{z}^{\downarrow}})$
(4)

More precisely, there should be terms for the measurement device too, when
truly considering the entire system. $\rho_{spin}$ describes only the
subsystem (i.e. the quantum ideal gas) sans measurement device, i.e. a state
with the measurement device traced out – this is in line with von Neumann’s
focus on the entropy changes due to changes in the subsystem (von Neumann
1955, 358–379). I follow Henderson (2003) and H$\&$S in talking about the
system’s state as though I have already traced the measurement device out
whenever measurement is involved.

Stages three and four are where the particles are (reversibly) separated
according to their spin states by a semi-permeable wall into two sides of the
box, each with volume $V$.666This semi-permeable wall can be assumed to be a
black box which reversibly separates particles to different sides based on
their different orthogonal/disjoint states; see (von Neumann 1955, 367–370)
for discussion. I follow everyone in the debate in accepting this assumption.
As a result of this separation, we in effect double the mixture’s volume. The
gas expands to fill up volume $V$ on each side.

Stage five involves an isothermal and quasi-static compression of the mixture
so that we return to a total volume $V$ (effectively halving the volume on
each side of the box), while pressure on both sides becomes equal.
Importantly, due to this compression, S${}_{\text{TD}}$ decreases due to the
decrease in volume.

Stage six brings all the particles into the pure spin state
$\ket{\psi_{x}^{\uparrow}}$ quasi-statically and without work done, while
stage seven removes the semi-permeable wall, such that the system returns to
its original state.

Fig. 1: from top to bottom, stage one to stage seven, as described by H$\&$S.

Now consider how $S_{\text{VN}}$ and $S_{\text{TD}}$ change across the various
stages. Stage seven ends with the body of gas having the same thermodynamic
state (same $V$, same $P$, and constant $T$) as stage one. Furthermore, all
the thermodynamic transformations performed were reversible, and removing the
wall alone does no additional work. Thus, the system at stage one must have
the same thermodynamic entropy as stage seven, i.e. $\Delta S_{\text{TD}}$ =
0, since S${}_{\text{TD}}$ depends only on the initial and final state of the
system. $\Delta S_{\text{VN}}=0$ from stage one to seven too, since the system
is in the same state in both the first and seventh stages.

Since stage six does not involve thermodynamic transformations, there is no
change in S${}_{\text{TD}}$. Likewise, the transformation of $\rho_{spin}$ to
$\ket{\psi_{x}^{\uparrow}}$ here does not change S${}_{\text{VN}}$ as the
transformation can be performed unitarily. This is possible as a result of our
separation of the gases to different sides of the box according to their spin-
eigenstates - given this, we can perform unitary operations on each side of
the box (or perform the more general measurement procedure recommended by (von
Neumann 1955, 365-367)), to transform them into the same state as stage one.
Both unitary transformations and von Neumann’s procedure do not increase
S${}_{\text{VN}}$, and so there is no change in S${}_{\text{VN}}$ at stage six
as a result.

There are no changes in S${}_{\text{TD}}$ or S${}_{\text{VN}}$ in stages three
and four. While there is an increase in the gas’s volume, as noted above, from
$V$ to $2V$, and hence an accompanying increase in S${}_{\text{TD}}$ by
$n.R.log\textit{ }2$,777Here, $n$ refers to the number of moles of gas in the
system, and $R$ is the gas constant. there is also a compensating change in
the thermodynamic entropy of mixing888Henderson (2003) explains the mixing
entropy, describing the mixing of different gases, crisply: “After separation,
each separated gas occupies the original volume $V$ alone. To return to the
mixture, each gas is compressed to a volume $c_{i}V$ (where $c$ is the
concentration of the $i^{th}$ gas). The compression requires work
$W=-n.k.T\sum_{i}\textit{ }c_{i}\textit{ }log\textit{ }c_{i}$ to be invested,
and the entropy of the gas is reduced by $\Delta S=-n.k.\sum_{i}\textit{
}c_{i}\textit{ }log\textit{ }c_{i}$. An increase in entropy of the same amount
must then be associated with the mixing step of removing the partitions. This
is the ‘mixing entropy’.” (Henderson 2003, 292) Separation simply results in a
decrease in entropy of the same amount.,999Tim Maudlin raised the following
objection to the applicability of the entropy of mixing in this context when a
version of this paper was presented at a summer school. Mixing should have a
thermodynamic effect only when differences between the gases are already
assumed to be thermodynamically relevant: for example, mixing differently
colored gases should not have a thermodynamic effect unless the difference in
color is thermodynamically relevant. It is, however, not clear whether the
difference in spin is a thermodynamically relevant one, and might amount to
begging the question. This is a good point, but one that I am setting aside
for now, since everyone in the debate accepts the assumption that separating
the gases here decreases the entropy of mixing. As we shall see later, a more
fundamental issue arises with using the entropy of mixing in the ‘single
particle’ case. by $-n.R.log\textit{ }2$ which exactly compensates this
increase in S${}_{\text{TD}}$.101010see (H$\&$S 2006, 157, fn. 4). Since the
particles are in orthogonal spin states at this stage, there are no quantum
effects (e.g. ‘collapse’ effects) from simply filtering the gases with the
semi-permeable walls, and hence S${}_{\text{VN}}$ does not change
either.111111This is argued for in (von Neumann 1955, 370–376).

However, importantly, there is a decrease in S${}_{\text{TD}}$ in stage five,
of $-n.R.log\textit{ }2$ due to the isothermal compression and decrease in
volume. Yet, nowhere else is there any further change in S${}_{\text{TD}}$. We
have to account for why the overall change in S${}_{\text{TD}}$ from the first
to the seventh stages is 0.

