# Is there a maximum quantum advantage in sensing?

This is a rather conceptual question.

Quantum sensing takes advantage of entanglement (and other quantum properties such as squeezing) to get variances that scale much better than the ones one can obtain by following any classical strategy.

My question is about quantum fisher information: imagine you have a well-defined estimation problem. You have a probe to sense a target. The problem may or may not include losses, and may or may not be noisy. You divide your strategies in two: the ones with and the ones without entanglement. The first are quantum strategies and the second are classical strategies. In order to compare in a fair way, resources (e.g. number of particles per probe) need to match between classical and quantum strategies. Now, in order to see if there is a "quantum advantage", we compute the quantum Fisher information for both strategies, $$H_C$$ for classical and $$H_Q$$ for quantum, and define their ratio: $$A=\frac{H_Q}{H_C}.$$ Now, if $$A>1$$ we claim there is a quantum advantage.

The question is whether $$A$$ is bounded or not, i.e. if there is a maximum advantage for any estimation problem.

In quantum illumination, for instance, it seems that the best you can achieve is a factor of 4 between these two, i.e. $$A=4$$ for the target detection problem. But to my knowledge this does not prove that there isn't another strategy that, with the same resources, can outperform quantum illumination.

I have heard, however, that this factor 4 is indeed a fundamental limit, but have not found a proof for this.

I'm not sure if we have a general answer to this question without further definitions. (i.e., what type of quantity is to be estimated and, more specifically, how does it couple to the probe). However, I can provide an answer in one common estimation context.

Let's suppose we want to estimate a parameter $$\theta$$. This parameter is coupled to $$N$$ single-particle Hamiltonians: $$H = \theta \sum_{i=1}^N \hat{h}_i.$$ We suppose that a state, $$\left| \psi \right\rangle$$ is evolved under this Hamiltonian and then measured using any POVM we want to extract the quantity $$\theta$$. We call our strategy "classical" if the state is separable, and "quantum" if the state is entangled. In this case, we can show that the classical Fisher information, $$F_C$$, is upper-bounded by $$N$$ times the maximum single-particle Fisher information. $$F_C \leq N F_1$$

This makes sense -- the classical strategy can be seen as "repeating" the same measurement many times across several separated states, and the Fisher information is linear in the number of measurements made. Now, what's $$F_1$$, the single-particle Fisher information? It turns out this can be related to a quantity called the operator semi-norm of the Hamiltonian, the difference between the largest and smallest eigenvalue of the Hamiltonian. So we can say: $$F_C \leq N \| h \|_s$$

Meanwhile, a quantum strategy can use an entangled state -- specifically, something that looks like the GHZ state for spins, where if $$\left| \lambda_\mathrm{min} \right\rangle$$ is the smallest eigenvalue of $$h$$ and $$\left| \lambda_\mathrm{max} \right\rangle$$ is the largest one, the probe state is $$\left| \psi \right\rangle = \frac{1}{\sqrt{2}} \left( \left| \lambda_\mathrm{max} \right\rangle^{\otimes N} + \left| \lambda_\mathrm{min} \right\rangle^{\otimes N} \right)$$ In this case, it can be shown that the quantum Fisher information is the largest possible, and is proportional to: $$F_Q = N^2 \| h \|_s.$$ So here your $$A$$ would be proportional to the number of entangled particles, $$N$$.

In spectroscopy/atomic clocks this is known as the Heisenberg limit and generally held to be the maximum possible enhancement due to entanglement, but this might not be true if you can use multiparticle Hamiltonians. Note you'll often see it expressed as an enhancement by a factor of $$\sqrt{N}$$ because people write it in terms of the standard deviation rather than the variance.

• Thank you for your reply. Could you try to explain the factor of 4 (pdfs.semanticscholar.org/c844/…) between $F_Q$ and $F_C$ that arises in quantum illumination? It seems that if $N$ is the number of entangled particles (which in quantum illumination is 2), the maximum advantage you can get is 2? – MCV Jun 14 '19 at 15:21
• Could you specify where and which factor of 4 you mean? Fig 2 shows the quantum advantage capping out at 2. – zeldredge Jun 14 '19 at 15:37
• Sorry, I pasted the wrong reference. In this work journals.aps.org/prl/pdf/10.1103/PhysRevLett.101.253601 you can see that a factor of 4 is obtained. – MCV Jun 16 '19 at 18:57
• @MCV this is an improvement in the "error probability exponent," which is related to the improvement in a binary discrimination task, so it's a different problem than what I outlined above -- parameter estimation described by a Fisher information -- and the two numbers can't be directly compared to my knowledge. Unfortunately I'm not familiar enough with this problem to tell you whether there's a hard upper limit on that. – zeldredge Jun 17 '19 at 12:32

When thinking of quantum states as resources for computation, one must consider the dimension of the system. Systems with more degrees of freedom have greater capacity for computation. In the case of quantum illumination (QI), the dimensionality of the composite system (i.e. d=dS dI where dS and dI are the dimensions of the signal and idler subsystems, respectively) determines the capacity to distinguish signal from noise. This is important since the returning signal and the surrounding noise do not have orthogonal support (except in the limit as d goes to infinity). If conventional illumination has access to dS to distinguish signal from noise, QI has access to some effective dimension d'S to distinguish signal from noise. This effective dimension is bounded by [dS,d] where dS implies no advantage gained. Therefore, the advantage of QI, and possibly quantum sensing in general, is limited by the dimension of the entangled state used and one's access to the idler dimension to increase the effective signal. It has been shown that the Bell gives the maximum advantage of QI.