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The holographic encoding interpretation from the Bekenstein-Hawking entropy is due to it scaling with the area $S_{BH} = A/4$ of the surface, rather than the volume $V$. While I am aware that most quantum systems have volumetric entanglement entropy, I cannot really find any 3 dimensional result for the von Neumann entropy, which one should compare with $S_{BH}$. It does seem like they scale with the "volume" though, for instance the one dimensional XXZ Heisenberg chain and Hubbard model scale with the length of the chain. There seems to be a post A measure of entanglement which distinguishes between classical and quantum correlations which claims that this is true, but gives no reference.

The reason for the perhaps provocative question in the title is that for classical systems, such as the ideal gas one finds an entropy $$S= N \left(\ln V +\frac{3}{2}\ln\left(\frac{m U}{3 \pi^2}\right)+5\frac{1-\ln N}{2}\right),$$ which I cannot see how one would interpret as a volume law. Here $\hbar = 1$. It is extensive in the particle number $N$ however. If we look at the Bekenstein-Hawking formula for a Schwarzschild black hole, then we find an entropy which goes like mass squared. I would call such an entropy super extensive, since doubling the number of particles which would form our black hole, more than doubles the entropy.

All in all, I would just like to hear thoughts on how the area law is supposed to be counter-intuitive or even to correspond to a reduction in complexity based on comparing and interpreting other entropies.

Note: I am not looking for an explanation based on how a quantum system can overshoot this bound, hence the need for quantum gravity, e.g., Lecture 6 Holographic Principle of Hong Liu. I am well aware of such arguments.

I am specifically looking for an understanding from a volume and area law perspective in the von Neumann entropy if one exists.

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If I have a cubic lattice of spin 1/2 particles in a volume $V$ with lattice spacing $a$ then the number of spin states is $2^{V/a^3}$, and the entropy is $V/a^3\ln(2)$. I guess I'm saying the temperature is high or the particles don't have any magnetic moment so they have no desire to align their spins. That's the simplest example I can think of where the entropy is proportional to volume.

Not sure from your answer whether or not you were aware of this, but the exciting thing about black hole entropy isn't just "black holes have this entropy wow cool." It's that black holes have the maximum possible entropy - or rather if a volume of space has an entropy higher than the black hole entropy for a black hole of that size, that volume of space instead turns into a black hole. https://en.wikipedia.org/wiki/Bekenstein_bound

So I think what surprises people about this being the maximum entropy possible in our universe is that people imagine the universe can contain information proportional to its volume. That I could take a gigantic region of space, and put bits into that space... for example bits stored in the form of the spin states of spin 1/2 particles. But the Bekenstein bound tells us that actually, the maximum information possible in a volume of space is proportional to its area. If I tried to construct this giant lattice, depending on the lattice spacing, I would always eventually make a black hole instead of my giant volume-scaling information storage system when it reached some radius.

This gives rise to hints of a "holographic principle" - that maybe all the information of our universe can be stored on the surface of the universe. And that perhaps the 3d nature of our universe is an illusion arising from something that is only happening on the surface of the universe.

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  • $\begingroup$ Thank you for your answer, I like your first example, essentially when the temperature is high, the von Neumann entropy is just proportional to the degrees of freedom, so a nice general statement. A question however is about what happens when temperature is in some intermediate regime, and whether one can actually prove something similar happens. Yes, I am aware of the bound, that's what I was referring to when I said "how a quantum system can overshoot this bound". $\endgroup$ Commented Mar 22, 2023 at 22:23
  • $\begingroup$ Yeah... this is currently a hole in my answer. I didn't realize this at first - but my answer actually relies on being at very high temperature, and that's kind of lame. I may revisit later but for now I'd call this a half answer. $\endgroup$
    – AXensen
    Commented Mar 22, 2023 at 22:27

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