As von Neumann argues, only one possibility remains. While S${}_{\text{TD}}$
remains constant in stage two, notice that there was an increase in
S${}_{\text{VN}}$, of $-N.k.-log\textit{ }2=\frac{N.R}{N_{A}}.log\textit{
}2=n.R.log\textit{ }2$,121212$N$ is the total number of particles: since each
particle is assumed to be non-interacting and independent from others under
the ideal gas assumption, their entropies are additive. $N_{A}$ is Avogadro’s
number. as a result of the spin measurement. This is equivalent to the change
of S${}_{\text{TD}}$ in stage five. The state of each particle changes from a
pure state
$\frac{1}{\sqrt{2}}(\ket{\psi_{z}^{\uparrow}}+\ket{\psi_{z}^{\downarrow}})$ to
a mixed state represented by $\rho_{spin}$, and hence S${}_{\text{VN}}$ for
the gas increases on the whole. In order to ensure that entropic changes are
consistent, von Neumann thinks that we should accept S${}_{\text{VN}}$’s
contribution to S${}_{\text{TD}}$ in this context, where both quantum effects
and thermodynamical considerations are at play. Without accepting
S${}_{\text{VN}}$ in our entropic accounting, we end up with a violation of
thermodynamics since we have a reversible thermodynamic cycle with non-zero
change in S${}_{\text{TD}}$, contra the Second Law. In other words, we should
accept that S${}_{\text{VN}}$ corresponds to S${}_{\text{TD}}$.

Furthermore, the correspondence of S${}_{\text{VN}}$ and S${}_{\text{TD}}$ in
this context can be defended from another perspective, apart from
considerations about consistency from the thermodynamic perspective:
consistent accounting from the perspective of quantum mechanics also demands
correspondence. This is simply a change in perspective with regards to the
thought-experiment, but, to my knowledge, this argument has not been
explicitly made in the literature, thus underselling the case for
correspondence in von Neumann’s thought experiment.

Instead of arguing for correspondence by considering thermodynamic
consistency, i.e. ensuring that $\Delta S_{\text{TD}}$ = 0 throughout the
cycle, we can also consider consistency from the quantum mechanical
perspective. We started and ended with the same spin state, and so it should
be the case that $\Delta S_{\text{VN}}=0$ throughout the cycle. Yet, there is
an inconsistency: if we only consider the increase of S${}_{\text{VN}}$ in
stage two as a result of measurement, we should end in stage seven with an
increase in S${}_{\text{VN}}$, not $\Delta S_{\text{VN}}=0$. As described,
there is nowhere else in the thought-experiment where S${}_{\text{VN}}$
changes. However, there is a decrease in S${}_{\text{TD}}$ in stage five due
to the thermodynamic process of isothermal compression, exactly balancing out
the increase in S${}_{\text{VN}}$. Hence, we can ensure consistency, i.e. that
$\Delta S_{\text{VN}}=0$, only by taking S${}_{\text{VN}}$ to correspond to
S${}_{\text{TD}}$. In other words, just as the thermodynamic accounting of
S${}_{\text{TD}}$ is consistent only if we consider S${}_{\text{VN}}$, the
quantum entropic accounting of S${}_{\text{VN}}$ is also consistent only if we
consider S${}_{\text{TD}}$. Consistency from a quantum mechanical perspective
also demands correspondence between S${}_{\text{VN}}$ and S${}_{\text{TD}}$.

Though the debate has largely focused only on how the thought-experiment
demonstrates one direction of correspondence, of S${}_{\text{VN}}$ to
S${}_{\text{TD}}$ as a result of thermodynamical considerations, the
correspondence demonstrated by this thought-experiment in fact goes both ways.
Of course, since von Neumann was focused on demonstrating the thermodynamic
nature of $S_{\text{VN}}$ (specifically the irreversibility of measurement),
rather than the quantum nature of S${}_{\text{TD}}$, it was natural that he
chose to approach it the way he did.

##  4 Hemmo and Shenker’s Arguments

H$\&$S disagree with von Neumann’s argument, and criticize it by considering
three cases: the single-particle case, the finite but large $N$ particles
case, and the infinite particles case.

###  4.1 Single Particle Case - Thermodynamics

H$\&$S first consider von Neumann’s argument in the single particle case (see
Fig. 2). They claim that the argument does not go through here, since
S${}_{\text{TD}}$ actually remains constant, contrary to our thought-
experiment’s description. In other words, using thermodynamical
considerations, they find that S${}_{\text{VN}}$ should not be included in our
accounting for S${}_{\text{TD}}$.

Here’s their argument. Consider the stages where there are entropic changes.
In stage two when the spin measurement was performed, S${}_{\text{VN}}$
increases as before, since it tracks the change of the particle’s spin state
from pure to mixed.

Contrariwise, S${}_{\text{TD}}$ does not change in stage five (isothermal
quasi-static compression) nor anywhere else (this will be important later).
After stage two, the single particle is in either the
$\ket{\psi_{z}^{\uparrow}}$ state or the $\ket{\psi_{z}^{\downarrow}}$ state.
After stages three and four, with the expansion and separation via semi-
permeable wall, there is a particle only in one side of the box, and not the
other. We make an S${}_{\text{TD}}$-conserving location
measurement131313Prunkl (ms) claims that the location measurement leads to a
violation of the Second Law. If true, this makes H$\&$S’s argument even more
problematic. Here, for the sake of argument, I assume that the location
measurement is unproblematic. to figure out which side of the box is empty and
which side the particle is at, so as to compress the box against the empty
side. The compression is then performed as per before. However, this
compression does not decrease S${}_{\text{TD}}$:141414As an anonymous reviewer
rightfully notes, the location measurement is important for ensuring
$\Delta$S${}_{\text{TD}}=0$ here. Without the location measurement, we might
end up compressing in the wrong direction against the side with the gas,
rather than the empty vacuum - this will have thermodynamic effects since we
are doing work on the gas. However, the H$\&$S set-up emphasizes the location
measurement, and I will play along for the sake of argument. to restore the
volume of the ‘gas’ to $V$ no work needs to be done, since we are compressing
against vacuum. Since there is a change in S${}_{\text{VN}}$ in this cycle,
but no change in S${}_{\text{TD}}$, the apparent answer, in order to do our
entropic accounting, is to ignore, not incorporate, S${}_{\text{VN}}$ into
S${}_{\text{TD}}$. Hence S${}_{\text{VN}}$ does not correspond to
S${}_{\text{TD}}$.

Fig. 2: from top to bottom, stage one to stage seven for the single particle
case as described by H$\&$S.

Their analysis is problematic. Though their ultimate point in this analysis –
that $S_{\text{TD}}$ fails to corresponds to $S_{\text{VN}}$ – still holds, it
does not hold in the way they claim. In fact, the way it fails suggests to us
that we should disregard the single particle case.

