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Assume we have a quantum state $$ |\psi\rangle = \sum_{j=0}^{2^n -1} \psi_j |j\rangle $$ where $\psi_j \in \mathbb R \ \forall j$.

My knowledge of the HKP shadow tomography protocol is limited, but it seems like given a set of $M$ observables, one can estimate their expectation values with $O(\log M)$ measurements. If that's the case, if we let our set of observables be $\{|0\rangle \langle 0|, |1\rangle \langle 1|, \dots, |2^n-1\rangle \langle 2^n -1 |\}$, wouldn't we be able to do full tomography on $|\psi\rangle$ using only $O(n)$ measurements? I'm certain I'm missing something here but I'm not sure what. Is it the fact that the amplitudes can be exponentially small and the accuracy becomes the limiting factor?

confusion
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1 Answers1

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You can't get the amplitudes, they are not physical. Your proposed set of observables will give you the probabilities instead, which is not enough for tomography. But in principle you're correct, you can use a set of observables ($O(4^n)$ many) to do "tomography" using $O(n)$ measurements.

And your intuition is also correct: The accuracy is the limiting factor. There are two problems: 1) You may be estimating very small quantities with additive precision. 2) You will estimate certain matrix coefficients of your state up to an additive error, but you should be interested in a more meaningful error measure, such as the trace distance. Converting errors will introduce a dimensional factor which will make the error even worse.

Point 2) also explains why you are not bypassing sampling complexity lower bounds from state tomography in this way. See also Lower bounds on the number of measurements outcomes required for quantum state tomography for the mentioned lower bounds and Using Classical Shadow to predict quantum state's fidelity has nothing to do with the dimension of the density matrix? for the resolution of that apparent contradiction, in particular the comments to my answer there.

Markus Heinrich
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