# von Neumann entropy and quantum version of thermodynamic entropy

Smitarani Mishra, Shaon Sahoo

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Smitarani Mishra, Shaon Sahoo. von Neumann entropy and quantum version of thermodynamic entropy. 2024. hal-04850007

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Distributed under a Creative Commons Attribution 4.0 International License
von Neumann entropy and quantum version of thermodynamic entropy

Smitarani Mishra and Shaon Sahoo*

Department of Physics, Indian Institute of Technology, Tirupati

(Dated: December 19, 2024)

The debate whether the von Neumann (VN) entropy is a suitable quantum version of the ther-
modynamic (TH) entropy has a long history. In this regard, we briefly review some of the main
reservations about the VN entropy and explain that the objections about its time-invariance and
subadditivity properties can be either avoided or addressed convincingly. In a broader context, we
analyze here whether and when the VN entropy is the same or equivalent to the TH entropy. For
a thermalized isolated or open system, the VN entropy for an appropriately chosen density oper-
ator is the same as the quantum version of, respectively, the Boltzmann or Gibbs entropy (these
latter entropies are equivalent to the TH entropy for large thermalized systems). Since the quantum
thermalization is essentially defined for a subsystem of a much larger system, it is important to
investigate if the VN entropy of a subsystem is equivalent to its TH entropy. We here show that
the VN entropy of a small subsystem of a large thermalized system is proportional to the quantum
statistical entropy (Boltzmann’s version) of the subsystem. For relevant numerical results, we take
a one-dimensional spin-1/2 chain with next-nearest neighbor interactions.

# I. INTRODUCTION

Since the development of quantum mechanics, there
has been an enormous progress in our understanding
of how the quantum many-body systems behave. This
progress saw some foundational questions being asked
and at-least partially answered. One of such questions is
on whether and how the statistical mechanics works for
the quantum many-body systems[1-3]. This gives rise
to the important field of quantum statistical mechanics
(QSM). The thermodynamic (TH) entropy is a very im-
portant state function which helps us understand how a
system approaches equilibrium besides playing a crucial
role in studying equilibrium properties of a system [4].
The usefulness of the TH entropy naturally calls for its
appropriate quantum counterpart in QSM.

The TH entropy, as defined by Clausius in the con-
text of second law of thermodynamics, is a state function
which takes the maximum value for a system in the equi-
librium under fixed macroscopic conditions [4, 5]. For an
isolated system, this entropy never decreases when the
system undergoes some process. Thus any entropy de-
fined in the similar spirit for a quantum system must be
a state function and should increase (or remain constant)
when the system undergoes a process.

Boltzmann, and later Gibbs, showed us how the TH
entropy can be calculated from the microscopic details
using statistical approaches [4, 6]. For isolated sys-
tem, Boltzmann gave formula (actually Planck first wrote
down the explicit expression for the Boltzmann entropy)
for the TH entropy in terms of the volume of the relevant
phase space (or the number of associated microstates).
For the open system, Gibbs on the other hand, using
his ensemble theory, provided the formula for the TH

entropy in terms for probability distribution function in
phase space. The quantum versions of these statistical
entropies (Boltzmann's and Gibbs') are somewhat intu-
itively clear [5, 7]: for an isolated quantum system, the
entropy is given in terms of the dimension of the relevant
subspace of the system's Hilbert space while for an open
quantum system, the entropy can be calculated from the
partition function (which depends on the eigenvalues of
the appropriate Hamiltonian of the system). Question,
however, remains on the quantum version of the TH en-
tropy for a part (subsystem) of a quantum system. We
will address this issue separately in this work.

From the thermodynamical considerations, von Neu-
mann (VN) gave a formula for entropy of a quantum
ensemble (representing a given mixed state) [8]. This en-
tropy, known as VN entropy, attracted a lot of attention
and criticism in regards to its connection to the TH en-
tropy. We show in this work that, for isolated and open
quantum system or for small part of a large quantum
system, the VN entropy is equivalent to the TH entropy
if the quantum system is large and in thermal equilib-
rium. We largely exclude from our discussions the quan-
tum systems which do not thermalize or systems not in
equilibrium.

It is worth mentioning here the logical steps we fol-
low in this work while establishing the said equivalence
between the entropies and finding the conditions under
which the equivalence works. First, identify under what
conditions the TH entropy is equivalent to the statisti-
cal entropy (either Boltzmann's or Gibbs' version), then
reinterpret the statistical entropy in terms of the quan-
tum variables (in Boltzmann's case: dimension of rel-
evant subspace in place of volume of accessible phase-
space, and in Gibbs' case: energy eigenvalues in place of
continuous classical energy). In the last step of under-
standing the equivalence, one checks if the VN entropy
of an appropriate density matrix is equivalent to the
quantum version of the statistical entropy (i.e. the rein-

<footnote>* Corresponding author.
shaon@iittp.ac.in</footnote>
interpreted statistical entropy in terms of quantum variables). To better understand the equivalence, one should also check if the VN entropy inherits the important properties of the equilibrium TH entropy. Accordingly, whereever appropriate, we briefly deliberate, if the VN entropy is a state function, additive and whether it increases for isolated system due to an irreversible process.

