技术控

    今日:41| 主题:49409
收藏本版 (1)
最新软件应用技术尽在掌握

[其他] The Quantum Version of the One-Time Pad is Teleportation

[复制链接]
蛋撻,大叔 发表于 2016-10-3 18:55:14
83 2

立即注册CoLaBug.com会员,免费获得投稿人的专业资料,享用更多功能,玩转个人品牌!

您需要 登录 才可以下载或查看,没有帐号?立即注册

x
For whatever reason, one of the topics I keep coming back to on this blog is "ways to think about quantum teleportation". I've talked about how teleportation is    like storing bandwidth, how if it was any more efficient you could    send unbounded amounts of data without sending a single bit, and how you can    split it into copy-sending and dis-entangling.  
  Well, no reason to stop now!
  In this post I'm going to describe Alice and Bob performing a one-time pad cipher, then I'm going to describe Alice and Bob performing a quantum teleportation. See if you notice any similarities.
  One-Time Pad

  Alice wants to send a bit $M$ to Bob. This is a bit tricky, since Eve is listening in on the line and Alice doesn't want Eve to learn $M$.
  Fortunately, Alice and Bob have a shared random bit S that Eve doesn't know. Instead of talking directly about $M$, Alice and Bob can talk about how $M$ relates to $S$.
  So Alice compares $M$ to $S$. If they're different, she yells out "Hey Bob! $M$ is the opposite of $S$! Flip $S$ to get $M$!". If they're the same, she instead yells out "Hey Bob! $M$ is the same as $S$! Just use $S$ as your copy of $M$!".
  Bob hears Alice yelling, does or doesn't flip $S$ based on what Alice said, and now he has $M$. With Eve none the wiser.
  Quantum Teleportation

  Alice wants to send a qubit $M$ to Bob. This is a bit tricky, since their communication channel is classical and would decohere $M$ if used.
  Fortunately, Alice and Bob have a shared EPR pair $S$. Instead of talking directly about $M$, Alice and Bob can talk about how $M$ relates to $S$.
  (The EPR pair $S$ has even X-parity and even Z-parity. If Alice and Bob were to both measure the X axis of their respective part of the pair, they'd get the same answer. Ditto for the Z axis.)
  Alice compares $M$ to $S_A$ along the Z axis (she measures the Z-parity of $M$ vs $S_A$). If they're different, she yells out "Hey Bob! Along the Z axis, $M$ is the opposite of $S$! Flip $S_B$ around the X axis to fix that!". If they're the same, she instead yells out "Hey Bob! Along the Z axis, $M$ is the same as $S$! No need to do a flip yet!".
  Alice also compares $M$ to $S$ along the X axis. If they're different, she yells out "Hey Bob! Along the $X$ axis, $M$ is opposite to $S$! Fix $S_B$ with a half turn around the Z axis!". If they're the same, she instead yells out "Hey Bob! Along the $X$ axis, $M$ is the same as $S$! You're good to go!".
  Bob hears Alice yelling, does or doesn't flip $S$ around the X and Z axes based on what Alice said, and now he has $M$.
  Similarities

  As I'm sure you noticed from the writing getting a bit repetitive, the one-time pad scenario sounded almost identical to the quantum teleportation scenario. The only real difference is that teleportation had to do two compare-and-flips instead of one.
  I hope that makes it clear why I say quantum teleportation is basically just the quantum version of a one-time pad cipher. (    Side note: and it's all thanks to      X-parity and Z-parity being compatible, unlike X-value and Z-value. Y-parity is also compatible with the other two, but there's no need to involve it in the protocol because the third axis' parity is implied by the other two.    )  
  One use for this strong analogy between quantum teleportation and one-time pads is seeing through some of the silly things people say about quantum teleportation. For example, sometimes people say that, at least when the quantum teleportation doesn't require Bob to flip his qubit at all, teleportation moves $M$ from Alice to Bob    instantaneously(or even retrocausally).  
  Of course what they mean is that, when you focus on just the 25% of runs where no corrections were needed, Bob's analysis of his qubit will match up with the qubit Alice wanted to send. Even if Bob does the analysis before Alice performs the parity measurements necessary to actually do the teleportation.
  That might sound a bit mind-bending... but really it's just an effect of the post-selection we did by focusing on a special 25% of the runs. The exact same thing happens with the one-time pad. If you focus on the 50% of runs where no corrections were needed, i.e. where $S=M$, Bob's analysis of his copy of $S$ will match up with $M$ even if he does it before Alice compared $M$ to $S$. For reasons that I hope are    incredibly obvious.  
  (    Rant: This is      thereason I say post-selection is misleading, or even cheating. Post-selection creates weirdness out of nothing, even in the classical world. But it's an artefact of the analysis, not anything useful in the real world. And it can be tricky to spot post-selection-weird hiding amongst proper quantum-weird. Endless confusion results.    )  
  Another interesting, if less satisfying, example of taking a "weird" fact about teleportation and applying it to one-time pads is "moving infinite information". Sometimes people say that, because quantum teleportation moves qubits and the state of a qubit is defined by continuous amplitudes, teleportation uses just 2 bits of communication to move a continuous infinity of detail.
  The one-time pad also "moves infinite information". But the "moved information" is in terms of (for example) Eve's beliefs about the various bits. Alice yells out a single bit, and a continuous parameter that described Alice (Eve's inferred chance that Alice's bit is ON) suddenly applies to Bob.
  ... I'm not sure what to think about that particular correspondence. It quickly leads into the territory of Bell inequalities and philosophy.
  Summary

  Have a shared secret. Compare the shared secret to a message you want to send. Yell out the result of the comparison. That's how both the one-time pad and quantum teleportation are done.
友荐云推荐




上一篇:Android 刷量技术揭秘(一):工具篇
下一篇:The best way to map a Java 1.8 Optional entity attribute with JPA and Hibernate
酷辣虫提示酷辣虫禁止发表任何与中华人民共和国法律有抵触的内容!所有内容由用户发布,并不代表酷辣虫的观点,酷辣虫无法对用户发布内容真实性提供任何的保证,请自行验证并承担风险与后果。如您有版权、违规等问题,请通过"联系我们"或"违规举报"告知我们处理。

华哥电商 发表于 2016-10-24 18:36:54
楼主你好。。新人。混眼熟。顺便骗点经验。到手~拍拍屁股走人~
回复 支持 反对

使用道具 举报

段利斯 发表于 2016-11-17 05:24:25
缺乏基情了!
回复 支持 反对

使用道具 举报

*滑动验证:
您需要登录后才可以回帖 登录 | 立即注册

本版积分规则

我要投稿

推荐阅读

扫码访问 @iTTTTT瑞翔 的微博
回页顶回复上一篇下一篇回列表手机版
手机版/CoLaBug.com ( 粤ICP备05003221号 | 文网文[2010]257号 )|网站地图 酷辣虫

© 2001-2016 Comsenz Inc. Design: Dean. DiscuzFans.

返回顶部 返回列表