Von Neumann Entropy: Temperature & Info Explained

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Thermodynamic entropy is defined through the Gibbs equilibrium state and is related to temperature via the inverse temperature parameter beta. In contrast, von Neumann entropy applies to all quantum states, represented by a formula that does not inherently connect to temperature outside of thermal equilibrium. The discussion highlights that while thermodynamic entropy correlates with Shannon entropy, von Neumann entropy serves as its quantum counterpart. The relationship between these entropies and information theory is significant, as both entropies measure information encoded in probability distributions. Overall, the connection between von Neumann entropy and physical temperature is limited to thermal equilibrium states.
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What's the difference between thermodynamic entropy and von Neumann entropy? In particular, how is temperature related to the von Neumann entropy?
Also, what has information got to do with these two entropies?
 
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touqra said:
What's the difference between thermodynamic entropy and von Neumann entropy? In particular, how is temperature related to the von Neumann entropy?
Also, what has information got to do with these two entropies?
Hi, thermodynamic entropy is the entropy of the Gibbs equilibrium state
s := exp( - beta H)/trace(exp( - beta H)/) where beta is to be interpreted as the inverse temperature and H as the Hamiltonian. The Von Neumann entropy equals - trace (s log s). It is just that this last formula makes sense for ALL states (ie. semi positive definite matrices with trace equal to unity). Therefore Von Neumann entropy does NOT relate in general to any meaningful notion of temperature.

Cheers,

Careful
 
Thermodynamic entropy is the same as Shannon Entropy (see http://en.wikipedia.org/wiki/Shannon_entropy) , and it is a measure of how much information is encoded in a probability distribution. The relationship between Information Theory and Thermodynamics indicated by the equality of Shannon and Boltzmann entropy were beautifully described in the papers of Jaynes (see http://en.wikipedia.org/wiki/Edwin_Jaynes) .

The von Neumann entropy is the QM analogous and its relation to Quantum Information Theory is explained in a very simple way in a paper by Plenio and Vitelli that you can find in

http://arxiv.org/abs/quant-ph/0103108
 
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Alamino said:
Thermodynamic entropy is the same as Shannon Entropy (see http://en.wikipedia.org/wiki/Shannon_entropy) , and it is a measure of how much information is encoded in a probability distribution. The relationship between Information Theory and Thermodynamics indicated by the equality of Shannon and Boltzmann entropy were beautifully described in the papers of Jaynes (see http://en.wikipedia.org/wiki/Edwin_Jaynes) .

The question touqra posed was how Von Neumann entropy relates to a *physical* temperature. There exists no good physical notion of temperature for a general state which is not a thermal equilibrium (ie. Gibbs) state AFAIK. Therefore, the statement that thermodynamic entropy equals Von Neumann entropy is extremely misleading to say the very least.
 
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I don't remembered saying "that thermodynamic entropy equals Von Neumann entropy"... I said "analogous", what can be esily seen by the similarity of both formulas. And I remembered that he also asked about the relation of both entropies with respect to information, which was what my post was about.

Anyway, the last paper I indicated talks about the erasure of quantum information by thermalization and indicates where temperature enters in this matter (particualrly, look at equation (47)).
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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