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How can I interpret a quantum circuit that results in the same state for the initialization $\newcommand{\ket}[1]{|#1\rangle}\newcommand{\bra}[1]{\langle #1|}\ket{+}$ and $\ket{-}$?

For example, the circuit consisting of a CNOT where the control qubit is measured gives the same result for both initialization states $\ket{+}$ and $\ket{-}$.

In both cases $$\ket{\psi'}=\frac{1}{2}\left(\ket{\psi}\bra{\psi} + X\ket{\psi}\bra{\psi}X^\dagger\right)$$

This is just a simple example. In general the circuit could have multiple ancilla qubits initialized to $\ket{+}$ or $\ket{-}$ and an arbitrary $n$-qubit input state.

Are there any general statements I can make about such circuits?

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  • $\begingroup$ Are you only interested about 2-qubits circuits? $\endgroup$
    – Tristan Nemoz
    Commented May 16 at 20:44
  • $\begingroup$ @TristanNemoz No, I'm not only interested in 2-qubit circuits. I even consider larger circuits that have multiple ancilla qubits initialized to $|+\rangle$ or $|-\rangle$. $\endgroup$
    – upe
    Commented May 16 at 21:22

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The way to think about the circuit you showed (and it can generalise to some multi-qubit cases) is that when you measure, you collapse the qubit into a particular state, 0 or 1. You can now run the circuit backwards in time, and it's as if the qubit was in that state at the various steps. So, if you get a 1 measurement outcome, in this case it's like the top qubit was in the $|1\rangle$ state when it controlled the controlled-not, and so that determines what happens to the target qubit. (This works particularly well because the measurement basis and the basis of the control coincide.) So, all that really matters is that for the different possible input states, they have identical probabilities for giving the (tracked-back) measurement outcomes, i.e. $$ |\langle 0|+\rangle|^2=|\langle 0|-\rangle|^2. $$

By that token, for your depicted circuit, arbitrary families of states $\alpha|0\rangle+\beta|1\rangle$ for fixed $|\alpha|^2$ should all give the same output.

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  • $\begingroup$ But is this a general property of circuits with that property? $\endgroup$ Commented May 17 at 18:24
  • $\begingroup$ @NorbertSchuch I wouldn't want to claim so, but it does cover a family of cases beyond just two-qubit. I imagine one could find other ways of constructing circuits with a suitable property (I haven't thought about it). $\endgroup$
    – DaftWullie
    Commented May 20 at 6:16
  • $\begingroup$ I guess one could try to make sth like this more precise by showing (all of this not thought through carefully) that if I can input both + and -, I can as well input a mixture of + or -, or equiv a mixture of any other ONB ... and then try to see that this implies that the corresponding state is treated classically by the circuit. But since the argument works for any bases (and rightfully so), this "propagate the meas. back" can at least occur in any basis which does not induce a bias towards the pre-mixture state (i.e. any meas. in the YZ plane). $\endgroup$ Commented May 20 at 11:16
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One way of seeing this is that the difference between the circuits is a $Z$-gate applied to the first qubit. This won't affect the measurement, and $Z$-gates are not 'transposed' by the CNOT gate to the target qubit, so the circuit results must be identical.

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To see why $|+\rangle$ and $|-\rangle$ states are not the same consider $|\psi\rangle = |i\rangle$ and the measurement outputs $|1\rangle$.

enter image description here

enter image description here

The two states only differ by a global phase, but they are not the same.

Instead, the reason why the two initialisations are logically equivalent resides in the measurement. Apply the deferred measurement principle in reverse to notice that the first qubit works like a classical 50/50 bit.

enter image description here

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  • $\begingroup$ What program creates such a graphics? $\endgroup$ Commented May 19 at 7:16
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    $\begingroup$ Quirk by Craig Gidney $\endgroup$ Commented May 19 at 7:51

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