Noisy Quantum Communication

Author:

Marty Scholes

[email protected]

Last Modified:

2007.09.12

Disclaimer

I am not a physicist. I could best be described as a “backyard physicist,” which is to say that I have read a lot of material on the web and am fascinated by quantum mechanics and quantum computing. Anything you read here could be wrong.

References

Some good information on quantum mechanics and quantum computing can be found at the following links:

• http://www.toqc.com/

• http://en.wikipedia.org/wiki/Qubit

• http://tph.tuwien.ac.at/~oemer/qcl.html

• http://www.pha.jhu.edu/~javalab/qubit/qubit.html

Overview

Most material available will suggest that entangled decoherence cannot be used as a way to transmit information. The reason is as follows. Imagine two qubits, entangled and in a state of superposition such that when one qubit is measured, the other qubit will immediately decohere to the same value. When the first qubit is measured, it will yield a completely random zero or one, and the second qubit will yield the same value. Measuring the second qubit before the first qubit is measured will have the same effect, producing a random zero or one on both qubits. Conventional wisdom is that no information can be transmitted in this fashion because the same effect is produced regardless of which qubit is measured first. Furthermore, physics tell us once the qubits are separated, there is nothing that can be done to one qubit in isolation which will affect the other qubit, other than the previously mentioned measurement.

Thesis

This document will show a way that communication may take place using separated pairs of entangled qubits. The message passed is “noisy” in that there is a quantifiable probability of error. While not in the scope of this document, numerous error correcting codes exist which can reduce the error rate to any chosen level.

The Setup

Using physicists favorite imaginary people, Alice and Bob, assume the following takes place:

• A pair of qubits are prepared in superposition such that there is a 50% probability that they will both be zero and a 50% probability that they will both be one. Call these qubits Q_A and Q_B. For simplicity sake, keep the phase angle at zero.

• Give Q_A to Alice and Q_B to Bob. Separate Alice and Bob by an arbitrary distance.

The Communication

Alice can communicate a bit to Bob at a predetermined time by doing the following:

• Before the predetermined time, Alice either measures Q_A to send Bob a value of TRUE or does not measure Q_A to send Bob a value of FALSE.

• After the predetermined time, Bob performs a Hadamard transformation on Q_B and then measures it. Based on what the measurement of Q_B yields, Bob knows with some certainty, whether or not Alice has sent a TRUE or a FALSE.

Step by Step

Let's examine the states of the qubits based on the scenarios of whether or not Alice has measured her qubit.

Alice has measured her qubit.

At this point both qubits will have collapsed into identical states. Both qubits will either be pure zero or pure one. Since this is a truly random operation, either outcome is equally likely, but both qubits will have the same value.

When Bob performs a Hadamard operation on his qubit, it will become a superposition of zero and one. Once Bob measures his qubit, he will see either a zero or one with equal probability. In other words, he has learned nothing.

Alice has NOT measured her qubit.

At this point both qubits are still in superposition and a measurement will yield a zero or one with equal probability. Fortunately, Bob does not measure his qubit yet.

When Bob performs a Hadamard operation on his qubit, it will move to a state of pure zero. When Bob measures the qubit, it will always show a zero with 100% probability.

The Net Effect

If Alice measures her qubit, Bob will read either a zero or a one. If Alice does not measure her qubit, Bob will always measure a zero.

If Bob measures a one, then he knows with 100% probability that Alice has measured her qubit.

If Bob measures a zero, then he knows nothing about what Alice did.

In short, Bob can learn some information with 100% certainty based on what Alice did.

Closing Thoughts

As stated, I lack the training and qualification to say any of this with authority. I am open to feedback from those more learned that I. Please feel free to drop me an e-mail with any feedback and insight.

Marty