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The Von Neumann entropy of a QM system, as far as I understand it, is a measure of the information a particular observer has about that system. Is this definition of entropy directly related to heat transfer in a sense analogous to the classical viewpoint where $\Delta S\geq\frac{Q}{T}$ ?

Note that I'm not asking about the equivalence of the mathematical form. My question is whether the amount of information one has affects the evolution of the system i.e., would the physics between two truly identical states (assuming that's even possible) be different if the level of knowledge of its starting conditions was different between the two?

In the classical sense, information is an accounting tool and it doesn't affect how the state evolves over time. Also in the classical sense, entropy may be discussed among physicists and amateurs like me as akin to the amount of knowledge one has of the state but, as I just mentioned, physics doesn't care about how much information there is. Information is not a physical quantity, but entropy clearly is.

This Nature News article suggests there isn't an academic consensus answering the question I posed. I want to see what this community has to say.

http://www.nature.com/news/battle-between-quantum-and-thermodynamic-laws-heats-up-1.21720

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    $\begingroup$ Though I'm not an information theorist, I imagine that the Von Neumann entropy actually refers to the maximum amount of information a particular observer can have, not necessarily the amount that the observer actually has. (This is similar to other information-theoretic definitions of entropy, which detail the minimum amount of information required to communicate a given message.) In this way, the actual behavior of the observer becomes irrelevant. $\endgroup$
    – probably_someone
    Commented Nov 7, 2017 at 20:47

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While, it's not your question, it's worth point out. The Von-Neumann Entropy is the definition of thermo dynamic entropy for a canonical ensemble. Thus if you use the canonical-ensemble density-matrix in the Von-Neumann Entropy you will get the thermodynamic Entropy of the system:

$ -tr\left<\rho ln(\rho)\right>=tr\left<e^{-\beta H}/Z(\beta H-lnZ)\right>=S $

My question is whether the amount of information one has affects the evolution of the system i.e., would the physics between two truly identical states (assuming that's even possible) be different if the level of knowledge of its starting conditions was different between the two? My question is whether the amount of information one has affects the evolution of the system i.e., would the physics between two truly identical states (assuming that's even possible) be different if the level of knowledge of its starting conditions was different between the two?

If the initial density matrices are different, then time time evolution will be different. So if two density matrices have different von-neumann entropy, they must be different and have different evolutions.

Ignorance In Measurement

Now what your really interested in, does "knowing something about the system change it's evolution?" To understand this, you have to think about how you'll know something about a system. To lean something about the system you have to measure it by entangling it with another system. This process of entangling is physical and thus has physical consequences.

Suppose the entanglement process entangles your system states $\left|n\right>_s$ with your measurement device states $\left|n\right>_m$. Then the combined state will look something like

$ \sum_n c_n\left|n\right>_s \otimes \left|n\right>_m $

with a pure state density matrix

$ \rho = \sum_{n,n'}c_n c^*_{n'} \left|n\right>_s \otimes \left|n\right>_m \left<n'\right|_s \otimes \left<n'\right|_m $

Making a measurement in your measurement device is a projection onto the state you observe: $P_M = \left|M\right>\left<M\right|$. Suppose $\left|M\right>=\sum_l M_l \left|l\right>_m$ This measurement will collapse your system into a mixed state:

$ \rho \rightarrow \sum_{l,l'}c_{l}M_lc^*_{l'}M^*_{l'}\left|l\right>_s\left<l'\right|_s $

Different measurements will collapse to different mixed states with different entropies, and will evolve differently.

Ignorance In Chaotic Motion

Essentially, by acquiring information about a system, we change it's state and subsequent evolution. This sort of information is kinda different from what we consider in our loss of information in statistical mechanics, we usually attribute this to chaotic evolution and your intuition from classical mechanics applies here. If you start a quantum system in the exact same state, it will evolve to the exact same state. But we can't predict what this state will be, here the entropy is describing how well we can predict the final state. This sort of knowledge doesn't effect the evolution of the system just like in classical mechanics.

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    $\begingroup$ yes, but only when you remain in a canonical-ensemble density-matrix (close to equilibrium). Otherwise you will have trouble defining a temperature. You maybe able to define a local or "reduced-density-matrix" from tracing out "the non-local" degrees of freedom, then this might give you a density matrix of that form to get a local temperature. There maybe other ways of generalizing heat for quantum systems out of equilibrium that make use of the Von Neumann Entropy, but far from equilibrium temperature doesn't make sense at all so It wouldn't be related to SdT $\endgroup$
    – Shane P Kelly
    Commented Nov 8, 2017 at 15:36
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    $\begingroup$ Yes you interpert me correctly. My point about non-equilibrium, is a system isn't considered to be at a true equilibrium unless it has the density matrix of a canonical-ensemble. There are interesting exceptions based on how equilibrium is defined. But it is true that a system in equilibrium with a large enough bath will have a canonical-ensemble density matrix and the change in VN entropy will relate to Q/T. $\endgroup$
    – Shane P Kelly
    Commented Nov 9, 2017 at 0:21
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    $\begingroup$ Thanks a lot. Another question, what is the relationship between the canonical-ensemble density matrix (with a large enough bath) and information? My intuition tells me information is minimized. As an aside, I get confused when physicists use the term "information". It implies QM systems don't evolve unless you know something about them. That can't be right. $\endgroup$
    – SuchDoge
    Commented Nov 9, 2017 at 16:06
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    $\begingroup$ Information measures are usually talking about how many bits you need to represent or describe something. For a probability distribution, it tells you something about how many different outcomes their are. For a density matrix the different outcomes corresponds to different pure states the system may be in. Just like in classical mechanics, you don't maximize your multiplicity( minimize information), you maximize a free energy which maximizes multiplicity while trying to minimize energy at the same time. $\endgroup$
    – Shane P Kelly
    Commented Nov 10, 2017 at 0:20
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    $\begingroup$ Information is kinda a weird idea when applied to physical systems because your not trying to send a signal, your trying to describe a system or distribution. So it's always better to think of VN entropy as describing the multiplicity of possible states(this is the best interpretation both classically and in quantum mechanics). There are two things that are special about quantum mechanics: 1) this uncertainty can be inherent to the system, rather then just a lack of knowledge, 2) measurement can affects this uncertainty. A system will evolve wether you know something about it or not. $\endgroup$
    – Shane P Kelly
    Commented Nov 10, 2017 at 0:31

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