We organize our paper in the following way. In the next section (Sec. II), we review the thermodynamic and statistical entropies, and subsequently, discuss the suitable quantum versions of these statistical entropies. In Sec. III, we discuss von Neumann entropy and some reservations about the entropy while it comes to representing the thermodynamic entropy. Next, in Sec. IV, we check for three different cases (isolated system, open system and subsystem) if the von Neumann entropy is equivalent to the thermodynamic entropy. We conclude our work in Sec. V.

## II. THERMODYNAMIC AND STATISTICAL ENTROPIES

In this section we first briefly review the contexts and definitions of the thermodynamic (TH) entropy and the statistical (ST) entropies (both Boltzmann's and Gibbs' versions). Next we try to understand under what conditions these entropies (TH and ST) are equivalent. At the end this section, we discuss the reinterpretations of the statistical entropies in terms of the quantum variables.

### A. Entropy: from Clausius to Boltzmann and Gibbs

The Clausius theorem or the Clausius inequality states that [4], for any cyclic transformation or thermodynamic process, $\oint \frac{dq}{T} \le 0$, where $dq$ is the heat increment received by the system at temperature $T$. Here the equality sign works only for a reversible cycle and this helps us to define a thermodynamic (TH) state function, called entropy. Streaky speaking, this state function (thermodynamic entropy, $S_{TH}$) is well defined only for a thermodynamic system in equilibrium. If a system in equilibrium at temperature $T$ receives $dq$ amount of heat reversibly and subsequently reaches an equilibrium again, then the difference in entropy between these two equilibrium phases would be $dS_{TH} = \frac{dq}{T}$.

The Boltzmann's version of statistical entropy is defined by the number of accessible microstates $\Omega$ of a system (with a fixed energy). For a gaseous system, this number is proportional to the phase-space volume available to the system in the given macrostate. Here the statistical entropy is defined as $S_{BN} = k_B \ln \Omega$ with $k_B$ being the Boltzmann (BN) constant [4, 7, 9]. By dividing the accessible phase-space as per the values of the well-defined macrovariables, one can in-fact calculate the BN entropy for each such division [5, 7]. This way one can

even make the BN entropy useful in studying a system approaching equilibrium.

The statistical entropy as per Gibbs (GB), for a given macrostate, is defined as $S_{GB} = -k_B \int \rho(x) \ln \rho(x) dx$, where $dx$ is an appropriate (symmetrized) volume element of the phase-space and $\rho(x)$ is the density or probability distribution function describing (the ensemble of) the system [7]. In principle, one can apply this formula to calculate entropy of a system in non-equilibrium (described by a suitable distribution $\rho$). This entropy, interestingly, does not increase (or change) due to the Liouville evolution (the Hamiltonian dynamics). There are many suggestions on how one can resolved this issue and make the Gibbs entropy increase for a system approaching equilibrium [7]. We will again come back to this issue in Sec. III C.

In the following subsection we discuss under what conditions the TH entropy and the statistical entropy (BN or GB entropy) are equivalent.

### B. Conditions of equivalence between thermodynamic and statistical entropy

To understand the conditions under which the equivalence works, we outline here a method to establish such equivalence. A common way of showing the equivalence between the TH and BN entropies of a system with a given energy is to bring the system with energy $E_1$ in contact with another system with energy $E_2$ [6]. Before the contact, the systems were in thermal equilibrium and after the contact they will again come to a common equilibrium via exchange of heat. Although the systems exchange heat, total energy ($E = E_1 + E_2$) of the composite system remains constant; here one ignores the interaction energy between the systems. In the physical setup to establish the equivalence, the main premise is that a thermodynamic system in equilibrium with a given energy can be in any of a certain number of microstates with an equal probability (the principle of equal a priori probabilities). Here the number of accessible microstates is a function of the system's energy. If $\Omega_1(E'_1)$ ($\Omega_2(E'_2)$) is the number of microstates of first (second) system with energy $E'_1$ ($E'_2$), then at this energy distribution, the number of microstates of the composite system would be $\Omega(E'_1) = \Omega_1(E'_1)\Omega_2(E - E'_1)$. Here we assume that the two systems can independently be in any of their accessible microstates (this assumption is related to our assumption that the interaction energy between the two systems can be ignored). The total number of microstates of the composite system with energy $E$ is $M(E) = \sum_{E'_1} \Omega(E'_1)$, where the sum is over all possible $E'_1$ values. As per the premise, the composite system in equilibrium is equally likely to be in any of the $M(E)$ number of microstates. When the systems are thermodynamically large, $M(E)$ will be almost equal to $\Omega(\bar{E}_1)$, i.e., $M(E) \approx \Omega(\bar{E}_1)$, where $\bar{E}_1$ maximizes the function $\Omega(E'_1)$. This means that the composite system, after
settling in a new equilibrium state, will spend most of its
time in the microstates corresponding to the energy dis-
tribution $\overline{E}_1$ and $E - \overline{E}_1$ for the first and second systems,
respectively.

We stress here that, when the system is in non-
equilibrium, it would not be equally probable to be in
different microstates (at a given time, a subset of mi-
crostates may be more probable than the other sub-
sets). Only in equilibrium, as per the premise, the sys-
tem is equally probable to be in any of the accessible mi-
crostates. Now since in equilibrium the composite system
is equally probable to be any of the $M(E)$ microstates,
which is approximated by $\Omega(\overline{E}_1)$, a condition for the
new equilibrium between the systems can be established
by a maximization condition of $\Omega(E'_1)$ (or, in practice,
$\ln\Omega(E'_1)$). After comparing this condition with a stan-
dard thermodynamic relation (namely, $\frac{1}{T} = (\frac{\partial S}{\partial E})_{N,V}$),
one infers that the thermodynamic entropy of the (first)
system is equal to the BN entropy, i.e., $k_B \ln \Omega_1(\overline{E}_1)$.

