I recently learned about the Entropy of entanglement and now asking myself, what is the difference between the classical entropy and the entropy of entanglement? Is there even any or is entropy really just a measure of how entangled a system is? I thought I understood what entanglement and entropy is and now I'm confused as hell.
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Possible duplicate https://physics.stackexchange.com/questions/227094/what-is-entanglement-entropy-and-all-those-stories-about-counting – Mr Anderson Jul 09 '21 at 12:24
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1You might be interested in my answer to this question: https://physics.stackexchange.com/questions/263197/is-information-entropy-the-same-as-thermodynamic-entropy In particular, I claim (controversially) that you can perfectly well think of classical entropy as a strict subset of entanglement entropy, applied to small systems near thermal equilibrium. – Rococo Jul 09 '21 at 13:43
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@Rococo your claim is indeed controversial: since we use the same formula, the two entropies will be given by the same expressionwhen written for the same system. They however mean different things. In a sense, one can interpret thermal entropy as entanglement entropy to say that a thermal state is not entangled - this is a mathematically rigorous, but physically trivial statement. – Roger V. Jul 09 '21 at 14:07
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what's your understanding of "entropy of entanglement"? Because the way they are usually defined, entropy of entanglement and "classical entropy" are rather different objects. Given a bipartite state, its "entropy of entanglement" is the (von Neumann) entropy of the reduced state. The "classical entropy", or rather, its quantum counterpart (called von Neumann entropy), would be the entropy of the state itself, definable as the entropy of its eigenvalues. – glS Jul 12 '21 at 16:36
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I have added a reference to Jaynes' article, where different "entropies" are discussed - thinking of them as the same thing often leads to misconceptions and incorrect results. – Roger V. Jul 26 '21 at 09:03
2 Answers
If you are a believer of the "church of the larger Hilbert space", whose central belief is that any mixed quantum state can always be seen as part of a larger pure state (whether you believe in this is, to quite some extent, a matter of interpretation rather than hard facts), then: Yes, every mixed state is part of a larger entangled pure state, and then, its entropy is a measure of its entanglement with the other part of that pure state.
Some more details can also be found in this older answer of mine (thanks to @Rococo for digging this out!).

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regarding purification (the process referred to here), see Norbert's nice earlier answer as well:
https://physics.stackexchange.com/questions/424007/is-purification-physicaly-meaningful/424017
– Rococo Jul 10 '21 at 16:09 -
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TL; DR: the basic formula is the same, but the content and the domain of application are totally different. Entropy in statistical physics is not a measure of entanglement.
Let us consider what is proposed in the Wikipedia article linked as von Neuman entanglement entropy:
- we start with a two-component system, described by the density matrix $$\rho_{AB}=|\Psi_{AB}\rangle\langle\Psi_{AB}|$$
- we trace this density matrix over one of the components: $$\rho_A=Tr[\rho_{AB}]$$
- we calculate the entropy using the expression for the Shannon's entropy ($S=-\sum_ip_i\log p_i$) generalized for the use with the density matrix: $$S(\rho_A)=-Tr[\rho_A\log \rho_A]$$
This is a rather special situation, as compared to (equilibrium) statistical mechnaics where the partition function has a well-defined form (e.g., in the canonical ensemble): $$ \rho=e^{-\beta H} $$ Moreover, in most cases we will not even use this expression, but rather define the entropy as a logarithm of the phase volume occupied by the system (assuming that the system has an equal probability of being in each state): $$S=k_B\log\Omega.$$ Finally, in thermodynamics the entropy is introduced via $$ dS = \frac{\delta Q}{T}. $$
To recapitulate: although the basic formula for calculating the entropy is the same (except for thermodynamic approach), the content is different:
- in statistical physics we take the statistical matrix of the system of many degrees of freedom, which has a well-nown form (at least in equilibrium). In particular, this entropy is never zero, because equilibrium statistical ensemble is not in a pure state.
- the entanglement entropy can be calculated for any system with two quantum numbers, even for one particle. No assumptions about the size of the system or the particular form of the density matrix are made. This entropy takes zero value in a pure state, which si why it can be used as a measure of entanglement.
Update:
@Rococo has brought up in the comments this article, where both concepts are used: a complex entangled system is thermalized, and the entanglement entropy becomes thermodynamic entropy (as the entanglement vanishes).
Update 2
Jaynes in his article The minimum entropy production principle points out at the confusions resulting from using word entropy for quantities that actually differ mathematically or have different physical content:
By far the most abused word in science is "entropy". Confusion over the different meanings of this word, already serious 35 years ago, reached disaster proportions with the 1948 advent of Shannon's information theory, which not only appropriated the same word for a new set of meanings; but even worse, proved to be highly relevant to statistical mechanics.
Jaynes himself distinguishes at least 6 different types of entropy.

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5What this misses IMO is that there are, at a minimum, cases in which the entanglement entropy and thermal entropy of a (sub)system can be explicitly shown to coincide: https://arxiv.org/abs/1603.04409 . As such, I don't think it is correct to say that their 'domain[s] of application are totally different.' – Rococo Jul 09 '21 at 13:46
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@Rococo thank you for bringing this to my attention - I included it in the answer. – Roger V. Jul 09 '21 at 13:54
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@Rococo note that they still mean different things: entanglement entropy measures whether the system is entangled, which is never the case of a system in thermal equilibrium. In the experiment that you cited one becomes the other in a course of a non-reversible process. – Roger V. Jul 09 '21 at 14:03
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2Hi Roger- thanks. Without getting into a long discussion, I will just say that I'm not sure why you say that a system in thermal equilibrium is never entangled. In an isolated quantum system, thermalization and entanglement between different subsystems are exactly the same process, resulting in the equivalence of the two entropies as discussed in the paper above. A thermal state in this context is a highly entangled state. – Rococo Jul 09 '21 at 14:25
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See also this question, and in particular the associated Quanta article and its links: https://physics.stackexchange.com/questions/248488/what-are-the-phenomena-responsible-for-irreversible-increase-in-entropy/ – Rococo Jul 09 '21 at 14:25
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@Rococo (grand-)canonical distribution is a mixed state - this is statistical mechanics. One can calculate the entropy and convince oneself that it is not zero. Thermalization is a loss of entanglement - they are not "the same process", in fact - entanglement is not even a process, but a characteristic of a state. – Roger V. Jul 09 '21 at 14:31
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2If you describe quantum mechanics in a pure state + unitary evolution picture, thermalization is the process of interacting and thereby getting entangled with the environment, such that in the end, the reduced state of the system is described by a Gibbs state. – Norbert Schuch Jul 10 '21 at 13:43
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1@NorbertSchuch I see what you mean. But thermalization also implies taking thermodynamic limit. Otherwise, it doesn't make sense to talk about thermodynamic entropy. Btw, could you expand your answer? - I liked the joke, but I would be interested to learn more about the debate that you are referring to. – Roger V. Jul 10 '21 at 14:40
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@RogerVadim I'm not an expert on quantum thermodynamics, but there are ways to define thermal equilibrium for finite (small) systems. In any case, the point remains that if you believe (or take the standpoint) that the world is described by pure states & unitary evolution, then the equilibrium reached by a system + environment will still be a pure state, and thus, the entropy of the system in equilibrium measures its entanglement with the environment. (While "equilbrium" for a finite system in principle is time-average, for a large enough bath, this quickly approaches an ensemble average.) – Norbert Schuch Jul 10 '21 at 16:54
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@RogerVadim Which joke? The "church of the larger Hilbert space"? – Norbert Schuch Jul 10 '21 at 16:55