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Imagine two eigenstates of a system $|0\rangle$ and $|1\rangle$, and suppose you manage to prepare your system in the superposition $|\psi_{in}\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$. After some time, the system evolves naturally to the state $|\psi_{out}\rangle = (|0\rangle + e^{i\phi}|1\rangle)/\sqrt{2}$. The probability of the output being the same as the input is $p(\phi) = |\langle \psi_{in}|\psi_{out} \rangle|^2$.

I'm reading a paper that claims we can estimate this quantity with a statistical error (meaning variance) of $\Delta^2p(\phi) = \langle \psi_{out}| \left( |\psi_{in}\rangle \langle \psi_{in}| \right)^2 |\psi_{out}\rangle - p^2(\phi)$. Can anyone tell me where this expression comes from? Maybe I'm missing something obvious, but it's not clear how this relates to any of the usual expressions I know for variance or standard deviation.

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The variance of an operator $\hat A$ is $$\langle \psi| \hat A ^2 |\psi\rangle - \langle \psi | \hat A |\psi\rangle^2 , $$ as in statistics, $\overline{A^2}-\bar {A}^2$. You have used this in the uncertainty principle.

Your operator $\hat A$ here, however, is a projector, $P=|\psi_{in}\rangle \langle \psi_{in}|=P^2$, yielding $p-p^2$.

So $(\tfrac{1}{2}\sin \phi)^2$. Which paper?

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    $\begingroup$ Oh wow... can't believe I didn't see why the projector made sense as the "operator" here. Thanks for this. The paper is "Quantum-Enhanced Measurements:Beating the Standard Quantum Limit," Giovannetti, Lloyd, and Maccone, Science 306 (2004). $\endgroup$ – flevinBombastus Apr 15 at 20:56

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