From the above discussion, we conclude that the TH
entropy and the BN entropy of an isolated system is
equivalent ($S_{TH} \equiv S_{BN}$) if the system is large (so that
the temperature $T$ can be defined and $M(E)$ can be ap-
proximated by $\Omega(\overline{E}_1)$) and the system is in thermal equi-
librium (so that the principle of equal a priori probabil-
ities becomes applicable).

While establishing equivalence between the TH en-
tropy and GB entropy ($S_{TH} \equiv S_{GB}$), one first finds out
the equilibrium probability distribution function $\rho_{eq}(x)$.
For that, one considers a similar setup as discussed above:
a target system and another system (called, reservoir)
which is in this case considered to be much larger com-
pared to the first system [6]. Under the same condi-
tions as mentioned above (i.e., the system is large, in-
teraction energy between the systems is ignorable and
both the systems are in thermal equilibrium), one finds
that, for a microstate denoted by $x$, $\rho_{eq}(x) = \frac{1}{Q}e^{-\beta H_1(x)}$,
where $H_1(x)$ is the energy of the first system correspond-
ing to the microstate $x$ and $Q$ is the partition func-
tion, $Q = \int e^{-\beta H_1(x)} dx$. This distribution helps us to
get the average energy (internal energy, $U$) of the sys-
tem in terms of the partition function $Q$. Comparing
this expression with a standard thermodynamic relation
(namely, $U = \left[\frac{\partial(A/T)}{\partial(1/T)}\right]_{N,V}$), one can establish a connec-
tion between the thermodynamic (Helmholtz) free energy
$A$ and the partition function $Q$. With this connection,
it is not very difficult to show that the TH entropy of
the system is equal to its GB entropy in equilibrium,
i.e., $-k_B \int \rho_{eq}(x) \ln \rho_{eq}(x) dx$ [6]. The conditions under
which this equality works are what we just mentioned
above.

C. Quantum version of statistical entropy

It is not straightforward how one can reinterpret the
BN or GB entropy in terms of quantum variable(s). This

is because, one of the conditions for equivalence, namely
the principle of equal *a priori* probabilities, needs re-
examination. To understand this, we recall that, a clas-
sical system moves from one microstate to another due
to its time evaluation (each point in a system's equal-
energy trajectory is a microstate and they are statis-
tically equally probable). On the other hand, what is
a microstate for a quantum system? It is tempting to
think of the eigenstates of the system's Hamiltonian as
the microstates, as one can then refer to all the eigen-
states withing a narrow energy window to introduce the
principle of equal probabilities for a quantum system.
The main problem with this assumption is that the sys-
tem does not move from one eigenstate to another due to
its time evaluation. In the following we discuss how one
may resolve this problem.

If we consider an isolated large quantum system with
size *N* (the 'size' can be interpreted as the number of
particles or constituents of the system), the average gap
between any two consecutive eigenvalues will be expo-
nentially small with the system size. We note that the
(average) density of states is proportional to the Hilbert
space dimension (*D*) and the average gap (*δ*) is inversely
proportional to the density of states. Since *D* ∝ *e*<sup>*cN*</sup>,
we have *δ* ∝ *e*<sup>-*cN*</sup>, where *c* is a positive constant. Thus,
if *N* is large enough, it will be practically impossible to
prepare the quantum system in a pure state. In other
words, because of our said inability, we only can study a
system which can be in any of the many closely spaced
eigenstates with some probability distribution. Here we
may apply the principle of equal *a priori* probabilities
and assume the probability distribution to be uniform
in the equilibrium (because, we do not have any specific
reason to believe that one state will be preferred over the
others in the same energy range).

There are other ways one can justify this equal proba-
bilities principle for a quantum system. For example, due
to some (weak) interaction in past, currently the isolated
system can be in an entangled state with its environment.
This effectively puts the isolated system in a mixed state,
i.e., the system can be in any of many microstates (eigen-
states) corresponding to a particular energy scale. The
associated probability distribution, as per the principle,
can be taken to be uniform. Yet another way to jus-
tify the principle is the following. Let $H_0$ be the main
Hamiltonian of the noninteracting system and $H'$ rep-
resents the effective (weak) interaction among the par-
ticles (taken to be perturbation to the main Hamilto-
nian). Hence the total Hamiltonian describing the sys-
tem is $H = H_0 + H'$. Now if we take eigenstates of $H_0$ to
be the microstates, the system will make transitions from
one to other microstates during time evaluation due the
effect of the interaction $H'$. If the effective interaction is
taken to be weak, all the eigenstates of $H_0$ from a narrow
energy window can be assumed to be equally probable,
enabling us to use the results of statistical mechanics.