For the single particle case, they claim that “… [S${}_{\text{TD}}$] is null
throughout the experiment.” (H$\&$S 2006, 162) This then allows them to claim
that thermodynamic accounting for S${}_{\text{TD}}$ is consistent only if we
did not consider S${}_{\text{VN}}$. This then supports their claim that
S${}_{\text{VN}}$ does not correspond to S${}_{\text{TD}}$ since adding
S${}_{\text{VN}}$ into the thermodynamic accounting actually renders the
otherwise consistent calculations inconsistent.

They are right to say that the stage five compression (after location
measurement) has no thermodynamic effect because we are compressing against
vacuum: no work needs to be done, and so $\Delta$S${}_{\text{TD}}=0$ for stage
five. However, I claim that $\Delta$S${}_{\text{TD}}\neq 0$ for the single
particle case overall, because $\Delta$S${}_{\text{TD}}\neq 0$ in stages three
and four in this context.

As far as I can tell, H$\&$S did not analyze stage three and four, i.e. the
isothermal expansion and separation, in terms of the single particle case at
all. Rather, they seem to have assumed that $\Delta$S${}_{\text{TD}}=0$ in
these stages as with the original case of the macroscopic gas.151515Prunkl
(ms) appears to do the same. However, this assumes that there is both a change
in entropy of $n.R.log\textit{ }2$ due to isothermal expansion and a change in
the entropy of mixing of $-n.R.log\textit{ }2$ due to separation, as they say
so themselves for the original case: “The increase of thermodynamic entropy
due to the volume increase $\Delta$S = $\frac{1}{T}\int P\text{ }dV$ is
exactly compensated by the decrease of thermodynamic mixing entropy $\Delta$S
= $\sum w_{k}\textit{ }ln\textit{ }w_{k}$ (where $w_{k}$ is the relative
frequency of molecules of type k) due to the separation.” (H$\&$S 2006, 157,
fn. 4, emphasis mine)

In the single particle case, it makes sense that isothermal expansion should
still increase S${}_{\text{TD}}$, since the single particle ‘gas’ is expanding
against a piston and doing work. However, it does not make physical sense to
speak of the entropy of mixing here at all, since there is no separation of
gases in the single particle case. The entropy of mixing is explicitly defined
for systems where different gases are separated from/mixed with one another
via semi-permeable walls, but a single particle cannot be separated from/mixed
with itself. The quote above makes this conceptual point explicit: by H$\&$S’s
own lights, the relative frequency of a single particle is simply unity (and
null for particles of other types), so the entropy of mixing is $1\textit{
}ln\textit{ }1=0$. There is no thermodynamic entropy of mixing in the single
particle case.

Discounting the entropy of mixing, however, we find that
$\Delta$S${}_{\text{TD}}=n.R.log\textit{ }2\neq 0$ for stages three and four,
and hence for the entire process, contrary to H$\&$S’s claim. Interestingly,
correspondence does fail to obtain between S${}_{\text{TD}}$ and
S${}_{\text{VN}}$, since $\Delta$S${}_{\text{TD}}+$
$\Delta$S${}_{\text{VN}}=2n.R.log\textit{ }2\neq 0$, despite the process being
reversible ex hypothesi: incorporating S${}_{\text{VN}}$ into thermodynamic
accounting violates the Second Law.

However, on this new analysis, we gain some clarity as to why the single
particle case is problematic. While it is true that incorporating
S${}_{\text{VN}}$ into the thermodynamic accounting violates the Second Law,
S${}_{\text{TD}}$ accounting by itself also violates the Second Law (contrary
to H$\&$S). Even without considering S${}_{\text{VN}}$,
$\Delta$S${}_{\text{TD}}\neq 0$ despite the process being reversible.
Thermodynamic accounting is inconsistent here no matter what we do, which
suggests that the reversible process they described for the single particle
case is thermodynamically unsound: if so, any argument H$\&$S make in this
context may be disregarded.

The upshot: I agree with H$\&$S that correspondence fails for the single
particle case, but not why it fails. It is not because the process they
described is already thermodynamically consistent without taking
S${}_{\text{VN}}$ into account. Rather, it is because the process is already
thermodynamically inconsistent anyway.

In recent work, John Norton argued that thermodynamically reversible processes
for single-particle systems are impossible in principle, which might explain
why the process described by H$\&$S is thermodynamically unsound: it was not
justified to assume the process was reversible for a single particle system.
For Norton, a reversible process is “loosely speaking, one whose driving
forces are so delicately balanced around equilibrium that only a very slight
disturbance to them can lead the process to reverse direction. Because such
processes are arbitrarily close to a perfect balance of driving forces, they
proceed arbitrarily slowly while their states remain arbitrarily close to
equilibrium states.” (Norton 2017, 135) Norton notes that these thermodynamic
equilibrium states are balanced not because there are no fluctuations, but
because these fluctuations are negligible for macroscopic systems. However,
fluctuations relative to single-particle systems are large, and generally
prevent these systems from being in equilibrium states at any point of the
process, rendering reversible processes impossible in the single particle
case. (Norton 2017, 135) If reversible processes are impossible for single
particle systems in general, then it should come as no surprise that the
particular single particle reversible process used by H$\&$S is likewise
thermodynamically unsound, as my analysis above suggests. If so, H$\&$S’s
claim that correspondence fails in this process is simply besides the point,
since this process is not thermodynamic at all.

Since any reversible process cannot be realized for single particle systems in
general, the issue seems not to be with any particular process per se, but
with the single particle case simpliciter. To my knowledge, no one prior to
H$\&$S discussed von Neumann’s experiment in terms of a single particle; von
Neumann (1955), Peres (1990, 2002), Shenker (1999) and Henderson (2003) all
explicitly or implicitly assume a large (or infinite) number of particles.
This is for good reason. As H$\&$S acknowledge, and as we have seen: “The case
of a single particle is known to be problematic as far as arguments in
thermodynamics are concerned”. (H$\&$S 2006, 158) Matter in phenomenological
thermodynamics is assumed to be continuous.161616See Compagner (1989) for a
discussion of the so-called ‘continuum limit’ as a counterpart to the
thermodynamic limit in phenomenological thermodynamics. A ‘gas’ composed of
one particle can be many things, but it is surely not continuous in any
commonly accepted sense. In other words, it is just not clear whether the
domain of thermodynamics should apply to the single-particle case at all.

