michael_nielsen

The shared state (|00>+|11>)/sqrt 2 is symmetric, so the original circuit is equivalent to one where we move the CNOT target to the third qubit:pic.twitter.com/Dnn9AzBvN8

We can measure the second qubit as early as we like, without changing the output from the circuit. In fact, we can do the measurement right at the beginning of the protocol, giving:pic.twitter.com/obUVj5snWf

We can now ignore the second qubit - it no longer has any impact on the output we're interested in. We can also commute the X^x past the CNOT target, to get:pic.twitter.com/AqM7vyjC48

I don't know of any way of computing the output here, except to do the algebra:pic.twitter.com/KQ4FnQgP0C

Here's how to analyze your final circuit (with a slight abuse of time...). What you're really saying is that the final state before the measurement has overlap 1/√2 with |z〉|ψ〉.pic.twitter.com/0T9ERPGELO

The only interaction between the two qubits is the CNOT gate. What if we twisted the CNOT gate around and run the gates on the first qubit backwards (i.e., we reverse time)? In other words, starting with |z〉|0〉 we would like to get overlap 1/√2 with |ψ^*〉|ψ〉.pic.twitter.com/8iWOoK87Fb