Fisher information of parametric channel

Question

Suppose $\Phi_\theta$ is a quantum channel whose action can be written for any state $\rho\in \mathcal S(\mathcal H_S)$ in the Stinespring representation as $\Phi_\theta(\rho)= \text{Tr}_E(U_\theta (\rho_S\otimes \tau_E) U_\theta^\dagger)$ for some unitary $U_\theta$ living on the total space $S\otimes E$. Suppose also that, if $a = (a_1, b_1,...,a_n, b_n)^t$ is the vector of the combined modes describing the system $S$ (so that the initial state of the system $\rho_S$ can be written as some function $\rho_S=f(a, a^\dagger)$), such unitary is described by its action on them $U_\theta a^\dagger U_\theta^\dagger=u_\theta a^\dagger$ for some matrix $u_\theta$.

The quantum Fisher information of a state $\rho_\theta$ is generally defined as $H(\theta) = \text{Tr}(\partial_\theta \rho_\theta L_\theta)$, where $L_\theta$ is the SLD operator. More explicitly, there exists the following formula $$H(\theta) = 2\int_0^{+\infty} dt \ e^{-\rho_\theta t}\partial_\theta \rho_\theta e^{-\rho_\theta t}$$ which only involves $\rho_\theta$ and its derivatives. Assuming knowledge of $u_\theta$, is there a simple way to express the QFI for a channel such as $\Phi_\theta$, in terms of any initial state $\rho$?

Answer 1

Imagine you have access to the auxiliary mode $\tau_E$ and want to find the quantum Fisher information of that scenario. The true QFI will be upper bounded by that, because the information in $\rho_\theta$ is strictly less than the information in the joint system.

Since $U_\theta(\rho_S\otimes\tau_E)U_\theta^\dagger\equiv |\Psi_\theta\rangle\langle\Psi_\theta|$ is pure by construction ($\rho_S=|\psi\rangle\langle \psi|$ and $\tau_E=|0\rangle\langle 0|$ and $|\Psi_\theta\rangle\equiv U_\theta|\psi\rangle\otimes |0\rangle\equiv U_\theta |\Psi_0\rangle$), its QFI is found to be $$H(\theta)=4(\langle \partial_\theta\Psi_\theta|\partial_\theta\Psi_\theta\rangle-|\langle \Psi_\theta|\partial_\theta\Psi_\theta\rangle|^2)=4(\langle\Psi_\theta|G^2|\Psi_\theta\rangle-\langle\Psi_\theta|G|\Psi_\theta\rangle^2)=4(\langle\psi|\otimes\langle 0|G^2|\psi\rangle\otimes|0\rangle-\langle\psi|\otimes\langle 0|G|\psi\rangle\otimes|0\rangle^2),$$ where the last equality used unitary invariance of the variance of the generator $G\equiv -i (\partial U_\theta/\partial \theta) U_\theta^\dagger$. In this case, $U_\theta$ is an SU($2n$) operation represented by a $2n\times 2n$ matrix $u_\theta$ and so $G$ will be a linear combination of the generators of SU($2n$). The coefficients of the linear combination may depend on $\theta$, depending on the form of $u_\theta$. One set of generators contains all of the generators of the form (cf wiki) $c_i c_j^\dagger +c_i^\dagger c_j$ ($i\neq j$), $\mathrm{i}(c_i c_j^\dagger -c_i^\dagger c_j)$ ($i\neq j$), and $c_1^\dagger c_1 -c_j^\dagger c_j$ ($j>1$) for $\mathbf{c}=(a_1,b_1,\cdots, a_n,b_n)$. Since $G$ is a linear combination of those, its variance is given by the sum of all of the covariances of those generators. The generators with odd $i$ and $j$ deal with variances with respect to the original state $|\psi\rangle$ and depend on both $|\psi\rangle$ and the form of $U_\theta=\exp(i G)$. (The form of $u_\theta$ told us that $G$ was made from generators of SU($2n$).) The generators with even $i$ and/or $j$ all have their expectation values vanish, because all of the $b$ modes start in their vacuum states, except for the final $n-1$ generators that retain the term $c_1^\dagger c_1=a_1^\dagger a_1$ that can have nonzero expectation value. As for their second-order moments, many of them vanish but not all, because for example $\langle (c_i c_2^\dagger +c_i^\dagger c_2)(c_j c_2^\dagger +c_j^\dagger c_2)\rangle=\langle (c_i^\dagger c_2)(c_j c_2^\dagger)\rangle=\langle c_i^\dagger c_j\rangle$ when mode 2 begins in the vacuum state.

All that work, just for an upper bound to the QFI. It turns out that the QFI is actually equivalent to the [minimum QFI over all purifications] (https://doi.org/10.1038/nphys1958), so we can say (I think, maybe there is more to the optimization) $$H_\theta=\min_{|\phi\rangle}4(\langle\psi|\otimes\langle \phi|G^2|\psi\rangle\otimes|\phi\rangle-\langle\psi|\otimes\langle \phi|G|\psi\rangle\otimes|\phi\rangle^2),$$ where now we enjoy none of the simplifications from above because $|\phi\rangle$ is not necessarily the vacuum state for the $b$ modes.