As Myrvold (2011) notes, Maxwell also made a similar claim with regards to
phenomenological thermodynamics in general; it does not and should not hold in
the single particle case. On his view, the laws of phenomenological
thermodynamics, notably the Second Law, must be continually violated on small
scales:

> If we restrict our attention to any one molecule of the system, we shall
> find its motion changing at every encounter in a most irregular manner.
>
> If we go on to consider a finite number of molecules, even if the system to
> which they belong contains an infinite number, the average properties of
> this group, though subject to smaller variations than those of a single
> molecule, are still every now and then deviating very considerably from the
> theoretical mean of the whole system, because the molecules which form the
> group do not submit their procedure as individuals to the laws which
> prescribe the behaviour of the average or mean molecule.
>
> Hence the second law of thermodynamics is continually being violated, and
> that to a considerable extent, in any sufficiently small group of molecules
> belonging to a real body. As the number of molecules in the group is
> increased, the deviations from the mean of the whole become smaller and less
> frequent […] (Maxwell 1878, 280)

The Second Law, and hence phenomenological thermodynamics, should not be
expected to hold true universally in small scale cases, and especially not in
the single-particle case. Von Neumann and everyone else in the debate should
have recognized this point. Why, then, should it matter that the thought-
experiment succeeds or fails in this case? Phenomenological thermodynamics
does not apply to single-particle cases. There is thus no profit in trying to
establish correspondence between S${}_{\text{VN}}$ and S${}_{\text{TD}}$ in
this case. Indeed, if we took seriously Maxwell’s claim that the Second Law
fails at small scales, a failure of thermodynamic entropic accounting might
even be expected; it does not rule out the possible thermodynamic nature of
S${}_{\text{VN}}$ even though the sum of S${}_{\text{VN}}$ and
S${}_{\text{TD}}$ might be inconsistent with the Second Law. In short, it is
not clear why the single-particle case is relevant to the discussion at hand.

H$\&$S’s reasoning is untenable, because they fail to respect the context of
phenomenological thermodynamics by bringing it into a context where it is not
expected to hold. Instead, it seems more appropriate that the single-particle
case is precisely beyond the purview of classical thermodynamics, requiring an
analogue that only corresponds to classical thermodynamics at the appropriate
scales and limits. We may then take S${}_{\text{VN}}$ to be the analogue of
S${}_{\text{TD}}$ in this case, only approximating S${}_{\text{TD}}$ as the
system in question approaches the context suitable for traditional
thermodynamic analysis. If so, we may see von Neumann as merely demonstrating
that S${}_{\text{VN}}$ corresponds, not at all domains but in the domain where
thermodynamics is taken to hold, to S${}_{\text{TD}}$.

###  4.2 Single Particle Case Redux - Statistical Mechanics and Information

Given the foregoing discussion, H$\&$S might insist that S${}_{\text{VN}}$
fails to correspond to S${}_{\text{TD}}$ even when take into account a more
relevant domain for single particles – statistical mechanics.

After directly arguing that S${}_{\text{VN}}$ does not correspond to
S${}_{\text{TD}}$ (H$\&$S 2006, 162–165), they further argue that
S${}_{\text{VN}}$ does not correspond to information entropy (more on this
below) in the single-particle case. Prima facie, this should seem irrelevant
to von Neumann’s argument, which was to establish the correspondence of the
thermodynamic S${}_{\text{TD}}$ and quantum S${}_{\text{VN}}$: why should
information entropy’s failure to correspond with S${}_{\text{VN}}$ be a worry
at all?

Here’s one plausible worry, on a charitable reading. If information entropy
corresponds to S${}_{\text{TD}}$, and H$\&$S shows that S${}_{\text{VN}}$
fails to correspond to information entropy, then we might conclude,
indirectly, that S${}_{\text{VN}}$ does not correspond to S${}_{\text{TD}}$
after all.171717Caveat: I am not committed to the information entropy’s
relationship to thermodynamics. One may, like Earman and Norton (1998, 1999),
be skeptical that information entropy is related to S${}_{\text{TD}}$ at all,
in which case H$\&$S’s argument here is simply irrelevant. This argument
assumes that information entropy does correspond to S${}_{\text{TD}}$, an
assumption H$\&$S seem to hold as well: this is in line with the so-called
‘subjectivist’ view of statistical mechanics.181818Notably, see Jaynes (1957).
Furthermore, my above argument against the misapplication of phenomenological
thermodynamics does not seem to apply here, since this argument is being made
in the context of statistical mechanics and its particle picture, with no
commitment to phenomenological thermodynamics.

However, H$\&$S do not do much to motivate the linkage between information
entropy and S${}_{\text{TD}}$; indeed, in their words, “a linkage between the
Shannon information and thermodynamic entropy has not been established”
(H$\&$S 2006, 164). Without this link, the failure of correspondence between
the information entropy and S${}_{\text{VN}}$ appears, at best, irrelevant to
the correspondence between S${}_{\text{TD}}$ and S${}_{\text{VN}}$.
Nevertheless, I will take a charitable view here and assume that there is a
correspondence between information entropy and S${}_{\text{TD}}$, for the sake
of assessing their argument. Here’s a plausible (if arguable) sketch: if one
were a subjectivist like Jaynes (1957), one might take the Gibbs entropy in
statistical mechanics to be a special case of the information entropy. After
all, both have the form:

$-\sum_{i}p_{i}\text{ }ln\text{ }p_{i}$ (5)

with i being the number of possible states with associated probabilities of
occurring $p_{i}$, with the Gibbs entropy being multiplied by an additional
Boltzmann’s constant $k$.191919Using the so-called Planck units, where $k=1$,
Gibbs entropy and information entropy are then formally equivalent. We know
that statistical mechanics corresponds to phenomenological thermodynamics at
the thermodynamic limit so we can think of the Gibbs entropy, and hence
information entropy, as corresponding to S${}_{\text{TD}}$. I take this to be
in line with what H$\&$S have in mind: “to the extent that the Shannon
information underwrites the thermodynamic entropy, it does so via statistical
mechanics” (2006, 165). Assuming that the above picture is plausible, a
failure of correspondence between S${}_{\text{VN}}$ and the information
entropy provides evidence against the correspondence between S${}_{\text{VN}}$
and S${}_{\text{TD}}$.

