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The amount of information or entropy that can be contained in a region of space is given by the Bekenstein Bound:

$I \leq \frac{2\pi RE} {hc \ln2}$

However, recent publications have shown information encoding in multiple dimensions, up to 10 dimensions on a single photon(Kues et al.)

What is the relationship between photon encoding in multiple dimensions and the Bekenstein Bound? And more broadly, the computational limitations of our universe?

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  • $\begingroup$ I know nothing about Bekenstein bound, but not that the "multiple dimensions" of the experiment you are referring to are not spatial dimensions. Using the same definition of "dimension", a single photon is in general infinite-dimensional, as it it characterised by continuous degrees of freedom such as the spatial profile of its wavefunction. $\endgroup$
    – glS
    Commented Feb 7, 2019 at 10:58
  • $\begingroup$ I suppose I should have clarified, I am aware that photon encoding is obviously not in spatial dimensions(otherwise string theorists would have went crazy), a big part of what I want to know is what effect these results(if any) have on the computation limits of the universe which is usually cited at about $10^{120}$ operations if we consumed all mass-energy with perfect efficiency and proportional storage. $\endgroup$
    – hisairnessag3
    Commented Feb 8, 2019 at 9:34

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I think this link will be useful, particularly when he begins discussing the 5th argument. The answer as to why quantum computing doesn't violate the Bekenstein Bound is that if we have $n$ two dimensional entangled quantum systems, qubits, (the argument works similarly with 10 dimensional systems, except we work with a base of 10 instead of 2) there is a theorem, the Holevo Theorem which gives the amount of reliable classical bits that can be recovered from the system, namely $n$ classical bits. The system requires $2^n$ complex variables to describe, so given an $n$ sufficiently large, we could certainly make it look like we are violating the Bekenstein Bound, e.g. by having a system which requires $2^{10000}$ continuous complex variables to completely specify, but we couldn't "cash the system out". That is we couldn't recover even a minuscule fraction of that information from the system. Thus when we consider the number of classical bits stored in a quantum system, it still only scales linearly with the number of quantum bits, and thus we can't violate the Bekenstein Bound.

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  • $\begingroup$ Hmmm, you mention that classical and quantum information storage will scale linearly(if I am not taking you out of context to harshly). However, Freeman Dyson made an argument, in a discussion with Krauss on "can life exist forever"(goo.gl/Ab5hVg), that the amount of information you can store in the classical phase space is $N \log L$ vs quantum $N$ $\endgroup$
    – hisairnessag3
    Commented Feb 11, 2019 at 17:18
  • $\begingroup$ Can you give a specific time in the video and/or give what $L$ is. $\endgroup$
    – Physical Mathematics
    Commented Feb 11, 2019 at 22:30
  • $\begingroup$ I would watch the whole thing;) but 28:50 abouts $\endgroup$
    – hisairnessag3
    Commented Feb 11, 2019 at 22:36
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    $\begingroup$ What he's talking about in his talk involved a classical system with variable volume. Usually when we consider a classical system with $n$ bits we are considering other things to be constant. Plus the whole $\log L$ thing isn't really pertinent to the Bekenstein bound issue which really only is an issue if you were able to recover all the quantum amplitudes of a quantum system, as the number of such amplitude grows with $2^n$, $n$ being the number of qubits. $\endgroup$
    – Physical Mathematics
    Commented Feb 11, 2019 at 23:20

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