Once we have agreement on quantum version of the
principle of equal *a priori* probabilities, it is not diffi-
cult to come up with quantum versions of the statistical entropies. To get the quantum version of the BN (qBN) entropy, we prepare a system in such a way that its energy lies in the narrow energy shell ($E, E + \Delta E$], where $\Delta E$ is arbitrary but macroscopically small and microscopically large [5]. Let {$E_n$} be the set of eigenvalues of the system Hamiltonian and {|n⟩} be the corresponding set of eigenkets. As per the premise, the system will be in any state $|n\rangle$ with equal probability if $E_n \in (E, E + \Delta E]$. Subsequently, guided by our understanding of the BN entropy, we can write its quantum version ($S_{qBN}$) in the following way:

$$S_{BN} \rightarrow S_{qBN} = k_B \ln D, \quad (1)$$

where $D$ is the number of eigenvalues within the shell. It may be noted here that one can subdivide the energy shell into different macrostates (subspaces), with one of them having the largest number of microstates (the largest dimension) [5, 7]. For a thermodynamically large system, the dimension of this largest subspace, which corresponds to the thermal equilibrium, is essentially equal to the dimension of the shell ($D$).

To get the quantum version of Gibbs (qGB) entropy, in place of continuous energy, we use energy eigenvalues of the system's Hamiltonian to calculate the partition function. Here we assume that, the system's interaction with the reservoir is weak (it had interaction with the reservoir in the past) so that we still can effectively describe the dynamics of the system by its Hamiltonian. As before, let {$E_n$} be the set of eigenvalues of the system Hamiltonian. In the spirit of Gibbs distribution for classical system, we can define the probability to find the system with energy $E_n$ as $p_n = \frac{1}{Q}e^{-\beta E_n}$, where $Q$ is the partition function, $Q = \sum_n e^{-\beta E_n}$. Here $\beta$ denotes the inverse temperature. Subsequently, we can write the quantum version of the GB entropy in the following way:

$$S_{GB} \rightarrow S_{qGB} = -k_B \sum_n p_n \ln p_n. \quad (2)$$

### III. VON NEUMANN ENTROPY

While reformulating some of the major aspects of statistical mechanics so that it can be used in studying quantum systems, von Neumann provided a formula for entropy of a quantum system in a mixed state [8]. If the density matrix $\rho$ represents the mixed state of a system (not to be confused with the probability distribution function denoted by the same symbol), the entropy of the system, according to von Neumann, is

$$S_{VN} = -k_B \mathrm{Tr}(\rho \ln \rho). \quad (3)$$

This formula attracted both the applause and criticism alike [10–14]. In the following two subsections, we first discuss two different perspectives to understand how one can arrive at the formula and what it actually quantifies.

Next we highlight some of the major criticisms on the applicability of the formula and explain our views that many of such criticisms can be avoided or are unfounded.

#### A. As basis-independent representation of quantum statistical entropy

One way to understand the VN entropy is to view it as the basis-independent representation of the quantum statistical entropy of a thermodynamic system in equilibrium. The VN entropy for an appropriate density matrix can give us the qBN or qGB entropy. In fact, while deriving the expression of the thermodynamic (TH) entropy of a subsystem, Landau also came up with the same expression appears in Eq. 3 as the basis-independent representation of the quantum statistical entropy [9]. We note that, if we take $\rho = \frac{1}{d}I$, where $d$ is the dimension of the system's relevant subspace (of the total Hilbert space) and $I$ is the identity matrix, then it is easy to check that $S_{VN} = -k_B \mathrm{Tr}(\rho \ln \rho) = -k_B \ln d = S_{qBN}$. On the other hand, if we take $\rho = \frac{e^{-\beta H}}{\mathrm{Tr}(e^{-\beta H})}$, where $H$ the system Hamiltonian and $\beta$ is the inverse temperature, then it is difficult to verify that $S_{VN} = S_{qGB}$.

#### B. As entropy of thermodynamic ensemble

Sometimes we may not know in which (pure) state our quantum system is in; von Neumann introduced the concept of the density matrix ($\rho$) to describe the physics of the system [8]. The density matrix is a statistical mixture of many pure states; for example,

$$\rho = \sum_i p_i |\psi_i\rangle \langle\psi_i|, \quad (4)$$

where the system is expected to be found in state $|\psi_i\rangle$ with probability $p_i$ (clearly, $\sum_i p_i = 1$). To calculate the entropy of a system in state $\rho$, von Neumann considered an ensemble, similar to the Gibbsian ensemble for a classical system. The ensemble representing $\rho$ in Eq. 4 will have $Np_i$ number of copies in the state $|\psi_i\rangle$, where $N$ is the total number of copies in the ensemble.

While following the Gibbsian way, one immediately faces problems in defining entropy of the system in the state $\rho$. First of all, the decomposition in Eq. 4 is not unique and, then, the pure states {|$\psi_i$⟩} in a decomposition may not be in general distinguishable (as they are not in general orthonormal). Through thermodynamic considerations, von Neumann proposed a formula (as given in Eq. 3) to quantify entropy of a system in state $\rho$. Without going through the details of his argument, we briefly discuss here the two main features of the VN entropy. First we distinguish two different processes: (1) time evaluation of the system, and (2) measurement on the system. If $\rho(0)$ is the density matrix at time $t = 0$ then the density matrix at the $t$ is given
by $\rho(t) = U(t)\rho(0)U^{\dagger}(t)$, where $U(t)$ is the time evaluation operator. It is not difficult to see that the VN entropy is invariant under this time evaluation process, i.e., $S_{VN}(t) = S_{VN}(0)$. von Neumann showed that a measurement, in general, irreversibly changes the density matrix to another density matrix $\rho \rightarrow \rho'$, where

$$ \rho' = \sum_n \langle \phi_n | \rho | \phi_n \rangle P_n. \quad (5) $$

We may here note that the set of states $\{|\phi_n\rangle\}$ are the eigenkets of the observable for which one is performing the measurement. The subtleties of degeneracy can be found in the von Neumann's work [8]. He showed that the entropy of a system, in general, increases (or remains constant but never decreases) due to the measurement process, i.e., $S_{VN}(\rho') \ge S_{VN}(\rho)$. If we identify the first process (time evaluation) and the second process (measurement) with the thermodynamic reversible and irreversible processes, respectively, then these results are consistent with the thermodynamic laws. Before we discuss, in the next subsection, some of the difficulties with the VN entropy formula, we try to intuitively understand the formula below.