Their argument comes into two parts. Ignoring S${}_{\text{TD}}$ for the time
being (which does not change throughout the cycle for the single-particle case
– see $\S$4.1), they claim that we can consider the stage five location
measurement to be a decrease in information entropy of $ln$ $2$, as a result
of learning information about which one of two parts of the box the particle
is in. On first glance, this seems to resolve the arithmetic inconsistency in
entropic accounting: $ln$ 2 is exactly the increase in S${}_{\text{VN}}$ as a
result of the spin state changing from a pure state
$\ket{\psi_{x}^{\uparrow}}$ to the mixed state $\rho_{spin}$ in stage two. In
other words, for both the information and von Neumann entropy’s accounting to
be correct (i.e. net change of zero across the cycle), we must consider
S${}_{\text{VN}}$ as corresponding to information entropy. Now, since
information entropy also corresponds, ex hypothesi, to S${}_{\text{TD}}$, we
have an indirect argument for the correspondence of S${}_{\text{VN}}$ to
S${}_{\text{TD}}$.

However, H$\&$S claim that this argument fails for collapse interpretations,
i.e. interpretations of quantum mechanics on which a superposed quantum state
ontologically collapses into a pure state upon measurement (either precisely
or approximately).202020On GRW-type approaches, though, collapse occurs with
or without measurement, but measurement increases the likelihood of collapse,
roughly speaking. They allow that, on no-collapse interpretations, e.g.
Bohmian or many-worlds interpretations, the location measurement in stage five
does not decrease S${}_{\text{VN}}$, since the state of the system never
changes in light of measurements, and so the above argument goes through.

Let us see what they could mean by this claim by following the state of the
particle through the cycle. At stage two, everyone agrees that the state of
the system is $\rho_{spin}$ following the z-spin measurement;
S${}_{\text{VN}}$ increases by $ln$ 2\. At this point, the particle’s position
degrees of freedom remain independent from its spin degrees of freedom, as per
our ideal gas assumption, though we might assume the particle starts out on
the left half of the box, with the mixture of position states $\rho_{pos}(L)$
with ‘L’ representing the left side. (Consider Fig. 1 but with only one
particle). Following the semipermeable wall’s filtering at stages three and
four, the location of the particle becomes classically correlated with the
spin. Let’s say that the semipermeable wall sends $\ket{\psi_{z}^{\uparrow}}$
particles to the left, represented by $\rho_{pos}(L)$, and
$\ket{\psi_{z}^{\downarrow}}$ particles to the right, represented by
$\rho_{pos}(R)$. As such, the (mixed) state of the particle is now:

$\rho_{particle}=\frac{1}{2}\bigg{(}\ket{\psi_{z}^{\uparrow}}\bra{\psi_{z}^{\uparrow}}\otimes\rho_{pos}(L)+\ket{\psi_{z}^{\downarrow}}\bra{\psi_{z}^{\downarrow}}\otimes\rho_{pos}(R)\bigg{)}$
(6)

For no-collapse interpretations, H$\&$S agree that the state of the particle
stays the same as above after the location measurement in stage five. We
perform the compression in stage five and remove the partition at the end of
stage six, thereby removing the classical correlations between position and
spin. No further change in either information entropy or S${}_{\text{VN}}$
occurs, and hence the correspondence goes through (H$\&$S 2006, 164) - the
spin state remains mixed until unitarily transformed into a pure state and
completing the cycle.

For collapse interpretations, they claim that the location measurement
decreases S${}_{\text{VN}}$ by $ln$ 2, because, on collapse interpretations,
the state of the particle upon the measurement, depending on which side the
particle is found, becomes:

$\rho_{particle}=\begin{cases}\ket{\psi_{z}^{\uparrow}}\bra{\psi_{z}^{\uparrow}}\otimes\rho_{pos}(L)\\\
\ket{\psi_{z}^{\downarrow}}\bra{\psi_{z}^{\downarrow}}\otimes\rho_{pos}(R)\end{cases}$
(7)

The spin state of the system here effectively goes from being a mixed state to
a pure state as a result of this measurement: S${}_{\text{VN}}$ decreases by
$ln$ 2\. Summing up the entropy changes, there was a decrease of $ln$ 2 in
information entropy, and a net change of zero for S${}_{\text{VN}}$ as a
result of the increase in stage two and the decrease in stage five. Overall,
then, the change is not zero but $-ln$ $2$; our accounting has gone awry, and
there is a failure of correspondence between S${}_{\text{VN}}$ and information
entropy. If this is right, S${}_{\text{VN}}$ does not correspond to
S${}_{\text{TD}}$.

However, I think that H$\&$S are wrong to claim that S${}_{\text{VN}}$
decreases following the location measurement for collapse interpretations. As
Prunkl (Prunkl ms, 11–12) notes, there is an inconsistency here. Everyone,
including H$\&$S, agrees that the spin state of the particle is mixed – not
pure – after stage two’s spin measurement, even on collapse interpretations
(H$\&$S 2006, 160). In that case, why does the particle’s spin become pure
after the location measurement?

I think this results from a confusion over the nature of mixed states. In
particular, they seem to have adopted what Hughes (Hughes 1992, $\S$5.4,
$\S$5.8) call the “ignorance interpretation” of mixed states, confusing what I
call classical and quantum ignorance. They seem to be assuming that mixed
states simply represents classical ignorance, i.e. the lack of knowledge about
a particular system: a system represented by a mixed state really is in a pure
state, but we know not which. This is why the location measurement is supposed
to reveal to us the pure state of this system (by revealing which side it is
on and hence the correlated spin state) and hence ‘wash away’ our classical
ignorance of the real state of the system - post-measurement, we know exactly
which pure state this system is in, unlike pre-measurement; hence
S${}_{\text{VN}}$ decreases.

