7
$\begingroup$

What advantages does the phase sensitivity estimation obtained from the QFI give us compared to say, the phase sensitivity estimation obtained via the calculus of error propagation?

I'm getting interested in quantum-enhanced metrology and have come across the quantum Fisher information (QFI) as a measure of how much a quantum state $|\Psi(\theta)\rangle$ changes with respect to some variable, for example, the phase accumulated during an interferometer, θ. This is interesting as it provides a means to estimate the phase sensitivity of the interferometer given by $$\Delta\theta=\frac{1}{\sqrt{F_Q}},\qquad\qquad\qquad(1.1)$$ where $F_Q$ is the QFI and for pure states can be written as $$F_Q=4\left(\langle\Psi'|\Psi'\rangle-\left|\langle\Psi'|\Psi\rangle\right|^2\right),\qquad\qquad(1.2)$$ where $|\Psi'\rangle=\tfrac{d}{d\theta}|\Psi\rangle$ and $|\Psi\rangle$ the output state. Now, I'm also aware of other phase sensitivity estimations such as the formula derived via the calculus of error propagation $$\Delta\theta=\frac{\langle\Delta O\rangle}{\left|\frac{d\langle O\rangle}{d\theta}\right|},\qquad\qquad\qquad(1.3)$$ where $\langle\Delta O\rangle$ is the standard deviation and $O$ some Hermitian operator normally describing some measurement such as, the population difference between the two output arms of the interferometer.

My question is, why or when would I prefer one method over the other? My current understanding draws me to the form of both equations, (1.2) has no dependence on the measurement process whilst (1.3) does. This implies that a measurement procedure explained with $O$ might not be appropriate to obtain the degree of sensitivity given by (1.1), so comparing the two can tell you whether your measurement procedure is optimal?

$\endgroup$

1 Answer 1

2
$\begingroup$

The Classical Fisher Information (CFI) is a measure regarding how much a probability distribution changes with respect to some parameter, $\theta$. Whilst the quantum Fisher Information (QFI) is a measure regarding how much a quantum state changes with respect to some parameter, again here I have used $\theta$. In parameter estimation theory there is an important result giving a lower bound to the variance of an estimator $\Theta(\epsilon)$, where an estimator is defined as a function associating a measurement value $\epsilon$ with an estimate of the parameter $\theta$. This lower bound is called the Cramer-Rao lower bound, and for unbiased estimators, $\tfrac{\partial\Theta}{\partial\theta}=1$ is given by $$\left(\Delta\theta_{CR}\right)^2=\frac{1}{F(\theta)},\qquad\qquad\qquad(2.1)$$ where $F(\theta)$ is the CFI, which in its most general form can be written as $$F(\theta)=\sum_\epsilon\frac{1}{P(\epsilon|\theta)}\left(\frac{\partial P(\epsilon|\theta)}{\partial\theta}\right)^2,\qquad\qquad\qquad(2.2)$$ where $P(\epsilon|\theta)$ is the probability to obtain the result $\epsilon$ given the parameter has the value $\theta$. Now, the QFI is derived by maximising Eq.(2.2) overall generalised measurements in quantum mechanics; these measurements are those expressed with the positive-operator-values-measure (POVM). Indeed, after maximising Eq.(2.2) and simplifying the resulting expression for pure states, i.e. $\rho_\theta=|\Psi\rangle\langle\Psi|$, one obtains Eq.(1.2) for the QFI given in the original post. This now leads to a new lower bound called the Quantum Cramer-Rao lower bound and is given by $$\left(\Delta\theta_{CR}\right)^2\geq\left(\Delta\theta_{QCR}\right)^2=\frac{1}{F_Q},\qquad\qquad\qquad(2.3)$$ where $F_{Q}$ is the QFI. Indeed, this result provides us with an optimised lower bound for the variance in the estimation of $\theta$ provided one uses a corresponding optimal measurement which can be verified by using the method of error propagation suggested in Eq.(1.3) or the CFI and comparing the results. In my experience thus far, the advantage of using the QFI seems to be in the simplicity of computing it compared to Eq.(1.3). This provides a quick way to evaluate the input state's ability to estimate the parameter $\theta$, which can then be used as a base when including the estimate given that a measurement has now been performed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.