As in Sec. II C, if a pure quantum state is taken to be a microstate, then the qBN entropy of a system in a pure quantum state is zero. In this case, as we see from Eq. 3, the VN entropy is also zero. When the system is not in pure state, rather in a statistical mixture of pure states (represented by $\rho$), a formula for entropy must quantify the amount of “mixedness” (or statistical impurity) in the state $\rho$, similar to how the BN entropy quantifies the amount of “uncertainty” of a system to be found in a microstate. If we now see the decomposition in Eq. 4, it would suggest that the corresponding entropy of the system is $S_{SN}(\{p_i\}) = -k_B \sum_i p_i \ln p_i$, which is the Shannon entropy for the probability distribution $\{p_i\}$. Unfortunately, the decomposition of a mixed state is not unique, hence this way of defining entropy is unacceptable. One way to circumvent this problem is to consider all possible decompositions of $\rho$ and take the one for which $S_{SN}(\{p_i\})$ is minimum. The advantage of defining entropy this way is that the entropy is now unique for a given $\rho$. In fact it can be proved that the VN entropy of a given $\rho$ corresponds to the aforementioned minimum [15, 16]:

$$ S_{VN}(\rho) = \inf \left\{ S_{SN}(\{p_i\}) : \rho = \sum_i p_i |\psi_i\rangle \langle\psi_i| \right\}. \quad (6) $$

Another way to understand the VN entropy is as follows. If we take a set of orthonormal states to be a set of distinguishable microstates, we can define an entropy from the probabilities associated with these microstates. If $\{|\phi_n\rangle\}$ is a set of orthonormal states, the entropy one can define is $S_{SN}(\{w_i\}) = -k_B \sum_n w_n \ln w_n$, where $w_n = \langle\phi_n|\rho|\phi_n\rangle$. Unfortunately, again, this entropy is not unique as it depends on the set of orthonormal states (microstates). To understand how to

avoid this problem, we first define a new density matrix $\rho' = \sum_n w_n |\phi_n\rangle\langle\phi_n|$, where $w_n = \langle\phi_n|\rho|\phi_n\rangle$. We may note that $|\phi_n\rangle$ is an eigenket of $\rho'$ with eigenvalue $w_n$. This density matrix ($\rho'$) is in the same form as the one defined in Eq. 5. But for this updated density matrix after measurement ($\rho \rightarrow \rho'$), von Neumann showed that $-\mathrm{Tr}(\rho' \ln \rho') \ge -\mathrm{Tr}(\rho \ln \rho)$ [8]. Thus we see that the VN entropy corresponds to the minimum of $S_{SN}(\{w_i\})$, which we obtain when we choose the microstates as the eigenkets of the given density matrix $\rho$. Accordingly,

$$ S_{VN}(\rho) = \inf \left\{ S_{SN}(\{w_i\}) : w_i = \langle \phi_i | \rho | \phi_i \rangle \right\}, \quad (7) $$

where it is understood that $\langle\phi_i|\phi_j\rangle = \delta_{ij}$.

### C. Criticism against VN entropy: Are we always asking legitimate questions?

Over the century since von Neumann's work, the entropy after his name found a lot of applications. But along the way it attracted a few major criticisms on its applicability and meaning. Here we address two main objections while trying to interpret the VN entropy as the thermodynamic (TH) entropy.

a. *Time invariance of VN entropy:* The first of these objections is that the VN entropy of an isolated quantum system does not change (more specifically, increase) as the system evolves with time. As seen earlier, $S_{VN}(t) = S_{VN}(0)$ as $\rho(0) \rightarrow \rho(t)$ under the time evaluation operator $U(t)$. It is known that, for an isolated system, the TH entropy increases as the system approaches thermal equilibrium. Does not the VN entropy apply to systems in non-equilibrium?

At the outset, we stress that the above issue arises only when the system is not in thermal equilibrium. In the thermal equilibrium, as discussed earlier in Sec. III A, the VN entropy and the TH entropy are equivalent (under certain general conditions). Before we discuss two different approaches to resolve the problem, we first note that the VN entropy is the quantum-analog of the Gibbs (GB) entropy [17]. Hence, some of the major criticisms faced by the GB entropy are also applicable to the VN entropy. For example, the GB entropy of a system is time invariant since the “shape” of the probability distribution function does not change under Hamiltonian evolution of the system [7]. There are two different ways one can tackle this time-invariance problem. First, one considers that the VN entropy formula needs modification. Accordingly, new entropy formula is proposed which increases as the system evolves with time (this entropy increases and finally saturates in equilibrium). The observational entropy is one such popular alternative to the VN entropy [14, 18]. Second, instead of considering that the VN entropy formula is wrong, one may consider that, while the formula is correct, the time evaluation of density matrix $\rho$ is not correctly described by von Neumann-
Liouville equation. We prefer the second option for the
following reasons.