However, as Hughes (Hughes 1992, 144/150) argues, this interpretation of mixed
states – as representing classical ignorance about which pure state a
particular system is in – cannot be the right interpretation of all mixed
states. To begin, a mixed state can be decomposed in non-unique ways in
general. Here’s a simple example: a mixed state representing a mixture of
$\ket{\psi_{z}^{\uparrow}}$ and $\ket{\psi_{z}^{\downarrow}}$ can also
represent a mixture of $\ket{\psi_{x}^{\uparrow}}$ and
$\ket{\psi_{x}^{\downarrow}}$ and so on. If we insist that a mixed state
represent our classical ignorance about the real state of a particular system,
then we end up having to say that a system’s state is really both either
$\ket{\psi_{z}^{\uparrow}}$ or $\ket{\psi_{z}^{\downarrow}}$, and either
$\ket{\psi_{x}^{\uparrow}}$ or $\ket{\psi_{x}^{\downarrow}}$. Of course, this
is impossible given quantum mechanics. The defender of the classical ignorance
interpretation might insist that we simply pick one pair of possible pure
states but not both at once. In general, however, there’s no way to do that
non-arbitrarily given some density matrix. Furthermore, this problem only
worsens when we consider that there are usually more than just two ways to
decompose a density matrix - a principled choice based on the mixed state
alone is not feasible. The mixed state cannot be a representation of classical
ignorance.

Instead, to paraphrase Hughes (Hughes 1992, 144–145), mixed states should be
(minimally) interpreted as such: if we prepared in the same way an ensemble of
systems, each described with the same mixed state, i.e. a mixture of pure
states with certain weights, then the relative frequency of any given
measurement outcome from the ensemble is exactly what we would get if the
ensemble comprised of various ‘sub-ensembles’ each in one of the pure states
in the mixture, with the relative frequency of each sub-ensemble in the
ensemble given by the corresponding weights.

In other words, the sort of quantum ignorance relevant in the right
interpretation of mixed states is not whether we are ignorant about the real
state of this particular system, but whether we are ignorant about the
measured states of an ensemble of identically prepared systems like this one.
If this is right, quantum ignorance cannot be ‘washed away’ upon measurement
of a single system unlike the sort of ignorance H$\&$S were implicitly
assuming, and it seems like this quantum ignorance is precisely what remains
after the location measurement.

This was roughly Henderson’s (2003) criticism against Shenker (1999), which is
why it is puzzling that H$\&$S (2006) commit the same mistake:

> This preparation produces the pure states [$\ket{\psi_{z}^{\downarrow}}$]
> and [$\ket{\psi_{z}^{\uparrow}}$] with equal probabilities. In a particular
> trial, the observer may take note of the measurement result, and he
> therefore discovers that he has say a [$\ket{\psi_{z}^{\uparrow}}$]. If he
> applies a projective measurement in the
> [$\\{\ket{\psi_{z}^{\uparrow}},\ket{\psi_{z}^{\downarrow}}\\}$] basis, he
> could predict that he will measure [$\ket{\psi_{z}^{\uparrow}}$]. However,
> this does not mean that, if someone handed him another state prepared in the
> same way, he could again predict that the outcome of his measurement would
> be [$\ket{\psi_{z}^{\uparrow}}$]. In this sense the observer does not know
> the state of the system which is being prepared, and it is because of this
> ignorance that the state is mixed. Looking at the measurement result does
> not remove the fact that there is a probability distribution over the
> possible outcomes. (Henderson 2003, 294)

This applies to the location measurement in stage five too: measuring the
location of the particle in this case does not change the state of the
particle from a mixed one to a pure one even on collapse interpretations.
Firstly, it seems quite irrelevant whether we adopt a collapse or no-collapse
interpretation, because the collapse mechanism applies to superposed pure
states, not statistical mixtures. If anything, collapse had already happened
in the stage two measurement procedure, yet everyone including H$\&$S (H$\&$S
2006, 160) accepts that the system is in a mixed state after stage two even
for collapse interpretations. More importantly, there remains a probability
distribution over the states of the particle as a result of stage two’s spin
measurement, even after the location measurement. Given an ensemble of
particles prepared from stages one to five in the same way, we are still not
be able to predict with certainty whether an ensemble of particles would all
be measured on the left or right sides of the box (and hence all spin-up or
spin-down) as a result of the mixed state resulting from stage two, only that
half of the ensembles will be on the left and the other half will be on the
right. Quantum ignorance remains – the system remains in a mixed state even
after the location measurement, as:

$\rho_{particle}=\frac{1}{2}\bigg{(}\ket{\psi_{z}^{\uparrow}}\bra{\psi_{z}^{\uparrow}}\otimes\rho_{pos}(L)+\ket{\psi_{z}^{\downarrow}}\bra{\psi_{z}^{\downarrow}}\otimes\rho_{pos}(R)\bigg{)}$
(8)

This is exactly the state of the system in no-collapse interpretations, i.e.
quantum ignorance does not discern between collapse and no-collapse
interpretations. What has gone away is the classical ignorance that H$\&$S
(mistakenly) assumed was relevant for mixed states, ignorance about this
particular system’s state. By measuring the system’s location, we come to
learn of the correlations between location measurement and the particle’s
spin. This ignorance does not change the mixed state to a pure state: instead,
this loss of classical ignorance – gain in information – is represented as a
decrease in information entropy just as before, and this information is what
we use to perform the compression in stage five.

As a result, there is no additional decrease in S${}_{\text{VN}}$ in stage
five for collapse interpretations; the entropy accounting lines up after all,
as with no-collapse interpretations: the decrease in information entropy does
correspond to the increase in S${}_{\text{VN}}$, and so information entropy
does correspond to S${}_{\text{VN}}$ after all. H$\&$S’s argument does not
establish the failure of correspondence between S${}_{\text{VN}}$ and
S${}_{\text{TD}}$ via the failure of S${}_{\text{VN}}$ and information entropy
to correspond.

To sum up, their arguments in the single-particle case are either ill-
motivated and irrelevant to von Neumann and our discussion of correspondence
when considered in terms of phenomenological thermodynamics, or outright fails
when considered in the more relevant domain of (informational approaches to)
statistical mechanics. Either way, their argument does not support the failure
of correspondence between S${}_{\text{VN}}$ and S${}_{\text{TD}}$.212121Let me
briefly note that their argument in the two particles case fails for similar
reasons. On the one hand, from the perspective of phenomenological
thermodynamics, their argument is irrelevant: following Maxwell and others,
two particles do not a thermodynamic system make. On the other hand, in the
domain of statistical mechanics, the analysis in terms of information entropy
is irrelevant from non-informational views of statistical mechanics. From an
informational perspective, however, their argument rests again on the supposed
difference between collapse and no-collapse interpretations of mixed states.
Since this difference is non-existent, their argument likewise fails apart in
that case.

