Quantum teleportation achieved over 16 km

Recently, in this report, it’s been reported that a physics laboratory in China achieved a new distance record in quantum teleportation: 16 km. That’s quite a feat, considering that up until now, the max. distance had been a few hundred meters.

While working on a project with a team in the US, I was monitoring a system (before debugging it) that failed in a part that I wasn’t going to debug, involving a TCP/IP error. I quipped that “unless [they] are using TCP over quantum teleportation, the Heisenberg uncertainty principle doesn’t apply here” (i.e. it didn’t just break because I was looking at it). That got me thinking: how feasible would it be to implement TCP/IP over quantum teleportation, with the current state of affairs?

The answer is: completely infeasible. Here’s why:

To be moderately useful, an average TCP packet – say about 512 bytes in size (plus the TCP/IP/Ethernet overhead) should have at least a 75% chance of arriving to its destination. 75% is pretty bad, considering that that means that one in four packets would be re-transmitted at least once, but I did say “moderately useful”. That means that each bit in the package would have to have a change of at least 99.994% of arriving: you’d have to be able to send the Ethernet pre-amble (8 bytes), followed by a 64-byte Ethernet header, followed by a 20-byte (minimum) IPv4 header, followed by a 20-byte (minimum) TCP header, which is an overhead of 112 bytes (not counting the 12-byte inter-frame gap mandated for Ethernet), so to get to 99.994%, you apply this formula P = \sqrt[\frac{1}{8(B + O)}]{0.75} in which P is the required probability for a single bit, B is the size of the message in bytes (octets) and O is the size of the overhead in bytes (octets). With B = 512 and O = 112, P comes out as 0.99994 – 99.994%.

In the article, they reported an 89% accuracy for teleporting a quantum state. If we equate the quantum state to a bit, we can see immediately that that is well below the 99.994% target.

So, how good are the chances of a TCP packet reaching its destination? How far are we from the target that three out of four reaches the destination intact? Well, for a packet to reach the destination intact, each and every one of the bits in the packet has to reach its destination. With each bit having a probability p of arriving, the packet’s chances P are P = p^{8(B + O)}. With B = 512 and O = 112, that makes P = 0.89^{8(512 + 112)} which is 2.264 * 10^{-253} or, in english, a change of 4 in 1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 (that’s a one with 252 zeroes).

That’s pretty slim!

About rlc

Software Analyst in embedded systems and C++, C and VHDL developer, I specialize in security, communications protocols and time synchronization, and am interested in concurrency, generic meta-programming and functional programming and their practical applications. I take a pragmatic approach to project management, focusing on the management of risk and scope. I have over two decades of experience as a software professional and a background in science.
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