Like the TH entropy, the VN entropy is a state func-
tion - it depends only on ρ and does not depend on any
choice or preference of an observer. The VN entropy for-
mula correctly captures equilibrium thermodynamics of
a system. In fact, starting with the VN entropy formula,
one can derive the expression of the equilibrium Gibbs
distribution using the maximum entropy principle and
some other conditions [19]. All these point out that the
expression of the VN entropy is a suitable one to study
the thermodynamics of a quantum system; main chal-
lenge here is to figure out how one the the density matrix
evolves from a given initial state to the final one (rerep-
senting thermal equilibrium). In the present study, we
leave out any such detail discussion on how the density
matrix evolves.

b. *Subadditivity of VN entropy:* The second major objection against the VN entropy is that it obeys the subadditivity property while the TH entropy is additive (for a large uniform system where the interaction energy between its two parts can be ignored). The subadditivity property is stated as follows: if a system (AB) is in state ρ<sub>AB</sub>, and its parts A and B are respectively in states ρ<sub>A</sub> and ρ<sub>B</sub> (which are reduced density matrices of the state ρ<sub>AB</sub>), then S<sub>VN</sub>(ρ<sub>AB</sub>) ≤ S<sub>VN</sub>(ρ<sub>A</sub>) + S<sub>VN</sub>(ρ<sub>B</sub>) [20]. This implies that, if the system is in a pure state, its VN entropy will be zero but the sum of the VN entropies of its parts is positive (the sum can be zero in a very special case). This contradicts the notion of additivity of the TH entropy. We can circumvent this apparent contradiction in the following way.

As we noted earlier, the VN entropy is the quantum-
analog of the GB entropy. The GB entropy is applicable
for an open system which is in thermal equilibrium with
its bath. For such a thermodynamic system, its TH en-
tropy is equal to the TH entropies of its parts. Now
to verify if the VN entropy has this additivity property,
we first note that, the quantum thermalization is defined
with respect to a smaller subsystem of a large system
[1, 5, 21–23]. Here, a part of the system is thermalized
while the rest of the system works as its bath. There-
fore, instead of studying additivity property of the whole
quantum system, we should study it for a subsystem. It
is know that the VN entropy follows a volume law for
excited states (especially the states from the middle of
its spectrum) [24–27]. In fact, for our system (see Sec.
IV C), we see that the average VN entropy $\bar{S}_{VN}$ (aver-
aged over all the states from a narrow energy window
in the middle of the spectrum) is extensive and grows
linearly with subsystem size, as seen in Fig. 1. We,
therefore, infer here that the VN entropy also obeys the
additivity property if interpreted appropriately.

We finish this subsection by concluding that the major
criticisms against the VN entropy either can be avoided
or can be addressed convincingly if issues are analyzed
appropriately. In this regard, we concur with A. Peres
when he says that there should be no doubt that von Neu-

<figure><img src="image_2.png" /><figcaption>FIG. 1: In the middle of the spectrum, the average VN entropy (S̄<sub>VN</sub>) grows linearly with subsystem size. Here total system size is N = 16.</figcaption></figure>

mann's entropy is equivalent to the entropy of classical
thermodynamics [28].

IV. ENTROPY OF LARGE QUANTUM SYSTEMS

We already saw in Sec. II and Sec. III A that for
a thermalized large isolated or open quantum system,
the VN entropy is equivalent to the TH entropy (or to
an appropriate quantum statistical entropy). It is non-
trivial to check this hypothesis for a small subsystem of a
large quantum system. In this section we check, mainly
through numerical analysis, whether and how the VN en-
tropy is equivalent to the TH entropy for a subsystem.
For completeness and more clarity, we first briefly sum-
marize how the VN entropy is equal (or equivalent) to
the Th entropy for a thermalized isolated and open sys-
tem. We exclude from our discussion the systems which
are not in equilibrium or the systems which do not ther-
malize.

A. Isolated quantum system

As discussed in Sec. II C, an isolated quantum system
will be in a mixed state, for one or multiple reasons.
Using the principle of equal *a priori* probabilities, one can
show that the VN entropy is equal to the qBN entropy
(which in turn is equivalent to the TH entropy).

Assume that we prepare a system in such a way that
its energy lies in the narrow energy shell (E, E + ΔE],
where ΔE is arbitrary but macroscopically small and mi-
croscopically large [5]. Let {|n⟩} be the set of the eigen-
kets of the system Hamiltonian, and E_n be the eigenvalue
corresponding to |n⟩. The system will be in any state |n⟩
with equal probability if E_n ∈ (E, E + ΔE]. Accordingly,
<figure><img src="image_3.png" /><figcaption>FIG. 2: The subsystem VN entropy ($S_{VN}$) for individual eigenkets and $\ln(DOS)$ are plotted against the energy of the full system. Here total system size is 16 and subsystem size is 6.</figcaption></figure>

the thermodynamic entropy ($S_{TH}$), or equivalently, the quantum Boltmann entropy ($S_{qBN}$) is,

$$S_{TH} \equiv S_{qBN} = k_B \ln D, \qquad (8)$$

where $D$ is the number of eigenvalues within the shell. It may be noted here that one can subdivide the energy shell into different macrostates (subspaces), with one of them having the largest number of microstates (the largest dimension) [5, 7]. For a thermodynamically large system, the dimension of this largest subspace, which corresponds to the thermal equilibrium, is essentially equal to the dimension of the shell ($D$).