###  4.3 Finitely Many Particles

H$\&$S’s argument in the case of finitely many particles rests on the
assumption of equidistribution, i.e. that the particles will be equally
distributed across the left and right sides of the box after separation by the
semi-permeable wall.

Assuming equidistribution, the increase in S${}_{\text{VN}}$ given the spin
measurement in stage two is $Nln2$ (H$\&$S 2006, 169). Furthermore, the
decrease in thermodynamic entropy in the fourth stage is $Nln2$ as well. The
entropic accounting therefore seems to work out.

However, H$\&$S press further on the ‘rough’ nature of equidistribution when
$N$ is large but finite: they claim that the change in S${}_{\text{VN}}$ will
only only be $Nln2$ when $N$ is infinite, since equidistribution only truly
holds when $N\rightarrow\infty$. In other cases, S${}_{\text{VN}}$ will
strictly only approximate S${}_{\text{TD}}$, and hence S${}_{\text{VN}}$ and
S${}_{\text{TD}}$ combined will never be exactly zero; hence, “Von Neumann’s
argument goes through as an approximation” (H$\&$S 2006, 169). However, they
claim that this state of affairs suggest, instead, that von Neumann’s argument
strictly fails: “[…] since Von Neumann’s argument is meant to establish a
conceptual identity between the quantum mechanical entropy and thermodynamic
entropy, we think that such an implication is mistaken […] no matter how large
N may be, as long as it is finite, the net change of entropy throughout the
experiment will not be exactly zero.” (H$\&$S 2006, 169)

As I have already discussed in $\S$4.1, it is not clear to me that von
Neumann’s goal really was to establish strict identity (what they call
“conceptual identity”), i.e. correspondence between S${}_{\text{VN}}$ and
S${}_{\text{TD}}$ in all domains. Rather, it seems to be the establishing of
correspondence only in domains where S${}_{\text{TD}}$ is taken to hold. If
so, their argument here simply misses the point.

Furthermore, as is well-known, the particle analogue of thermodynamics,
statistical mechanics, become equivalent to phenomenological thermodynamics
only when $N=\infty$, viz. when $N$ arrives at the thermodynamic limit. As
such, to complain that S${}_{\text{VN}}$ does not match up to
S${}_{\text{TD}}$ outside of this domain is to demand the unreasonable, since
it is not clear that even statistical mechanics, the bona fide particle
analogue of thermodynamics, can satisfy this demand. Since S${}_{\text{VN}}$
approximates S${}_{\text{TD}}$ the same way statistical mechanical entropies
approximate S${}_{\text{TD}}$ (and becomes equivalent at $N=\infty$), and
physicists generally accept that statistical mechanics corresponds to
thermodynamics nevertheless, why should this problem of approximation be
particularly problematic for S${}_{\text{VN}}$? I think H$\&$S take too
seriously the notion of conceptual identity involved in von Neumann’s thought-
experiment to be strict equality, though I suspect a better way to understand
von Neumann’s strategy is to understand S${}_{\text{VN}}$ as an approximation
to S${}_{\text{TD}}$ that is more fundamental than S${}_{\text{TD}}$ in small
$N$ cases, but becomes part of the S${}_{\text{TD}}$ calculus in domains where
S${}_{\text{TD}}$ applies.

To have a case against S${}_{\text{VN}}$ as a quantum analogue of
S${}_{\text{TD}}$ in the case of finitely many particles, H$\&$S must explain
what exactly the problem is with approximations in this case, if it has worked
out so well for the case of statistical mechanics and thermodynamics. If not,
they might just be “taking thermodynamics too seriously’.222222See Callender
(2001).

One might say something stronger: unless they can justify why we cannot use
approximations at all in science, they do not have a case at all. As they note
themselves, S${}_{\text{TD}}$ is itself only on average approximately $-Nln2$
(H$\&$S 2006, 169), only being equal to $-Nln2$ when $N=\infty$. So, in fact,
the approximate quantity of S${}_{\text{VN}}$, $\sim Nln2$, exactly matches
the approximate quantity of S${}_{\text{TD}}$, $\sim-Nln2$, in the case of
finitely many particles. Unless there is something wrong with approximations
in physics in general, this, then, is in fact a case of S${}_{\text{VN}}$
corresponding to S${}_{\text{TD}}$, contrary to their argument.

###  4.4 Infinitely Many Particles

H$\&$S consider von Neumann’s argument in the infinite particles case in two
different ways: one as $N\rightarrow\infty$ and one as $N=\infty$. As they
rightly point out, the two cases are very different for calculations of
physical quantities.

Consider stage two and stage five in this context. H$\&$S emphasize that a
spin measurement is “a physical operation which takes place in time” (H$\&$S
2006, 170), which constrains what is physically possible.

For the case where $N\rightarrow\infty$, stage two is to be understood as a
succession of physical measurements where “we measure individual quantities of
each of the particles separately and only then count the relative frequencies”
(H$\&$S 2006, 170), before coming up with a density matrix describing this
state. In this case, as with the case described in $\S$4.3, S${}_{\text{VN}}$
approaches $Nln2$ as $N\rightarrow\infty$. Their complaint here consist of two
premises: one, that, as with $\S$4.3, S${}_{\text{VN}}$ never reaches $Nln2$
unless $N=\infty$. Two, that since measurements are physical measurements, we
can never perform an infinite series of these measurements, and so we can
never measure infinite particles. A fortiori the measurable S${}_{\text{VN}}$
can never arrive at $Nln2$, and so the entropic accounting is again supposed
to be inconsistent if we consider both S${}_{\text{VN}}$ and
S${}_{\text{TD}}$.

However, it is clear that their argument is moot given a clear understanding
of the sort of thermodynamics we are interested in (see $\S$4.3). While it is
true that S${}_{\text{VN}}$ will never reach $Nln2$, recall that
S${}_{\text{TD}}$ (or, more likely, one of its statistical mechanical
analogues, given the domain of finitely many particles merely approaching
$\infty$ rather than $N=\infty$) will likewise never reach $Nln2$. In other
words, it does not matter that we can never perform an infinite series of
these measurements, and hence never come to know of S${}_{\text{VN}}$ at the
thermodynamic limit, since we can likewise never have a thermodynamic entropy
equivalent to $Nln2$ unless we are at the thermodynamic limit. The two
entropies, then, in fact correspond in this case.

