Black-Hole Radiation Decoding Is Quantum Cryptography

Abstract

We propose to study equivalence relations between phenomena in high-energy physics and the existence of standard cryptographic primitives, and show the first example where such an equivalence holds. A small number of prior works showed that high-energy phenomena can be explained by cryptographic hardness. Examples include using the existence of one-way functions to explain the hardness of decoding black-hole Hawking radiation (Harlow and Hayden 2013, Aaronson 2016), and using pseudorandom quantum states to explain the hardness of computing AdS/CFT dictionary (Bouland, Fefferman and Vazirani, 2020).

In this work we show, for the former example of black-hole radiation decoding, that it also implies the existence of secure quantum cryptography. In fact, we show an existential equivalence between the hardness of black-hole radiation decoding and a variety of cryptographic primitives, including bit-commitment schemes and oblivious transfer protocols (using quantum communication). This can be viewed (with proper disclaimers, as we discuss) as providing a physical justification for the existence of secure cryptography. We conjecture that such connections may be found in other high-energy physics phenomena.

For the most up-to-date version of this work, please refer to https://arxiv.org/abs/2211.05491.

Z. Brakerski—Supported by the Israel Science Foundation (Grant No. 3426/21), and by the European Union Horizon 2020 Research and Innovation Program via ERC Project REACT (Grant 756482).

A Proof of Estimation Lemma

Lemma A.1

(Lemma 2.5, Restated). Let $n \in {\mathbb N}$ and let $(\varGamma _1, \ldots , \varGamma _n)$ be an arbitrary set of events over some probability space. Define $\varUpsilon _i = \bigwedge _{j=1}^i \varGamma _j$ and let $\varUpsilon _0$ be the universal event. Denote $\epsilon = \Pr [\varUpsilon _n]$. Finally denote $\alpha _i = \Pr [\varGamma _i | \varUpsilon _{i-1}]$. Then for all $\delta \in (0,1)$

$$\begin{aligned} \Pr _{i \in [n]}\left[ \alpha _i > 1- \frac{\ln (1/\epsilon )}{\delta n}\right] \ge 1- \delta . \end{aligned}$$

(16)

Proof

By definition it holds that

$$\begin{aligned} \alpha _i = \frac{\Pr [\varGamma _i \wedge \varUpsilon _{i-1}]}{\Pr [\varUpsilon _{i-1}]} = \frac{\Pr [\varUpsilon _{i}]}{\Pr [\varUpsilon _{i-1}]} . \end{aligned}$$

(17)

Then by a telescopic product it holds that

$$\begin{aligned} \prod _{i=1}^n \alpha _i = \Pr [\varUpsilon _n] = \epsilon , \end{aligned}$$

(18)

or alternatively

$$\begin{aligned} \frac{\ln (1/\epsilon )}{n} = \mathop {{\mathbb E}}_{i \in [n]}\left[ \ln (1/\alpha _i)\right] . \end{aligned}$$

(19)

We can apply Markov’s inequality (since $\ln (1/\alpha _i)$ is positive) to conclude that for all $\delta \in (0,1)$:

$$\begin{aligned} \Pr _i \left[ \ln (1/\alpha _i) \ge \frac{\ln (1/\epsilon )}{\delta n} \right] \le \delta . \end{aligned}$$

(20)

Therefore with probability at least $1-\delta$ over i, we have that $\ln (1/\alpha _i) < \frac{\ln (1/\epsilon )}{\delta n}$, which implies that $1/\alpha _i < e^{\frac{\ln (1/\epsilon )}{\delta n}}$. That is

$$\begin{aligned} \alpha _i > e^{-\frac{\ln (1/\epsilon )}{\delta n}} \ge 1 - \frac{\ln (1/\epsilon )}{\delta n} . \end{aligned}$$

(21)

This concludes the proof of the lemma.