Now to calculate the corresponding von Neumann entropy ($S_{VN}$) of the isolated quantum system, we note that the system is in a mixed state, represented by the following density matrix,

$$\rho = \sum_{E_n \in (E, E+\Delta E]} p_n |n\rangle \langle n|, \qquad (9)$$

where $p_n$ is the probability to find the system in state $|n\rangle$. Since $\Delta E$ is macroscopically small, all the accessible microstates (here, eigenkets) will essentially have almost the same energy. Application of the principle of equal a priori probabilities suggests that $p_n = \frac{1}{D}$. As $S_{VN} = -k_B \text{Tr}(\rho \ln \rho)$, we need the eigenvalues of $\rho$ to efficiently compute $S_{VN}$. It is clear from Eq. 9 that $|n\rangle$ is an eigenket of $\rho$ with eigenvalue $\frac{1}{D}$, i.e., $\rho|n\rangle = \frac{1}{D}|n\rangle$ for any energy eigenket from the shell. This implies that

$$S_{VN} = -k_B \text{Tr}(\rho \ln \rho) = k_B \ln D. \qquad (10)$$

We see from Eqs. 8 and 10 that, for an isolated quantum system, the VN entropy is the same as the TH entropy

($S_{VN} = S_{qBN} \equiv S_{TH}$), provided that the system is large and in thermal equilibrium. The detail discussion on the conditions under which this equivalence works can be found in Sec. II B.

## B. Weakly interacting open quantum system

Here we consider an open system which is in contact with the thermal bath. Although the system exchanges heat with the bath, we assume that the interaction with the bath is so weak that the energy spectrum of the system is more or less unaffected by the interaction, and the dynamics of the quantum system can still be described (at-least effectively) by its Hamiltonian.

As before, let $\{E_n\}$ be the set of eigenvalues of the system Hamiltonian with $|n\rangle$ being the eigenket corresponding to eigenvalue $E_n$. In the spirit of Gibbs distribution for classical system, we can define the probability to find the system with energy $E_n$ as $p_n = \frac{1}{Q}e^{-\beta E_n}$, where $Q$ is the partition function, $Q = \sum_n e^{-\beta E_n}$. Here $\beta$ denotes the inverse temperature. We can now get the TH entropy, quantified by the quantum Gibbs entropy, as,

$$S_{TH} \equiv S_{qGB} = -k_B \sum_n p_n \ln p_n. \qquad (11)$$

To find the VN entropy of the open system, one takes the following density operator: $\rho = \frac{e^{-\beta H}}{\text{Tr}(e^{-\beta H})}$, where the inverse temperature can be fixed by solving the equation $\text{Tr}(He^{-\beta H}) = U$ with $U$ being the (given) internal or average energy. In fact, for overwhelmingly large number of the pure states (of the system-bath combination), it can
<figure><img src="image_4.png" /><figcaption>FIG. 3: The average VN entropy ($\bar{S}_{VN}$) for subsystem is plotted against $\ln(DOS)$ across the energy spectrum (separately for the left and the right halves of the spectrum). Our system size is 16 and subsystem size is 6.</figcaption></figure>

be quite generally shown that, a weakly interacting open quantum system is very well described by a (reduced) density matrix which is equivalent to the known Gibbsian distribution [29, 30]. As a result, the VN entropy of the open system is,

$$S_{VN} = -k_B \mathrm{Tr}(\rho \ln \rho) = -k_B \sum_n p_n \ln p_n, \quad (12)$$

where the expression of $p_n$ is the same as the one appears in the formula for $S_{qGB}$ in Eq. 11. We now see from Eqs. 11 and 12 that the VN entropy is the same as the TH entropy ($S_{VN} = S_{qBN} \equiv S_{TH}$) for a weakly interacting open quantum system in thermal equilibrium. The detail discussion on the conditions under which this equivalence works can be found in Sec. II B.

### C. Subsystem of a quantum system

As mentioned in the beginning of this section (Sec. IV), it is non-trivial to verify if the VN entropy is equivalent to the TH entropy for a subsystem (of a much larger quantum system in thermal equilibrium). We analyze this possibility here.

We consider a large composite quantum system AB in a state described by the density matrix $\rho$. We also assume that the subsystem A is macroscopically large but much smaller compare to its counterpart B. We here show, mainly by the numerical analysis, that the VN entropy of the subsystem A is proportional to its TH entropy.

To first calculate the TH entropy of a subsystem, we again assume that the total system is prepared with an energy lying in $(E, E + \Delta E]$. As discussed earlier, the

system will be in a mixed state $\rho$, as appear in Eq. 9. The TH entropy (qBN entropy) of the full system is $S_{TH} \equiv S_{qBN} = k_B \ln D$, as in Eq. 8, where the D is the number of eigenvalues of the system lying in the aforementioned energy shell. Now since the TH entropy is additive, we can assume that the TH entropy of the subsystem (denoted as $S_{TH}^{sb}$) is proportion to $k_B \ln D$ or to the $k_B \ln(DOS)$, where DOS is the density-of-states of the full system at the energy scale E. Accordingly,

$$S_{TH}^{sb} \propto k_B \ln(DOS), \quad (13)$$

where the proportionality constant depends on the size of the subsystem in comparison to the size of the full system.