What of the second case? Here, H$\&$S concede that “arithmetically Von
Neumann’s argument goes through at the infinite limit” (H$\&$S 2006, 172),
which makes sense because, as I have insisted so far, von Neumann’s strategy
was never to demonstrate the strict identity of S${}_{\text{VN}}$ and
S${}_{\text{TD}}$, i.e. the correspondence of S${}_{\text{VN}}$ and
S${}_{\text{TD}}$ in all domains. Instead, it was to show that
S${}_{\text{VN}}$ corresponds to S${}_{\text{TD}}$ only in the domain where
phenomenological thermodynamics hold, in all other cases merely approximating
S${}_{\text{TD}}$ in large $N$ cases or replacing it altogether (in e.g.
single-particle cases). I maintain that H$\&$S’s main mistake was to confuse
the domain where phenomenological thermodynamics hold, with domains where they
do not hold.

H$\&$S complain that “[…] real physical systems are finite. This means that
Von Neumann’s argument does not establish a conceptual identity between the
Von Neumann entropy and thermodynamic entropy of physical systems. Identities
of physical properties mean that the two quantities refer to the same
magnitude in the world.” (H$\&$S 2006, 172) In line with what I have said in
$\S$4.1, it seems that there was no physically meaningful theoretical term in
phenomenological thermodynamics that could refer to some quantity in the
single-particle case, which was why von Neumann needed to come up with a new
measure of entropy to begin with. Furthermore, extending a concept to a new
domain does not require strict identity, as we have seen and understood for a
long time in the case of statistical mechanics and phenomenological
thermodynamics.

As Peres (2002) summarizes: “There should be no doubt that von Neumann’s
entropy. . . is equivalent to the entropy of classical thermodynamics. (This
statement must be understood with the same vague meaning as when we say that
the quantum notions of energy, momentum, angular momentum, etc., are
equivalent to the classical notions bearing the same names).” (Peres 2002,
174) ‘Equivalence’ here should not be understood in terms of strict (or
conceptual) identity i.e. correspondence at all domains. Rather, we should
understand equivalence loosely as correspondence in the suitable domains of
application, and successful extension of old concepts in these domains to new
domains. As Peres noted above, ‘equivalence’ should be understood in the
context of discovery, where one is trying to develop new concepts which are
analogous to old ones in different domains. For von Neumann, we have a theory
(phenomenological thermodynamics) that is well-understood, but also another
theory (quantum mechanics) that we want to understand in light of the former
theory. Finding correspondence provides us with ways to extend concepts from
the original theory to the new theory: for example, with S${}_{\text{VN}}$ we
may now define ‘something like’ S${}_{\text{TD}}$ whereas before there was no
way to talk about these cases. The same goes for statistical mechanics: by
finding a correspondence between e.g. temperature to mean kinetic energy in
the thermodynamic limit, we can extend the notion of ‘something like’
temperature beyond its original domain into systems with small numbers of
particles, whereas before there was, again, no way to talk about these cases.

I see nothing wrong in these cases in the context of discovery. We should give
up a strong and untenable notion of conceptual identity in this context. If
so, H$\&$S’s objection loses much bite.

They further claim that “the fact that the behavior of the two quantities
coincides approximately for a very large number of particles is not enough,
because in any ensemble of finite gases there are systems in which the
identity will not be true. This means that in a real experiment the Von
Neumann entropy is not identical with the thermodynamic entropy.” (H$\&$S
2006, 172) This again reveals a confusion between phenomenological and
statistical thermodynamics. If they want to talk about particles at all, it
seems they must adopt some form of statistical mechanical picture with
microscopic variables, given phenomenological thermodynamics’ emphasis on
purely macroscopic variables like volume or temperature. Yet, if so, they must
recognize that thermodynamic entropy S${}_{\text{TD}}$ is in general not
strictly identical to statistical mechanical entropy, e.g. the Gibbs entropy
or information entropy (briefly discussed in $\S$4.2) either. Their complaint
about approximate coincidences not being enough for (the relevant sort of)
equivalence thus weakens significantly, especially since they must assume some
such equivalence (which cannot be strict identity) to even talk about
particles within the context of phenomenological thermodynamics to begin with.
Furthermore, statistical mechanics is evidently empirically successful in
explaining and predicting traditionally thermodynamic phenomena despite this
‘non-equivalence’ – it is not clear why this ‘non-equivalence’ should matter
if, for all practical purposes, statistical mechanics is the conceptual
successor of thermodynamics. Of course, if they could come up with a
principled reason why approximations should not be allowed period, while
accounting for statistical mechanics’ empirical success in accounting for
thermodynamic behavior, then this could change. As of now, I see no such
argument forthcoming.

##  5 Conclusion, and Some Open Questions

Given the above, I hope to have shown that H$\&$S’s argument against the
correspondence of S${}_{\text{VN}}$ and S${}_{\text{TD}}$ – to my knowledge
the only one in the philosophical literature – fails to hold in all three
cases considered ($\S$4.1 – $\S$4.4), as a result of their misunderstanding
about domains where phenomenological thermodynamics should hold and domains
where it should not. This is compounded with misunderstandings about the role
of approximations and the relevant interpretation of density matrices and
ignorance in quantum mechanics. I conclude that their argument fails on the
whole; the correspondence holds for now.

Of course, even if H$\&$S’s claims were debunked, this does not yet amount to
a positive argument for the equivalence between von Neumann entropy and
thermodynamic entropy. Even assuming correspondence, correspondence does not
entail equivalence. However, the former does provide good prima facie reasons
to believe the latter, especially given the novel take on correspondence I
provided in the end of $\S$3: we can accept the correspondence based on
thermodynamic considerations about the Second Law and S${}_{\text{TD}}$
accounting, but also based on quantum mechanical considerations about
S${}_{\text{VN}}$ accounting. The correspondence supports a ‘two-way street’ –
equivalence – between S${}_{\text{TD}}$ and S${}_{\text{VN}}$.

While I hope to have conclusively refuted H$\&$S’s argument, this is but the
beginning of further inquiry into questions arising from this supposed
correspondence. Amidst the tangle of entropies, there remains much more
housekeeping to be done for philosophers of physics.

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