Decoherence

Why opening the box breaks the spell

Quantum computers can answer questions like "what are the factors of this number?"  If the numbers are large enough these are questions that classical computers could never have the resources to answer.  For example you could ask it "what numbers divide 621405631250025693248096949484035695232" and it might respond "1932840132984031923 divides 621405631250025693248096949484035695232" as this is one of the divisors.

Except not yet.  At the moment quantum computers are pretty puny and only have a few qubits each.  A quantum computer with 4 qubits might be able to tell you what the divisors of 15 are, but it couldn't go any higher without needing more qubits.  And the problem that is making it difficult to build bigger, better, quantum computers is decoherence.

Before describing decoherence, let's look at how a quantum computer works.  In the following diagram points on the paper represent orthoganal states of the system.  But because I fully believe in The Universal Wavefunction, these states are not just states of the quantum computer - they are states of the entire universe.  (And why not?  Nothing in the physics actually tells us that the laws of quantum mechanics only applies to the very small!)  The state $\lvert s \rangle$ represents the start of computation, e.g. you've just loaded the question "what numbers divide 621405631250025693248096949484035695232".  The state $\lvert \checkmark \rangle$ represents a state of the universe in which computation has completed and is showing a correct answer to the question.  (Imagine a scientist in a white coat staring at a printout and smiling!)  The states $\lvert \times \rangle$ represent various states of the universe in which computation has completed but produced incorrect results. (Scientists are staring at the printouts and frowning.)

There are multiple paths from $\lvert s \rangle$ to $\lvert \checkmark \rangle$ represented by dotted lines.  These represent multiple different possible evolutions of the internal state of the machine.  Essentially, the machine guesses the solution infinitely many times and checks whether the guess is correct. Each one of those dotted lines represents a single guess.  Now, according to Feynman in his book QED: The Strange Theory of Light and Matter whenever you have two states the probability for one to evolve into the other is given by the square of the length of a complex number called the probability amplitude. And this probability amplitude is the sum of multiple probability amplitudes - one for each possible path between the two states.  The way our machine works is that it checks its guess and if it's correct it makes sure the phase of that particular path is some fixed value, say zero.  So the probability amplitudes for paths leading to $\lvert \checkmark \rangle$ all have the same phase and therefore interfere constructively!  On the other hand, if the guess is wrong the machine makes sure the phase for the path is random.  This means that the paths to $\lvert \times \rangle$ (shown in the diagram below) all interfere destructively, leading to a very low resultant probability amplitude for going from $\lvert s \rangle$ to $\lvert \times \rangle$.

So far, we have shown that: a) The likelihood of $\lvert \checkmark \rangle$ (a state of the universe containing a correct answer on a printout and historical evidence of state $\lvert s \rangle$) is very high; and b) The likelihood of $\lvert \times \rangle$ (a state of the universe containing an incorrect answer on a printout and historical evidence of state $\lvert s \rangle$) is very low.  So where does decoherence come into the picture (and ruin everything)?  Well, let's suppose that some evidence of the path taken by the machine leaks out.  What does this even mean?  It means that somehow you have recorded what happened inside the box and saved the result.  But this means that we are no longer talking about a single state $\lvert \checkmark \rangle$ - there is now one state where the records show one path was taken, another where the records show that a 2nd path was taken, another for a 3rd path, and so on.  All of these states correspond to a correct answer by the machine, but they are different states corresponding to a correct answer by the machine.  The picture below shows this situation.  The 3 paths shown are no longer paths between the same two states and as such their probability amplitudes cannot be added.  This means that there is no longer any constructive interference.  Similarly for the states in which the answer shown by the machine is incorrect, but this time it is the destructive interference which is missing.

Conclusion

States corresponding to worlds where you've set up the machine, waited patiently, and have a correct result are far more likely than states corresponding to worlds where you've set up the machine, waited patiently, and have an incorrect result.  But states corresponding to worlds where you've set up the machine, peeked, and have a correct result are just as likely as states corresponding to worlds where you've set up the machine, peeked, and have an incorrect result.

The problem of decoherence, is the problem of how on earth to avoid peeking.