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On the power of weak measurements in separating quantum states. (English) Zbl 1317.81017

Summary: We investigate the power of weak measurements in the framework of quantum state discrimination. First, we define and analyze the notion of weak consecutive measurements. Our main result is a convergence theorem whereby we demonstrate when and how a set of consecutive weak measurements converges to a strong measurement. Second, we show that for a small set of consecutive weak measurements, long before their convergence, one can separate close states without causing their collapse. We thus demonstrate a tradeoff between the success probability and the bias of the original vector towards collapse. Next, we use post-selection within the two-state vector formalism and present the non-linear expansion of the expectation value of the measurement device’s pointer to distinguish between two predetermined close vectors.

MSC:

81P15 Quantum measurement theory, state operations, state preparations
81P50 Quantum state estimation, approximate cloning

References:

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[29] \[|\cos (g\hat{X})-i \langle \hat{A}\rangle_w \sin (g\hat{X})|^2= |\cos (g\hat{X})+b \sin (g\hat{X})|^2= \cos^2(g\hat{X})+b^2\sin^2(g\hat{X})+b\sin (2g\hat{X})= (1-b^2)\cos^2(g\hat{X})+ b^2+b\sin (2g\hat{X})=(1-b^2)( \cos (2g\hat{X})+1)/2 +b^2+b\sin (2g\hat{X})=\frac{(1-b^2)}{2}\cos (2g\hat{X})+\frac{(1+b^2)}{2}+b\sin (2g\hat{X})\]|cos(gX^)-i⟨A^⟩wsin(gX^)|2=|cos(gX^)+bsin(gX^)|2=cos2(gX^)+b2sin2(gX^)+bsin(2gX^)=(1-b2)cos2(gX^)+b2+bsin(2gX^)=(1-b2)(cos(2gX^)+1)/2+b2+bsin(2gX^)=(1-b2)2cos(2gX^)+(1+b2)2+bsin(2gX^)
[30] Use integration by parts on \[\int e^{-\frac{x^2}{2\sigma^2}} \cos (kx) dx\]∫e-x22σ2cos(kx)dx
[31] Note that \[\cos (\eta ) = e^{-2(g\sigma )^2}\] cos(η)=e-2(gσ)2, and \[\sin (\eta )=\sqrt{1-e^{-4(g\sigma )^2}}\] sin(η)=\sqrt{1-e-4(gσ)2}, therefore \[\frac{\sin (\eta )2g \sigma^2}{e^{2(g\sigma )^2}-\cos (\eta )}= \frac{\sqrt{1-e^{-4(g\sigma )^2}} 2g\sigma^2}{e^{2(g\sigma )^2}(1-e^{-4(g\sigma )^2})}\] sin(η)2gσ2e2(gσ)2-cos(η)=\sqrt{1-e-4(gσ)2}2gσ2e2(gσ)2(1-e-4(gσ)2)
[32] Use integration by parts on \[\int e^{-\frac{x^2}{2\sigma^2}}[\cos (2kx)-2kx\sin (2kx)]dx\]∫e-x22σ2[cos(2kx)-2kxsin(2kx)]dx and the previous formula for \[\langle \hat{X}\sin (2g\hat{X})\rangle_{in} \]⟨X^sin(2gX^)⟩in
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