<p>The real reason of conservation of information in quantum mechanics is that the von Neumann entropy <span class="math-container">$S(\rho) = -Tr(\rho \log_2 \rho)$</span> of an isolated quantum system with density operator <span class="math-container">$\rho$</span> does not change in time. This happens because <span class="math-container">$S(\rho)$</span> only depends on the spectrum of <span class="math-container">$\rho$</span>, and the unitarity of time evolution preserves the spectrum.</p>

<p>If you start out with a perfectly certain state it will evolve unitarily, and remain perfectly certain - but realistically, there are going to be subspaces you did not measure and their uncertainty are going to "leak out" and get mixed with the state. This is basically <a href="https://en.wikipedia.org/wiki/No-hiding_theorem" rel="nofollow noreferrer">the no hiding theorem</a> which states that when your system interacts with the outside world and decoheres the information about the original state ends up on the outside and cannot be recovered by operating on just the internal state (but it is indeed conserved for the whole system). </p>

<p>Note that none of this depends on uncertainty per se nor the issue of what measurement does, but rather that the deterministic effects of the Schrödinger equation/unitary evolution makes things dependent on degrees of freedom you have little control over.</p>

CC BY-SA 4.0
2019-05-07T16:32:59.397Z
2019-05-07T16:32:59.397Z
Anders Sandberg
<p>I read about Hawking and Leonard Susskind and their debates about whether information is destroyed or not if it falls into a black hole.  It was resolved in favor of information always being conserved, preserved, whatever (i.e. not destroyed).</p>

<p>I see how this could be consistent with determinism.  However, how is this in keeping with quantum physics, when there is so much randomness.  It seems that at any moment, all matter, energy, even every atom in our bodies, is indeterminate and "scrambled" somewhat.  Even if you could pinpoint or describe every atom and its position perfectly in your body at one moment, by the next moment there is uncertainty.  So how is this compatible with conservation of information?</p>

CC BY-SA 4.0
2019-05-06T16:16:07.227Z
S = kLnW. S increases over time. Shannon: https://en.wikipedia.org/wiki/Information_theory Sounds to me like something is screwy here.
Wayfaring Stranger
CC BY-SA 4.0
2019-05-06T19:37:24.700Z
Not sure if OP's definition of information is the same was what is being "conserved": https://en.wikipedia.org/wiki/Black_hole_information_paradox
Morrison Chang
<black-hole><astrophysics>
Law of Conservation of Information?
yoda
# Law of Conservation of Information?

(Asked by: username_0 on 2019-05-06)

I read about Hawking and Leonard Susskind and their debates about whether information is destroyed or not if it falls into a black hole. It was resolved in favor of information always being conserved, preserved, whatever (i.e. not destroyed).

I see how this could be consistent with determinism. However, how is this in keeping with quantum physics, when there is so much randomness. It seems that at any moment, all matter, energy, even every atom in our bodies, is indeterminate and "scrambled" somewhat. Even if you could pinpoint or describe every atom and its position perfectly in your body at one moment, by the next moment there is uncertainty. So how is this compatible with conservation of information?

**Tags:** black-hole, astrophysics

> **Comments:**
> **username_1**: S = kLnW. S increases over time. Shannon: https://en.wikipedia.org/wiki/Information_theory Sounds to me like something is screwy here.  
> **username_2**: Not sure if OP's definition of information is the same was what is being "conserved": https://en.wikipedia.org/wiki/Black_hole_information_paradox  

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## 1 answers

### Answer by username_3 on 2019-05-07. Score: 1

The real reason of conservation of information in quantum mechanics is that the von Neumann entropy $S(\rho) = -Tr(\rho \log\_2 \rho)$ of an isolated quantum system with density operator $\rho$ does not change in time. This happens because $S(\rho)$ only depends on the spectrum of $\rho$, and the unitarity of time evolution preserves the spectrum.

If you start out with a perfectly certain state it will evolve unitarily, and remain perfectly certain - but realistically, there are going to be subspaces you did not measure and their uncertainty are going to "leak out" and get mixed with the state. This is basically [the no hiding theorem](https://en.wikipedia.org/wiki/No-hiding_theorem) which states that when your system interacts with the outside world and decoheres the information about the original state ends up on the outside and cannot be recovered by operating on just the internal state (but it is indeed conserved for the whole system).

Note that none of this depends on uncertainty per se nor the issue of what measurement does, but rather that the deterministic effects of the Schrödinger equation/unitary evolution makes things dependent on degrees of freedom you have little control over.


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