Now to calculate the VN entropy of the subsystem, we first need to calculate the reduced density matrix of the subsystem $\rho_A = \mathrm{Tr}_B(\rho)$, where $\rho$ is the density matrix of the full system, as given in Eq. 9 (with $p_n = \frac{1}{D}$). Denoting the reduced density matrix corresponding to the eigenket $|n\rangle$ by $\rho_A^{(n)}$, where $\rho_A^{(n)} = \mathrm{Tr}_B(|n\rangle\langle n|)$, we have

$$\rho_A = \frac{1}{D} \sum_n \rho_A^{(n)}. \quad (14)$$

Since the subsystem is in the mixed state $\rho_A$, we now can calculate the VN entropy of the subsystem directly from $\rho_A$, i.e,

$$S_{VN}^{sb} = -k_B \mathrm{Tr}(\rho_A \ln \rho_A), \quad (15)$$

where $\rho_A$ is given by Eq. 14. For a system which thermalizes (or obeys the eigenstate thermalization hypothesis)
esis [21, 22]), we can also have

$$S_{VN}^{sb} = -k_B \mathrm{Tr}(\rho_A \ln \rho_A) \approx -\frac{k_B}{D} \sum_n \mathrm{Tr}(\rho_A^{(n)} \ln \rho_A^{(n)}). \quad (16)$$

This approximate equality works since, as per the hypothesis, the expectation value of a local operator (say, O) is almost the same for different eigenkets from a narrow energy window, i.e., Tr(ρ<sub>A</sub><sup>(n)</sup> O) ≈ Tr(ρ<sub>A</sub><sup>(m)</sup> O). This is possible if the set of eigenvalues of the reduced density matrix ρ<sub>A</sub><sup>(n)</sup> is almost the same for different n. This implies that Tr(ρ<sub>A</sub><sup>(n)</sup> ln ρ<sub>A</sub><sup>(n)</sup>) ≈ Tr(ρ<sub>A</sub><sup>(m)</sup> ln ρ<sub>A</sub><sup>(m)</sup>), which guarantees the approximate equality in Eq. 16. If we write $\bar{S}_{VN} = -\frac{k_B}{D} \sum_n \mathrm{Tr}(\rho_A^{(n)} \ln \rho_A^{(n)})$, then $S_{VN}^{sb} \approx \bar{S}_{VN}$ for a thermalized system.

We note that, if the system does not thermalize (or is integrable), then the VN entropy for the individual eigenkets (−k<sub>B</sub>Tr(ρ<sub>A</sub><sup>(n)</sup> ln ρ<sub>A</sub><sup>(n)</sup>)) can take widely different values. This in fact can be seen in Fig. 2, where the subsystem VN entropy for different eigenkets are plotted against the energy across the spectrum. We here mention that, for all our numerical results, we take the following one-dimensional (with open boundary condition) Heisenberg spin-1/2 chain with the next-nearest neighbor interactions:

$$H = J \left( \sum_{i=1}^{N-1} \vec{S}_i \cdot \vec{S}_{i+1} + \Delta_2 \sum_{i=1}^{N-2} S_i^z S_{i+2}^z \right). \quad (17)$$

In our calculations, we take J = 1 and the size of the system N = 16. This model system is integrable when Δ<sub>2</sub> = 0 and is non-integrable (hence thermalizes) when Δ<sub>2</sub> > 0 [31].

To make more sense of the results in Fig. 2, we plot the subsystem’s average VN entropy (S̄<sub>VN</sub>) against ln(DOS) (separately for the left and right halves of the spectrum). The results are shown in Fig. 3. We find that, for a thermalized system (Δ<sub>2</sub> > 0), its subsystem’s average

VN entropy (S̄<sub>VN</sub>) is proportional to ln(DOS) with a proportionality constant approximately equal to N<sub>sb</sub>/N (in our calculations, the subsystem size N<sub>sb</sub> = 6 and the full system size N = 16). This proportionality constant takes a different value (a smaller one) if the system is not thermalized. We take k<sub>B</sub> = 1 for all our numerical calculations of entropy. Thus our numerical results suggest that,

$$S_{VN}^{sb} \approx \bar{S}_{VN} \propto \ln(DOS). \quad (18)$$

From Eqs. 13 and 18, we conclude that, like for isolated and open systems, for a subsystem also the VN entropy is equivalent to the TH entropy (qBN entropy), i.e., S<sub>VN</sub><sup>sb</sup> ≡ S<sub>TH</sub><sup>sb</sup>.

It may be worth mentioning here that, the result that the subsystem’s average VN entropy is proportional to system’s ln(DOS), is earlier reported in [32] in a different context. We also note that, elsewhere [33], the thermodynamic entropy of a subsystem is defined as what we call here S<sub>VN</sub><sup>sb</sup>. As discussed in this work, thermodynamic (TH) entropy is appropriately calculated as the logarithm of DOS (since, S<sub>TH</sub> ≡ S<sub>qBN</sub> under some general conditions).

V. CONCLUSION

In this work we analyze whether and when the von Neumann (VN) entropy is equal or equivalent to the thermodynamic (TH) entropy. In reviewing the major objections about the VN entropy, we explain that the concerns about the time-invariance and subadditivity properties can be addressed convincingly. We show in this study that, for large thermalized isolated or open system, the VN entropy is equivalent to the TH entropy. Next, through numerical analysis, we also show that the VN entropy of a subsystem of a large thermalized system is equivalent to its TH entropy (qBN entropy).

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