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In the following paper by Harrow et al.: https://arxiv.org/pdf/1607.03236.pdf, they want to implement a measurement operator that is the average of a set of measurement operators. On page 9, right above theorem 9, they say:

One special case that will be important for us is when we have a sequence of projectors $\Lambda_i$, where $\Lambda_i$ corresponds to the two-outcome measurement $M_i = {\Lambda_i, I−\Lambda_i}$, and we would like to implement the POVM element $ \Lambda := \frac{1}{n} \sum_{j=1}^{n} \Lambda_j$. Define the projector $\Pi = \sum_{i=0}^{n-1} \Lambda_{i+1} \otimes (Q |{i} \rangle \langle{i}| Q^{-1})$ where $Q$ is the quantum Fourier transform on $\mathbb{Z}_n$. Given quantum circuits which implement each measurement $M_i$, it is easy to write down a circuit that implements the measurement ${\Pi, I − \Pi}$.

I don't see why this is easy. How do we write down this circuit?

Also, I do not see why we need the fourier transform for this implementation. Why can't we just have $|{0} \rangle \langle{0}| $ for the second register. I am guessing its necessary for the implementation but I do not see why.

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The point of the second register is to initialize it in some state with an equal probability of being found in any basis state $|i\rangle$. Then, conditioned on the state of the second register, one chooses which of the measurement procedures $M_i$ to perform on the state. Then the Fourier transform sets each of the projectors $\Lambda_i$ to have equal weight for the $|0\rangle$ state in the second register, and we can project onto that one to proceed. One could also project onto a different register state to avoid requiring $Q$ here.

Okay, so we know how to implement $M_i$, so we can do the projector $\Lambda_i$. Now construct a circuit that says "implement $\Lambda_i$ if the register is in state $|i\rangle$." That looks like $$\Pi^\prime=\sum_i \Lambda_i\otimes|i\rangle\langle i|.$$ Next, perform a Fourier transform on the second register and project the result into state $|0\rangle$. Also do that prior to $\Pi^\prime$, but with the inverse Fourier transform. That looks like, using their notation $\Delta=\mathbb{I}\otimes |0\rangle\langle 0|$, $$\Delta (\mathbb{I}\otimes Q)\Pi^\prime (\mathbb{I}\otimes Q^{-1})\Delta=\sum_i \Lambda_i\otimes |0\rangle\langle 0|Q|i\rangle\langle i|Q^{-1}|0\rangle\langle 0|.$$ Now, the purpose of the Fourier transform is to realize that $Q^{-1}=Q^\dagger$ and $|\langle 0| Q|i\rangle|=1/\sqrt{n}$ regardless of $i$. Then, we find that implementing first $\Delta$, then the inverse Fourier transform on the second register, then $\Pi^\prime$, then the Fourier transform on the second register, then $\Delta$ again enacts $$\Delta (\mathbb{I}\otimes Q)\Pi^\prime (\mathbb{I}\otimes Q^{-1})\Delta=\frac{1}{n}\sum_i \Lambda_i\otimes |0\rangle\langle 0|=\Lambda\otimes|0\rangle\langle 0|.$$ So if the second register is initialized in state $|0\rangle$ prior to all of these operations, the combined circuit will enact $\Lambda$ on the first register.

To recap: initialize the second register in state $|0\rangle$. Apply inverse Fourier transform $Q^{-1}$ on second register. Apply projector $\Lambda_i$ on first register conditioned on second register being in state $|i\rangle$ (this step requires the assumed knowledge of the circuits $M_i$ to implement such projectors, and the ability to choose which circuit to implement). Apply Fourier transform $Q$ to second register. Project second register onto state $|0\rangle$. You have enacted the desired transformation $\Lambda$ on the first register!

The same can be done with a replacement $\Lambda_i\to\mathbb{I}-\Lambda_i$ to lead to the overall replacement $\Lambda\to\mathbb{I}-\Lambda$.

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